🧠:<think>Okay, so I have this problem about a travel blogger visiting temples in India. There are two parts to the problem. Let me tackle them one by one.Starting with the first part: Each temple is modeled as a regular n-sided polygon inscribed in a circle of radius R. I need to derive a formula for the perimeter Pn in terms of n and R, and then figure out for which values of n (with n ≥ 3) the perimeter is maximized when R is constant.Alright, so a regular n-sided polygon inscribed in a circle. I remember that in a regular polygon, all sides are equal, and all central angles are equal. Since it's inscribed in a circle, each vertex lies on the circumference.First, I need to find the length of one side of the polygon. If I can find that, then the perimeter would just be n times that length.To find the length of one side, I can think of the polygon as being made up of n congruent isosceles triangles, each with two sides equal to the radius R and the included angle at the center being 2π/n radians. Because the full circle is 2π radians, and there are n sides, so each central angle is 2π/n.So, each triangle has sides of length R, R, and the base which is the side of the polygon. The base can be found using the Law of Cosines. The Law of Cosines states that for a triangle with sides a, b, and c, opposite angles A, B, and C respectively, c² = a² + b² - 2ab cos(C).In this case, the two sides are both R, and the included angle is 2π/n. So, the length of the base, which is the side of the polygon, let's call it s, would be:s² = R² + R² - 2*R*R*cos(2π/n)s² = 2R² - 2R² cos(2π/n)s² = 2R²(1 - cos(2π/n))Therefore, s = sqrt(2R²(1 - cos(2π/n))) = R*sqrt(2(1 - cos(2π/n)))Hmm, that seems right. Alternatively, I remember that 1 - cos(θ) can be written as 2 sin²(θ/2). So, substituting that in:s = R*sqrt(2 * 2 sin²(π/n)) = R*sqrt(4 sin²(π/n)) = R*2 sin(π/n)Oh, that's simpler! So, s = 2R sin(π/n). That makes sense because as n increases, the side length decreases, which aligns with intuition since the polygon becomes closer to the circle.So, the perimeter Pn is n times the side length:Pn = n * 2R sin(π/n) = 2nR sin(π/n)Alright, so that's the formula for the perimeter. Now, the next part is to determine for which values of n (n ≥ 3) the perimeter Pn is maximized, given R is constant.So, we need to find the value of n that maximizes Pn = 2nR sin(π/n). Since R is constant, we can ignore it for the purpose of maximization. So, we need to maximize f(n) = n sin(π/n) for n ≥ 3.Hmm, n is an integer greater than or equal to 3, but perhaps we can treat n as a continuous variable to find the maximum, and then check the integer values around that.Let me consider f(n) = n sin(π/n). Let's take the derivative with respect to n to find the critical points.But wait, n is an integer, so calculus might not directly apply, but sometimes treating it as a continuous variable can give us an idea.Let me set x = n, so f(x) = x sin(π/x). Let's compute the derivative f’(x):f’(x) = sin(π/x) + x * cos(π/x) * (-π/x²) = sin(π/x) - (π/x) cos(π/x)Set derivative equal to zero:sin(π/x) - (π/x) cos(π/x) = 0So, sin(π/x) = (π/x) cos(π/x)Divide both sides by cos(π/x):tan(π/x) = π/xSo, tan(θ) = θ, where θ = π/x.Hmm, the equation tan(θ) = θ has solutions at θ = 0, θ ≈ 4.4934, θ ≈ 7.7253, etc. But since θ = π/x, and x ≥ 3, so θ ≤ π/3 ≈ 1.0472 radians. So, the only solution in this range is θ = 0, but θ = 0 would mean x approaches infinity, which isn't practical.Wait, so perhaps the function f(x) = x sin(π/x) doesn't have a maximum in the domain x ≥ 3? Let's analyze the behavior.As x approaches infinity, sin(π/x) ≈ π/x, so f(x) ≈ x*(π/x) = π, which is a constant. So, as n increases, the perimeter approaches 2Rπ, which is the circumference of the circle.But for finite n, f(n) = n sin(π/n). Let's compute f(n) for n = 3,4,5,... and see how it behaves.Compute f(3) = 3 sin(π/3) ≈ 3*(√3/2) ≈ 2.598f(4) = 4 sin(π/4) ≈ 4*(√2/2) ≈ 2.828f(5) = 5 sin(π/5) ≈ 5*0.5878 ≈ 2.939f(6) = 6 sin(π/6) = 6*(0.5) = 3f(7) ≈ 7 sin(π/7) ≈ 7*0.4339 ≈ 3.037f(8) ≈ 8 sin(π/8) ≈ 8*0.3827 ≈ 3.061f(9) ≈ 9 sin(π/9) ≈ 9*0.3420 ≈ 3.078f(10) ≈ 10 sin(π/10) ≈ 10*0.3090 ≈ 3.090Wait, so as n increases, f(n) increases? But earlier, I thought as n approaches infinity, f(n) approaches π ≈ 3.1416. So, f(n) is increasing with n, approaching π.But wait, let's compute f(100): 100 sin(π/100) ≈ 100*(π/100 - (π/100)^3/6) ≈ π - (π^3)/(6*100^2) ≈ 3.1416 - 0.0005 ≈ 3.1411So, f(n) increases as n increases, approaching π from below. Therefore, the perimeter Pn = 2R f(n) increases as n increases, approaching 2Rπ.But wait, the problem says to determine for which values of n (n ≥ 3) the perimeter is maximized. But since as n increases, Pn increases and approaches 2Rπ, which is the circumference. So, technically, the maximum perimeter is achieved as n approaches infinity, but since n must be an integer ≥3, there is no maximum; it just keeps increasing as n increases.But that can't be, because in reality, as n increases, the polygon becomes closer to the circle, so the perimeter approaches the circumference, which is a fixed value. So, the perimeter is increasing with n, approaching 2πR.But the question is asking for which n the perimeter is maximized. Since n is an integer ≥3, and the perimeter increases with n, the maximum would be as n approaches infinity, but since n must be finite, the perimeter can be made arbitrarily close to 2πR by choosing a large enough n.But perhaps I'm misunderstanding. Maybe the problem is considering n as a real number? Or perhaps it's a trick question where the maximum is achieved as n approaches infinity.Wait, let's think again. The perimeter is Pn = 2nR sin(π/n). Let's see how this behaves as n increases.We can write sin(π/n) ≈ π/n - (π/n)^3/6 for large n. So,Pn ≈ 2nR [π/n - (π^3)/(6n^3)] = 2Rπ - (2Rπ^3)/(6n^2) = 2Rπ - (Rπ^3)/(3n^2)So, as n increases, the second term becomes negligible, and Pn approaches 2Rπ from below.Therefore, Pn is an increasing function of n, approaching 2Rπ as n tends to infinity.Therefore, for any finite n, Pn is less than 2Rπ, and as n increases, Pn increases. So, there is no maximum for finite n; it just keeps increasing. So, the perimeter is maximized as n approaches infinity, but since n must be an integer ≥3, the maximum isn't achieved at any finite n.But the question says "determine for which values of n (with n ≥ 3) the perimeter Pn is maximized given that R remains constant."Hmm, maybe I made a mistake earlier. Let me double-check.Wait, perhaps I should consider the behavior of f(n) = n sin(π/n). Let's compute f(n) for n=3,4,5,... and see if it's increasing.f(3) ≈ 3*(0.8660) ≈ 2.598f(4) ≈ 4*(0.7071) ≈ 2.828f(5) ≈ 5*(0.5878) ≈ 2.939f(6) ≈ 6*(0.5) = 3f(7) ≈ 7*(0.4339) ≈ 3.037f(8) ≈ 8*(0.3827) ≈ 3.061f(9) ≈ 9*(0.3420) ≈ 3.078f(10) ≈ 10*(0.3090) ≈ 3.090f(11) ≈ 11*(0.2817) ≈ 3.100f(12) ≈ 12*(0.2588) ≈ 3.106f(20) ≈ 20*sin(π/20) ≈ 20*0.1564 ≈ 3.128f(50) ≈ 50*sin(π/50) ≈ 50*0.0628 ≈ 3.140f(100) ≈ 100*sin(π/100) ≈ 100*0.0314 ≈ 3.1416So, yes, as n increases, f(n) increases, approaching π ≈ 3.1416. Therefore, Pn = 2R f(n) approaches 2Rπ.So, in terms of maximizing Pn, since it's increasing with n, the maximum would be as n approaches infinity. But since n must be an integer ≥3, there's no finite n that gives the maximum; it just keeps increasing.But the question is asking for which values of n the perimeter is maximized. Maybe it's a trick question, and the answer is that the perimeter increases without bound as n increases, so there's no maximum. But that doesn't make sense because the perimeter is approaching a limit, 2πR.Wait, no, it's not increasing without bound; it's approaching a limit. So, the perimeter is maximized in the limit as n approaches infinity, but for any finite n, it's less than 2πR.But the problem says "determine for which values of n (with n ≥ 3) the perimeter Pn is maximized." So, perhaps the answer is that the perimeter is maximized as n approaches infinity, but since n must be finite, the maximum isn't achieved at any specific n; it's just that Pn increases with n.Alternatively, maybe I made a mistake in the derivative approach earlier. Let me try again.We have f(n) = n sin(π/n). Let's consider n as a continuous variable and take the derivative.f(n) = n sin(π/n)f’(n) = sin(π/n) + n cos(π/n) * (-π/n²) = sin(π/n) - (π/n) cos(π/n)Set f’(n) = 0:sin(π/n) = (π/n) cos(π/n)Divide both sides by cos(π/n):tan(π/n) = π/nLet θ = π/n, so tan(θ) = θWe know that tan(θ) = θ has solutions at θ = 0, θ ≈ 4.4934, θ ≈ 7.7253, etc. But θ = π/n, and since n ≥3, θ ≤ π/3 ≈ 1.0472. So, the only solution in this interval is θ = 0, which corresponds to n approaching infinity.Therefore, the function f(n) = n sin(π/n) has its maximum as n approaches infinity. Hence, the perimeter Pn is maximized as n approaches infinity.But since n must be an integer ≥3, the perimeter can be made arbitrarily close to 2πR by choosing a sufficiently large n, but it never actually reaches 2πR.Therefore, the perimeter is maximized as n approaches infinity, but there's no finite n that achieves the maximum.Wait, but the problem says "determine for which values of n (with n ≥ 3) the perimeter Pn is maximized." So, perhaps the answer is that the perimeter increases without bound as n increases, but that's not correct because it approaches a limit.Alternatively, maybe I'm overcomplicating. Since Pn increases with n, the maximum perimeter is achieved as n approaches infinity, but for any finite n, it's less than 2πR. So, in terms of finite n, there's no maximum; it's unbounded above in the limit.But the problem might be expecting a specific answer, perhaps that the perimeter is maximized as n increases, so the maximum is achieved in the limit as n approaches infinity.Alternatively, maybe I made a mistake in the initial formula. Let me double-check.Perimeter of regular n-gon inscribed in circle of radius R:Each side length s = 2R sin(π/n)Perimeter Pn = n*s = 2nR sin(π/n)Yes, that's correct.So, Pn = 2nR sin(π/n). As n increases, sin(π/n) ≈ π/n - (π/n)^3/6, so Pn ≈ 2nR(π/n - (π^3)/(6n^3)) = 2Rπ - (2Rπ^3)/(6n^2) = 2Rπ - (Rπ^3)/(3n^2)So, as n increases, the second term becomes negligible, and Pn approaches 2Rπ from below.Therefore, Pn is increasing with n, approaching 2Rπ. So, the maximum perimeter is 2Rπ, achieved in the limit as n approaches infinity.But since n must be an integer ≥3, the perimeter can be made as close to 2Rπ as desired by choosing a large enough n, but it never actually reaches 2Rπ.Therefore, the perimeter is maximized as n approaches infinity, but there's no finite n that achieves the maximum.But the problem asks for "values of n (with n ≥ 3)" where the perimeter is maximized. So, perhaps the answer is that the perimeter is maximized as n approaches infinity, but since n must be finite, there's no maximum; it's unbounded above.Alternatively, maybe the problem expects us to recognize that as n increases, the perimeter increases, so the maximum is achieved as n increases without bound.But perhaps I'm overcomplicating. Maybe the answer is that the perimeter is maximized as n approaches infinity, but since n must be finite, the maximum isn't achieved at any specific n.Alternatively, perhaps the problem expects us to consider that the perimeter is maximized for the largest possible n, but since n can be any integer ≥3, there's no upper bound.Wait, but the problem says "determine for which values of n (with n ≥ 3) the perimeter Pn is maximized." So, perhaps the answer is that the perimeter increases with n, so it's maximized as n approaches infinity, but for any finite n, it's less than 2πR.Alternatively, maybe the problem expects us to consider that the perimeter is maximized when n is as large as possible, but since n can be any integer, there's no maximum.Wait, but in reality, n can't be infinite, so perhaps the answer is that the perimeter is maximized as n approaches infinity, but in practical terms, there's no finite n that gives the maximum.Hmm, I think I need to conclude that the perimeter Pn increases with n and approaches 2πR as n approaches infinity. Therefore, the perimeter is maximized as n approaches infinity, but for any finite n, it's less than 2πR.So, to answer the first part: The formula for the perimeter is Pn = 2nR sin(π/n), and the perimeter is maximized as n approaches infinity, meaning there's no finite n that achieves the maximum; it just increases without bound (but actually approaches a limit).Wait, no, it doesn't increase without bound; it approaches 2πR. So, the perimeter is maximized in the limit as n approaches infinity, but for any finite n, it's less than 2πR.So, perhaps the answer is that the perimeter is maximized as n approaches infinity, but since n must be finite, there's no specific n that gives the maximum; it's just that Pn increases with n.But the problem says "determine for which values of n (with n ≥ 3) the perimeter Pn is maximized." So, maybe the answer is that the perimeter is maximized as n approaches infinity, but since n must be finite, there's no maximum; it's unbounded above.Alternatively, perhaps the problem expects us to consider that the perimeter is maximized when n is as large as possible, but since n can be any integer, there's no maximum.Wait, but in reality, n can't be infinite, so perhaps the answer is that the perimeter is maximized as n approaches infinity, but for any finite n, it's less than 2πR.I think I need to wrap this up. So, the formula is Pn = 2nR sin(π/n), and the perimeter increases with n, approaching 2πR as n approaches infinity. Therefore, there's no finite n that maximizes the perimeter; it's just that as n increases, the perimeter gets closer to 2πR.Now, moving on to the second part: The blogger wants to find a central point C such that the sum of the squared Euclidean distances from C to each temple is minimized. This is the geometric median, but wait, actually, the geometric median minimizes the sum of distances, but the problem says sum of squared distances, which is the centroid or the mean.Wait, no, the centroid minimizes the sum of squared distances. The geometric median minimizes the sum of distances, but that's a different problem.So, given points (1,3), (4,5), (7,8), (10,12), find the point C = (x,y) that minimizes the sum of squared distances.The centroid (or geometric mean) is given by the average of the x-coordinates and the average of the y-coordinates.So, x = (1 + 4 + 7 + 10)/4 = (22)/4 = 5.5y = (3 + 5 + 8 + 12)/4 = (28)/4 = 7Therefore, the point C is (5.5, 7).Wait, let me verify that.Sum of x-coordinates: 1 + 4 = 5, 5 +7=12, 12+10=22Sum of y-coordinates: 3 +5=8, 8+8=16, 16+12=28Number of points: 4So, x = 22/4 = 5.5, y = 28/4 =7.Yes, that's correct.Therefore, the coordinates of point C are (5.5,7).So, to recap:1. The perimeter formula is Pn = 2nR sin(π/n), and the perimeter increases with n, approaching 2πR as n approaches infinity. Therefore, there's no finite n that maximizes the perimeter; it's just that as n increases, the perimeter gets closer to 2πR.2. The central point C that minimizes the sum of squared distances is the centroid, which is (5.5,7).I think that's it.</think>

🧠:<think>Okay, so I have this problem about a college student using a logistic growth model to spread awareness about racial injustice. There are two parts to the problem. Let me try to tackle them one by one.Problem 1: Calculating the Growth Rate ( k )The logistic growth model given is:[ A(t) = frac{L}{1 + e^{-k(t - t_0)}} ]We are told that:- The maximum level of awareness ( L ) is 100 units.- The initial level of awareness at ( t = 0 ) is 5 units.- The inflection point occurs at ( t = 3 ).We need to find the growth rate ( k ).First, let's recall what the logistic growth model represents. It models the growth of something that starts off slowly, then increases rapidly, and then slows down as it approaches the maximum level. The inflection point is where the growth rate is the highest, which in this case is at ( t = 3 ).Given that the inflection point is at ( t = 3 ), that means ( t_0 = 3 ). So, plugging that into the equation, we have:[ A(t) = frac{100}{1 + e^{-k(t - 3)}} ]Now, we also know that at ( t = 0 ), the awareness level is 5 units. So, let's plug ( t = 0 ) and ( A(0) = 5 ) into the equation:[ 5 = frac{100}{1 + e^{-k(0 - 3)}} ][ 5 = frac{100}{1 + e^{-3k}} ]Let me solve this equation for ( k ). First, multiply both sides by the denominator:[ 5(1 + e^{-3k}) = 100 ][ 5 + 5e^{-3k} = 100 ]Subtract 5 from both sides:[ 5e^{-3k} = 95 ]Divide both sides by 5:[ e^{-3k} = 19 ]Now, take the natural logarithm of both sides:[ ln(e^{-3k}) = ln(19) ][ -3k = ln(19) ]So, solving for ( k ):[ k = -frac{ln(19)}{3} ]Wait, that gives a negative value for ( k ). But in the logistic growth model, ( k ) is typically a positive growth rate. Did I make a mistake?Let me check my steps again.Starting from:[ 5 = frac{100}{1 + e^{-3k}} ]Multiply both sides by denominator:[ 5(1 + e^{-3k}) = 100 ][ 5 + 5e^{-3k} = 100 ]Subtract 5:[ 5e^{-3k} = 95 ]Divide by 5:[ e^{-3k} = 19 ]Yes, that's correct. Taking natural log:[ -3k = ln(19) ][ k = -frac{ln(19)}{3} ]Hmm, so ( k ) is negative. But in the logistic model, ( k ) is positive because it's the growth rate. Maybe I misapplied the model?Wait, let me think. The logistic model is:[ A(t) = frac{L}{1 + e^{-k(t - t_0)}} ]So, if ( t_0 = 3 ), then at ( t = 3 ), the exponent becomes zero, so ( A(3) = frac{L}{2} ). That makes sense because the inflection point is at half the maximum.But when ( t < t_0 ), the exponent is negative, so ( e^{-k(t - t_0)} ) becomes ( e^{-k(-something)} = e^{k(something)} ), which is positive. So, as ( t ) increases, the exponent becomes less negative, approaching zero.Wait, but in our case, when ( t = 0 ), the exponent is ( -k(-3) = 3k ). So, ( e^{3k} ). So, the denominator is ( 1 + e^{3k} ).So, plugging in ( t = 0 ):[ A(0) = frac{100}{1 + e^{3k}} = 5 ]So, that gives:[ frac{100}{1 + e^{3k}} = 5 ][ 1 + e^{3k} = 20 ][ e^{3k} = 19 ][ 3k = ln(19) ][ k = frac{ln(19)}{3} ]Ah! That makes sense. I think I messed up the sign earlier. Because when ( t = 0 ), ( t - t_0 = -3 ), so the exponent is ( -k(-3) = 3k ), which is positive. So, the equation becomes ( e^{3k} ), not ( e^{-3k} ). That was my mistake.So, correcting that:From ( A(0) = 5 ):[ 5 = frac{100}{1 + e^{3k}} ][ 1 + e^{3k} = 20 ][ e^{3k} = 19 ][ 3k = ln(19) ][ k = frac{ln(19)}{3} ]Calculating ( ln(19) ):I know that ( ln(16) = 2.7726 ) because ( e^{2.7726} approx 16 ). ( ln(20) approx 2.9957 ). So, ( ln(19) ) is somewhere between 2.7726 and 2.9957. Let me compute it more accurately.Using a calculator, ( ln(19) approx 2.9444 ). So,[ k = frac{2.9444}{3} approx 0.9815 ]So, approximately 0.9815 per unit time.Let me double-check:Compute ( e^{3k} ):If ( k = 0.9815 ), then ( 3k = 2.9445 ), and ( e^{2.9445} approx 19 ). Yes, that's correct.So, the growth rate ( k ) is approximately 0.9815.Problem 2: Determining the Social Networking Impact ( m )Now, the student modifies the model to include a social network effect, making the model:[ A(t) = frac{L}{1 + e^{-k(t - t_0)}} times e^{mt} ]We are told that ( L = 100 ), ( t_0 = 3 ), ( k ) is the previously calculated value (approximately 0.9815), and at ( t = 2 ), the awareness level is 30 units. We need to find ( m ).So, plugging in the known values:[ 30 = frac{100}{1 + e^{-0.9815(2 - 3)}} times e^{m times 2} ]First, let's compute the logistic part at ( t = 2 ):Compute the exponent:( 2 - 3 = -1 )So, exponent is ( -0.9815 times (-1) = 0.9815 )Thus,[ frac{100}{1 + e^{0.9815}} ]Compute ( e^{0.9815} ):Since ( e^{1} approx 2.71828 ), and 0.9815 is slightly less than 1, so ( e^{0.9815} approx 2.668 ) (using calculator: 0.9815 * ln(e) = 0.9815, so e^0.9815 ≈ 2.668).So,[ frac{100}{1 + 2.668} = frac{100}{3.668} approx 27.26 ]So, the logistic part at ( t = 2 ) is approximately 27.26.Therefore, the equation becomes:[ 30 = 27.26 times e^{2m} ]Divide both sides by 27.26:[ e^{2m} = frac{30}{27.26} approx 1.0999 ]Take natural logarithm of both sides:[ 2m = ln(1.0999) ]Compute ( ln(1.0999) ):We know that ( ln(1.1) approx 0.09531 ). Since 1.0999 is very close to 1.1, let's approximate:[ ln(1.0999) approx 0.09531 ]So,[ 2m approx 0.09531 ][ m approx 0.04765 ]So, approximately 0.04765 per unit time.Let me verify the calculations step by step to ensure accuracy.First, compute the logistic part at ( t = 2 ):[ frac{100}{1 + e^{-k(2 - 3)}} = frac{100}{1 + e^{k}} ]Wait, since ( k = frac{ln(19)}{3} approx 0.9815 ), so ( e^{k} approx e^{0.9815} approx 2.668 ). So, denominator is ( 1 + 2.668 = 3.668 ), so ( 100 / 3.668 ≈ 27.26 ). Correct.Then, ( 27.26 times e^{2m} = 30 ), so ( e^{2m} ≈ 30 / 27.26 ≈ 1.0999 ). Taking ln: ( 2m ≈ 0.0953 ), so ( m ≈ 0.04765 ). That seems correct.Alternatively, using more precise calculations:Compute ( e^{0.9815} ):Using a calculator: 0.9815Compute e^0.9815:We can use Taylor series or calculator approximation.But for precision, let's compute it:e^0.9815:We know that e^0.98 ≈ e^(1 - 0.02) = e / e^0.02 ≈ 2.71828 / 1.02020 ≈ 2.6645Similarly, e^0.9815 is slightly higher. Let's compute 0.9815 - 0.98 = 0.0015.So, e^0.9815 ≈ e^0.98 * e^0.0015 ≈ 2.6645 * 1.001501 ≈ 2.6645 + 0.004 ≈ 2.6685So, e^0.9815 ≈ 2.6685Thus, 1 + e^0.9815 ≈ 3.6685So, 100 / 3.6685 ≈ 27.26So, 27.26 * e^{2m} = 30So, e^{2m} = 30 / 27.26 ≈ 1.0999Compute ln(1.0999):We can use the approximation ln(1 + x) ≈ x - x^2/2 + x^3/3 - ... for small x.Here, x = 0.0999So, ln(1.0999) ≈ 0.0999 - (0.0999)^2 / 2 + (0.0999)^3 / 3Compute:0.0999 ≈ 0.1(0.0999)^2 ≈ 0.00998(0.0999)^3 ≈ 0.000997So,ln(1.0999) ≈ 0.0999 - 0.00998 / 2 + 0.000997 / 3≈ 0.0999 - 0.00499 + 0.000332≈ 0.0999 - 0.00499 = 0.09491 + 0.000332 ≈ 0.09524So, ln(1.0999) ≈ 0.09524Thus, 2m ≈ 0.09524So, m ≈ 0.04762So, approximately 0.0476So, rounding to four decimal places, m ≈ 0.0476Alternatively, if we use a calculator for ln(1.0999):ln(1.0999) ≈ 0.09531So, 2m = 0.09531 => m ≈ 0.047655So, approximately 0.0477Therefore, m is approximately 0.0477Summary of Calculations:1. For the first part, the growth rate ( k ) was calculated by using the initial condition and solving the logistic equation, resulting in ( k approx 0.9815 ).2. For the second part, the modified logistic model was used with the known value at ( t = 2 ) to solve for ( m ), resulting in ( m approx 0.0477 ).I think these are the correct values. Let me just recap the steps to ensure I didn't skip anything.Recap for Problem 1:- Given ( A(0) = 5 ), plug into the logistic model with ( t_0 = 3 ) and ( L = 100 ).- Solved for ( k ) by manipulating the equation, realizing that the exponent becomes positive because ( t < t_0 ).- Calculated ( k ) to be approximately 0.9815.Recap for Problem 2:- Modified model includes an exponential term ( e^{mt} ).- At ( t = 2 ), ( A(2) = 30 ).- Plugged in the known values, calculated the logistic part at ( t = 2 ) as approximately 27.26.- Solved for ( m ) by setting up the equation ( 30 = 27.26 times e^{2m} ), resulting in ( m approx 0.0477 ).Everything seems consistent. I don't see any errors in the reasoning now.Final Answer1. The growth rate ( k ) is boxed{dfrac{ln(19)}{3}}.2. The value of ( m ) is boxed{dfrac{ln(1.1)}{2}}.Wait, hold on. In the second part, I approximated ( m ) as 0.0477, which is approximately ( ln(1.1)/2 ). Because ( ln(1.1) approx 0.09531 ), so ( 0.09531 / 2 approx 0.04765 ). So, actually, ( m = frac{ln(1.1)}{2} ). That's a more precise way to write it instead of approximating.So, perhaps the exact value is ( m = frac{ln(1.1)}{2} ), since ( e^{2m} = 1.1 ), so ( 2m = ln(1.1) ), hence ( m = frac{ln(1.1)}{2} ).Yes, that's a better way to present it rather than the approximate decimal.Similarly, for the first part, ( k = frac{ln(19)}{3} ), which is exact.So, revising the final answers:1. ( k = frac{ln(19)}{3} )2. ( m = frac{ln(1.1)}{2} )So, boxed:1. boxed{dfrac{ln(19)}{3}}2. boxed{dfrac{ln(1.1)}{2}}Final Answer1. The growth rate ( k ) is boxed{dfrac{ln(19)}{3}}.2. The value of ( m ) is boxed{dfrac{ln(1.1)}{2}}.</think>

🧠:<think>Alright, so I've got this problem here where I need to decrypt a message that's been encrypted using a combination of number theory and linear algebra. It's a two-part problem, and I need to tackle each part step by step. Let me try to break it down.First, the message is encoded in a 3x3 matrix M, where each element is a prime number between 2 and 29. My task is to determine the determinant of this matrix. Since the primes are randomly selected, I need to pick nine primes within that range and construct the matrix. Then, I have to calculate the determinant.Okay, so let me start by selecting nine prime numbers between 2 and 29. The primes in that range are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. I need to choose nine of these. Let me just randomly pick them:Let's see, 2, 3, 5, 7, 11, 13, 17, 19, 23. Hmm, that's nine primes. Wait, 29 is also a prime, but I only need nine, so maybe I'll exclude 29 for now.So, my matrix M will be a 3x3 matrix with these primes. Let me arrange them in a way that seems random. Maybe:First row: 2, 3, 5Second row: 7, 11, 13Third row: 17, 19, 23Let me write that down:M = [2 3 5][7 11 13][17 19 23]Alright, now I need to calculate the determinant of this matrix. The determinant of a 3x3 matrix can be calculated using the rule of Sarrus or the general formula. I think the general formula is more straightforward here.The determinant of a 3x3 matrix:|a b c||d e f||g h i|is a(ei - fh) - b(di - fg) + c(dh - eg).So, applying this to our matrix M:a = 2, b = 3, c = 5d = 7, e = 11, f = 13g = 17, h = 19, i = 23So, determinant = 2*(11*23 - 13*19) - 3*(7*23 - 13*17) + 5*(7*19 - 11*17)Let me compute each part step by step.First, compute 11*23: 11*20=220, 11*3=33, so 220+33=253Then, 13*19: 13*20=260, minus 13=247So, 11*23 - 13*19 = 253 - 247 = 6Next, compute 7*23: 7*20=140, 7*3=21, so 140+21=161Then, 13*17: 10*17=170, 3*17=51, so 170+51=221So, 7*23 - 13*17 = 161 - 221 = -60Then, compute 7*19: 7*20=140, minus 7=13311*17: 10*17=170, 1*17=17, so 170+17=187So, 7*19 - 11*17 = 133 - 187 = -54Now, plug these back into the determinant formula:2*(6) - 3*(-60) + 5*(-54)Compute each term:2*6 = 12-3*(-60) = +1805*(-54) = -270So, determinant = 12 + 180 - 27012 + 180 = 192192 - 270 = -78So, the determinant of matrix M is -78.Wait, but determinants can be negative, right? So, that's okay.But let me double-check my calculations because sometimes I might have messed up somewhere.First part: 11*23=253, 13*19=247, 253-247=6. That seems correct.Second part: 7*23=161, 13*17=221, 161-221=-60. Correct.Third part: 7*19=133, 11*17=187, 133-187=-54. Correct.Then, 2*6=12, -3*(-60)=180, 5*(-54)=-270.12 + 180 = 192, 192 - 270 = -78. Yep, that seems right.So, determinant k is -78.But in modular arithmetic, we usually deal with positive numbers, so maybe I should take the absolute value? Or perhaps not, because the sign matters in modular inverses. Wait, no, in modular arithmetic, negative numbers can be converted to positive equivalents by adding the modulus. So, if we have -78 mod 31, we can compute it as 31 - (78 mod 31). Let me see.But before that, moving on to part 2.The second layer of encryption involves modular arithmetic. The determinant k is used as the key in the modular inverse operation under modulo 31. So, I need to find the modular inverse of k modulo 31.First, let's note that k is -78. Since we're working modulo 31, let's compute k mod 31.Compute 78 divided by 31: 31*2=62, 78-62=16. So, 78 mod 31 is 16. Therefore, -78 mod 31 is equivalent to -16 mod 31. To make it positive, add 31: -16 +31=15. So, -78 ≡15 mod31.Therefore, k ≡15 mod31.Now, we need to find the modular inverse of 15 modulo31. That is, find an integer x such that 15x ≡1 mod31.To find the modular inverse, we can use the Extended Euclidean Algorithm.Let me recall how that works. The algorithm finds integers x and y such that ax + by = gcd(a,b). If a and b are coprime, then gcd(a,b)=1, and x is the modular inverse of a modulo b.So, let's apply the Extended Euclidean Algorithm to 15 and 31.First, divide 31 by15:31 = 2*15 +1So, 1 =31 -2*15Therefore, 1 ≡ -2*15 mod31Which implies that -2 is the inverse of 15 modulo31. But we usually express the inverse as a positive number between 1 and 30.So, -2 mod31 is 29. Therefore, 15*29 ≡1 mod31.Let me verify that: 15*29=435. Now, divide 435 by31: 31*14=434, so 435-434=1. Therefore, 435≡1 mod31. Correct.Therefore, the modular inverse of 15 modulo31 is29.But wait, hold on. Let me make sure I didn't make a mistake earlier.Wait, in the first step, I had k = -78, which is equivalent to15 mod31. Then, the inverse of15 is29. So, the modular inverse is29.But let me double-check the determinant calculation again because sometimes I might have messed up.Wait, determinant was -78. Then, -78 mod31: 31*2=62, 78-62=16, so 78≡16 mod31, so -78≡-16 mod31, which is15 mod31. Correct.So, k≡15 mod31, inverse is29.But wait, the problem says: "If the determinant k does not have an inverse modulo 31, explain why and provide the smallest prime number that could replace an element in the original matrix M to ensure that the new determinant does have an inverse modulo31."In this case, k=15, which is coprime with31 because31 is prime and15<31, so gcd(15,31)=1. Therefore, 15 does have an inverse modulo31, which is29. So, in this case, we don't need to do anything else.But just to be thorough, what if the determinant had a common factor with31? Since31 is prime, the determinant must be a multiple of31 to not have an inverse. But in our case, determinant was-78, which is not a multiple of31. So, it's invertible.Therefore, the modular inverse is29.Wait, but let me think again. The determinant was-78, which is equivalent to15 mod31, and15 is invertible. So, everything is fine.But just to make sure, let me try another approach. Maybe I can compute the inverse of15 mod31 using another method, like trial and error.We need to find x such that15x ≡1 mod31.Let's compute15*1=15 mod31=1515*2=30 mod31=3015*3=45 mod31=45-31=1415*4=60 mod31=60-31=2915*5=75 mod31=75-62=1315*6=90 mod31=90-93= -3 mod31=2815*7=105 mod31=105-93=1215*8=120 mod31=120-93=2715*9=135 mod31=135-124=1115*10=150 mod31=150-155= -5 mod31=2615*11=165 mod31=165-155=1015*12=180 mod31=180-186= -6 mod31=2515*13=195 mod31=195-186=915*14=210 mod31=210-217= -7 mod31=2415*15=225 mod31=225-217=815*16=240 mod31=240-248= -8 mod31=2315*17=255 mod31=255-248=715*18=270 mod31=270-279= -9 mod31=2215*19=285 mod31=285-279=615*20=300 mod31=300-279=2115*21=315 mod31=315-310=515*22=330 mod31=330-310=2015*23=345 mod31=345-341=415*24=360 mod31=360-341=1915*25=375 mod31=375-341=34 mod31=315*26=390 mod31=390-341=49 mod31=1815*27=405 mod31=405-391=1415*28=420 mod31=420-408=1215*29=435 mod31=435-434=1Ah, there we go. 15*29=435≡1 mod31. So, x=29 is indeed the inverse.So, that confirms it.Therefore, the modular inverse of k modulo31 is29.But just to make sure, let me recap:1. I constructed a 3x3 matrix with primes 2,3,5,7,11,13,17,19,23.2. Calculated determinant as-78.3. Converted -78 mod31 to15.4. Found inverse of15 mod31 is29.So, everything checks out.But wait, just to be thorough, what if the determinant had been a multiple of31? Then, it wouldn't have an inverse. So, in that case, I would need to adjust the matrix.But in this case, determinant is-78, which is15 mod31, which is invertible.Therefore, the answer for part2 is29.But let me think about the second part again. It says: "If the determinant k does not have an inverse modulo31, explain why and provide the smallest prime number that could replace an element in the original matrix M to ensure that the new determinant does have an inverse modulo31."So, in this case, since k=15 has an inverse, we don't need to do anything. But just to explore, what if k was a multiple of31? Then, it wouldn't have an inverse because gcd(k,31)=31, which is not1. So, in that case, we would need to adjust the matrix.But in our case, it's not necessary.However, just for practice, let me consider a scenario where determinant is a multiple of31. Suppose determinant was31, then k=31≡0 mod31, which doesn't have an inverse.In that case, we would need to change one element in the matrix to make the determinant not a multiple of31.The smallest prime number is2. So, perhaps replacing a prime in the matrix with2 could adjust the determinant.But in our case, since determinant is-78, which is15 mod31, which is invertible, we don't need to do that.Therefore, the answers are:1. Determinant is-78.2. Modular inverse is29.But let me check if the problem wants the determinant as a positive number or not. Since determinant can be negative, but in modular arithmetic, we usually express it as a positive equivalent. So, -78 mod31 is15, which is what we used.But the problem says: "the determinant you found in part1 is to be used as the key k in the modular inverse operation under modulo31." So, k is-78, but in modular arithmetic, we can represent it as15 mod31.Therefore, the key k is15, and its inverse is29.But wait, actually, in modular arithmetic, the key is k mod31, which is15. So, the inverse is29.Therefore, the answers are:1. Determinant is-78.2. Modular inverse is29.But let me make sure that the determinant is indeed-78. Let me recalculate the determinant.Given matrix:[2 3 5][7 11 13][17 19 23]Compute determinant:2*(11*23 -13*19) -3*(7*23 -13*17) +5*(7*19 -11*17)Compute each minor:11*23=253, 13*19=247, so 253-247=67*23=161, 13*17=221, so 161-221=-607*19=133, 11*17=187, so 133-187=-54So, determinant=2*6 -3*(-60) +5*(-54)=12 +180 -270=192-270=-78. Correct.So, yes, determinant is-78.Therefore, the answers are:1. Determinant is-78.2. Modular inverse is29.But let me think again about the second part. The problem says: "the determinant you found in part1 is to be used as the key k in the modular inverse operation under modulo31." So, k is-78, but in modular arithmetic, we can represent it as15 mod31. So, the inverse is29.Therefore, the final answers are:1. Determinant is-78.2. Modular inverse is29.But let me check if the problem expects the determinant to be positive. Sometimes, determinants are reported as absolute values, but in this case, it's just the determinant, which can be negative.Alternatively, maybe I should have taken the absolute value before applying modulo. Let me see.If I take |k|=78, then 78 mod31=16, so inverse of16 mod31.Wait, but the problem says "the determinant you found in part1 is to be used as the key k". So, k is-78, which is equivalent to15 mod31.Therefore, the inverse is29.Alternatively, if I take absolute value, k=78, which is16 mod31, inverse of16 mod31.Wait, let me compute that as well.Compute inverse of16 mod31.Using Extended Euclidean Algorithm:31 =1*16 +1516=1*15 +115=15*1 +0So, gcd=1.Backwards:1=16 -1*15But 15=31 -1*16So, 1=16 -1*(31 -1*16)=2*16 -1*31Therefore, 2*16 ≡1 mod31. So, inverse of16 is2.But wait, that contradicts with earlier. So, if k was78, which is16 mod31, inverse is2.But in our case, k is-78≡15 mod31, inverse is29.Therefore, the answer depends on whether we take k as-78 or78.But the problem says: "the determinant you found in part1 is to be used as the key k". So, determinant is-78, so k=-78.But in modular arithmetic, we can represent negative numbers as their positive equivalents by adding the modulus.So, -78 mod31=15, as we computed earlier.Therefore, k=15 mod31, inverse is29.Therefore, the answer is29.But just to make sure, let me compute 15*29 mod31.15*29=435435 divided by31: 31*14=434, so 435-434=1.Therefore, 15*29≡1 mod31. Correct.Therefore, the modular inverse is29.So, to summarize:1. The determinant of matrix M is-78.2. The modular inverse of k=-78 modulo31 is29.Therefore, the answers are:1. Determinant: -782. Modular inverse:29But let me check if the problem expects the determinant to be positive. If so, then |k|=78, which is16 mod31, inverse is2. But the problem didn't specify to take absolute value, so I think we should stick with k=-78, which is15 mod31, inverse29.Therefore, the final answers are:1. Determinant: -782. Modular inverse:29But let me think again about the matrix. Maybe I should have chosen different primes to make the determinant positive? Or does it not matter?No, the determinant can be negative. The problem didn't specify to make it positive, so I think -78 is correct.Therefore, I think I've solved the problem correctly.</think>

🧠:<think>Okay, so I have this problem about a car collector's vintage vehicle with a faulty engine control system. The system is described by a nonlinear differential equation, and I need to solve two parts. Let me try to tackle them one by one.Problem 1: Solving the Differential EquationThe equation given is:[ frac{dA(t)}{dt} = k_1 A(t) - k_2 f(A(t)) ]With ( f(A(t)) = A(t)^2 ), ( k_1 = 5 ), and ( k_2 = 3 ). So plugging these in, the equation becomes:[ frac{dA}{dt} = 5A - 3A^2 ]Hmm, this looks like a first-order ordinary differential equation (ODE). It's nonlinear because of the ( A^2 ) term. I remember that equations of the form ( frac{dA}{dt} = f(A) ) can sometimes be solved by separation of variables. Let me try that.So, rewrite the equation as:[ frac{dA}{dt} = 5A - 3A^2 ]Which can be written as:[ frac{dA}{5A - 3A^2} = dt ]Now, I need to integrate both sides. Let me handle the left side first. The integral of ( frac{1}{5A - 3A^2} dA ). This seems like a rational function, so partial fractions might work here.Let me factor the denominator:( 5A - 3A^2 = A(5 - 3A) )So, I can express ( frac{1}{A(5 - 3A)} ) as partial fractions:Let ( frac{1}{A(5 - 3A)} = frac{C}{A} + frac{D}{5 - 3A} )Multiplying both sides by ( A(5 - 3A) ):1 = C(5 - 3A) + D(A)Now, let's solve for C and D. Let me plug in A = 0:1 = C(5 - 0) + D(0) => 1 = 5C => C = 1/5Next, plug in A = 5/3 to eliminate C:1 = C(5 - 3*(5/3)) + D*(5/3) => 1 = C(5 - 5) + (5/3)D => 1 = 0 + (5/3)D => D = 3/5So, the partial fractions are:[ frac{1}{5A} + frac{3}{5(5 - 3A)} ]Therefore, the integral becomes:[ int left( frac{1}{5A} + frac{3}{5(5 - 3A)} right) dA = int dt ]Let me compute each integral separately.First integral: ( frac{1}{5} int frac{1}{A} dA = frac{1}{5} ln|A| + C_1 )Second integral: ( frac{3}{5} int frac{1}{5 - 3A} dA ). Let me make a substitution here. Let u = 5 - 3A, then du = -3 dA, so dA = -du/3.So, the integral becomes:( frac{3}{5} times left( -frac{1}{3} right) int frac{1}{u} du = -frac{1}{5} ln|u| + C_2 = -frac{1}{5} ln|5 - 3A| + C_2 )Putting it all together, the left side integral is:[ frac{1}{5} ln|A| - frac{1}{5} ln|5 - 3A| + C ]Where C is the constant of integration. The right side integral is:[ int dt = t + C' ]So, combining both sides:[ frac{1}{5} ln|A| - frac{1}{5} ln|5 - 3A| = t + C ]Multiply both sides by 5 to simplify:[ ln|A| - ln|5 - 3A| = 5t + C' ]Combine the logs:[ lnleft| frac{A}{5 - 3A} right| = 5t + C' ]Exponentiate both sides to eliminate the logarithm:[ left| frac{A}{5 - 3A} right| = e^{5t + C'} = e^{C'} e^{5t} ]Let me denote ( e^{C'} ) as another constant, say K, which can be positive or negative. So:[ frac{A}{5 - 3A} = K e^{5t} ]Now, solve for A:Multiply both sides by (5 - 3A):[ A = K e^{5t} (5 - 3A) ]Expand the right side:[ A = 5K e^{5t} - 3K e^{5t} A ]Bring all terms with A to the left:[ A + 3K e^{5t} A = 5K e^{5t} ]Factor out A:[ A (1 + 3K e^{5t}) = 5K e^{5t} ]Solve for A:[ A = frac{5K e^{5t}}{1 + 3K e^{5t}} ]This is the general solution. Let me rewrite it for clarity:[ A(t) = frac{5K e^{5t}}{1 + 3K e^{5t}} ]Alternatively, I can write this as:[ A(t) = frac{5}{3 + (5/(3K)) e^{-5t}} ]But maybe it's better to leave it in the exponential form. Let me see if I can express it differently. Let me denote ( C = 5K ), then:[ A(t) = frac{C e^{5t}}{1 + (3C/5) e^{5t}} ]But perhaps it's clearer to just keep it as:[ A(t) = frac{5K e^{5t}}{1 + 3K e^{5t}} ]Where K is a constant determined by initial conditions.Behavior as t approaches infinityNow, I need to determine the behavior of A(t) as t approaches infinity.Looking at the general solution:[ A(t) = frac{5K e^{5t}}{1 + 3K e^{5t}} ]As t becomes very large, the exponential terms dominate. Let me factor out ( e^{5t} ) from numerator and denominator:[ A(t) = frac{5K e^{5t}}{e^{5t}(3K + e^{-5t})} = frac{5K}{3K + e^{-5t}} ]As t approaches infinity, ( e^{-5t} ) approaches zero. So:[ lim_{t to infty} A(t) = frac{5K}{3K} = frac{5}{3} ]So, regardless of the initial condition (as long as K ≠ 0), the air-fuel ratio A(t) approaches 5/3 as t becomes large.Wait, but let me think about this. If K is zero, then A(t) would be zero for all t, but that's a trivial solution. So, assuming K ≠ 0, which is the case if the initial condition A(0) ≠ 0.So, the system tends to a steady state of 5/3 as time goes to infinity.Problem 2: Introducing a PerturbationNow, the collector wants to introduce a time-dependent perturbation ( g(t) = sin(omega t) ). So, the original differential equation is modified to include this perturbation.The original equation was:[ frac{dA}{dt} = 5A - 3A^2 ]So, adding the perturbation, the new equation becomes:[ frac{dA}{dt} = 5A - 3A^2 + sin(omega t) ]We need to find the particular solution when ( omega = 2 ), and analyze the stability.Hmm, so this is a nonhomogeneous nonlinear differential equation. Solving nonlinear ODEs with forcing terms can be tricky. I might need to use methods like perturbation techniques or look for particular solutions.But since the equation is nonlinear, finding an exact solution might not be straightforward. Maybe I can consider a linearization around the steady state found in part 1, which is 5/3.Let me denote the perturbed solution as ( A(t) = A_s + delta(t) ), where ( A_s = 5/3 ) is the steady state, and ( delta(t) ) is a small perturbation.Substituting into the differential equation:[ frac{d}{dt}(A_s + delta) = 5(A_s + delta) - 3(A_s + delta)^2 + sin(2t) ]Since ( A_s ) is a constant, its derivative is zero:[ frac{ddelta}{dt} = 5A_s + 5delta - 3(A_s^2 + 2A_s delta + delta^2) + sin(2t) ]Simplify term by term:First, compute 5A_s:5*(5/3) = 25/3Then, 5δ remains as is.Next, expand the quadratic term:-3(A_s^2 + 2A_s δ + δ^2) = -3A_s^2 - 6A_s δ - 3δ^2So, putting it all together:[ frac{ddelta}{dt} = 25/3 + 5delta - 3A_s^2 - 6A_s δ - 3δ^2 + sin(2t) ]Compute constants:Compute 25/3 - 3A_s^2:A_s = 5/3, so A_s^2 = 25/9Thus, 25/3 - 3*(25/9) = 25/3 - 25/3 = 0So, the constant terms cancel out.Now, collect terms with δ:5δ - 6A_s δ = δ(5 - 6*(5/3)) = δ(5 - 10) = -5δAnd the nonlinear term: -3δ^2So, the equation becomes:[ frac{ddelta}{dt} = -5delta - 3delta^2 + sin(2t) ]This is still a nonlinear equation because of the δ² term. However, if the perturbation δ is small, maybe we can neglect the δ² term as a first approximation. Let me consider that.Assuming δ is small, so δ² is negligible, then the equation approximates to:[ frac{ddelta}{dt} ≈ -5δ + sin(2t) ]This is a linear nonhomogeneous ODE, which is easier to solve.The equation is:[ frac{ddelta}{dt} + 5δ = sin(2t) ]This is a linear first-order ODE. The integrating factor method can be used.The integrating factor (IF) is:[ IF = e^{int 5 dt} = e^{5t} ]Multiply both sides by IF:[ e^{5t} frac{ddelta}{dt} + 5 e^{5t} δ = e^{5t} sin(2t) ]The left side is the derivative of (δ e^{5t}):[ frac{d}{dt} (δ e^{5t}) = e^{5t} sin(2t) ]Integrate both sides:[ δ e^{5t} = int e^{5t} sin(2t) dt + C ]Now, compute the integral ( int e^{5t} sin(2t) dt ). I remember that integrals of the form ( int e^{at} sin(bt) dt ) can be solved using integration by parts twice and then solving for the integral.Let me denote I = ( int e^{5t} sin(2t) dt )Let u = sin(2t), dv = e^{5t} dtThen, du = 2 cos(2t) dt, v = (1/5) e^{5t}So, integration by parts gives:I = uv - ∫ v du = (1/5) e^{5t} sin(2t) - (2/5) ∫ e^{5t} cos(2t) dtNow, let me compute the remaining integral, let me call it J = ( int e^{5t} cos(2t) dt )Again, integration by parts:Let u = cos(2t), dv = e^{5t} dtThen, du = -2 sin(2t) dt, v = (1/5) e^{5t}So, J = uv - ∫ v du = (1/5) e^{5t} cos(2t) + (2/5) ∫ e^{5t} sin(2t) dtNotice that the integral now is I again. So, J = (1/5) e^{5t} cos(2t) + (2/5) ISubstitute J back into the expression for I:I = (1/5) e^{5t} sin(2t) - (2/5) [ (1/5) e^{5t} cos(2t) + (2/5) I ]Expand:I = (1/5) e^{5t} sin(2t) - (2/25) e^{5t} cos(2t) - (4/25) IBring the (4/25) I term to the left:I + (4/25) I = (1/5) e^{5t} sin(2t) - (2/25) e^{5t} cos(2t)Factor I:(29/25) I = (1/5) e^{5t} sin(2t) - (2/25) e^{5t} cos(2t)Multiply both sides by 25/29:I = (5/29) e^{5t} sin(2t) - (2/29) e^{5t} cos(2t) + CSo, going back to the equation:δ e^{5t} = I + C = (5/29) e^{5t} sin(2t) - (2/29) e^{5t} cos(2t) + CDivide both sides by e^{5t}:δ(t) = (5/29) sin(2t) - (2/29) cos(2t) + C e^{-5t}So, the general solution for δ(t) is:δ(t) = (5/29) sin(2t) - (2/29) cos(2t) + C e^{-5t}Therefore, the particular solution for A(t) is:A(t) = A_s + δ(t) = 5/3 + (5/29) sin(2t) - (2/29) cos(2t) + C e^{-5t}The term C e^{-5t} is the transient solution, which decays to zero as t increases, assuming C is a constant determined by initial conditions.Stability AnalysisNow, to analyze the stability of the solution. The homogeneous solution is C e^{-5t}, which decays exponentially to zero because the exponent is negative. The particular solution is the steady-state response, which is a sinusoidal function with amplitude determined by the coefficients.Since the transient term dies out, the system approaches the particular solution as t increases. The particular solution is a bounded oscillation, so the system is stable. Specifically, it's asymptotically stable because any initial perturbation decays away, leaving the system to oscillate around the steady state 5/3.However, I should note that this analysis assumes that the perturbation δ is small, so the nonlinear term δ² was neglected. If δ is not small, the nonlinear term could affect the stability. But for small perturbations, the system remains stable.Summary of Findings1. The general solution for A(t) without perturbation is:[ A(t) = frac{5K e^{5t}}{1 + 3K e^{5t}} ]As t approaches infinity, A(t) approaches 5/3.2. When introducing a perturbation ( sin(2t) ), the particular solution is:[ A(t) = frac{5}{3} + frac{5}{29} sin(2t) - frac{2}{29} cos(2t) + C e^{-5t} ]The solution is stable because the transient term decays, leaving a bounded oscillation around the steady state.Final Answer1. The general solution is ( boxed{A(t) = dfrac{5K e^{5t}}{1 + 3K e^{5t}}} ) and as ( t to infty ), ( A(t) ) approaches ( boxed{dfrac{5}{3}} ).2. The particular solution is ( boxed{A(t) = dfrac{5}{3} + dfrac{5}{29} sin(2t) - dfrac{2}{29} cos(2t) + C e^{-5t}} ) and the solution is stable.</think>