🧠:<think>Okay, so I have these two math problems to solve, both related to fractals and geometry. Let me take them one at a time.Starting with the first problem: It's about a Sierpinski tetrahedron, which is a 3D fractal created by recursively removing smaller tetrahedrons from the original solid. The question is asking for a formula for the total surface area after n iterations, given the original side length a.Hmm, I remember that for 2D fractals like the Sierpinski triangle, the surface area (or in 2D, the area) changes in a predictable way with each iteration. Maybe I can apply similar reasoning here.First, let me recall how a Sierpinski tetrahedron is constructed. Starting with a regular tetrahedron, each iteration involves dividing each tetrahedron into four smaller tetrahedrons, each with 1/4 the volume of the original. Then, the central inverted tetrahedron is removed. So, each iteration replaces each tetrahedron with three smaller ones.Wait, but in terms of surface area, how does this affect the total area? Each time we remove a tetrahedron, we're taking away some volume, but we're also exposing new surfaces. So, the surface area might actually increase.Let me think about the surface area of a regular tetrahedron. The surface area of a regular tetrahedron with side length a is 4 times the area of one face. Each face is an equilateral triangle with area (√3/4)a². So, total surface area is 4*(√3/4)a² = √3 a².Now, when we perform the first iteration, we divide the original tetrahedron into four smaller tetrahedrons, each with side length a/2. The central one is removed, so we're left with three smaller tetrahedrons. But, when we remove the central one, we expose three new faces on each of the remaining tetrahedrons.Wait, actually, each face of the central tetrahedron that was removed was previously internal, so removing it adds new surface area. Each face of the central tetrahedron is an equilateral triangle with side length a/2. Since the central tetrahedron has four faces, but each face is shared with one of the surrounding tetrahedrons. So, removing the central tetrahedron would expose three new faces, each of area (√3/4)(a/2)².But actually, each of the three surrounding tetrahedrons had one face covered by the central tetrahedron. So, removing the central one would expose three new faces, each of area (√3/4)(a/2)². So, the total new surface area added is 3*(√3/4)(a²/4) = (3√3/16)a².But wait, the original surface area was √3 a². After the first iteration, we have three smaller tetrahedrons, each with surface area √3 (a/2)² = √3 a² /4. So, three of them would have a total surface area of 3*(√3 a² /4) = (3√3/4)a². But we also added the new exposed faces, which is (3√3/16)a². So, the total surface area after the first iteration is (3√3/4 + 3√3/16)a².Let me compute that: 3√3/4 is equal to 12√3/16, so adding 3√3/16 gives 15√3/16 a². Hmm, that seems a bit less than the original. But wait, that can't be right because we should be increasing the surface area.Wait, maybe I made a mistake in the reasoning. Let's think again.Each original face is divided into four smaller faces, each of side length a/2. When we remove the central tetrahedron, we are removing one of the four smaller tetrahedrons, which was attached to the center of each face. So, each face of the original tetrahedron is now a larger triangle with a smaller triangle removed from the center. So, each original face is now a hexagon? Or is it still a triangle?Wait, no, actually, each face is divided into four smaller triangles, and the central one is removed. So, each face becomes a sort of star with three smaller triangles and the central one missing. So, each original face contributes three smaller faces, each of area (√3/4)(a/2)².But originally, each face had area (√3/4)a². After the first iteration, each face is replaced by three smaller faces, each of area (√3/4)(a/2)². So, the area per face becomes 3*(√3/4)(a²/4) = (3√3/16)a². Since there are four original faces, the total surface area becomes 4*(3√3/16)a² = (12√3/16)a² = (3√3/4)a².Wait, but that's the same as the total surface area of the three smaller tetrahedrons. So, perhaps the total surface area after the first iteration is (3√3/4)a², which is less than the original √3 a². But that contradicts the intuition that removing the central tetrahedron should expose new surfaces, thus increasing the total surface area.Hmm, maybe I'm confusing the way the surface area changes. Let me think differently.Each time we remove a tetrahedron, we are taking away some volume, but we are also adding new surfaces. So, the original surface area is √3 a². When we remove the central tetrahedron, we are removing a volume, but we are also exposing three new faces of the central tetrahedron. Each of these new faces is an equilateral triangle with side length a/2.So, each face has area (√3/4)(a/2)² = (√3/16)a². There are three such new faces, so the total added surface area is 3*(√3/16)a² = (3√3/16)a².But, we also have the original surface area, but parts of it have been altered. Wait, actually, the original surface area is still there, except that each face now has a hole in the center. However, the hole is covered by the three new faces we've added. So, perhaps the total surface area is the original surface area plus the added surface area from the removal.Wait, but the original surface area is still present, except that each face is now a larger triangle with a smaller triangle missing. So, the area lost from each face is (√3/4)(a/2)² = √3 a² /16. Since there are four faces, the total area lost is 4*(√3 a² /16) = √3 a² /4.But we added three new faces, each of area √3 a² /16, so total added area is 3*(√3 a² /16) = 3√3 a² /16.So, the total surface area after the first iteration is original surface area minus area lost plus area added.Original surface area: √3 a².Minus area lost: √3 a² /4.Plus area added: 3√3 a² /16.So, total surface area: √3 a² - √3 a² /4 + 3√3 a² /16.Let me compute that:√3 a² is 16√3 a² /16.Minus √3 a² /4 is 4√3 a² /16.Plus 3√3 a² /16.So, 16√3/16 - 4√3/16 + 3√3/16 = (16 - 4 + 3)√3/16 = 15√3/16 a².Wait, so after the first iteration, the surface area is 15√3/16 a², which is less than the original √3 a². That seems counterintuitive because we added new surfaces. Maybe my approach is wrong.Alternatively, perhaps each iteration increases the surface area. Let me think about the scaling factor.In 2D, the Sierpinski triangle has a surface area that increases by a factor of 3/2 each iteration. Maybe in 3D, it's similar but with a different scaling factor.Wait, in 3D, each iteration replaces each tetrahedron with three smaller ones, each scaled down by a factor of 1/2. So, the surface area of each smaller tetrahedron is (1/2)^2 = 1/4 of the original. So, three of them would have a total surface area of 3*(1/4) = 3/4 of the original. But we also added new surfaces from the removed central tetrahedron.Wait, the central tetrahedron had a surface area of √3 (a/2)^2 = √3 a² /4. But when we remove it, we expose three new faces, each of area √3 (a/2)^2 /4 = √3 a² /16. So, three of them add up to 3√3 a² /16.So, the total surface area after the first iteration is 3/4 of the original surface area plus 3√3 a² /16.But the original surface area was √3 a², so 3/4 of that is (3√3/4)a². Adding 3√3/16 a² gives (12√3/16 + 3√3/16) = 15√3/16 a², same as before.Hmm, so it seems that the surface area is decreasing after the first iteration, which is confusing because we added new surfaces. Maybe the issue is that the original surface area is being subdivided, and only parts of it remain, while the new surfaces are added.Wait, perhaps the correct way to model this is to consider that each iteration, the surface area is multiplied by a factor. Let me see.After the first iteration, the surface area is 15√3/16 a². Let me compute the ratio: 15/16 ≈ 0.9375. So, it's slightly less than the original.But in 2D, the Sierpinski triangle's area decreases by a factor of 3/4 each iteration, but the perimeter increases. Maybe in 3D, the surface area behaves differently.Wait, perhaps I'm overcomplicating. Let me think recursively.Let S(n) be the total surface area after n iterations.At each iteration, each tetrahedron is replaced by three smaller tetrahedrons, each with 1/4 the surface area of the original. So, the total surface area contributed by the smaller tetrahedrons is 3*(1/4) S(n-1).But, when we remove the central tetrahedron, we expose three new faces. Each face has an area equal to the area of a face of the small tetrahedron, which is (√3/4)(a/2)^2 = √3 a² /16. Since we have three such faces per removed tetrahedron, and each iteration removes one tetrahedron per original tetrahedron, the total added surface area is 3*(√3 a² /16).But wait, actually, each iteration removes one central tetrahedron from each existing tetrahedron. So, the number of new faces added is 3 per tetrahedron. So, if at iteration n, there are 3^n tetrahedrons, each removal adds 3 new faces. So, the total added surface area is 3^n * 3*(√3 a² /16).Wait, that might not be correct because the side length decreases with each iteration.Wait, perhaps it's better to model the surface area as a geometric series.At each iteration, the surface area is multiplied by a factor. Let me see.After the first iteration, S(1) = (3/4) S(0) + 3*(√3 (a/2)^2 /4). Wait, S(0) is √3 a².So, S(1) = (3/4)√3 a² + 3*(√3 a² /16) = (3√3/4 + 3√3/16) a² = (12√3/16 + 3√3/16) a² = 15√3/16 a².Similarly, for the second iteration, each of the three tetrahedrons from the first iteration will undergo the same process. So, each will contribute (3/4) of their surface area plus 3*(√3 (a/4)^2 /4).Wait, because the side length is now a/2, so the next iteration would have side length a/4.Wait, let me think again. At each iteration, the side length is halved, so the surface area of each new small tetrahedron is (1/2)^2 = 1/4 of the previous one.So, the surface area contributed by the smaller tetrahedrons is 3*(1/4) S(n-1).But we also add new surfaces. Each removal adds 3 new faces, each of area (√3/4)(a/2^n)^2.Wait, maybe it's better to express the total surface area as S(n) = (3/4) S(n-1) + 3*(√3/4)(a/2^n)^2 * number of removals.But the number of removals at each iteration is equal to the number of tetrahedrons from the previous iteration, which is 3^{n-1}.Wait, this is getting complicated. Maybe I should look for a pattern.Let me compute S(0) = √3 a².S(1) = 15√3/16 a² ≈ 0.9375√3 a².Now, let's compute S(2). Each of the three tetrahedrons from S(1) will undergo the same process.So, each contributes (3/4) of its surface area plus 3*(√3/4)(a/4)^2.Wait, the side length at the second iteration is a/4, so the area of each new face is (√3/4)(a/4)^2 = √3 a² /64.Each removal adds 3 such faces, so per tetrahedron, it's 3*(√3 a² /64).But how many tetrahedrons are there at iteration 1? It's 3^1 = 3.So, the total added surface area at iteration 2 is 3 * 3*(√3 a² /64) = 9√3 a² /64.Meanwhile, the surface area from the smaller tetrahedrons is (3/4) S(1) = (3/4)*(15√3/16 a²) = 45√3/64 a².So, S(2) = 45√3/64 a² + 9√3/64 a² = 54√3/64 a² = 27√3/32 a² ≈ 0.84375√3 a².Hmm, so S(0) = √3 a², S(1) ≈ 0.9375√3 a², S(2) ≈ 0.84375√3 a².Wait, that seems like the surface area is decreasing, which contradicts the idea that we're adding surfaces. Maybe my approach is wrong.Alternatively, perhaps the surface area is increasing because each iteration adds more surfaces, but the scaling factor is such that the total surface area converges to a finite limit.Wait, let's think about the scaling. Each iteration, the number of tetrahedrons is multiplied by 3, and each has a surface area scaled by (1/4). So, the total surface area from the smaller tetrahedrons is 3*(1/4) S(n-1).But we also add new surfaces. Each removal adds 3 new faces, each of area (√3/4)(a/2^n)^2.Wait, maybe the total surface area after n iterations is S(n) = (3/4)^n * √3 a² + sum_{k=1}^n 3^k * 3*(√3/4)(a/2^k)^2.Wait, let me try to express it as a recurrence relation.At each step, S(n) = (3/4) S(n-1) + 3*(√3/4)(a/2^n)^2 * 3^{n-1}.Wait, because at each iteration, we have 3^{n-1} tetrahedrons, each contributing 3 new faces of area (√3/4)(a/2^n)^2.So, S(n) = (3/4) S(n-1) + 3^{n} * (√3/4)(a^2 /4^n).Simplify that:S(n) = (3/4) S(n-1) + (√3/4) * 3^n * a² /4^n.Which is S(n) = (3/4) S(n-1) + (√3 a² /4) * (3/4)^n.Now, this is a linear recurrence relation. Let me write it as:S(n) = (3/4) S(n-1) + C*(3/4)^n, where C = √3 a² /4.This is a nonhomogeneous linear recurrence. The solution will be the sum of the homogeneous solution and a particular solution.The homogeneous equation is S(n) = (3/4) S(n-1), which has the solution S_h(n) = K*(3/4)^n.For the particular solution, since the nonhomogeneous term is C*(3/4)^n, we can try a particular solution of the form S_p(n) = D*n*(3/4)^n.Plugging into the recurrence:D*n*(3/4)^n = (3/4) [D*(n-1)*(3/4)^{n-1}] + C*(3/4)^n.Simplify:D*n*(3/4)^n = (3/4) D*(n-1)*(3/4)^{n-1} + C*(3/4)^n.Divide both sides by (3/4)^{n-1}:D*n*(3/4) = D*(n-1) + C*(3/4).Multiply both sides by 4/3:D*n = (4/3) D*(n-1) + C.Simplify:D*n = (4D/3)(n - 1) + C.Expand:D*n = (4D/3)n - 4D/3 + C.Bring all terms to left:D*n - (4D/3)n + 4D/3 - C = 0.Factor n:D*(1 - 4/3) n + 4D/3 - C = 0.Simplify:D*(-1/3) n + 4D/3 - C = 0.This must hold for all n, so the coefficient of n must be zero:- D/3 = 0 => D = 0.But that's not possible because then the particular solution would be zero, which doesn't help. So, perhaps my guess for the particular solution is incorrect. Maybe I need to try a different form.Alternatively, since the nonhomogeneous term is of the same form as the homogeneous solution, I might need to multiply by n. Wait, I already tried that. Maybe I need to try a different approach.Alternatively, let's solve the recurrence using the method for linear recurrences.The general solution is S(n) = S_h(n) + S_p(n).We have S_h(n) = K*(3/4)^n.For S_p(n), since the nonhomogeneous term is C*(3/4)^n, and since 3/4 is a root of the homogeneous equation, we need to multiply by n. So, let's assume S_p(n) = D*n*(3/4)^n.Plugging into the recurrence:D*n*(3/4)^n = (3/4) [D*(n-1)*(3/4)^{n-1}] + C*(3/4)^n.Simplify:D*n*(3/4)^n = (3/4) D*(n-1)*(3/4)^{n-1} + C*(3/4)^n.Divide both sides by (3/4)^{n-1}:D*n*(3/4) = D*(n - 1) + C*(3/4).Multiply both sides by 4/3:D*n = (4/3) D*(n - 1) + C.Expand:D*n = (4D/3)n - 4D/3 + C.Bring all terms to left:D*n - (4D/3)n + 4D/3 - C = 0.Factor n:D*(1 - 4/3) n + 4D/3 - C = 0.Simplify:D*(-1/3) n + 4D/3 - C = 0.This must hold for all n, so the coefficient of n must be zero:- D/3 = 0 => D = 0.But again, D=0, which doesn't help. This suggests that my particular solution guess is incorrect. Maybe I need to try a different form, like S_p(n) = D*(3/4)^n.But that's the homogeneous solution. Alternatively, since the nonhomogeneous term is of the same form, perhaps I need to use a different method.Alternatively, let's write the recurrence as:S(n) - (3/4) S(n-1) = C*(3/4)^n.This is a linear nonhomogeneous recurrence. The solution can be written as the sum of the homogeneous solution and a particular solution.The homogeneous solution is S_h(n) = K*(3/4)^n.For the particular solution, since the nonhomogeneous term is C*(3/4)^n, and since 3/4 is a root of the homogeneous equation, we can try S_p(n) = D*n*(3/4)^n.Plugging into the equation:D*n*(3/4)^n - (3/4) D*(n-1)*(3/4)^{n-1} = C*(3/4)^n.Simplify:D*n*(3/4)^n - (3/4) D*(n-1)*(3/4)^{n-1} = C*(3/4)^n.Factor out (3/4)^{n-1}:(3/4)^{n-1} [D*n*(3/4) - (3/4) D*(n - 1)] = C*(3/4)^n.Simplify inside the brackets:(3/4)^{n-1} [ (3D/4) n - (3D/4)(n - 1) ] = C*(3/4)^n.Factor out 3D/4:(3/4)^{n-1} * (3D/4) [n - (n - 1)] = C*(3/4)^n.Simplify inside the brackets:(3/4)^{n-1} * (3D/4) [1] = C*(3/4)^n.So,(3D/4) * (3/4)^{n-1} = C*(3/4)^n.Simplify:(3D/4) * (3/4)^{n-1} = C*(3/4)^n.Divide both sides by (3/4)^{n-1}:3D/4 = C*(3/4).So,3D/4 = C*(3/4).Multiply both sides by 4/3:D = C.Since C = √3 a² /4, then D = √3 a² /4.Therefore, the particular solution is S_p(n) = (√3 a² /4) * n*(3/4)^n.So, the general solution is:S(n) = K*(3/4)^n + (√3 a² /4) * n*(3/4)^n.Now, apply the initial condition. At n=0, S(0) = √3 a².So,√3 a² = K*(3/4)^0 + (√3 a² /4)*0*(3/4)^0.Simplify:√3 a² = K*1 + 0 => K = √3 a².Therefore, the solution is:S(n) = √3 a²*(3/4)^n + (√3 a² /4)*n*(3/4)^n.Factor out √3 a²*(3/4)^n:S(n) = √3 a²*(3/4)^n [1 + n/4].So,S(n) = √3 a²*(3/4)^n (1 + n/4).Alternatively, we can write this as:S(n) = √3 a²*(3/4)^n ( (4 + n)/4 ) = √3 a²*(3/4)^n*(4 + n)/4.Simplify:S(n) = √3 a²*(4 + n)/4 * (3/4)^n.Alternatively,S(n) = √3 a²*(4 + n)/4 * (3/4)^n.But let me check this formula with n=0:S(0) = √3 a²*(4 + 0)/4 * (3/4)^0 = √3 a²*(4/4)*1 = √3 a². Correct.For n=1:S(1) = √3 a²*(4 +1)/4*(3/4)^1 = √3 a²*(5/4)*(3/4) = √3 a²*(15/16). Which matches our earlier result.For n=2:S(2) = √3 a²*(4 +2)/4*(3/4)^2 = √3 a²*(6/4)*(9/16) = √3 a²*(3/2)*(9/16) = √3 a²*(27/32). Which also matches our earlier result.So, the formula seems correct.Therefore, the total surface area after n iterations is:S(n) = √3 a² * (4 + n)/4 * (3/4)^n.Alternatively, we can write it as:S(n) = √3 a² * (n + 4)/4 * (3/4)^n.Simplifying further, we can write:S(n) = √3 a² * (n + 4) * (3/4)^n /4.But perhaps it's better to leave it as:S(n) = √3 a² * (3/4)^n * (1 + n/4).Alternatively, factor out the 1/4:S(n) = √3 a² * (3/4)^n * (4 + n)/4.So, that's the formula for the total surface area after n iterations.Now, moving on to the second problem.The poet is modeling light reflection using the complex function f(z) = z² + c, starting from z₀ = 0. We need to determine the conditions on c such that the sequence z_{n+1} = f(z_n) does not escape to infinity. This is related to the Mandelbrot set, where c is in the Mandelbrot set if the sequence remains bounded.So, the condition is that the sequence remains bounded, i.e., does not escape to infinity. The Mandelbrot set is defined as the set of complex numbers c for which the function f(z) = z² + c does not escape to infinity when iterated from z₀ = 0.The classic result is that c is in the Mandelbrot set if and only if the sequence remains bounded. It's known that if at any point |z_n| > 2, then the sequence will escape to infinity. So, a common test is to check if |z_n| exceeds 2; if it does, c is outside the Mandelbrot set.But the exact conditions are more nuanced. The boundary of the Mandelbrot set is given by the set of points c where the sequence does not escape, but it's not simply |c| ≤ 2, because some points inside |c| ≤ 2 still escape, and some outside might not, but actually, it's known that if |c| > 2, the sequence will escape, but the converse isn't true.Wait, actually, it's known that if |c| > 2, then the sequence will escape to infinity. So, the Mandelbrot set is contained within the disk of radius 2 centered at the origin. But not all points inside |c| ≤ 2 are in the Mandelbrot set; some do escape.But the exact condition is that c is in the Mandelbrot set if the sequence does not escape to infinity. So, the condition is that for all n, |z_n| ≤ 2. But this is not a simple condition to express without iteration.However, for the purpose of this problem, I think the answer is that c must lie within the Mandelbrot set, which is the set of complex numbers c for which the sequence z_{n+1} = z_n² + c remains bounded when starting from z₀ = 0. Equivalently, c is in the Mandelbrot set if the sequence does not escape to infinity.But perhaps the problem expects a more specific condition. For example, in some cases, if |c| ≤ 1/4, the sequence remains bounded, but that's only for certain regions.Wait, actually, the main cardioid of the Mandelbrot set is given by c = (1/2 - 1/2 e^{iθ})², which is the set of c where the sequence remains bounded for certain θ. But that's more detailed.Alternatively, the condition can be expressed in terms of the magnitude of c. It's known that if |c| > 2, the sequence escapes. For |c| ≤ 2, it might or might not escape, depending on c.But perhaps the problem is expecting the condition that |c| ≤ 2, but that's not sufficient, as some points inside |c| ≤ 2 escape.Alternatively, the condition is that c is in the Mandelbrot set, which is defined by the boundedness of the sequence. So, the condition is that c is such that the sequence z_{n+1} = z_n² + c does not escape to infinity when starting from z₀ = 0.In the context of the fractal structure of the tetrahedron, perhaps the reflection patterns correspond to points in the Mandelbrot set, meaning that the light reflection remains bounded, creating intricate patterns similar to the fractal structure of the tetrahedron.So, the condition is that c is in the Mandelbrot set, i.e., the sequence does not escape to infinity.But to express this more formally, the condition is that for all n ≥ 0, |z_n| ≤ 2, where z_{n+1} = z_n² + c and z₀ = 0.Alternatively, using the escape radius of 2, if |z_n| > 2 for any n, then c is outside the Mandelbrot set.So, the condition is that for all n, |z_n| ≤ 2.But since this is a fractal, the boundary is complex, and there's no simple closed-form condition for c, but it's known that if |c| > 2, then the sequence escapes.Therefore, the necessary condition is |c| ≤ 2, but it's not sufficient. However, for the purpose of this problem, perhaps the answer is that c must satisfy |c| ≤ 2, but with the caveat that some points within this disk still escape.But in the context of the problem, the poet is noticing a pattern in the reflection, so perhaps the condition is that c is in the Mandelbrot set, meaning the sequence remains bounded.So, to answer the question: The condition on c is that the sequence z_{n+1} = z_n² + c remains bounded, i.e., does not escape to infinity when starting from z₀ = 0. This is equivalent to c being in the Mandelbrot set.In the context of the fractal structure of the tetrahedron, this means that the reflection patterns created by the light correspond to points in the Mandelbrot set, resulting in intricate, non-escaping trajectories that mirror the self-similar structure of the Sierpinski tetrahedron.So, summarizing:1. The total surface area after n iterations is S(n) = √3 a² * (3/4)^n * (1 + n/4).2. The condition on c is that it lies within the Mandelbrot set, i.e., the sequence z_{n+1} = z_n² + c does not escape to infinity when starting from z₀ = 0.</think>

🧠:<think>Alright, so I've got this problem about Mr. Thompson, a retired school principal who's tracking his caloric intake and expenditure. The goal is to figure out when his net caloric balance is within -500 to 500 calories over a 24-hour period. Let me try to break this down step by step.First, let's understand what net caloric balance means. It's the difference between his caloric intake and expenditure at any given time. So, mathematically, that would be:Net Balance ( N(t) = I(t) - E(t) )Given the functions:( I(t) = 250 sinleft(frac{pi t}{12}right) + 2000 )( E(t) = 150 cosleft(frac{pi t}{6}right) + 1800 )So, substituting these into the net balance equation:( N(t) = 250 sinleft(frac{pi t}{12}right) + 2000 - left(150 cosleft(frac{pi t}{6}right) + 1800right) )Simplify that:( N(t) = 250 sinleft(frac{pi t}{12}right) - 150 cosleft(frac{pi t}{6}right) + 200 )So, the net balance is a function that combines sine and cosine terms. The challenge is to find the times ( t ) within 0 to 24 hours where ( N(t) ) is between -500 and 500.Hmm, okay. So, part a is asking for the time intervals when ( -500 leq N(t) leq 500 ). To approach this, I think I need to solve the inequalities:1. ( N(t) leq 500 )2. ( N(t) geq -500 )Which translates to:1. ( 250 sinleft(frac{pi t}{12}right) - 150 cosleft(frac{pi t}{6}right) + 200 leq 500 )2. ( 250 sinleft(frac{pi t}{12}right) - 150 cosleft(frac{pi t}{6}right) + 200 geq -500 )Simplify both inequalities:For the first inequality:( 250 sinleft(frac{pi t}{12}right) - 150 cosleft(frac{pi t}{6}right) leq 300 )For the second inequality:( 250 sinleft(frac{pi t}{12}right) - 150 cosleft(frac{pi t}{6}right) geq -700 )So, now we have two inequalities to solve:1. ( 250 sinleft(frac{pi t}{12}right) - 150 cosleft(frac{pi t}{6}right) leq 300 )2. ( 250 sinleft(frac{pi t}{12}right) - 150 cosleft(frac{pi t}{6}right) geq -700 )This seems a bit complicated because it's a combination of sine and cosine functions with different periods. Let me see if I can express them with the same argument or find a way to combine them.First, note the arguments of the sine and cosine functions:- Sine has ( frac{pi t}{12} ), which is a period of ( frac{2pi}{pi/12} = 24 ) hours.- Cosine has ( frac{pi t}{6} ), which is a period of ( frac{2pi}{pi/6} = 12 ) hours.So, the sine function has a period of 24 hours, and the cosine has a period of 12 hours. That means the cosine function completes two cycles in the same time the sine completes one.Hmm, maybe we can express both functions in terms of the same variable. Let me set ( theta = frac{pi t}{12} ). Then, ( frac{pi t}{6} = 2theta ). So, the functions become:( N(t) = 250 sin(theta) - 150 cos(2theta) + 200 )That might make it easier to handle. So, substituting ( theta = frac{pi t}{12} ), which implies ( t = frac{12theta}{pi} ). Since ( t ) ranges from 0 to 24, ( theta ) will range from 0 to ( 2pi ).So, now, ( N(theta) = 250 sin(theta) - 150 cos(2theta) + 200 )We can use a double-angle identity for cosine: ( cos(2theta) = 1 - 2sin^2(theta) ). Let me substitute that in:( N(theta) = 250 sin(theta) - 150 (1 - 2sin^2(theta)) + 200 )Simplify:( N(theta) = 250 sin(theta) - 150 + 300 sin^2(theta) + 200 )Combine constants:( N(theta) = 300 sin^2(theta) + 250 sin(theta) - 150 + 200 )( N(theta) = 300 sin^2(theta) + 250 sin(theta) + 50 )So, now we have a quadratic in terms of ( sin(theta) ). Let me denote ( x = sin(theta) ), so:( N(x) = 300x^2 + 250x + 50 )Now, we can write the inequalities in terms of ( x ):1. ( 300x^2 + 250x + 50 leq 500 )2. ( 300x^2 + 250x + 50 geq -500 )Simplify both:First inequality:( 300x^2 + 250x + 50 - 500 leq 0 )( 300x^2 + 250x - 450 leq 0 )Second inequality:( 300x^2 + 250x + 50 + 500 geq 0 )( 300x^2 + 250x + 550 geq 0 )So, now we have two quadratic inequalities:1. ( 300x^2 + 250x - 450 leq 0 )2. ( 300x^2 + 250x + 550 geq 0 )Let me solve each inequality separately.Starting with the first inequality:( 300x^2 + 250x - 450 leq 0 )Divide all terms by 50 to simplify:( 6x^2 + 5x - 9 leq 0 )Now, solve ( 6x^2 + 5x - 9 = 0 )Using quadratic formula:( x = frac{-5 pm sqrt{25 + 216}}{12} )( x = frac{-5 pm sqrt{241}}{12} )Calculate ( sqrt{241} approx 15.524 )So,( x = frac{-5 + 15.524}{12} approx frac{10.524}{12} approx 0.877 )( x = frac{-5 - 15.524}{12} approx frac{-20.524}{12} approx -1.710 )So, the roots are approximately 0.877 and -1.710.Since this is a quadratic opening upwards (coefficient 6 is positive), the inequality ( 6x^2 + 5x - 9 leq 0 ) holds between the roots. So, ( x in [-1.710, 0.877] ). However, ( x = sin(theta) ), which must be between -1 and 1. So, the valid interval is ( x in [-1, 0.877] ).Now, moving to the second inequality:( 300x^2 + 250x + 550 geq 0 )Again, divide by 50:( 6x^2 + 5x + 11 geq 0 )Solve ( 6x^2 + 5x + 11 = 0 )Discriminant: ( 25 - 264 = -239 )Since discriminant is negative, the quadratic never crosses zero and is always positive because the coefficient of ( x^2 ) is positive. So, this inequality is always true for all real x.Therefore, the only restriction comes from the first inequality, which gives ( x in [-1, 0.877] ). But since ( x = sin(theta) ), we have:( -1 leq sin(theta) leq 0.877 )But since ( sin(theta) ) can't be less than -1, the lower bound is automatically satisfied. So, the critical condition is ( sin(theta) leq 0.877 ).So, we need to find all ( theta ) in [0, 2π] where ( sin(theta) leq 0.877 ).The values of ( theta ) where ( sin(theta) = 0.877 ) can be found in the first and second quadrants.Calculate ( theta = arcsin(0.877) approx arcsin(0.877) approx 1.065 ) radians (since sin(1.065) ≈ 0.877). The other solution in [0, 2π] is ( pi - 1.065 approx 2.076 ) radians.So, ( sin(theta) leq 0.877 ) is true for ( theta in [0, 1.065] cup [2.076, 2pi] ).Wait, actually, no. Because sine is increasing from 0 to π/2, then decreasing from π/2 to π, etc. So, when ( sin(theta) leq 0.877 ), it's true except between the two points where ( sin(theta) = 0.877 ). So, actually, ( sin(theta) leq 0.877 ) is true for ( theta in [0, 1.065] cup [2.076, 2pi] ).Wait, no, actually, when ( sin(theta) ) is above 0.877, it's between 1.065 and 2.076. So, the regions where ( sin(theta) leq 0.877 ) are ( [0, 1.065] ) and ( [2.076, 2pi] ).Therefore, ( theta in [0, 1.065] cup [2.076, 2pi] ).But we need to translate this back to ( t ). Remember ( theta = frac{pi t}{12} ), so:For ( theta in [0, 1.065] ):( t in [0, frac{12}{pi} times 1.065] approx [0, 4.07] ) hours.For ( theta in [2.076, 2pi] ):( t in [frac{12}{pi} times 2.076, frac{12}{pi} times 2pi] approx [8.06, 24] ) hours.So, putting it all together, the net caloric balance is within the desired range when ( t in [0, 4.07] cup [8.06, 24] ).But wait, let's double-check this. Because the quadratic inequality gave us ( x leq 0.877 ), which translates to ( sin(theta) leq 0.877 ). So, the regions where ( sin(theta) ) is less than or equal to 0.877 are indeed the intervals we found.However, I should also consider that ( sin(theta) ) is negative in certain intervals, but since ( x ) can be as low as -1, but our quadratic inequality only restricts ( x leq 0.877 ). So, the regions where ( sin(theta) ) is negative are automatically included because ( x ) is less than 0.877.Wait, but in our earlier step, we had ( x in [-1, 0.877] ), but since ( x = sin(theta) ), which naturally is between -1 and 1, so the only restriction is ( sin(theta) leq 0.877 ). So, the times when ( sin(theta) leq 0.877 ) are indeed the intervals we found.But let me think again. The net balance ( N(t) ) is within -500 to 500 when ( N(t) leq 500 ) and ( N(t) geq -500 ). We found that the second inequality is always true, so the only restriction is ( N(t) leq 500 ), which happens when ( sin(theta) leq 0.877 ).Therefore, the net balance is within the desired range during the intervals when ( t in [0, 4.07] ) and ( t in [8.06, 24] ).But wait, let me verify this with specific times. For example, at t=0:( I(0) = 250 sin(0) + 2000 = 2000 )( E(0) = 150 cos(0) + 1800 = 150*1 + 1800 = 1950 )( N(0) = 2000 - 1950 = 50 ) which is within the range.At t=4.07:Calculate ( theta = pi*4.07/12 ≈ 1.065 ) radians, so ( sin(theta) ≈ 0.877 )So, ( N(t) = 300*(0.877)^2 + 250*(0.877) + 50 )Calculate:( 300*(0.769) ≈ 230.7 )( 250*(0.877) ≈ 219.25 )Sum: 230.7 + 219.25 + 50 ≈ 500So, at t=4.07, N(t)=500, which is the upper limit.Similarly, at t=8.06:( theta = pi*8.06/12 ≈ 2.076 ) radians, which is in the second quadrant where sine is still positive but decreasing.Wait, but ( sin(2.076) ≈ sin(pi - 1.065) ≈ 0.877 ). So, same value.So, N(t) at t=8.06 is also 500.But wait, what about t=12?At t=12:( I(12) = 250 sin(π*12/12) + 2000 = 250 sin(π) + 2000 = 0 + 2000 = 2000 )( E(12) = 150 cos(π*12/6) + 1800 = 150 cos(2π) + 1800 = 150*1 + 1800 = 1950 )( N(12) = 2000 - 1950 = 50 ) which is within the range.Wait, but according to our earlier intervals, t=12 is within [8.06,24], so it should be within the range, which it is.What about t=6?At t=6:( I(6) = 250 sin(π*6/12) + 2000 = 250 sin(π/2) + 2000 = 250*1 + 2000 = 2250 )( E(6) = 150 cos(π*6/6) + 1800 = 150 cos(π) + 1800 = 150*(-1) + 1800 = 1650 )( N(6) = 2250 - 1650 = 600 ) which is above 500. So, it's outside the desired range.But according to our intervals, t=6 is between 4.07 and 8.06, which is excluded. So, that makes sense.Similarly, at t=10:( I(10) = 250 sin(π*10/12) + 2000 = 250 sin(5π/6) + 2000 = 250*(0.5) + 2000 = 125 + 2000 = 2125 )( E(10) = 150 cos(π*10/6) + 1800 = 150 cos(5π/3) + 1800 = 150*(0.5) + 1800 = 75 + 1800 = 1875 )( N(10) = 2125 - 1875 = 250 ) which is within the range. Wait, but t=10 is within [8.06,24], so it should be included, which it is.Wait, but earlier I thought t=6 is excluded, which it is, and t=10 is included, which it is. So, seems consistent.But let me check another point, say t=16:( I(16) = 250 sin(π*16/12) + 2000 = 250 sin(4π/3) + 2000 = 250*(-√3/2) + 2000 ≈ 250*(-0.866) + 2000 ≈ -216.5 + 2000 ≈ 1783.5 )( E(16) = 150 cos(π*16/6) + 1800 = 150 cos(8π/3) + 1800 = 150 cos(2π/3) + 1800 = 150*(-0.5) + 1800 = -75 + 1800 = 1725 )( N(16) = 1783.5 - 1725 ≈ 58.5 ) which is within the range.So, t=16 is within [8.06,24], and N(t)=58.5, which is fine.Wait, but what about t=18:( I(18) = 250 sin(π*18/12) + 2000 = 250 sin(3π/2) + 2000 = 250*(-1) + 2000 = 1750 )( E(18) = 150 cos(π*18/6) + 1800 = 150 cos(3π) + 1800 = 150*(-1) + 1800 = 1650 )( N(18) = 1750 - 1650 = 100 ) which is within the range.So, seems consistent.But wait, let's check t=20:( I(20) = 250 sin(π*20/12) + 2000 = 250 sin(5π/3) + 2000 = 250*(-√3/2) + 2000 ≈ -216.5 + 2000 ≈ 1783.5 )( E(20) = 150 cos(π*20/6) + 1800 = 150 cos(10π/3) + 1800 = 150 cos(4π/3) + 1800 = 150*(-0.5) + 1800 = -75 + 1800 = 1725 )( N(20) = 1783.5 - 1725 ≈ 58.5 ) which is within the range.Okay, seems consistent.But wait, what about t=24:( I(24) = 250 sin(π*24/12) + 2000 = 250 sin(2π) + 2000 = 0 + 2000 = 2000 )( E(24) = 150 cos(π*24/6) + 1800 = 150 cos(4π) + 1800 = 150*1 + 1800 = 1950 )( N(24) = 2000 - 1950 = 50 ) which is within the range.So, all these test points seem to confirm that the intervals [0,4.07] and [8.06,24] are correct.But let me think about the quadratic approach again. We transformed the problem into a quadratic in ( x = sin(theta) ), found that ( x leq 0.877 ), which translates to ( sin(theta) leq 0.877 ). Then, we found the corresponding ( theta ) intervals, which gave us the t intervals.However, I should also consider that ( N(t) ) could potentially dip below -500, but our second inequality showed that it's always above -500 because the quadratic is always positive. So, the only constraint is the upper bound of 500.But wait, let me check if that's the case. Suppose ( N(t) ) could go below -500. Let's see.Wait, in our quadratic, we had ( N(x) = 300x^2 + 250x + 50 ). The minimum value of this quadratic occurs at ( x = -b/(2a) = -250/(2*300) = -250/600 ≈ -0.4167 ). Plugging this back into N(x):( N(-0.4167) = 300*(0.1736) + 250*(-0.4167) + 50 ≈ 52.08 - 104.17 + 50 ≈ -2.09 )So, the minimum value of N(t) is approximately -2.09, which is well above -500. Therefore, the net balance never goes below -500, so the second inequality is always satisfied. Therefore, the only constraint is ( N(t) leq 500 ), which happens when ( sin(theta) leq 0.877 ), leading to the intervals we found.Therefore, the time intervals during a 24-hour period when Mr. Thompson's net caloric balance is within -500 to 500 calories are approximately from 0 to 4.07 hours and from 8.06 hours to 24 hours.But let me express these times more precisely. Since 4.07 hours is approximately 4 hours and 4 minutes (0.07*60 ≈ 4.2 minutes), and 8.06 hours is approximately 8 hours and 4 minutes.So, the intervals are approximately:- From midnight (0 hours) to 4:04 AM- From 8:04 AM to midnight (24 hours)But since the problem is about a 24-hour period, we can express the intervals as [0, 4.07] and [8.06, 24].However, to be more precise, let's calculate the exact times when ( sin(theta) = 0.877 ).We had ( theta = arcsin(0.877) ≈ 1.065 ) radians and ( theta = π - 1.065 ≈ 2.076 ) radians.So, converting back to t:For ( theta = 1.065 ):( t = frac{12}{pi} * 1.065 ≈ frac{12 * 1.065}{3.1416} ≈ frac{12.78}{3.1416} ≈ 4.07 ) hours.For ( theta = 2.076 ):( t = frac{12}{pi} * 2.076 ≈ frac{24.912}{3.1416} ≈ 8.06 ) hours.So, the exact times are approximately 4.07 hours and 8.06 hours.Therefore, the net caloric balance is within the desired range from 0 to approximately 4.07 hours and from approximately 8.06 hours to 24 hours.But to express this in a more standard time format, let's convert 4.07 hours to hours and minutes:0.07 hours * 60 ≈ 4.2 minutes, so 4 hours and 4 minutes.Similarly, 8.06 hours is 8 hours and 3.6 minutes, approximately 8 hours and 4 minutes.So, the intervals are:- From 12:00 AM to 4:04 AM- From 8:04 AM to 12:00 AM (next day)But since we're considering a 24-hour period, it's from 0 to 4.07 hours and from 8.06 hours to 24 hours.Now, moving on to part b: Calculate the total net caloric balance over a 24-hour period and verify if the average net caloric balance per hour meets Mr. Thompson's goal.First, the total net caloric balance is the integral of N(t) from t=0 to t=24.So,( text{Total} = int_{0}^{24} N(t) dt = int_{0}^{24} [250 sin(frac{pi t}{12}) - 150 cos(frac{pi t}{6}) + 200] dt )We can split this integral into three parts:1. ( 250 int_{0}^{24} sin(frac{pi t}{12}) dt )2. ( -150 int_{0}^{24} cos(frac{pi t}{6}) dt )3. ( 200 int_{0}^{24} dt )Let's compute each integral separately.First integral:( I_1 = 250 int_{0}^{24} sinleft(frac{pi t}{12}right) dt )Let ( u = frac{pi t}{12} ), so ( du = frac{pi}{12} dt ), which means ( dt = frac{12}{pi} du ). When t=0, u=0; t=24, u=2π.So,( I_1 = 250 * frac{12}{pi} int_{0}^{2pi} sin(u) du )The integral of sin(u) from 0 to 2π is:( int_{0}^{2pi} sin(u) du = [-cos(u)]_{0}^{2pi} = (-cos(2π) + cos(0)) = (-1 + 1) = 0 )So, ( I_1 = 250 * frac{12}{pi} * 0 = 0 )Second integral:( I_2 = -150 int_{0}^{24} cosleft(frac{pi t}{6}right) dt )Let ( v = frac{pi t}{6} ), so ( dv = frac{pi}{6} dt ), which means ( dt = frac{6}{pi} dv ). When t=0, v=0; t=24, v=4π.So,( I_2 = -150 * frac{6}{pi} int_{0}^{4pi} cos(v) dv )The integral of cos(v) from 0 to 4π is:( int_{0}^{4pi} cos(v) dv = sin(v) bigg|_{0}^{4pi} = sin(4π) - sin(0) = 0 - 0 = 0 )So, ( I_2 = -150 * frac{6}{pi} * 0 = 0 )Third integral:( I_3 = 200 int_{0}^{24} dt = 200 * [t]_{0}^{24} = 200 * (24 - 0) = 4800 )Therefore, the total net caloric balance is:( text{Total} = I_1 + I_2 + I_3 = 0 + 0 + 4800 = 4800 ) calories.Wait, that can't be right. Because if the total is 4800 calories over 24 hours, the average per hour is 200 calories. But earlier, we saw that at t=0, N(t)=50, and at t=12, N(t)=50, so the average should be around 50 calories per hour, not 200.Wait, I must have made a mistake in the integral calculation.Wait, let's re-examine the integrals.First, N(t) = 250 sin(πt/12) - 150 cos(πt/6) + 200So, integrating term by term:1. Integral of 250 sin(πt/12) dt from 0 to 24:As before, substitution u = πt/12, du = π/12 dt, dt = 12/π du. Limits from 0 to 2π.Integral becomes 250 * (12/π) ∫ sin(u) du from 0 to 2π = 250*(12/π)*(-cos(u)) from 0 to 2π = 250*(12/π)*(-cos(2π) + cos(0)) = 250*(12/π)*(-1 + 1) = 0.Correct.2. Integral of -150 cos(πt/6) dt from 0 to24:Substitution v = πt/6, dv = π/6 dt, dt = 6/π dv. Limits from 0 to 4π.Integral becomes -150*(6/π) ∫ cos(v) dv from 0 to4π = -150*(6/π)*(sin(v)) from 0 to4π = -150*(6/π)*(sin(4π) - sin(0)) = -150*(6/π)*(0 - 0) = 0.Correct.3. Integral of 200 dt from 0 to24:200*(24 -0) = 4800.So, total is indeed 4800 calories.Wait, but that seems high because the average is 200 calories per hour, but earlier, at t=0 and t=12, N(t)=50, which is much lower.Wait, perhaps I made a mistake in the expression for N(t). Let me double-check.Original N(t) = I(t) - E(t) = [250 sin(πt/12) + 2000] - [150 cos(πt/6) + 1800] = 250 sin(πt/12) - 150 cos(πt/6) + 200.Yes, that's correct.So, integrating N(t) over 24 hours gives 4800 calories. Therefore, the average net caloric balance per hour is 4800 /24 = 200 calories per hour.But wait, that contradicts our earlier observation that at t=0 and t=12, N(t)=50, which is much lower than 200.Wait, perhaps the average is 200, but the function fluctuates around that average. Let me think.Wait, the average value of a function over an interval is the integral divided by the interval length. So, if the integral is 4800 over 24 hours, the average is indeed 200 calories per hour.But looking back at N(t), the constant term is 200, and the other terms are oscillating around zero. So, the average of the oscillating terms is zero, leaving the average net balance as 200 calories per hour.But wait, that seems contradictory because when I calculated N(t) at specific times, it was 50, 600, etc. But the average is 200.Wait, but 200 is the average, so some times it's above, some below. For example, at t=6, N(t)=600, which is above average, and at t=12, N(t)=50, which is below average.But the average is 200, which is within the desired range of -500 to 500. However, the problem states that Mr. Thompson wants his net balance to never exceed 500 and never go below -500. The average is 200, which is within the range, but the question is whether the average meets his goal.Wait, the average is 200, which is within the range, but the problem is about the instantaneous net balance, not the average. The average being 200 doesn't necessarily mean that the instantaneous values are always within the range. However, in this case, we've already determined the intervals when N(t) is within the range, and the average is 200, which is within the range.But the question is: "verify if the average net caloric balance per hour meets Mr. Thompson's goal of maintaining his net caloric balance within the specified range."Wait, the average is 200, which is within -500 to 500, so it meets the goal. However, the instantaneous values sometimes exceed 500, but the average is within the range.But wait, the problem says "maintaining his net caloric balance within the specified range." I think it refers to the instantaneous balance, not the average. So, even though the average is within the range, the instantaneous values sometimes go above 500, which violates his goal.But in part a, we found that the net balance is within the range except between 4.07 and 8.06 hours. So, the average is 200, which is within the range, but the instantaneous values sometimes exceed 500.Therefore, the average meets the goal, but the instantaneous values do not always meet the goal.But the question is: "verify if the average net caloric balance per hour meets Mr. Thompson's goal of maintaining his net caloric balance within the specified range."So, the average is 200, which is within -500 to 500, so it meets the goal. However, the instantaneous values sometimes exceed 500, which means his net balance is not always within the range, but the average is.But the question is specifically about the average, so the answer is yes, the average meets the goal.But let me make sure. The problem says: "verify if the average net caloric balance per hour meets Mr. Thompson's goal of maintaining his net caloric balance within the specified range."So, the average is 200, which is within -500 to 500, so it meets the goal. Therefore, the answer is yes.But wait, the total net caloric balance is 4800, so the average is 200 per hour, which is within the desired range. So, the average meets the goal.However, the instantaneous values sometimes exceed 500, but the question is about the average, so the answer is yes.So, summarizing:a. The net caloric balance is within -500 to 500 during [0, 4.07] and [8.06, 24] hours.b. The total net caloric balance is 4800 calories over 24 hours, so the average is 200 calories per hour, which is within the desired range.But wait, the problem says "verify if the average net caloric balance per hour meets Mr. Thompson's goal of maintaining his net caloric balance within the specified range."So, the average is 200, which is within -500 to 500, so it meets the goal. Therefore, the answer is yes.But I should also note that while the average is within the range, the instantaneous values sometimes exceed 500, which means his net balance is not always within the range, but the average is.But the question is specifically about the average, so the answer is yes.So, to wrap up:a. The time intervals are approximately from 0 to 4.07 hours and from 8.06 hours to 24 hours.b. The total net caloric balance is 4800 calories, with an average of 200 calories per hour, which is within the desired range.But wait, let me double-check the integral calculation because earlier I thought the average should be around 50, but it's actually 200.Wait, N(t) = 250 sin(πt/12) - 150 cos(πt/6) + 200The average value of sin and cos over their periods is zero, so the average of N(t) is just the constant term, which is 200. So, that's correct.Therefore, the average is 200, which is within the range.So, the final answers are:a. The net caloric balance is within the desired range during the intervals [0, 4.07] and [8.06, 24] hours.b. The total net caloric balance is 4800 calories, with an average of 200 calories per hour, which is within the desired range.</think>

🧠:<think>Alright, so I have this problem about a clumsy person navigating a 10x10 grid with obstacles. There are two parts: first, calculating the collision probability at a specific point, and second, finding the optimal path from (1,1) to (10,10) that minimizes the total collision probability. Let me try to break this down step by step.Starting with part 1: I need to find the probability P(5,5). The formula given is P(x,y) = [d(x,y)/S] * D, where d(x,y) is the Manhattan distance from (1,1) to (x,y), S is the total number of grid cells, and D is the density of obstacles.First, let me recall what Manhattan distance is. It's the sum of the absolute differences of their coordinates. So, from (1,1) to (5,5), the Manhattan distance d(5,5) should be |5-1| + |5-1| = 4 + 4 = 8. Got that.Next, S is the total number of grid cells. Since it's a 10x10 grid, that should be 10*10 = 100 cells. So S = 100.D is the density of obstacles, which is the number of obstacles divided by S. They told us there are 15 obstacles, so D = 15/100 = 0.15.Now, plugging these into the formula: P(5,5) = (8 / 100) * 0.15. Let me compute that. 8 divided by 100 is 0.08, and 0.08 multiplied by 0.15 is... 0.012. So, P(5,5) is 0.012, or 1.2%.Wait, let me double-check. 8/100 is 0.08, and 0.08*0.15 is indeed 0.012. Yep, that seems right.Moving on to part 2: The person is moving from (1,1) to (10,10) and wants to minimize the total collision probability. I need to formulate an optimization problem for this and find the optimal path and its total probability.Hmm, so the total collision probability would be the sum of P(x,y) for each cell (x,y) along the path. Since each step from one cell to another contributes the probability of the destination cell, right? Or is it the starting cell? Wait, the problem says "at any given position (x,y)", so I think it's the cell they're currently in. So, when moving from (x,y) to (x+1,y) or (x,y+1), the collision probability at (x,y) is added to the total.But wait, actually, when moving from (1,1), the first cell is (1,1), so do we include that? The starting point. Similarly, when moving to (10,10), do we include that? I think so, because the person is present in each cell they pass through.So, the total collision probability is the sum of P(x,y) for each cell (x,y) along the path, including the starting and ending points.Given that, the optimization problem is to find a path from (1,1) to (10,10) that minimizes the sum of P(x,y) over all cells in the path.In a grid, the person can only move right or up, right? Since it's a 10x10 grid, moving from (1,1) to (10,10) requires moving 9 steps right and 9 steps up, in some order. So, the total number of steps is 18, and the number of cells visited is 19 (including both start and end).But wait, actually, in a grid, moving from (1,1) to (10,10) would involve moving right 9 times and up 9 times, so the path length is 18 moves, visiting 19 cells. So, the total collision probability is the sum of P(x,y) for each of these 19 cells.But how do we model this as an optimization problem? It seems like a shortest path problem where each edge has a weight, but in this case, each node (cell) has a weight, which is P(x,y). So, we need to find the path from (1,1) to (10,10) that minimizes the sum of the node weights along the path.This is similar to the shortest path problem in graphs where nodes have weights instead of edges. One way to handle this is to convert the node weights into edge weights. For each node (x,y), we can split it into two nodes: (x,y)_in and (x,y)_out, connected by an edge with weight P(x,y). Then, all incoming edges to (x,y) go to (x,y)_in, and all outgoing edges from (x,y) go from (x,y)_out. This way, traversing through (x,y) incurs the weight P(x,y).Alternatively, since the grid is a directed acyclic graph (DAG) where you can only move right or up, we can use dynamic programming to compute the minimum total collision probability.Let me think about the dynamic programming approach. Let’s define dp[x][y] as the minimum total collision probability to reach cell (x,y). Then, the recurrence relation would be:dp[x][y] = P(x,y) + min(dp[x-1][y], dp[x][y-1])Because to reach (x,y), you can come from either (x-1,y) or (x,y-1), whichever gives the smaller total probability.The base cases would be dp[1][1] = P(1,1), since that's the starting point. For the first row (y=1), you can only come from the left, so dp[x][1] = dp[x-1][1] + P(x,1). Similarly, for the first column (x=1), dp[1][y] = dp[1][y-1] + P(1,y).Once we fill out the dp table, the value at dp[10][10] will be the minimum total collision probability.But wait, let me make sure. Each cell's probability is added once when you enter it, right? So, the starting cell (1,1) is added once, and each subsequent cell is added once as you move into it. So, the DP approach correctly accumulates the sum.Now, to compute this, I need to know P(x,y) for all cells. But in the problem statement, it's given that P(x,y) = [d(x,y)/S] * D. We already know S=100, D=0.15. So, for each cell (x,y), d(x,y) is |x-1| + |y-1|, since it's the Manhattan distance from (1,1).So, P(x,y) = (|x-1| + |y-1|)/100 * 0.15.Simplifying, P(x,y) = 0.15*(x + y - 2)/100, since |x-1| + |y-1| = (x-1) + (y-1) = x + y - 2.Wait, actually, no. Because |x-1| is x-1 for x >=1, which it always is since x starts at 1. Similarly for y. So, yes, d(x,y) = x + y - 2.Therefore, P(x,y) = 0.15*(x + y - 2)/100.So, for each cell (x,y), P(x,y) = 0.0015*(x + y - 2).So, for example, P(1,1) = 0.0015*(1 + 1 - 2) = 0.0015*0 = 0. That makes sense because the starting point has distance 0, so no collision probability.Similarly, P(2,1) = 0.0015*(2 + 1 - 2) = 0.0015*1 = 0.0015.P(1,2) is the same.P(2,2) = 0.0015*(2 + 2 - 2) = 0.0015*2 = 0.003.And so on.So, the collision probability increases as you move away from (1,1). So, the farther you are from (1,1), the higher the collision probability.Therefore, to minimize the total collision probability, the person should take the path that stays as close as possible to (1,1) for as long as possible. But since they have to reach (10,10), which is far away, the optimal path would be the one that increases x and y as slowly as possible, but since they have to reach (10,10), it's a trade-off.Wait, actually, since the collision probability increases with both x and y, the optimal path would be the one that minimizes the sum of (x + y) along the path. Because P(x,y) is proportional to (x + y - 2), so minimizing the sum of (x + y) would minimize the total collision probability.In a grid where movement is only right or up, the path that minimizes the sum of (x + y) is the one that moves diagonally as much as possible, but since we can't move diagonally, we have to alternate between right and up moves in a way that keeps x and y as balanced as possible.Wait, actually, in such cases, the minimal sum path is the one that moves right and up in a balanced way, i.e., moving right and up alternately to keep x and y as close as possible. This is similar to the shortest path in terms of Manhattan distance, but here we are minimizing the sum of (x + y) along the path.Alternatively, since each step right increases x by 1, and each step up increases y by 1, the sum of (x + y) over the path is equivalent to the sum of the positions visited. So, to minimize this sum, we need to keep x and y as small as possible for as long as possible.But since we have to reach (10,10), which is the maximum x and y, we have to increase both x and y. However, the path that increases x and y as slowly as possible would be the one that moves right and up in a way that x and y are as balanced as possible.Wait, actually, in terms of sum of (x + y), moving right and up in any order would result in the same total sum because each right move increases x by 1, and each up move increases y by 1. So, regardless of the order, the total sum of x's and y's would be the same.Wait, is that true? Let me think. Suppose we have two paths: one that goes all the way right first, then up, and another that goes up first, then right. The sum of x's for the right-first path would be 1+2+...+10 for x, and 1 for y until x=10, then y increases. Similarly, the up-first path would have y increasing first.But actually, the sum of x's and y's over the entire path would be the same because you have to cover all x from 1 to 10 and all y from 1 to 10, but in different orders. Wait, no, that's not correct because in the path, you don't cover all x and y, just the ones along the path.Wait, no, actually, each path from (1,1) to (10,10) with only right and up moves will pass through 19 cells, each with x from 1 to 10 and y from 1 to 10, but not all combinations. So, the sum of (x + y) over the path depends on the specific cells visited.But is there a way to make the sum as small as possible? Intuitively, moving in a way that x and y increase together would keep (x + y) smaller for more steps, whereas moving all the way in one direction first would have some steps with higher (x + y).Wait, let's consider two extreme paths:1. Path A: Move all the way right to (10,1), then all the way up to (10,10). The cells visited would be (1,1), (2,1), ..., (10,1), (10,2), ..., (10,10). The sum of (x + y) for this path would be:For x from 1 to 10, y=1: sum = (1+1) + (2+1) + ... + (10+1) = sum_{x=1 to 10} (x + 1) = sum_{x=1 to 10} x + sum_{x=1 to 10} 1 = 55 + 10 = 65.Then, for y from 2 to 10, x=10: sum = (10+2) + (10+3) + ... + (10+10) = sum_{y=2 to 10} (10 + y) = 9*10 + sum_{y=2 to 10} y = 90 + (54 - 1) = 90 + 53 = 143. Wait, sum from 2 to 10 is (10*11)/2 - 1 = 55 -1 = 54. So, 9*10 + 54 = 90 + 54 = 144.Wait, no, actually, for each y from 2 to 10, it's 10 + y, so sum is 9*10 + sum_{y=2 to 10} y = 90 + (54 - 1) = 90 + 53 = 143? Wait, no, sum from y=2 to 10 is 2+3+...+10 = (10*11)/2 - 1 = 55 -1 = 54. So, 9*10 + 54 = 144.So total sum for Path A: 65 + 144 = 209.2. Path B: Move diagonally as much as possible, i.e., alternate right and up moves. For example, (1,1) -> (2,1) -> (2,2) -> (3,2) -> (3,3) -> ... -> (10,10). Let's compute the sum for this path.Each step alternates between increasing x and y. So, starting at (1,1):(1,1): 1+1=2(2,1): 2+1=3(2,2): 2+2=4(3,2): 3+2=5(3,3): 3+3=6...(10,9): 10+9=19(10,10): 10+10=20So, how many terms are there? From (1,1) to (10,10), it's 19 cells. Let's list the (x+y) values:2, 3, 4, 5, 6, ..., 19, 20.Wait, from 2 to 20, that's 19 terms. The sum is the sum from 2 to 20 inclusive.Sum = (20*21)/2 - 1 = 210 -1 = 209.Wait, that's the same as Path A? That can't be right.Wait, no, actually, in Path A, the sum was 209, and in Path B, the sum is also 209. So, does that mean all paths have the same sum of (x + y)? That seems counterintuitive.Wait, let me check again. For Path A, moving all right then up, the sum was 65 + 144 = 209.For Path B, moving diagonally, the sum is 2 + 3 + 4 + ... + 20. The sum from 2 to 20 is (20*21)/2 - 1 = 210 -1 = 209.So, both paths have the same total sum of (x + y). Hmm, that suggests that all paths from (1,1) to (10,10) with only right and up moves have the same total sum of (x + y). Is that true?Wait, let's think about it. Each path must consist of 9 right moves and 9 up moves, in some order. Each right move increases x by 1, each up move increases y by 1. The sum of x's and y's over the path is the sum of all x's plus the sum of all y's.But for each path, the sum of x's is the same as the sum of y's? No, wait, no. Wait, each path has the same number of right and up moves, but the order affects the individual sums.Wait, no, actually, the total sum of x's over the path is the sum of x-coordinates of all cells visited. Similarly for y's.But regardless of the path, the sum of x's is the same because each path must pass through each x from 1 to 10 exactly once in the x direction, but no, that's not true. Each path passes through each x from 1 to 10, but the number of times each x is visited depends on the path.Wait, actually, no. Each path must move from x=1 to x=10, so each x from 1 to 10 is visited exactly once in the x direction, but in the y direction, each y from 1 to 10 is visited exactly once as well. Wait, no, that's not correct.Wait, in a grid path from (1,1) to (10,10) moving only right and up, each x from 1 to 10 is visited exactly once in the x direction, but each y from 1 to 10 is visited exactly once in the y direction. But actually, no, because when you move right, you stay in the same y, and when you move up, you stay in the same x.Wait, let me think differently. The number of times each x is visited is equal to the number of steps in the y direction at that x. Similarly, the number of times each y is visited is equal to the number of steps in the x direction at that y.But regardless, the sum of x's over the entire path is the sum of x coordinates for each cell visited. Similarly for y's.But regardless of the path, the sum of x's plus the sum of y's over the path is equal to the sum of (x + y) for each cell, which we saw is 209 for both paths. So, does that mean that all paths have the same total sum of (x + y)?Wait, that seems to be the case because the sum of (x + y) over the path is equal to the sum of x's plus the sum of y's, and regardless of the path, the total sum is fixed.Wait, let me test with a smaller grid. Let's say 2x2 grid, moving from (1,1) to (2,2).Path 1: Right then Up: (1,1), (2,1), (2,2). Sum of (x + y): (2) + (3) + (4) = 9.Path 2: Up then Right: (1,1), (1,2), (2,2). Sum of (x + y): (2) + (3) + (4) = 9.Same sum.Another example: 3x3 grid, from (1,1) to (3,3).Path 1: R, R, U, U: (1,1), (2,1), (3,1), (3,2), (3,3). Sum: 2 + 3 + 4 + 5 + 6 = 20.Path 2: U, U, R, R: (1,1), (1,2), (1,3), (2,3), (3,3). Sum: 2 + 3 + 4 + 5 + 6 = 20.Path 3: R, U, R, U: (1,1), (2,1), (2,2), (3,2), (3,3). Sum: 2 + 3 + 4 + 5 + 6 = 20.So, indeed, all paths have the same total sum of (x + y). Therefore, in our original problem, regardless of the path taken, the total sum of (x + y) is fixed at 209. Therefore, the total collision probability, which is proportional to the sum of (x + y - 2), would also be fixed.Wait, but that can't be right because the problem says to find the path that minimizes the total collision probability. If all paths have the same total, then any path is equally good. But that contradicts the problem statement.Wait, maybe I made a mistake in interpreting the collision probability. Let me go back.The collision probability at position (x,y) is P(x,y) = [d(x,y)/S] * D.d(x,y) is the Manhattan distance from (1,1), which is (x + y - 2). So, P(x,y) = (x + y - 2)/100 * 0.15.So, the total collision probability is the sum over the path of P(x,y) = 0.15/100 * sum over the path of (x + y - 2).Which is 0.0015 * sum over the path of (x + y - 2).But sum over the path of (x + y - 2) = sum over the path of (x + y) - 2*number_of_cells.Number of cells in the path is 19, so sum = (sum of x + y) - 38.But we saw that sum of x + y is 209, so total collision probability is 0.0015*(209 - 38) = 0.0015*171 = 0.2565.Wait, but that would mean that regardless of the path, the total collision probability is the same, which is 0.2565.But the problem says to find the path that minimizes the total collision probability, implying that it's not the same for all paths. So, where is the mistake?Wait, perhaps I misapplied the formula. Let me check again.P(x,y) = [d(x,y)/S] * D.d(x,y) is Manhattan distance from (1,1), which is (x-1) + (y-1) = x + y - 2.So, P(x,y) = (x + y - 2)/100 * D.But D is the density, which is 15/100 = 0.15.So, P(x,y) = (x + y - 2)/100 * 0.15 = 0.0015*(x + y - 2).Therefore, the total collision probability is sum over the path of 0.0015*(x + y - 2) = 0.0015*(sum over the path of (x + y) - 2*19).Sum over the path of (x + y) is 209, as before.So, total collision probability = 0.0015*(209 - 38) = 0.0015*171 = 0.2565.So, regardless of the path, the total collision probability is 0.2565.But the problem says to find the path that minimizes this total. If it's the same for all paths, then any path is optimal.But that seems contradictory. Maybe I misunderstood the problem.Wait, perhaps the collision probability is per step, not per cell. Let me re-read the problem."the probability P(x,y) that the person collides with an obstacle at any given position (x,y) on the grid depends on both the Manhattan distance from the starting point (1,1) and the density of obstacles in the grid.""Calculate the probability of collision at position (5,5)...""the clumsy person moves from (1,1) to (10,10) following a path that minimizes the total collision probability."So, it's the total collision probability along the path. If each cell has a collision probability, and the person is in each cell along the path, then the total collision probability is the sum of P(x,y) for each cell in the path.But if all paths have the same sum, then the total is fixed. Therefore, any path is equally good.But the problem says to "formulate an optimization problem that represents this scenario. What is the path the person should take, and what is the total collision probability along this path?"This suggests that there is a unique optimal path, which contradicts my earlier conclusion.Wait, perhaps I made a mistake in assuming that the sum of (x + y) is the same for all paths. Let me verify with a smaller grid.Take a 2x2 grid, moving from (1,1) to (2,2).Path 1: Right then Up: (1,1), (2,1), (2,2). Sum of (x + y): 2 + 3 + 4 = 9.Path 2: Up then Right: (1,1), (1,2), (2,2). Sum of (x + y): 2 + 3 + 4 = 9.Same sum.Another example: 3x3 grid, from (1,1) to (3,3).Path 1: R, R, U, U: (1,1), (2,1), (3,1), (3,2), (3,3). Sum: 2 + 3 + 4 + 5 + 6 = 20.Path 2: U, U, R, R: (1,1), (1,2), (1,3), (2,3), (3,3). Sum: 2 + 3 + 4 + 5 + 6 = 20.Path 3: R, U, R, U: (1,1), (2,1), (2,2), (3,2), (3,3). Sum: 2 + 3 + 4 + 5 + 6 = 20.So, indeed, in smaller grids, the sum is the same. Therefore, in the 10x10 grid, the sum of (x + y) over any path from (1,1) to (10,10) is fixed.Therefore, the total collision probability is fixed, regardless of the path. So, any path is equally good in terms of total collision probability.But the problem says to find the path that minimizes the total collision probability, which suggests that my initial assumption is wrong.Wait, perhaps the collision probability is not additive? Or perhaps it's multiplicative? Or maybe it's the probability of at least one collision, which would be 1 - product of (1 - P(x,y)).But the problem says "total collision probability", which is ambiguous. It could mean the sum of probabilities, treating each collision as independent, which would be the expected number of collisions. Alternatively, it could mean the probability of at least one collision, which is 1 - product of (1 - P(x,y)).But in the first part, it's asking for the probability at (5,5), which is a single value, so it's likely that the total is the sum.But if the sum is fixed, then the problem is ill-posed. Therefore, perhaps I misinterpreted the formula.Wait, let me re-examine the formula: P(x,y) = [d(x,y)/S] * D.d(x,y) is the Manhattan distance from (1,1), which is x + y - 2.S is the total number of grid cells, which is 100.D is the density, which is 15/100 = 0.15.So, P(x,y) = (x + y - 2)/100 * 0.15.Therefore, P(x,y) = 0.0015*(x + y - 2).So, for each cell, the collision probability is proportional to its distance from (1,1). Therefore, cells farther away have higher collision probabilities.But if the total sum over the path is fixed, then all paths have the same total collision probability. Therefore, the problem might have a different interpretation.Wait, perhaps the collision probability is the probability of colliding at each step, not at each cell. So, when moving from (x,y) to (x+1,y) or (x,y+1), the probability of collision is based on the next cell.In that case, the total collision probability would be the sum of P(next cell) for each move.But in that case, the starting cell (1,1) wouldn't contribute, and the ending cell (10,10) would contribute only if you stop there.Wait, but the problem says "at any given position (x,y)", so it's the probability of colliding while being at (x,y). So, it's per cell, not per move.Therefore, the total collision probability is the sum over all cells visited of P(x,y).But as we saw, this sum is fixed for all paths.Therefore, the problem might have a different formula.Wait, perhaps the formula is P(x,y) = d(x,y) * D / S, which is what it says: P(x,y) = [d(x,y)/S] * D.But if D is the density, which is 15/100, then P(x,y) = (d(x,y) * 15)/100^2.Wait, no, the formula is [d(x,y)/S] * D, which is (d(x,y)/100) * (15/100) = d(x,y)*15 / 10000.Wait, that would make P(x,y) = (x + y - 2)*15 / 10000.But in that case, the total collision probability would be sum over the path of (x + y - 2)*15 / 10000 = 15/10000 * sum over the path of (x + y - 2).Which is 0.0015 * sum over the path of (x + y - 2).But as before, sum over the path of (x + y - 2) = sum over the path of (x + y) - 2*19 = 209 - 38 = 171.So, total collision probability = 0.0015 * 171 = 0.2565.Therefore, regardless of the path, the total collision probability is 0.2565.Therefore, the optimization problem is trivial because all paths have the same total collision probability.But the problem says to "formulate an optimization problem that represents this scenario. What is the path the person should take, and what is the total collision probability along this path?"This suggests that the total collision probability is not fixed, which contradicts my calculations.Wait, perhaps the collision probability is not additive. Maybe it's the probability of colliding at least once along the path, which would be 1 - product of (1 - P(x,y)) for each cell in the path.In that case, the total collision probability would be 1 - product of (1 - P(x,y)).Since P(x,y) is small (each is at most 0.0015*(10 + 10 - 2) = 0.0015*18 = 0.027), the product would be approximately 1 - sum of P(x,y), but slightly less.But in that case, the total collision probability would still be approximately the same for all paths, since the sum is fixed.Wait, but if we consider the exact value, 1 - product of (1 - P(x,y)), then the path with cells that have lower P(x,y) would have a higher product, thus lower total collision probability.Wait, no, because 1 - product would be higher if the product is lower. So, if the product is lower, 1 - product is higher. Therefore, to minimize the total collision probability, we need to maximize the product of (1 - P(x,y)), which would be achieved by minimizing the sum of P(x,y), but since the sum is fixed, it's the same.Wait, no, actually, the product depends on the individual terms. For example, if you have two cells with P1 and P2, the product is (1 - P1)(1 - P2) = 1 - P1 - P2 + P1P2. So, the product is less than 1 - P1 - P2, but the difference depends on P1P2.Therefore, if you have cells with higher P(x,y), the product decreases more, leading to a higher total collision probability (1 - product). Therefore, to minimize the total collision probability, you want to minimize the sum of P(x,y) and also minimize the product of (1 - P(x,y)).But since the sum is fixed, the product is also fixed? No, because the product depends on the distribution of P(x,y). For example, if you have two cells with P=0.01 each, the product is (0.99)^2 = 0.9801, whereas if you have one cell with P=0.02, the product is 0.98, which is the same. Wait, no, 0.99*0.99 = 0.9801, which is less than 0.98. So, actually, the product is lower when you have more small probabilities.Wait, let me compute:Case 1: Two cells with P=0.01 each: product = 0.99 * 0.99 = 0.9801.Case 2: One cell with P=0.02: product = 0.98.So, 0.9801 > 0.98, meaning that the product is higher in Case 1, leading to a lower total collision probability (1 - 0.9801 = 0.0199 vs 1 - 0.98 = 0.02).Therefore, distributing the probabilities into more smaller terms leads to a lower total collision probability.Therefore, to minimize the total collision probability (1 - product), the person should take a path that has as many small P(x,y) as possible, i.e., stay as close to (1,1) as possible, thus minimizing the individual P(x,y) terms.Therefore, the optimal path would be the one that minimizes the maximum P(x,y) along the path, or spreads out the probabilities into more smaller terms.But in our case, since all paths have the same sum of P(x,y), but different distributions, the path that stays closer to (1,1) for as long as possible would have more cells with smaller P(x,y), leading to a higher product of (1 - P(x,y)), thus lower total collision probability.Therefore, the optimal path is the one that moves diagonally, keeping x and y as small as possible for as long as possible.In other words, the optimal path is the one that alternates between moving right and up, keeping x and y as balanced as possible.Therefore, the path would be moving right and up alternately, resulting in the minimal possible maximum P(x,y) along the path.But to confirm, let's consider two paths:Path A: All right first, then up.Path B: Alternating right and up.Compute the product of (1 - P(x,y)) for both paths.But since it's a 10x10 grid, computing this manually would be tedious. However, the principle holds that spreading out the probabilities into more smaller terms results in a higher product, thus lower total collision probability.Therefore, the optimal path is the one that keeps x and y as balanced as possible, i.e., moves right and up alternately.So, the path would be: (1,1) -> (2,1) -> (2,2) -> (3,2) -> (3,3) -> ... -> (10,9) -> (10,10).This path ensures that x and y increase together, keeping the individual P(x,y) as small as possible for as long as possible.Therefore, the total collision probability is 1 - product of (1 - P(x,y)) along this path.But since the problem asks for the total collision probability, and it's likely referring to the sum, which is fixed, but given the earlier contradiction, perhaps the intended answer is that all paths have the same total collision probability, so any path is optimal.But given the problem's wording, it's more likely that the total collision probability is the sum, which is fixed, so any path is equally good. However, if we consider the probability of at least one collision, then the optimal path is the one that stays closer to (1,1), i.e., the diagonal path.But since the problem didn't specify, and given that in part 1 it's asking for P(5,5), which is a single cell, I think the intended interpretation is that the total collision probability is the sum of P(x,y) along the path, which is fixed. Therefore, any path is equally good, and the total collision probability is 0.2565.But the problem says to "formulate an optimization problem that represents this scenario. What is the path the person should take, and what is the total collision probability along this path?"If the total is fixed, then any path is optimal, but perhaps the problem expects us to recognize that the total is fixed and thus any path is optimal.Alternatively, perhaps I misapplied the formula. Let me check again.Wait, the formula is P(x,y) = [d(x,y)/S] * D.But d(x,y) is the Manhattan distance from (1,1), which is x + y - 2.S is 100.D is 15/100 = 0.15.So, P(x,y) = (x + y - 2)/100 * 0.15 = 0.0015*(x + y - 2).Therefore, the total collision probability is sum over the path of 0.0015*(x + y - 2).Which is 0.0015*(sum of (x + y) - 2*19).Sum of (x + y) is 209, so 209 - 38 = 171.Total collision probability = 0.0015*171 = 0.2565.Therefore, regardless of the path, the total collision probability is 0.2565.Therefore, the optimization problem is trivial, and any path is optimal.But the problem says to "formulate an optimization problem that represents this scenario. What is the path the person should take, and what is the total collision probability along this path?"So, perhaps the answer is that any path is optimal, and the total collision probability is 0.2565.But the problem might expect us to consider that the total is fixed, so the answer is that any path is optimal, with total collision probability 0.2565.Alternatively, perhaps the problem expects us to consider that the total collision probability is the sum of P(x,y), which is fixed, so any path is optimal.But to be thorough, let me consider that perhaps the collision probability is not additive, but rather the probability of colliding at least once, which would be 1 - product of (1 - P(x,y)).In that case, the total collision probability would be 1 - product over the path of (1 - P(x,y)).Since P(x,y) is small, this would be approximately equal to the sum of P(x,y), but slightly less.But in that case, the path that minimizes the total collision probability would be the one that minimizes the product, which would be the path with the smallest possible individual P(x,y).But since the sum is fixed, the product is also fixed? No, because the product depends on the distribution.Wait, for example, if you have two cells with P=0.01 each, the product is 0.99*0.99=0.9801, so the total collision probability is 1 - 0.9801=0.0199.If you have one cell with P=0.02, the product is 0.98, so total collision probability is 0.02.Therefore, the total collision probability is higher when you have fewer, larger P(x,y). Therefore, to minimize the total collision probability, you want to spread out the probabilities into more smaller terms.Therefore, the optimal path is the one that stays as close to (1,1) as possible, thus having more cells with smaller P(x,y), leading to a higher product and thus lower total collision probability.Therefore, the optimal path is the one that moves diagonally, keeping x and y as balanced as possible.But since the problem didn't specify whether the total collision probability is the sum or the probability of at least one collision, it's ambiguous.However, given that part 1 asks for the probability at a single cell, and part 2 asks for the total along the path, it's more likely that the total is the sum, which is fixed.Therefore, the answer is that any path is optimal, and the total collision probability is 0.2565.But the problem says to "formulate an optimization problem that represents this scenario. What is the path the person should take, and what is the total collision probability along this path?"So, perhaps the intended answer is that the path is any path, and the total collision probability is 0.2565.But to be safe, I'll proceed with that.So, summarizing:1. P(5,5) = 0.012.2. The optimization problem is to minimize the sum of P(x,y) along the path from (1,1) to (10,10), which is fixed at 0.2565, so any path is optimal.But perhaps the problem expects us to recognize that the total is fixed, so any path is optimal.Alternatively, if considering the probability of at least one collision, the optimal path is the one that stays closest to (1,1), i.e., the diagonal path, but the total collision probability would still be approximately 0.2565.But given the problem's wording, I think the intended answer is that the total collision probability is fixed, so any path is optimal, and the total is 0.2565.Therefore, the final answers are:1. P(5,5) = 0.012.2. Any path is optimal, with total collision probability 0.2565.But to express 0.2565 as a fraction, it's 171/666.666..., but perhaps better to leave it as 0.2565.Alternatively, 0.2565 is 2565/10000, which simplifies to 513/2000.But 513 divided by 2000 is 0.2565.So, perhaps expressing it as 513/2000.But the problem might expect a decimal.Alternatively, perhaps I made a mistake in the calculation.Wait, 0.0015 * 171 = 0.2565.Yes, that's correct.So, the total collision probability is 0.2565.Therefore, the answers are:1. 0.012.2. Any path, total collision probability 0.2565.But the problem says "what is the path the person should take", implying a specific path, so perhaps I was wrong earlier, and the total collision probability is not fixed.Wait, perhaps the formula is different. Maybe P(x,y) is the probability of colliding when moving into (x,y), not when being in (x,y). So, the starting cell (1,1) doesn't contribute, and the ending cell (10,10) does.In that case, the total collision probability would be the sum of P(x,y) for all cells except (1,1), or including (10,10).But the problem says "at any given position (x,y)", so it's when being at (x,y). Therefore, including (1,1) and (10,10).But regardless, the sum is fixed.Wait, perhaps the formula is different. Maybe P(x,y) is the probability of colliding when moving from (x,y) to the next cell. So, it's based on the current cell's distance.In that case, the total collision probability would be the sum of P(x,y) for all cells except the last one.But in that case, the sum would still be fixed, because each path has the same number of moves, and the sum of P(x,y) over the starting cells would still be fixed.Wait, no, because the last cell (10,10) wouldn't contribute if you're only summing over the moves.But the problem says "at any given position (x,y)", so it's about being at (x,y), not moving into it.Therefore, the total collision probability is the sum over all cells visited, including (1,1) and (10,10).Therefore, the sum is fixed, and any path is optimal.Therefore, the answer is:1. P(5,5) = 0.012.2. Any path is optimal, with total collision probability 0.2565.But to express 0.2565 as a fraction, it's 513/2000.Alternatively, perhaps the problem expects us to express it as a decimal.So, final answers:1. 0.012.2. Any path, total collision probability 0.2565.But the problem says "what is the path the person should take", so perhaps it's expecting a specific path, which would be the one that stays closest to (1,1), i.e., the diagonal path.But given that the total collision probability is fixed, I think the answer is that any path is optimal.But to be thorough, perhaps the problem expects us to recognize that the total is fixed, so any path is optimal.Therefore, the final answers are:1. boxed{0.012}2. The person can take any path, and the total collision probability is boxed{0.2565}.</think>

🧠:<think>Okay, so I have this problem about a system of differential equations that Jack Gilbert Rentschler worked on. I need to solve two sub-problems. Let me start with the first one.Sub-problem 1: Show that the system has a steady-state solution and find the general form of (x(t)) and (y(t)) assuming (x(0) = x_0) and (y(0) = y_0).Alright, so the system is given by:[begin{cases}frac{dx(t)}{dt} = -a x(t) + b y(t) frac{dy(t)}{dt} = c x(t) - d y(t)end{cases}]First, I remember that a steady-state solution means that the derivatives are zero. So, I can set (frac{dx}{dt} = 0) and (frac{dy}{dt} = 0) and solve for (x) and (y).Setting the derivatives to zero:1. ( -a x + b y = 0 )2. ( c x - d y = 0 )So, from equation 1: ( -a x + b y = 0 ) => ( b y = a x ) => ( y = frac{a}{b} x )From equation 2: ( c x - d y = 0 ). Substitute ( y ) from equation 1 into this:( c x - d left( frac{a}{b} x right) = 0 )Simplify:( c x - frac{a d}{b} x = 0 )Factor out ( x ):( x left( c - frac{a d}{b} right) = 0 )So, either ( x = 0 ) or ( c - frac{a d}{b} = 0 ).If ( x = 0 ), then from equation 1, ( y = 0 ). So, the trivial solution is ( x = 0 ), ( y = 0 ).But if ( c - frac{a d}{b} = 0 ), then ( c = frac{a d}{b} ). In that case, the equations are dependent, and we can't find a unique non-trivial solution. So, unless ( c = frac{a d}{b} ), the only steady-state solution is the trivial one.Wait, but the problem says "show that the system has a steady-state solution." So, maybe they just mean the trivial solution? Or perhaps they consider the general solution approaching a steady state as ( t to infty ).Hmm, maybe I should think about the general solution of the system.To find the general solution, I can write this system in matrix form:[begin{pmatrix}frac{dx}{dt} frac{dy}{dt}end{pmatrix}=begin{pmatrix}-a & b c & -dend{pmatrix}begin{pmatrix}x yend{pmatrix}]Let me denote the matrix as ( M = begin{pmatrix} -a & b c & -d end{pmatrix} ).To solve this linear system, I need to find the eigenvalues and eigenvectors of matrix ( M ).The characteristic equation is ( det(M - lambda I) = 0 ).So,[det begin{pmatrix}-a - lambda & b c & -d - lambdaend{pmatrix}= 0]Calculating the determinant:( (-a - lambda)(-d - lambda) - b c = 0 )Expanding:( (a + lambda)(d + lambda) - b c = 0 )Multiply out:( a d + a lambda + d lambda + lambda^2 - b c = 0 )So,( lambda^2 + (a + d)lambda + (a d - b c) = 0 )This is a quadratic equation in ( lambda ). The solutions are:( lambda = frac{ - (a + d) pm sqrt{(a + d)^2 - 4(a d - b c)} }{2} )Simplify the discriminant:( D = (a + d)^2 - 4(a d - b c) = a^2 + 2 a d + d^2 - 4 a d + 4 b c = a^2 - 2 a d + d^2 + 4 b c )Which is ( D = (a - d)^2 + 4 b c )Since ( a, b, c, d ) are positive constants, ( D ) is positive because ( (a - d)^2 geq 0 ) and ( 4 b c > 0 ). Therefore, the eigenvalues are real and distinct.So, the system has two real eigenvalues, say ( lambda_1 ) and ( lambda_2 ), and corresponding eigenvectors ( mathbf{v}_1 ) and ( mathbf{v}_2 ).Therefore, the general solution is:[begin{pmatrix}x(t) y(t)end{pmatrix}= C_1 e^{lambda_1 t} mathbf{v}_1 + C_2 e^{lambda_2 t} mathbf{v}_2]Where ( C_1 ) and ( C_2 ) are constants determined by initial conditions.Now, depending on the eigenvalues, the behavior of the solution will change. Since ( a, b, c, d ) are positive, let's analyze the eigenvalues.The trace of the matrix ( M ) is ( -a - d ), which is negative. The determinant is ( a d - b c ). If ( a d > b c ), then the determinant is positive, and since the trace is negative, both eigenvalues are negative. If ( a d < b c ), determinant is negative, so one eigenvalue is positive, and the other is negative.Wait, but regardless, the system will have a steady-state solution as ( t to infty ) if all eigenvalues have negative real parts. If both eigenvalues are negative, then the solution will approach zero. If one eigenvalue is positive, the solution will blow up unless the initial condition has no component in that direction.But the problem says "show that the system has a steady-state solution." So, perhaps in the case when ( a d > b c ), the system is stable, and the solution tends to zero, which is the steady state. If ( a d < b c ), the system is unstable, but there's still a trivial steady state at zero.Alternatively, maybe the steady-state solution is non-trivial only when ( c = frac{a d}{b} ), but earlier, when setting derivatives to zero, we saw that only the trivial solution exists unless ( c = frac{a d}{b} ), which would make the equations dependent.Wait, perhaps I need to clarify. If ( c = frac{a d}{b} ), then the system is rank-deficient, and there are infinitely many steady-state solutions along a line. Otherwise, only the trivial solution.But the problem says "show that the system has a steady-state solution." So, regardless of the parameters, the trivial solution ( x = 0, y = 0 ) is always a steady state.So, perhaps the answer is that the trivial solution is always a steady-state, and the general solution is a combination of exponential functions based on the eigenvalues.Therefore, the general form of ( x(t) ) and ( y(t) ) is a linear combination of the eigenvectors multiplied by exponential functions of the eigenvalues times ( t ).But to write the general form explicitly, I need to find the eigenvalues and eigenvectors.Let me denote the eigenvalues as ( lambda_1 ) and ( lambda_2 ), given by:( lambda_{1,2} = frac{ - (a + d) pm sqrt{(a - d)^2 + 4 b c} }{2} )Let me compute the eigenvectors.For ( lambda_1 ):We solve ( (M - lambda_1 I) mathbf{v}_1 = 0 ).So,[begin{pmatrix}-a - lambda_1 & b c & -d - lambda_1end{pmatrix}begin{pmatrix}v_{1x} v_{1y}end{pmatrix}= 0]From the first equation:( (-a - lambda_1) v_{1x} + b v_{1y} = 0 )So,( v_{1y} = frac{a + lambda_1}{b} v_{1x} )Similarly, from the second equation:( c v_{1x} + (-d - lambda_1) v_{1y} = 0 )Substituting ( v_{1y} ):( c v_{1x} + (-d - lambda_1) left( frac{a + lambda_1}{b} v_{1x} right) = 0 )Simplify:( c v_{1x} - frac{(d + lambda_1)(a + lambda_1)}{b} v_{1x} = 0 )Factor out ( v_{1x} ):( left[ c - frac{(d + lambda_1)(a + lambda_1)}{b} right] v_{1x} = 0 )Since ( v_{1x} ) is not zero (eigenvector), the term in brackets must be zero.But wait, from the characteristic equation, we have:( lambda^2 + (a + d)lambda + (a d - b c) = 0 )So, for ( lambda = lambda_1 ):( lambda_1^2 + (a + d)lambda_1 + (a d - b c) = 0 )Which can be rearranged as:( a d - b c = - lambda_1^2 - (a + d)lambda_1 )So, substituting back into the bracket term:( c - frac{(d + lambda_1)(a + lambda_1)}{b} = c - frac{a d + a lambda_1 + d lambda_1 + lambda_1^2}{b} )But from the characteristic equation, ( a d - b c = - lambda_1^2 - (a + d)lambda_1 ), so ( a d = b c - lambda_1^2 - (a + d)lambda_1 )Substituting ( a d ) into the bracket term:( c - frac{(b c - lambda_1^2 - (a + d)lambda_1) + a lambda_1 + d lambda_1 + lambda_1^2}{b} )Simplify numerator:( b c - lambda_1^2 - (a + d)lambda_1 + a lambda_1 + d lambda_1 + lambda_1^2 = b c )So, the bracket term becomes:( c - frac{b c}{b} = c - c = 0 )Therefore, the equation is satisfied, which means our earlier expression for ( v_{1y} ) is correct.So, the eigenvector ( mathbf{v}_1 ) is proportional to ( begin{pmatrix} 1 frac{a + lambda_1}{b} end{pmatrix} )Similarly, for ( lambda_2 ), the eigenvector ( mathbf{v}_2 ) is proportional to ( begin{pmatrix} 1 frac{a + lambda_2}{b} end{pmatrix} )Therefore, the general solution is:[begin{pmatrix}x(t) y(t)end{pmatrix}= C_1 e^{lambda_1 t} begin{pmatrix} 1 frac{a + lambda_1}{b} end{pmatrix} + C_2 e^{lambda_2 t} begin{pmatrix} 1 frac{a + lambda_2}{b} end{pmatrix}]Alternatively, we can write this as:[x(t) = C_1 e^{lambda_1 t} + C_2 e^{lambda_2 t}][y(t) = C_1 e^{lambda_1 t} frac{a + lambda_1}{b} + C_2 e^{lambda_2 t} frac{a + lambda_2}{b}]So, that's the general form. The constants ( C_1 ) and ( C_2 ) are determined by the initial conditions ( x(0) = x_0 ) and ( y(0) = y_0 ).To find ( C_1 ) and ( C_2 ), we can set up the system:At ( t = 0 ):[x_0 = C_1 + C_2][y_0 = C_1 frac{a + lambda_1}{b} + C_2 frac{a + lambda_2}{b}]This is a linear system in ( C_1 ) and ( C_2 ). We can solve it using substitution or matrix methods.Let me denote ( mu_1 = frac{a + lambda_1}{b} ) and ( mu_2 = frac{a + lambda_2}{b} ), then the system becomes:[x_0 = C_1 + C_2][y_0 = C_1 mu_1 + C_2 mu_2]We can solve for ( C_1 ) and ( C_2 ):From the first equation: ( C_2 = x_0 - C_1 )Substitute into the second equation:( y_0 = C_1 mu_1 + (x_0 - C_1) mu_2 )Simplify:( y_0 = C_1 (mu_1 - mu_2) + x_0 mu_2 )Therefore,( C_1 = frac{y_0 - x_0 mu_2}{mu_1 - mu_2} )Similarly,( C_2 = x_0 - C_1 = x_0 - frac{y_0 - x_0 mu_2}{mu_1 - mu_2} = frac{x_0 (mu_1 - mu_2) - y_0 + x_0 mu_2}{mu_1 - mu_2} = frac{x_0 mu_1 - y_0}{mu_1 - mu_2} )So, substituting back ( mu_1 ) and ( mu_2 ):( C_1 = frac{y_0 - x_0 frac{a + lambda_2}{b}}{frac{a + lambda_1}{b} - frac{a + lambda_2}{b}} = frac{y_0 b - x_0 (a + lambda_2)}{(a + lambda_1) - (a + lambda_2)} = frac{y_0 b - x_0 (a + lambda_2)}{lambda_1 - lambda_2} )Similarly,( C_2 = frac{x_0 frac{a + lambda_1}{b} - y_0}{frac{a + lambda_1}{b} - frac{a + lambda_2}{b}} = frac{x_0 (a + lambda_1) - y_0 b}{(a + lambda_1) - (a + lambda_2)} = frac{x_0 (a + lambda_1) - y_0 b}{lambda_1 - lambda_2} )Therefore, the general solution with constants ( C_1 ) and ( C_2 ) determined by initial conditions is as above.So, summarizing, the system has a trivial steady-state solution at ( x = 0 ), ( y = 0 ). The general solution is a combination of exponential functions based on the eigenvalues, with coefficients determined by the initial conditions.Sub-problem 2: Given the initial conditions (x(0) = 1) and (y(0) = 0), and the constants (a = 2), (b = 1), (c = 3), and (d = 4), determine the values of (x(t)) and (y(t)) at (t = 2).Alright, so now we have specific values:( a = 2 ), ( b = 1 ), ( c = 3 ), ( d = 4 )Initial conditions: ( x(0) = 1 ), ( y(0) = 0 )We need to find ( x(2) ) and ( y(2) ).First, let's write down the system:[begin{cases}frac{dx}{dt} = -2 x + y frac{dy}{dt} = 3 x - 4 yend{cases}]We can solve this using the method from Sub-problem 1.First, find the eigenvalues of the matrix ( M = begin{pmatrix} -2 & 1 3 & -4 end{pmatrix} )The characteristic equation is:( det(M - lambda I) = 0 )So,[det begin{pmatrix}-2 - lambda & 1 3 & -4 - lambdaend{pmatrix}= 0]Calculating determinant:( (-2 - lambda)(-4 - lambda) - (1)(3) = 0 )Expand:( (2 + lambda)(4 + lambda) - 3 = 0 )Wait, actually, ( (-2 - lambda)(-4 - lambda) = (2 + lambda)(4 + lambda) ). Let's compute that:( (2 + lambda)(4 + lambda) = 8 + 6 lambda + lambda^2 )So,( 8 + 6 lambda + lambda^2 - 3 = 0 )Simplify:( lambda^2 + 6 lambda + 5 = 0 )Factor:( (lambda + 1)(lambda + 5) = 0 )So, eigenvalues are ( lambda_1 = -1 ) and ( lambda_2 = -5 )So, both eigenvalues are negative, which means the system is stable, and the solution will approach zero as ( t to infty ).Now, find the eigenvectors.For ( lambda_1 = -1 ):Solve ( (M - (-1) I) mathbf{v}_1 = 0 )So,[begin{pmatrix}-2 + 1 & 1 3 & -4 + 1end{pmatrix}=begin{pmatrix}-1 & 1 3 & -3end{pmatrix}]From the first row: ( -v_{1x} + v_{1y} = 0 ) => ( v_{1y} = v_{1x} )From the second row: ( 3 v_{1x} - 3 v_{1y} = 0 ) => same as first equation.So, eigenvector is any scalar multiple of ( begin{pmatrix} 1 1 end{pmatrix} )Similarly, for ( lambda_2 = -5 ):Solve ( (M - (-5) I) mathbf{v}_2 = 0 )So,[begin{pmatrix}-2 + 5 & 1 3 & -4 + 5end{pmatrix}=begin{pmatrix}3 & 1 3 & 1end{pmatrix}]From the first row: ( 3 v_{2x} + v_{2y} = 0 ) => ( v_{2y} = -3 v_{2x} )From the second row: same as the first row.So, eigenvector is any scalar multiple of ( begin{pmatrix} 1 -3 end{pmatrix} )Therefore, the general solution is:[begin{pmatrix}x(t) y(t)end{pmatrix}= C_1 e^{-t} begin{pmatrix} 1 1 end{pmatrix} + C_2 e^{-5 t} begin{pmatrix} 1 -3 end{pmatrix}]So,[x(t) = C_1 e^{-t} + C_2 e^{-5 t}][y(t) = C_1 e^{-t} - 3 C_2 e^{-5 t}]Now, apply initial conditions at ( t = 0 ):( x(0) = 1 = C_1 + C_2 )( y(0) = 0 = C_1 - 3 C_2 )So, we have the system:1. ( C_1 + C_2 = 1 )2. ( C_1 - 3 C_2 = 0 )Let's solve this.From equation 2: ( C_1 = 3 C_2 )Substitute into equation 1:( 3 C_2 + C_2 = 1 ) => ( 4 C_2 = 1 ) => ( C_2 = 1/4 )Then, ( C_1 = 3 * (1/4) = 3/4 )So, the particular solution is:[x(t) = frac{3}{4} e^{-t} + frac{1}{4} e^{-5 t}][y(t) = frac{3}{4} e^{-t} - frac{3}{4} e^{-5 t}]Now, we need to find ( x(2) ) and ( y(2) ).Compute ( x(2) ):( x(2) = frac{3}{4} e^{-2} + frac{1}{4} e^{-10} )Compute ( y(2) ):( y(2) = frac{3}{4} e^{-2} - frac{3}{4} e^{-10} )Let me compute these numerically.First, compute ( e^{-2} ) and ( e^{-10} ):( e^{-2} approx 0.135335283 )( e^{-10} approx 0.0000453999 )So,( x(2) = (3/4)(0.135335283) + (1/4)(0.0000453999) )Compute each term:( (3/4)(0.135335283) = 0.75 * 0.135335283 ≈ 0.101501462 )( (1/4)(0.0000453999) = 0.25 * 0.0000453999 ≈ 0.000011349975 )Add them together:( x(2) ≈ 0.101501462 + 0.000011349975 ≈ 0.101512812 )Similarly, ( y(2) = (3/4)(0.135335283) - (3/4)(0.0000453999) )Compute each term:( (3/4)(0.135335283) ≈ 0.101501462 )( (3/4)(0.0000453999) ≈ 0.000034049925 )Subtract:( y(2) ≈ 0.101501462 - 0.000034049925 ≈ 0.101467412 )So, rounding to, say, 6 decimal places:( x(2) ≈ 0.101513 )( y(2) ≈ 0.101467 )Alternatively, we can write them as fractions multiplied by exponentials, but since the question asks for the values, probably decimal approximations are acceptable.But let me double-check the calculations.First, ( e^{-2} ≈ 0.135335283 ), correct.( e^{-10} ≈ 0.0000453999 ), correct.Compute ( x(2) ):( (3/4)*0.135335283 = 0.75 * 0.135335283 ≈ 0.101501462 )( (1/4)*0.0000453999 ≈ 0.000011349975 )Sum ≈ 0.101512812, which is approximately 0.101513.Similarly, ( y(2) = 0.101501462 - 0.000034049925 ≈ 0.101467412 ), which is approximately 0.101467.So, rounding to six decimal places:( x(2) ≈ 0.101513 )( y(2) ≈ 0.101467 )Alternatively, if more precision is needed, but I think this is sufficient.Final AnswerSub-problem 1: The system has a steady-state solution at ( x = 0 ) and ( y = 0 ). The general solution is given by:[x(t) = C_1 e^{lambda_1 t} + C_2 e^{lambda_2 t}][y(t) = C_1 frac{a + lambda_1}{b} e^{lambda_1 t} + C_2 frac{a + lambda_2}{b} e^{lambda_2 t}]where ( lambda_1 ) and ( lambda_2 ) are the eigenvalues of the system matrix, and ( C_1 ) and ( C_2 ) are constants determined by initial conditions.Sub-problem 2: The values of ( x(t) ) and ( y(t) ) at ( t = 2 ) are approximately:[x(2) approx boxed{0.101513}][y(2) approx boxed{0.101467}]Final AnswerThe values of ( x(t) ) and ( y(t) ) at ( t = 2 ) are (boxed{0.1015}) and (boxed{0.1015}) respectively.Wait, hold on. The approximate values I got were 0.101513 and 0.101467. Rounding to four decimal places, that would be 0.1015 and 0.1015. But actually, 0.101513 is approximately 0.1015, and 0.101467 is approximately 0.1015 as well. So, both are approximately 0.1015.Alternatively, maybe more precise, but the question didn't specify the number of decimal places. Since the initial conditions are integers, perhaps two decimal places? Let me check:Compute ( x(2) ≈ 0.101513 ), which is approximately 0.1015, so 0.102 if rounded to three decimal places, or 0.10 if rounded to two.But in engineering, often three decimal places are used. So, 0.102 and 0.101.Wait, let me compute more accurately:( x(2) = (3/4)e^{-2} + (1/4)e^{-10} )Compute ( e^{-2} approx 0.1353352832366127 )Compute ( e^{-10} approx 0.00004539992976248552 )So,( (3/4)*0.1353352832366127 = 0.75 * 0.1353352832366127 ≈ 0.10150146242745953 )( (1/4)*0.00004539992976248552 ≈ 0.00001134998244062138 )Sum: 0.10150146242745953 + 0.00001134998244062138 ≈ 0.10151281240990015Similarly,( y(2) = (3/4)e^{-2} - (3/4)e^{-10} )Compute:( (3/4)*0.1353352832366127 ≈ 0.10150146242745953 )( (3/4)*0.00004539992976248552 ≈ 0.00003404994732186414 )Subtract:0.10150146242745953 - 0.00003404994732186414 ≈ 0.10146741248013767So, more precisely:( x(2) ≈ 0.1015128124 )( y(2) ≈ 0.1014674125 )Rounded to four decimal places:( x(2) ≈ 0.1015 )( y(2) ≈ 0.1015 )But actually, ( y(2) ≈ 0.101467 ) is closer to 0.1015 when rounded to four decimal places, but technically, 0.101467 is 0.1015 when rounded to four decimal places because the fifth decimal is 6, which rounds up.Wait, 0.101467 is:0.1014 (four decimal places) with the fifth being 6, so it rounds up to 0.1015.Similarly, ( x(2) ≈ 0.101513 ) is 0.1015 when rounded to four decimal places.So, both are approximately 0.1015.Therefore, the final answer is:( x(2) approx boxed{0.1015} )( y(2) approx boxed{0.1015} )But wait, looking back, the exact values are slightly different, but when rounded to four decimal places, both are 0.1015.Alternatively, if we keep more decimal places, they are approximately 0.101513 and 0.101467, which are very close but not exactly the same. However, depending on the precision required, they can both be approximated as 0.1015.Alternatively, perhaps the exact expressions are better:( x(2) = frac{3}{4} e^{-2} + frac{1}{4} e^{-10} )( y(2) = frac{3}{4} e^{-2} - frac{3}{4} e^{-10} )But since the question asks for the values at ( t = 2 ), it's likely expecting numerical approximations.So, to four decimal places, both are approximately 0.1015.Alternatively, if we compute more precisely:( x(2) ≈ 0.1015128124 ) ≈ 0.101513( y(2) ≈ 0.1014674125 ) ≈ 0.101467But if we need to box them, perhaps we can write them as:( x(2) approx boxed{0.1015} )( y(2) approx boxed{0.1015} )But to be precise, they are slightly different. However, given the closeness, it's reasonable to approximate both as 0.1015.Alternatively, if we compute with more precise exponentials:Using more decimal places for ( e^{-2} ) and ( e^{-10} ):( e^{-2} ≈ 0.1353352832366127 )( e^{-10} ≈ 0.00004539992976248552 )Compute ( x(2) ):( (3/4)*0.1353352832366127 = 0.75 * 0.1353352832366127 = 0.10150146242745953 )( (1/4)*0.00004539992976248552 = 0.00001134998244062138 )Sum: 0.10150146242745953 + 0.00001134998244062138 ≈ 0.10151281240990015Similarly, ( y(2) ):( (3/4)*0.1353352832366127 = 0.10150146242745953 )( (3/4)*0.00004539992976248552 = 0.00003404994732186414 )Subtract: 0.10150146242745953 - 0.00003404994732186414 ≈ 0.10146741248013767So, ( x(2) ≈ 0.1015128124 ) and ( y(2) ≈ 0.1014674125 )Rounded to six decimal places:( x(2) ≈ 0.101513 )( y(2) ≈ 0.101467 )But if we need to present them in boxes, perhaps we can write them as:( x(2) ≈ boxed{0.1015} )( y(2) ≈ boxed{0.1015} )Alternatively, if the question expects more precision, we can write:( x(2) ≈ boxed{0.101513} )( y(2) ≈ boxed{0.101467} )But since the initial conditions are integers and the constants are integers, it's likely that four decimal places are sufficient.So, I think the appropriate answer is:( x(2) ≈ boxed{0.1015} )( y(2) ≈ boxed{0.1015} )But to be precise, they are slightly different, but very close.Alternatively, perhaps the exact expressions are better, but the question asks for the values, so numerical approximations are expected.So, I think the final answer is:( x(2) approx boxed{0.1015} )( y(2) approx boxed{0.1015} )But to be accurate, since they are slightly different, maybe we should present them as:( x(2) approx boxed{0.1015} )( y(2) approx boxed{0.1015} )But actually, looking at the precise values:( x(2) ≈ 0.101513 ) which is approximately 0.1015( y(2) ≈ 0.101467 ) which is approximately 0.1015So, both can be rounded to 0.1015.Therefore, the final answer is:( x(2) approx boxed{0.1015} )( y(2) approx boxed{0.1015} )Final AnswerThe values of ( x(t) ) and ( y(t) ) at ( t = 2 ) are (boxed{0.1015}) and (boxed{0.1015}) respectively.</think>