🧠:<think>Alright, so I have this problem about a museum curator designing an exhibition space shaped like a right-angled triangular prism. The base is a right triangle with vertices at A(0,0), B(6,0), and C(0,8). The height of the prism is 15 units. There are two parts to the problem: first, calculating the volume of the prism, and second, finding the coordinates of point D, which is above the centroid of the triangular base, and then finding the distances from D to each vertex of the base.Starting with the first part: calculating the volume of the prism. I remember that the volume of a prism is given by the area of the base multiplied by the height. Since the base is a right-angled triangle, I can calculate its area easily.The coordinates of the vertices are A(0,0), B(6,0), and C(0,8). So, the legs of the right triangle are along the x-axis and y-axis. The length of AB is 6 units, and the length of AC is 8 units. The area of the triangle is (base * height)/2, which would be (6 * 8)/2. Let me compute that: 6 times 8 is 48, divided by 2 is 24. So, the area of the base is 24 square units.Now, the height of the prism is given as 15 units. So, the volume should be the area of the base times the height, which is 24 * 15. Let me calculate that: 24 times 15. Hmm, 20 times 15 is 300, and 4 times 15 is 60, so 300 + 60 is 360. So, the volume is 360 cubic units.Thinking about the significance of the geometric shape in contemporary art, a right-angled triangular prism can symbolize balance and symmetry. The right angle suggests a clear, structured form, while the prism adds depth, creating a three-dimensional aspect that can reflect the complexity and layers in modern art. The volume, being a measure of space, might represent the capacity or the immersive experience the exhibition aims to provide.Moving on to the second part: finding the coordinates of point D, which is directly above the centroid of the triangular base at a distance of 10 units along the height of the prism. Then, I need to calculate the distance from D to each vertex of the triangular base.First, I need to find the centroid of the triangular base. The centroid of a triangle is the average of its vertices' coordinates. So, for triangle ABC with vertices at A(0,0), B(6,0), and C(0,8), the centroid (let's call it G) can be found by averaging the x-coordinates and the y-coordinates.Calculating the x-coordinate of G: (0 + 6 + 0)/3 = 6/3 = 2.Calculating the y-coordinate of G: (0 + 0 + 8)/3 = 8/3 ≈ 2.6667.So, the centroid G is at (2, 8/3). Since the prism extends perpendicularly from the base, the z-coordinate for the base is 0, and the height is along the z-axis. Therefore, point D is directly above G, so its x and y coordinates are the same as G, and its z-coordinate is 10 units above the base.Hence, the coordinates of D are (2, 8/3, 10).Now, I need to find the distance from D to each vertex of the triangular base: A(0,0,0), B(6,0,0), and C(0,8,0).To find the distance between two points in 3D space, I can use the distance formula: distance = sqrt[(x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2].Starting with distance DA: from D(2, 8/3, 10) to A(0,0,0).Calculating the differences:x: 0 - 2 = -2y: 0 - 8/3 = -8/3z: 0 - 10 = -10Squaring each difference:(-2)^2 = 4(-8/3)^2 = 64/9(-10)^2 = 100Adding them up: 4 + 64/9 + 100.First, convert 4 and 100 to ninths to add them all together.4 = 36/9100 = 900/9So, 36/9 + 64/9 + 900/9 = (36 + 64 + 900)/9 = (1000)/9Therefore, the distance DA is sqrt(1000/9) = (sqrt(1000))/3.Simplify sqrt(1000): sqrt(100*10) = 10*sqrt(10). So, DA is (10*sqrt(10))/3 units.Next, distance DB: from D(2, 8/3, 10) to B(6,0,0).Differences:x: 6 - 2 = 4y: 0 - 8/3 = -8/3z: 0 - 10 = -10Squaring each:4^2 = 16(-8/3)^2 = 64/9(-10)^2 = 100Adding them: 16 + 64/9 + 100.Convert 16 and 100 to ninths:16 = 144/9100 = 900/9So, 144/9 + 64/9 + 900/9 = (144 + 64 + 900)/9 = (1108)/9Distance DB is sqrt(1108/9) = sqrt(1108)/3.Simplify sqrt(1108). Let's see if 1108 can be factored into squares.1108 divided by 4 is 277. 277 is a prime number, I believe. So, sqrt(1108) = sqrt(4*277) = 2*sqrt(277). Therefore, DB is (2*sqrt(277))/3 units.Now, distance DC: from D(2, 8/3, 10) to C(0,8,0).Differences:x: 0 - 2 = -2y: 8 - 8/3 = 16/3z: 0 - 10 = -10Squaring each:(-2)^2 = 4(16/3)^2 = 256/9(-10)^2 = 100Adding them: 4 + 256/9 + 100.Convert 4 and 100 to ninths:4 = 36/9100 = 900/9So, 36/9 + 256/9 + 900/9 = (36 + 256 + 900)/9 = (1192)/9Distance DC is sqrt(1192/9) = sqrt(1192)/3.Simplify sqrt(1192). Let's factor 1192.1192 divided by 4 is 298. 298 divided by 2 is 149, which is a prime number. So, sqrt(1192) = sqrt(4*298) = 2*sqrt(298). Therefore, DC is (2*sqrt(298))/3 units.Wait, let me double-check the calculations for DC:Wait, the y-coordinate difference was 8 - 8/3. 8 is 24/3, so 24/3 - 8/3 = 16/3. Squared is 256/9, which is correct.x difference squared is 4, z difference squared is 100. So, 4 + 256/9 + 100. Converting 4 and 100: 4 is 36/9, 100 is 900/9. 36 + 256 + 900 is 1192, so 1192/9. So, sqrt(1192)/3. 1192 factors: 4*298, as above. So, yes, 2*sqrt(298)/3.So, to recap, the distances are:DA: (10*sqrt(10))/3DB: (2*sqrt(277))/3DC: (2*sqrt(298))/3I should also verify if these distances make sense. Since D is above the centroid, which is closer to A than to B or C, I would expect DA to be the shortest distance, followed by DB and DC. Let me approximate the numerical values to check.Compute DA: sqrt(1000)/3 ≈ 31.6227766/3 ≈ 10.5409DB: sqrt(1108)/3 ≈ 33.3049/3 ≈ 11.1016DC: sqrt(1192)/3 ≈ 34.5254/3 ≈ 11.5085So, DA ≈10.54, DB≈11.10, DC≈11.51. That seems reasonable, as D is closer to A than to B or C.Wait, but let me think: the centroid is at (2, 8/3). So, in the base triangle, A is at (0,0), so the centroid is 2 units along x and 8/3 units along y from A. So, in 3D, D is 2 units in x, 8/3 in y, and 10 units up in z. So, the distance from D to A is sqrt(2^2 + (8/3)^2 + 10^2). Which is sqrt(4 + 64/9 + 100). Which is sqrt(1000/9). So, that's correct.Similarly, for DB: distance from D(2,8/3,10) to B(6,0,0). The differences are 4 in x, -8/3 in y, -10 in z. Squared: 16, 64/9, 100. Sum: 16 + 64/9 + 100 = (144 + 64 + 900)/9 = 1108/9. So, sqrt(1108)/3. Correct.And for DC: differences are -2 in x, 16/3 in y, -10 in z. Squared: 4, 256/9, 100. Sum: 4 + 256/9 + 100 = (36 + 256 + 900)/9 = 1192/9. sqrt(1192)/3. Correct.So, all the distances seem to be calculated correctly.Therefore, summarizing:1. The volume of the prism is 360 cubic units.2. The coordinates of point D are (2, 8/3, 10). The distances from D to A, B, and C are (10√10)/3, (2√277)/3, and (2√298)/3 units, respectively.I think that's all. I don't see any mistakes in my calculations, but let me just quickly recap:- Volume: Area of base (24) times height (15) is 360. Correct.- Centroid: ( (0+6+0)/3, (0+0+8)/3 ) = (2, 8/3). Correct.- Point D is 10 units above centroid, so z-coordinate is 10. So, (2, 8/3, 10). Correct.- Distances calculated using 3D distance formula, each time subtracting coordinates, squaring, adding, taking square root. All steps seem correct.Yes, I think I'm confident with these answers.Final Answer1. The volume of the prism is boxed{360} cubic units.2. The coordinates of point ( D ) are (left(2, dfrac{8}{3}, 10right)). The distances from ( D ) to each vertex are: - ( DA = dfrac{10sqrt{10}}{3} ) units, - ( DB = dfrac{2sqrt{277}}{3} ) units, - ( DC = dfrac{2sqrt{298}}{3} ) units.So, the final answers are:1. boxed{360}2. Coordinates of ( D ): (boxed{left(2, dfrac{8}{3}, 10right)}) Distances: - ( DA = boxed{dfrac{10sqrt{10}}{3}} ) - ( DB = boxed{dfrac{2sqrt{277}}{3}} ) - ( DC = boxed{dfrac{2sqrt{298}}{3}} )</think>

🧠:<think>Alright, so I have this problem where I need to analyze the revenue growth and market share dynamics of three toy companies: ToyCo, PlayInc, and FunToys. The goal is to figure out which company has the highest average annual growth rate and which one increased its market share the most over three years. Let me break this down step by step.Starting with the first part: Revenue Growth Analysis. Each company has its revenue modeled by an exponential function. I need to calculate the annual growth rate for each and determine which has the highest average over three years.First, let me recall that an exponential growth function is generally of the form ( R(t) = R_0 times e^{rt} ), where ( R_0 ) is the initial revenue, ( r ) is the growth rate, and ( t ) is time in years. The growth rate ( r ) is essentially the continuous growth rate. However, when we talk about annual growth rate, we might need to convert this into a more familiar percentage, perhaps using the formula for compound interest or something similar.Wait, actually, in finance, the continuous growth rate can be converted to an annual growth rate using the formula ( (e^r - 1) times 100% ). So, for each company, I can take their respective ( r ) values, plug them into this formula, and get the annual growth rate.Looking at the given functions:- ToyCo: ( R_T(t) = 50 times e^{0.05t} )- PlayInc: ( R_P(t) = 40 times e^{0.08t} )- FunToys: ( R_F(t) = 60 times e^{0.03t} )So, their continuous growth rates ( r ) are 0.05, 0.08, and 0.03 respectively.Calculating the annual growth rates:For ToyCo: ( e^{0.05} - 1 approx 1.05127 - 1 = 0.05127 ) or 5.127%For PlayInc: ( e^{0.08} - 1 approx 1.083287 - 1 = 0.083287 ) or 8.3287%For FunToys: ( e^{0.03} - 1 approx 1.03045 - 1 = 0.03045 ) or 3.045%So, comparing these, PlayInc has the highest annual growth rate at approximately 8.33%, followed by ToyCo at about 5.13%, and FunToys at around 3.05%.Wait, but the question says "average annual growth rate over the three years." Hmm, does that mean I need to compute the average of the growth rates each year, or is it just the same as the continuous growth rate since it's exponential?I think in exponential growth, the growth rate is constant, so the average annual growth rate over any period is just the same as the continuous growth rate converted to annual. So, I think my initial calculation is sufficient. Therefore, PlayInc has the highest average annual growth rate.Moving on to the second part: Market Share Dynamics. The total market size is given by ( M(t) = 200 times (1 + 0.06t) ). So, at any time t, the total market is 200 million dollars multiplied by (1 + 0.06t). That means the market is growing linearly over time.Each company's market share is proportional to its revenue. So, at any time t, the market share of each company is their revenue divided by the total market size.First, I need to find the market share at t = 3 years. Then, determine how the market share changes from t = 0 to t = 3. Finally, conclude which company had the most significant increase in market share.Let me structure this:1. Calculate each company's revenue at t = 0 and t = 3.2. Calculate the total market size at t = 0 and t = 3.3. Compute market shares at both times.4. Determine the change in market share for each company.5. Identify which company had the largest increase.Starting with step 1: Revenue at t = 0 and t = 3.For ToyCo:At t = 0: ( R_T(0) = 50 times e^{0} = 50 ) million.At t = 3: ( R_T(3) = 50 times e^{0.05 times 3} = 50 times e^{0.15} approx 50 times 1.1618 = 58.09 ) million.For PlayInc:At t = 0: ( R_P(0) = 40 times e^{0} = 40 ) million.At t = 3: ( R_P(3) = 40 times e^{0.08 times 3} = 40 times e^{0.24} approx 40 times 1.27125 = 50.85 ) million.For FunToys:At t = 0: ( R_F(0) = 60 times e^{0} = 60 ) million.At t = 3: ( R_F(3) = 60 times e^{0.03 times 3} = 60 times e^{0.09} approx 60 times 1.09417 = 65.65 ) million.Step 2: Total market size at t = 0 and t = 3.At t = 0: ( M(0) = 200 times (1 + 0.06 times 0) = 200 times 1 = 200 ) million.At t = 3: ( M(3) = 200 times (1 + 0.06 times 3) = 200 times (1 + 0.18) = 200 times 1.18 = 236 ) million.Step 3: Compute market shares.At t = 0:- ToyCo: ( frac{50}{200} = 0.25 ) or 25%- PlayInc: ( frac{40}{200} = 0.2 ) or 20%- FunToys: ( frac{60}{200} = 0.3 ) or 30%At t = 3:- ToyCo: ( frac{58.09}{236} approx 0.246 ) or 24.6%- PlayInc: ( frac{50.85}{236} approx 0.215 ) or 21.5%- FunToys: ( frac{65.65}{236} approx 0.278 ) or 27.8%Step 4: Determine the change in market share.For each company, subtract their market share at t = 0 from t = 3.- ToyCo: 24.6% - 25% = -0.4%- PlayInc: 21.5% - 20% = +1.5%- FunToys: 27.8% - 30% = -2.2%So, ToyCo's market share decreased by 0.4%, PlayInc's increased by 1.5%, and FunToys' decreased by 2.2%.Therefore, PlayInc had the most significant increase in market share over the three-year period.Wait, let me double-check my calculations because sometimes when dealing with percentages, it's easy to make a mistake.First, revenues:ToyCo at t=3: 50*e^0.15. e^0.15 is approximately 1.1618, so 50*1.1618=58.09. That seems correct.PlayInc at t=3: 40*e^0.24. e^0.24 is approximately 1.27125, so 40*1.27125=50.85. Correct.FunToys at t=3: 60*e^0.09. e^0.09≈1.09417, so 60*1.09417≈65.65. Correct.Total market at t=3: 200*(1+0.18)=236. Correct.Market shares at t=3:ToyCo: 58.09/236≈0.246 (24.6%). Correct.PlayInc: 50.85/236≈0.215 (21.5%). Correct.FunToys: 65.65/236≈0.278 (27.8%). Correct.Changes:ToyCo: 24.6 - 25 = -0.4%. Correct.PlayInc: 21.5 - 20 = +1.5%. Correct.FunToys: 27.8 - 30 = -2.2%. Correct.So, yes, PlayInc increased by 1.5%, which is the highest increase. ToyCo slightly decreased, and FunToys decreased more.Therefore, the conclusion is that PlayInc has the highest average annual growth rate and also the most significant increase in market share over the three years.But wait, let me think again. The market share is based on revenue. So, even though PlayInc has a higher growth rate, their market share increased because their revenue growth outpaced the market growth.The market itself is growing at 6% per year (since M(t) = 200*(1 + 0.06t)). So, the market size increases by 6% each year. But PlayInc's revenue is growing at 8.33% annually, which is higher than the market growth rate, so their market share should increase.Similarly, ToyCo's revenue grows at 5.13%, which is less than the market's 6%, so their market share decreases. FunToys grows at 3.05%, also less than 6%, so their market share decreases as well.This makes sense. So, the company with the highest growth rate above the market growth rate will gain market share, and PlayInc is the only one with a higher growth rate than the market.Therefore, my conclusion seems solid.Final Answer1. The company with the highest average annual growth rate is boxed{PlayInc}.2. The company with the most significant increase in market share is boxed{PlayInc}.</think>

🧠:<think>Okay, so I have this problem about a food influencer reviewing a student's fusion dish. There are two parts to it. Let me tackle them one by one.Starting with the first part: Determine the total exposure over the first 24 hours after the review is posted. The reach is modeled by the function ( R(t) = 5000 + 2000 sinleft(frac{pi t}{6}right) ). Hmm, so I need to find the total number of people reached over 24 hours. Wait, does that mean I need to integrate R(t) from t=0 to t=24? Because integration would give me the total exposure over that time period. Yeah, that makes sense. So, total exposure is the integral of R(t) dt from 0 to 24.Let me write that down:Total Exposure = ( int_{0}^{24} R(t) , dt = int_{0}^{24} left(5000 + 2000 sinleft(frac{pi t}{6}right)right) dt )Alright, so I can split this integral into two parts:( int_{0}^{24} 5000 , dt + int_{0}^{24} 2000 sinleft(frac{pi t}{6}right) , dt )Calculating the first integral: ( int_{0}^{24} 5000 , dt ). That's straightforward. The integral of a constant is just the constant times the interval length. So, 5000 * (24 - 0) = 5000 * 24.Let me compute that: 5000 * 24. 5000*20 is 100,000 and 5000*4 is 20,000, so total is 120,000.Okay, so the first part is 120,000.Now the second integral: ( int_{0}^{24} 2000 sinleft(frac{pi t}{6}right) , dt ). Hmm, I need to remember how to integrate sine functions. The integral of sin(ax) dx is (-1/a) cos(ax) + C. So, applying that here.Let me make a substitution. Let u = (π t)/6, so du/dt = π/6, which means dt = (6/π) du.So, substituting, the integral becomes:2000 * ∫ sin(u) * (6/π) duWhich is (2000 * 6 / π) ∫ sin(u) duThe integral of sin(u) is -cos(u) + C, so:(2000 * 6 / π) * (-cos(u)) evaluated from t=0 to t=24.But wait, I need to express the limits in terms of u. When t=0, u=0. When t=24, u=(π *24)/6 = 4π.So, plugging back in:(2000 * 6 / π) * [ -cos(4π) + cos(0) ]Compute cos(4π) and cos(0). Cos(4π) is 1 because cosine has a period of 2π, so 4π is two full periods. Cos(0) is also 1.So, substituting:(2000 * 6 / π) * [ -1 + 1 ] = (2000 * 6 / π) * 0 = 0.Wait, that's interesting. So the integral of the sine function over a full period is zero? That makes sense because the positive and negative areas cancel out over a full period. Since 24 hours is 4π in terms of the argument, which is two full periods (each period is 2π). So, yeah, the integral is zero.Therefore, the second integral contributes nothing to the total exposure.So, the total exposure is just 120,000 people.Wait, but that seems too straightforward. Let me double-check.The function R(t) = 5000 + 2000 sin(πt/6). So, it's a sine wave with amplitude 2000, oscillating around 5000. The average value of the sine function over a full period is zero, so the average reach is 5000. Therefore, over 24 hours, the total exposure should be 5000 * 24 = 120,000. Yep, that matches. So, that seems correct.Alright, so part 1 is 120,000.Moving on to part 2: The student's fusion creation has a quality score Q(x) modeled by a quadratic function ( Q(x) = -2x^2 + 12x + 20 ). We need to find the optimal number of ingredients x that maximizes the quality score and determine the maximum quality score.Quadratic functions have their maximum or minimum at the vertex. Since the coefficient of x^2 is negative (-2), this parabola opens downward, so the vertex is the maximum point.The formula for the x-coordinate of the vertex of a parabola given by ( ax^2 + bx + c ) is x = -b/(2a).So, here, a = -2, b = 12.So, x = -12/(2*(-2)) = -12/(-4) = 3.So, the optimal number of ingredients is 3.Now, to find the maximum quality score, plug x=3 back into Q(x):Q(3) = -2*(3)^2 + 12*(3) + 20.Calculating step by step:-2*(9) + 36 + 20 = -18 + 36 + 20.-18 + 36 is 18, plus 20 is 38.So, the maximum quality score is 38.Wait, let me double-check the calculation:-2*(3)^2 = -2*9 = -1812*3 = 36So, -18 + 36 = 1818 + 20 = 38. Yep, that's correct.Alternatively, another way to find the vertex is to complete the square. Let me try that to confirm.Starting with Q(x) = -2x^2 + 12x + 20.Factor out the coefficient of x^2 from the first two terms:Q(x) = -2(x^2 - 6x) + 20.Now, to complete the square inside the parentheses:Take half of -6, which is -3, square it: 9.So, add and subtract 9 inside the parentheses:Q(x) = -2[(x^2 - 6x + 9 - 9)] + 20Which is -2[(x - 3)^2 - 9] + 20Distribute the -2:Q(x) = -2(x - 3)^2 + 18 + 20Simplify:Q(x) = -2(x - 3)^2 + 38So, the vertex is at (3, 38), which confirms our earlier result.Therefore, the optimal number of ingredients is 3, and the maximum quality score is 38.I think that's solid. Both methods give the same result, so I feel confident.Final Answer1. The total exposure over the first 24 hours is boxed{120000} people.2. The optimal number of ingredients is boxed{3}, and the maximum quality score is boxed{38}.</think>

🧠:<think>Alright, so I've got this problem about the Volga River modeled by a parametric curve. It's divided into two sub-problems. Let me tackle them one by one.Starting with Sub-problem 1: I need to find the length of the river segment over the interval ( t in [0, T] ). The parametric equations are given as:- ( x(t) = a sin(bt) )- ( y(t) = c cos(dt) )- ( z(t) = e^{-frac{t^2}{2}} )And it's mentioned that ( b = d ). So, let's note that ( b = d ), which might simplify things later.First, I remember that the length of a parametric curve from ( t = 0 ) to ( t = T ) is given by the integral of the magnitude of the derivative of the position vector ( mathbf{r}(t) ). The formula is:[ L = int_{0}^{T} sqrt{left( frac{dx}{dt} right)^2 + left( frac{dy}{dt} right)^2 + left( frac{dz}{dt} right)^2} , dt ]So, I need to compute the derivatives ( x'(t) ), ( y'(t) ), and ( z'(t) ), square each, add them up, take the square root, and integrate from 0 to T.Let's compute each derivative step by step.1. ( x(t) = a sin(bt) ) So, ( x'(t) = a b cos(bt) )2. ( y(t) = c cos(dt) ) Since ( b = d ), let's denote ( b = d = k ) for simplicity. So, ( y(t) = c cos(kt) ) Then, ( y'(t) = -c k sin(kt) )3. ( z(t) = e^{-frac{t^2}{2}} ) So, ( z'(t) = e^{-frac{t^2}{2}} cdot (-t) = -t e^{-frac{t^2}{2}} )Now, let's square each derivative:1. ( (x'(t))^2 = (a b cos(bt))^2 = a^2 b^2 cos^2(bt) )2. ( (y'(t))^2 = (-c k sin(kt))^2 = c^2 k^2 sin^2(kt) )3. ( (z'(t))^2 = (-t e^{-frac{t^2}{2}})^2 = t^2 e^{-t^2} )Since ( b = k ), we can write ( (x'(t))^2 + (y'(t))^2 = a^2 b^2 cos^2(bt) + c^2 b^2 sin^2(bt) )So, the integrand becomes:[ sqrt{a^2 b^2 cos^2(bt) + c^2 b^2 sin^2(bt) + t^2 e^{-t^2}} ]Hmm, that looks a bit complicated. Let me see if I can factor out ( b^2 ) from the first two terms:[ sqrt{b^2 (a^2 cos^2(bt) + c^2 sin^2(bt)) + t^2 e^{-t^2}} ]I don't think this simplifies easily. Maybe we can write it as:[ sqrt{b^2 (a^2 cos^2(bt) + c^2 sin^2(bt)) + t^2 e^{-t^2}} ]But integrating this from 0 to T might not be straightforward. I wonder if there's a way to simplify this expression further or if it's a standard integral.Wait, let's consider the first part: ( a^2 cos^2(bt) + c^2 sin^2(bt) ). If ( a = c ), this would simplify to ( a^2 (cos^2(bt) + sin^2(bt)) = a^2 ). But since ( a ) and ( c ) are different constants, we can't do that. So, it remains as is.So, the integrand is:[ sqrt{b^2 (a^2 cos^2(bt) + c^2 sin^2(bt)) + t^2 e^{-t^2}} ]This seems like a non-trivial integral. I don't think it has an elementary antiderivative. Maybe we can express it in terms of elliptic integrals or something, but I'm not sure.Alternatively, perhaps the problem expects us to leave the answer in terms of an integral, rather than computing it explicitly. Let me check the problem statement again.It says, "Determine the length of the segment...". It doesn't specify whether to compute it explicitly or just set up the integral. Hmm.Looking back: It says "Determine the length...", which might imply that we need to compute it, but given the integrand, it's not obvious. Maybe there's a trick or substitution I can use.Wait, let's consider the expression inside the square root:( b^2 (a^2 cos^2(bt) + c^2 sin^2(bt)) + t^2 e^{-t^2} )Is there any way to combine these terms or find a substitution?Alternatively, perhaps the problem is designed so that when ( b = d ), some terms cancel or combine nicely. Let me see.Wait, ( x(t) = a sin(bt) ) and ( y(t) = c cos(bt) ). So, in the x-y plane, the projection of the river is an ellipse? Because ( x = a sin(bt) ) and ( y = c cos(bt) ). Let me see:If we write ( x = a sin(bt) ) and ( y = c cos(bt) ), then ( frac{x^2}{a^2} + frac{y^2}{c^2} = sin^2(bt) + cos^2(bt) = 1 ). So, the projection onto the x-y plane is indeed an ellipse with semi-major axis ( a ) and semi-minor axis ( c ) (or vice versa, depending on which is larger).But how does this help with the integral? Hmm.Wait, the derivative terms squared are ( a^2 b^2 cos^2(bt) + c^2 b^2 sin^2(bt) ). So, that's ( b^2 (a^2 cos^2(bt) + c^2 sin^2(bt)) ). Maybe we can write this as ( b^2 (a^2 (1 - sin^2(bt)) + c^2 sin^2(bt)) ) = ( b^2 (a^2 + (c^2 - a^2) sin^2(bt)) )So, that would be ( b^2 a^2 + b^2 (c^2 - a^2) sin^2(bt) )Hmm, not sure if that helps.Alternatively, if we let ( sin(bt) = s ), then ( cos(bt) = sqrt{1 - s^2} ), but that might complicate things more.Alternatively, perhaps we can write the expression as ( b^2 (a^2 cos^2(bt) + c^2 sin^2(bt)) = b^2 (a^2 + (c^2 - a^2) sin^2(bt)) )But still, integrating the square root of that plus ( t^2 e^{-t^2} ) is not straightforward.Wait, maybe the problem is designed so that the z-component is negligible? Or perhaps the river's elevation doesn't contribute much to the length? But the problem doesn't specify that, so I shouldn't assume.Alternatively, maybe the integral can be expressed in terms of known functions or constants, but I don't recall such integrals off the top of my head.Wait, let's consider the z-component: ( z'(t) = -t e^{-t^2/2} ), so ( (z'(t))^2 = t^2 e^{-t^2} ). The integral of ( sqrt{...} ) would include this term. Hmm.Alternatively, perhaps the problem expects us to recognize that the integral doesn't have an elementary form and to leave it as an integral expression. Let me check the problem statement again.It says, "Determine the length of the segment...", so maybe it's acceptable to present the integral expression as the answer.So, summarizing, the length ( L ) is:[ L = int_{0}^{T} sqrt{a^2 b^2 cos^2(bt) + c^2 b^2 sin^2(bt) + t^2 e^{-t^2}} , dt ]Alternatively, factoring out ( b^2 ):[ L = int_{0}^{T} sqrt{b^2 (a^2 cos^2(bt) + c^2 sin^2(bt)) + t^2 e^{-t^2}} , dt ]I think that's as far as we can go without more specific information about the constants or the interval T.Wait, but the problem mentions that ( T ) is the time it takes for a drop of water to travel from the source to the mouth of the river. So, perhaps ( T ) is related to the flow characteristics? But without more information, I don't think we can determine ( T ) explicitly.So, perhaps the answer is just the integral expression as above.Moving on to Sub-problem 2: Find the volume of the region swept by the river as it travels over ( t in [0, T] ), assuming the river maintains a constant width ( w ) perpendicular to its flow direction.Hmm, so the river is modeled as a parametric curve, and it has a constant width ( w ). So, the region swept out is like a "tube" or "cylinder" along the river's path, with radius ( w/2 ). But since it's a river, maybe it's more like a rectangular cross-section? Or perhaps it's a 2D width, so the volume would be the length of the river multiplied by the width and some height? Wait, but the river is in 3D space, so the swept volume would be a sort of extrusion along the curve.Wait, actually, in 3D, if the river has a constant width perpendicular to its flow, then the swept volume would be the surface area of the river's path times the width. But no, actually, it's more like a volume swept by a moving line segment of length ( w ) perpendicular to the direction of flow.Wait, perhaps it's better to model it as the river having a constant cross-sectional area, but the problem says "constant width ( w ) perpendicular to its flow direction". So, maybe the cross-section is a rectangle with width ( w ) and some height, but since elevation is given by ( z(t) ), perhaps the height is related to that? Hmm, not sure.Alternatively, maybe the river is approximated as a ribbon with width ( w ), so the volume swept is the integral over the path of the river of the width ( w ) times the differential arc length. Wait, that might make sense.Wait, in 3D, the volume swept by a curve with a constant width can be thought of as the surface area of the curve times the width. But actually, no, because the width is perpendicular to the flow direction, which is along the tangent of the curve.Wait, perhaps the volume is the integral over the curve of the width ( w ) times the differential arc length. So, ( V = int_{0}^{T} w , ds ), where ( ds ) is the differential arc length. But ( ds ) is the same as the integrand from Sub-problem 1. So, ( V = w cdot L ), where ( L ) is the length from Sub-problem 1.Wait, that seems too simplistic. Let me think again.If the river has a constant width ( w ) perpendicular to its flow direction, then at each point along the river, the cross-section is a line segment of length ( w ) perpendicular to the tangent vector. So, the volume swept out would be the integral over the curve of the area of the cross-section, which is ( w times ds ), where ( ds ) is the differential arc length.Wait, actually, no. The cross-sectional area would be ( w times h ), but if the width is ( w ) and the height is infinitesimal, then perhaps the volume is ( int w , ds ). But I'm not sure.Wait, maybe it's better to model it as a surface of revolution, but in this case, it's not a revolution, it's a translation of a line segment along the curve.In differential geometry, the volume swept by a moving segment can be calculated using the concept of parallel transport, but I'm not sure.Alternatively, perhaps the volume is the integral over the curve of the width ( w ) times the differential arc length, which would give ( V = w cdot L ). But that seems too simple, and I'm not sure if that's correct.Wait, let me think in terms of parameterization. The river is a curve ( mathbf{r}(t) ), and at each point, the river has a width ( w ) perpendicular to the flow direction, which is along ( mathbf{r}'(t) ). So, the cross-section is a line segment of length ( w ) in the plane perpendicular to ( mathbf{r}'(t) ).Therefore, the volume swept out would be the integral over the curve of the area of the cross-section, which is ( w times ds ), where ( ds ) is the differential arc length. So, ( V = int_{0}^{T} w , ds ).But ( ds = sqrt{(x'(t))^2 + (y'(t))^2 + (z'(t))^2} , dt ), which is exactly the integrand from Sub-problem 1. Therefore, ( V = w cdot L ), where ( L ) is the length from Sub-problem 1.Wait, but that would mean the volume is just the width times the length, which seems plausible if the cross-section is uniform. So, yes, that makes sense.Therefore, the volume ( V ) is:[ V = w cdot L = w int_{0}^{T} sqrt{a^2 b^2 cos^2(bt) + c^2 b^2 sin^2(bt) + t^2 e^{-t^2}} , dt ]So, in summary, for Sub-problem 1, the length is the integral expression, and for Sub-problem 2, the volume is the same integral multiplied by the width ( w ).But let me double-check if this makes sense. If the river were straight, say along the x-axis, with constant width ( w ), then the volume swept would indeed be the length times width times height, but in this case, since it's a 3D curve, the "height" is already accounted for in the differential arc length. Wait, no, because the width is perpendicular to the flow direction, which is along the tangent. So, the cross-section is a line segment of length ( w ) in the plane perpendicular to the tangent. Therefore, the area of the cross-section is ( w times ds ), but actually, no, the cross-sectional area would be ( w times text{differential length in the direction perpendicular} ). Hmm, maybe I'm confusing things.Wait, perhaps it's better to think of the volume as the integral over the curve of the area of the cross-section. Since the cross-section is a line segment of length ( w ) perpendicular to the tangent, the area would be ( w times ds ), but that would actually give a volume with units of length cubed, which is correct. Wait, no, ( ds ) is already a length, so ( w times ds ) would be area, but we need volume. Hmm, this is confusing.Wait, maybe I need to think in terms of a moving line segment. The volume swept by a moving line segment is given by the integral of the line segment's length times the differential movement. But in this case, the line segment is moving along the curve, so the volume would be the integral over the curve of the line segment's length times the differential arc length. So, ( V = int_{0}^{T} w , ds ), which is ( w cdot L ).Yes, that seems correct. So, the volume is just the width times the length of the river.Therefore, the answers are:Sub-problem 1: The length ( L ) is the integral from 0 to T of the square root of the sum of the squares of the derivatives.Sub-problem 2: The volume ( V ) is ( w ) times that integral.But let me write them formally.For Sub-problem 1:[ L = int_{0}^{T} sqrt{(a b cos(bt))^2 + (-c b sin(bt))^2 + (-t e^{-t^2/2})^2} , dt ]Simplifying inside the square root:[ L = int_{0}^{T} sqrt{a^2 b^2 cos^2(bt) + c^2 b^2 sin^2(bt) + t^2 e^{-t^2}} , dt ]And for Sub-problem 2:[ V = w cdot L = w int_{0}^{T} sqrt{a^2 b^2 cos^2(bt) + c^2 b^2 sin^2(bt) + t^2 e^{-t^2}} , dt ]I think that's the correct approach. I don't see a way to simplify the integral further without additional information, so these expressions are the answers.</think>