🧠:<think>Okay, so I have this problem about Neytiri's fan and their bioluminescent lights. It's divided into two parts. Let me tackle them one by one.Starting with part 1: It's about the Mandelbrot set, which I remember is a set of complex numbers for which a certain iterative function doesn't diverge. The function given is ( Z_{n+1} = Z_n^2 + C ), where ( C = 0.355 + 0.355i ). I need to find the set of initial values ( Z_0 = x + yi ) such that the sequence remains bounded after 100 iterations. Then, represent this set graphically on the complex plane.Hmm, okay. So, the Mandelbrot set is usually defined with ( C ) as the parameter, but here ( C ) is fixed, and we're varying ( Z_0 ). Wait, is that right? Or is it the other way around? Let me think.Wait, in the standard Mandelbrot set, ( C ) is the parameter, and ( Z_0 = 0 ). So, for each ( C ), you iterate starting from 0 and see if it stays bounded. But here, ( C ) is fixed, and we're varying ( Z_0 ). So, actually, this is more like the Julia set for the function ( f(Z) = Z^2 + C ). Julia sets are defined for each ( C ), and they consist of points ( Z_0 ) where the iteration doesn't escape to infinity.So, in this case, since ( C ) is fixed, we're looking at the Julia set for ( C = 0.355 + 0.355i ). The Julia set is the boundary between points that stay bounded and those that escape to infinity. So, the set of ( Z_0 ) that remain bounded is exactly the Julia set for this ( C ).But the problem says "the set of initial values ( Z_0 )" which remain bounded. So, it's the Julia set. Now, to represent this graphically, we would typically plot points in the complex plane where the iteration doesn't escape after a certain number of steps, say 100. Points that escape are colored based on how quickly they escape, and those that don't are usually colored black or another color.But since I can't actually compute this here, maybe I can describe the process. The Julia set for a quadratic function like this can be connected or disconnected, depending on whether ( C ) is in the Mandelbrot set. Wait, actually, if ( C ) is in the Mandelbrot set, the Julia set is connected; otherwise, it's a Cantor set.So, is ( C = 0.355 + 0.355i ) in the Mandelbrot set? Let me check. To determine if ( C ) is in the Mandelbrot set, we iterate ( Z_{n+1} = Z_n^2 + C ) starting from ( Z_0 = 0 ). If the magnitude of ( Z_n ) remains bounded, then ( C ) is in the set.Let me compute a few iterations:( Z_0 = 0 )( Z_1 = 0^2 + C = 0.355 + 0.355i )Compute the magnitude: |Z1| = sqrt(0.355² + 0.355²) ≈ sqrt(0.126 + 0.126) ≈ sqrt(0.252) ≈ 0.502Next iteration:( Z_2 = (0.355 + 0.355i)^2 + C )Let me compute ( (0.355 + 0.355i)^2 ):First, square the real part: 0.355² = 0.126Then, the cross terms: 2 * 0.355 * 0.355i = 2 * 0.126i = 0.252iThen, the imaginary part squared: (0.355i)^2 = -0.126So, adding them up: 0.126 + 0.252i - 0.126 = 0 + 0.252iThen, add C: 0 + 0.252i + 0.355 + 0.355i = 0.355 + (0.252 + 0.355)i = 0.355 + 0.607iCompute |Z2|: sqrt(0.355² + 0.607²) ≈ sqrt(0.126 + 0.368) ≈ sqrt(0.494) ≈ 0.703Next iteration:( Z_3 = (0.355 + 0.607i)^2 + C )Compute the square:Real part: 0.355² - 0.607² ≈ 0.126 - 0.368 ≈ -0.242Imaginary part: 2 * 0.355 * 0.607 ≈ 2 * 0.215 ≈ 0.430iSo, ( Z_3^2 = -0.242 + 0.430i )Add C: -0.242 + 0.430i + 0.355 + 0.355i = ( -0.242 + 0.355 ) + (0.430 + 0.355)i ≈ 0.113 + 0.785iCompute |Z3|: sqrt(0.113² + 0.785²) ≈ sqrt(0.0128 + 0.616) ≈ sqrt(0.6288) ≈ 0.793Next iteration:( Z_4 = (0.113 + 0.785i)^2 + C )Compute the square:Real part: 0.113² - 0.785² ≈ 0.0128 - 0.616 ≈ -0.603Imaginary part: 2 * 0.113 * 0.785 ≈ 2 * 0.0888 ≈ 0.1776iSo, ( Z_4^2 = -0.603 + 0.1776i )Add C: -0.603 + 0.1776i + 0.355 + 0.355i ≈ (-0.603 + 0.355) + (0.1776 + 0.355)i ≈ -0.248 + 0.5326iCompute |Z4|: sqrt((-0.248)^2 + 0.5326^2) ≈ sqrt(0.0615 + 0.2837) ≈ sqrt(0.3452) ≈ 0.587Hmm, it's oscillating. Let's do a few more iterations.( Z_5 = (-0.248 + 0.5326i)^2 + C )Compute square:Real part: (-0.248)^2 - (0.5326)^2 ≈ 0.0615 - 0.2837 ≈ -0.2222Imaginary part: 2 * (-0.248) * 0.5326 ≈ 2 * (-0.1318) ≈ -0.2636iSo, ( Z_5^2 = -0.2222 - 0.2636i )Add C: -0.2222 - 0.2636i + 0.355 + 0.355i ≈ ( -0.2222 + 0.355 ) + ( -0.2636 + 0.355 )i ≈ 0.1328 + 0.0914i|Z5| ≈ sqrt(0.1328² + 0.0914²) ≈ sqrt(0.0176 + 0.00835) ≈ sqrt(0.02595) ≈ 0.161Next iteration:( Z_6 = (0.1328 + 0.0914i)^2 + C )Compute square:Real part: 0.1328² - 0.0914² ≈ 0.0176 - 0.00835 ≈ 0.00925Imaginary part: 2 * 0.1328 * 0.0914 ≈ 2 * 0.01215 ≈ 0.0243iSo, ( Z_6^2 = 0.00925 + 0.0243i )Add C: 0.00925 + 0.0243i + 0.355 + 0.355i ≈ 0.36425 + 0.3793i|Z6| ≈ sqrt(0.36425² + 0.3793²) ≈ sqrt(0.1326 + 0.1439) ≈ sqrt(0.2765) ≈ 0.526Hmm, it's still oscillating but not growing beyond 0.7. Let me check a few more.( Z_7 = (0.36425 + 0.3793i)^2 + C )Compute square:Real part: 0.36425² - 0.3793² ≈ 0.1326 - 0.1439 ≈ -0.0113Imaginary part: 2 * 0.36425 * 0.3793 ≈ 2 * 0.1376 ≈ 0.2752iSo, ( Z_7^2 = -0.0113 + 0.2752i )Add C: -0.0113 + 0.2752i + 0.355 + 0.355i ≈ 0.3437 + 0.6302i|Z7| ≈ sqrt(0.3437² + 0.6302²) ≈ sqrt(0.118 + 0.397) ≈ sqrt(0.515) ≈ 0.718Next:( Z_8 = (0.3437 + 0.6302i)^2 + C )Compute square:Real part: 0.3437² - 0.6302² ≈ 0.118 - 0.397 ≈ -0.279Imaginary part: 2 * 0.3437 * 0.6302 ≈ 2 * 0.2167 ≈ 0.4334iSo, ( Z_8^2 = -0.279 + 0.4334i )Add C: -0.279 + 0.4334i + 0.355 + 0.355i ≈ 0.076 + 0.7884i|Z8| ≈ sqrt(0.076² + 0.7884²) ≈ sqrt(0.0058 + 0.6214) ≈ sqrt(0.6272) ≈ 0.792Hmm, similar to before. It's cycling between around 0.5 and 0.8. Let me check if it's growing beyond 2, which would mean it's escaping.But so far, after 8 iterations, it's still around 0.792. Let me do a few more.( Z_9 = (0.076 + 0.7884i)^2 + C )Compute square:Real part: 0.076² - 0.7884² ≈ 0.0058 - 0.6214 ≈ -0.6156Imaginary part: 2 * 0.076 * 0.7884 ≈ 2 * 0.0599 ≈ 0.1198iSo, ( Z_9^2 = -0.6156 + 0.1198i )Add C: -0.6156 + 0.1198i + 0.355 + 0.355i ≈ (-0.6156 + 0.355) + (0.1198 + 0.355)i ≈ -0.2606 + 0.4748i|Z9| ≈ sqrt((-0.2606)^2 + 0.4748²) ≈ sqrt(0.0679 + 0.2254) ≈ sqrt(0.2933) ≈ 0.5416Next:( Z_{10} = (-0.2606 + 0.4748i)^2 + C )Compute square:Real part: (-0.2606)^2 - (0.4748)^2 ≈ 0.0679 - 0.2254 ≈ -0.1575Imaginary part: 2 * (-0.2606) * 0.4748 ≈ 2 * (-0.1237) ≈ -0.2474iSo, ( Z_{10}^2 = -0.1575 - 0.2474i )Add C: -0.1575 - 0.2474i + 0.355 + 0.355i ≈ ( -0.1575 + 0.355 ) + ( -0.2474 + 0.355 )i ≈ 0.1975 + 0.1076i|Z10| ≈ sqrt(0.1975² + 0.1076²) ≈ sqrt(0.0390 + 0.0116) ≈ sqrt(0.0506) ≈ 0.225Hmm, it's fluctuating but not escaping. Maybe it's bounded? Or maybe it's just taking longer. Since the problem says after 100 iterations, I suppose we can assume that if it hasn't escaped by then, it's considered bounded.But wait, in the standard Mandelbrot set, if |Z_n| exceeds 2, it's guaranteed to escape. So, if after 100 iterations, |Z_n| is still less than or equal to 2, we consider it bounded.In our case, since C is fixed, and we're varying Z0, the Julia set is the boundary where points don't escape. So, the set of Z0 that remain bounded is the Julia set for this C.Graphically, it would be a fractal shape on the complex plane. Depending on whether C is in the Mandelbrot set, the Julia set is connected or not. Since our iterations didn't escape, maybe C is in the Mandelbrot set, so the Julia set is connected.But I'm not sure. Maybe I should check if C is in the Mandelbrot set. Wait, in the standard Mandelbrot set, C is in the set if starting from Z0=0, the iteration doesn't escape. So, if after many iterations, it's still bounded, then C is in the set.From our earlier iterations, starting at Z0=0, after 10 iterations, |Z10| is about 0.225, which is much less than 2. So, it's likely that C is in the Mandelbrot set, meaning the Julia set is connected.Therefore, the set of Z0 that remain bounded is the Julia set, which is a connected fractal. So, graphically, it would look like a connected shape, possibly with intricate detail.But since I can't compute all 100 iterations here, I can conclude that the set is the Julia set for C=0.355+0.355i, which is connected because C is in the Mandelbrot set.Moving on to part 2: Neytiri's fan synchronizes the lights' blinking patterns to a Fibonacci sequence. The nth light blinks according to the nth Fibonacci number. We need to express the sum of the squares of the first 20 Fibonacci numbers using closed-form expressions and then calculate the exact sum.Okay, so the sum S = F1² + F2² + ... + F20², where Fn is the nth Fibonacci number.I remember there's a formula for the sum of squares of Fibonacci numbers. Let me recall.I think the sum of the squares of the first n Fibonacci numbers is equal to Fn * Fn+1. Let me verify.Yes, the identity is:( sum_{k=1}^{n} F_k^2 = F_n F_{n+1} )So, for n=20, the sum would be F20 * F21.Therefore, the closed-form expression is F20 * F21.Now, I need to compute F20 and F21.Let me list the Fibonacci numbers up to F21.F1 = 1F2 = 1F3 = F2 + F1 = 1 + 1 = 2F4 = F3 + F2 = 2 + 1 = 3F5 = F4 + F3 = 3 + 2 = 5F6 = F5 + F4 = 5 + 3 = 8F7 = F6 + F5 = 8 + 5 = 13F8 = F7 + F6 = 13 + 8 = 21F9 = F8 + F7 = 21 + 13 = 34F10 = F9 + F8 = 34 + 21 = 55F11 = F10 + F9 = 55 + 34 = 89F12 = F11 + F10 = 89 + 55 = 144F13 = F12 + F11 = 144 + 89 = 233F14 = F13 + F12 = 233 + 144 = 377F15 = F14 + F13 = 377 + 233 = 610F16 = F15 + F14 = 610 + 377 = 987F17 = F16 + F15 = 987 + 610 = 1597F18 = F17 + F16 = 1597 + 987 = 2584F19 = F18 + F17 = 2584 + 1597 = 4181F20 = F19 + F18 = 4181 + 2584 = 6765F21 = F20 + F19 = 6765 + 4181 = 10946So, F20 = 6765 and F21 = 10946.Therefore, the sum S = 6765 * 10946.Let me compute that.First, compute 6765 * 10000 = 67,650,000Then, 6765 * 946.Wait, 10946 = 10000 + 946.So, 6765 * 10946 = 6765*(10000 + 946) = 6765*10000 + 6765*946Compute 6765*10000 = 67,650,000Now compute 6765*946.Let me break down 946 into 900 + 40 + 6.So, 6765*900 = 6765*9*100 = (60,885)*100 = 6,088,5006765*40 = 6765*4*10 = (27,060)*10 = 270,6006765*6 = 40,590Now, add them up:6,088,500 + 270,600 = 6,359,1006,359,100 + 40,590 = 6,399,690So, total 6765*946 = 6,399,690Now, add to 67,650,000:67,650,000 + 6,399,690 = 74,049,690Therefore, the sum S = 74,049,690.Wait, let me double-check the multiplication to be sure.Alternatively, I can compute 6765 * 10946 directly.But let me use another method.Compute 6765 * 10946:First, note that 10946 = 10946Compute 6765 * 10,000 = 67,650,000Compute 6765 * 946:Compute 6765 * 900 = 6,088,500Compute 6765 * 40 = 270,600Compute 6765 * 6 = 40,590Add them: 6,088,500 + 270,600 = 6,359,100; 6,359,100 + 40,590 = 6,399,690Add to 67,650,000: 67,650,000 + 6,399,690 = 74,049,690Yes, same result.Alternatively, I can use the formula for Fibonacci numbers in terms of Binet's formula, but since we have exact values, it's faster to compute directly.So, the sum is 74,049,690.Therefore, the closed-form expression is F20 * F21, which equals 6765 * 10946 = 74,049,690.Final Answer1. The set of initial values ( Z_0 ) forms the Julia set for ( C = 0.355 + 0.355i ), which is a connected fractal on the complex plane.2. The sum of the squares of the first 20 Fibonacci numbers is boxed{74049690}.</think>

🧠:<think>Okay, so I'm trying to figure out how to model the motion of these icons on a screen using parametric equations. The developer wants them to follow a smooth, sinusoidal wave pattern. The equations given are:[ x(t) = A sin(Bt + C) + D ][ y(t) = E cos(Ft + G) + H ]I need to determine the constraints on the constants A, B, C, D, E, F, G, and H so that the icons stay within the screen boundaries for all t between 0 and T. Also, the developer wants exactly 3 full oscillations both horizontally and vertically within time T, so I need to find B and F for that.Starting with part 1: Constraints on constants to keep icons within screen boundaries.First, the screen has width W and height H. So, the x-coordinate of the icon must satisfy 0 ≤ x(t) ≤ W for all t, and similarly, the y-coordinate must satisfy 0 ≤ y(t) ≤ H for all t.Looking at the x(t) equation: it's a sine function with amplitude A, shifted vertically by D. The sine function oscillates between -1 and 1, so when multiplied by A, it oscillates between -A and A. Then adding D shifts it to between D - A and D + A.To ensure x(t) stays within [0, W], the minimum value of x(t) should be at least 0, and the maximum should be at most W. So:D - A ≥ 0 --> D ≥ AD + A ≤ W --> D ≤ W - ASimilarly, for y(t): it's a cosine function with amplitude E, shifted vertically by H. The cosine function also oscillates between -1 and 1, so E*cos(...) oscillates between -E and E, then adding H shifts it to between H - E and H + E.So for y(t) to stay within [0, H]:H - E ≥ 0 --> H ≥ EH + E ≤ H --> Wait, that can't be right. Wait, H + E ≤ H? That would mean E ≤ 0, but E is an amplitude, so it should be positive. Hmm, maybe I made a mistake.Wait, the screen height is H, so y(t) must be between 0 and H. So:H - E ≥ 0 --> H ≥ EH + E ≤ H --> E ≤ 0, but E is positive, so this would imply E = 0, which can't be right because then there's no oscillation. Hmm, maybe I need to rethink.Wait, perhaps I misapplied the constraints. Let me consider the range of y(t):The cosine function varies between -1 and 1, so E*cos(...) varies between -E and E. Then adding H, it becomes between H - E and H + E.To have y(t) within [0, H], we need:H - E ≥ 0 --> H ≥ EH + E ≤ H --> E ≤ 0But E is positive, so the only way H + E ≤ H is if E = 0, which is not possible because then there's no vertical oscillation. So, maybe I need to adjust the equation.Wait, perhaps the vertical shift should be such that the midpoint is at H/2, so that the oscillation is symmetric around the center of the screen. So, if y(t) is E*cos(...) + H/2, then the maximum would be H/2 + E and the minimum would be H/2 - E. To ensure both are within [0, H], we need:H/2 + E ≤ H --> E ≤ H/2H/2 - E ≥ 0 --> E ≤ H/2So, E must be ≤ H/2. Similarly, for x(t), if we set D = W/2, then x(t) would oscillate between W/2 - A and W/2 + A. To keep within [0, W], we need:W/2 - A ≥ 0 --> A ≤ W/2W/2 + A ≤ W --> A ≤ W/2So, A must be ≤ W/2.Wait, but in the original equations, D and H are constants. So, perhaps D should be set to W/2 and H to H/2 to center the icons on the screen. Then, the amplitudes A and E must be ≤ W/2 and H/2 respectively.So, constraints:For x(t):A ≤ W/2D = W/2For y(t):E ≤ H/2H = H/2Wait, but H is the vertical shift, so maybe it's better to set H = H/2? Wait, no, H is the vertical shift, so if we set H = H/2, then y(t) would oscillate around H/2, which is the center of the screen. So, yes, that makes sense.So, to sum up, to keep the icons within the screen:For x(t):- Amplitude A must be ≤ W/2- Vertical shift D must be W/2For y(t):- Amplitude E must be ≤ H/2- Vertical shift H must be H/2Additionally, the phase shifts C and G can be any values, as they only affect the starting point of the sine and cosine waves, not the range. So, C and G can be any real numbers.But wait, what about B and F? They affect the frequency of the oscillations, but not the range. So, as long as A and E are within the above constraints, the icons will stay within the screen regardless of B and F. So, the constraints on B and F are not about the range, but about the number of oscillations.But part 1 is only about staying within screen boundaries, so B and F don't affect that, only the number of oscillations. So, for part 1, the constraints are:A ≤ W/2, D = W/2, E ≤ H/2, H = H/2.Wait, but in the original equations, H is the vertical shift, so if we set H = H/2, that would mean H is half of the screen height, which is correct. So, H is a constant, so maybe the vertical shift should be H = H/2, so that the center is at H/2.But in the original equation, y(t) = E cos(Ft + G) + H. So, H is the vertical shift. So, to center it, H should be H/2, but that would mean H is a variable, which is confusing because H is also the screen height. Maybe I should use a different symbol for the vertical shift. Let me denote the vertical shift as K instead of H to avoid confusion.Wait, in the original problem, the constants are A, B, C, D, E, F, G, and H. So, H is the vertical shift. So, to center the icons vertically, H should be set to H/2, where H is the screen height. So, H (constant) = H (screen height)/2.Similarly, D (constant) should be set to W/2, where W is the screen width.So, the constraints are:A ≤ W/2D = W/2E ≤ H/2H = H/2But wait, H is both the screen height and the vertical shift. That's confusing. Maybe the vertical shift should be denoted as, say, K, but in the problem, it's given as H. So, perhaps the vertical shift H (constant) must be equal to H (screen height)/2.So, in summary:To keep x(t) within [0, W]:A ≤ W/2D = W/2To keep y(t) within [0, H]:E ≤ H/2H (constant) = H (screen height)/2So, that's part 1.Now, part 2: The developer wants exactly 3 full oscillations both horizontally and vertically within time T.So, for x(t), which is A sin(Bt + C) + D, the period is 2π/B. To have 3 oscillations in time T, the period must be T/3. So:2π/B = T/3=> B = 6π/TSimilarly, for y(t), which is E cos(Ft + G) + H, the period is 2π/F. To have 3 oscillations in time T:2π/F = T/3=> F = 6π/TSo, both B and F should be 6π/T.Wait, but let me double-check.The number of oscillations is given by the frequency, which is 1/period. So, if we want 3 oscillations in time T, the period is T/3.For x(t), the period is 2π/B, so:2π/B = T/3=> B = 6π/TSimilarly for y(t):2π/F = T/3=> F = 6π/TYes, that seems correct.So, to summarize:1. Constraints:- A ≤ W/2- D = W/2- E ≤ H/2- H (constant) = H (screen height)/22. B = 6π/T and F = 6π/TI think that's it.</think>

🧠:<think>Alright, so I've got this problem where Alex is trying to model their mood fluctuations using a function ( f(t) = A sin(Bt + C) + D e^{-alpha t} ). The goal is to determine the constants ( A ), ( B ), ( C ), ( D ), and ( alpha ) based on some observations, and then compute the integral of this function over 30 days. Let me break this down step by step.First, let's list out the given information:1. Alex records mood every day at 8 AM for a week (7 days).2. The maximum mood value occurs every 3 days.3. At ( t = 0 ), the mood ( f(0) = D = 5 ).4. The mood decreases by 20% every day due to the exponential component ( D e^{-alpha t} ).5. The mood reaches a minimum value of -2 at ( t = 1.5 ) days.Starting with the constants:- From point 3, we know that ( D = 5 ). That's straightforward.Next, the exponential component decreases by 20% each day. So, the mood due to the exponential part at day ( t ) is 80% of the previous day's value. That means:( D e^{-alpha (t+1)} = 0.8 D e^{-alpha t} )Dividing both sides by ( D e^{-alpha t} ):( e^{-alpha} = 0.8 )Taking the natural logarithm of both sides:( -alpha = ln(0.8) )So,( alpha = -ln(0.8) )Calculating that:( ln(0.8) ) is approximately ( -0.2231 ), so ( alpha approx 0.2231 ) per day.Okay, so now we have ( D = 5 ) and ( alpha approx 0.2231 ).Moving on to the sine component. The function is ( A sin(Bt + C) ). We know that the maximum mood occurs every 3 days. The sine function has a period of ( 2pi / B ). Since the maximum occurs every 3 days, the period should be 6 days because the sine function reaches its maximum every half-period. Wait, actually, let me think about that.Wait, no. The maximum of a sine function occurs every period. But in this case, Alex notices that the maximum mood value occurs every 3 days. So, does that mean the period is 3 days? Or is it half the period?Wait, the sine function reaches its maximum every ( pi ) radians, which is half a period. So, if the maximum occurs every 3 days, that would correspond to half the period. Therefore, the full period is 6 days.So, the period ( T = 6 ) days. The period of the sine function is ( 2pi / B ), so:( 2pi / B = 6 )Solving for ( B ):( B = 2pi / 6 = pi / 3 approx 1.0472 ) per day.Alright, so ( B = pi / 3 ).Now, we need to find ( A ) and ( C ). We have two pieces of information left: the mood at ( t = 0 ) is 5, and the mood reaches a minimum of -2 at ( t = 1.5 ) days.First, let's use ( t = 0 ):( f(0) = A sin(B*0 + C) + D e^{-alpha*0} = A sin(C) + D = 5 )We know ( D = 5 ), so:( A sin(C) + 5 = 5 )Subtracting 5 from both sides:( A sin(C) = 0 )So, either ( A = 0 ) or ( sin(C) = 0 ). But ( A ) can't be zero because we have a sine component, so ( sin(C) = 0 ). Therefore, ( C ) must be an integer multiple of ( pi ). Let's denote ( C = kpi ), where ( k ) is an integer.Now, moving on to the other condition: the mood reaches a minimum of -2 at ( t = 1.5 ) days.So, ( f(1.5) = A sin(B*1.5 + C) + D e^{-alpha*1.5} = -2 )We can plug in the known values:( A sin((pi/3)*1.5 + C) + 5 e^{-0.2231*1.5} = -2 )Calculating each part:First, ( (pi/3)*1.5 = (pi/3)*(3/2) = pi/2 ). So, the argument of the sine function is ( pi/2 + C ).Since ( C = kpi ), the argument becomes ( pi/2 + kpi ).The sine of ( pi/2 + kpi ) is either 1 or -1, depending on whether ( k ) is even or odd.Because ( sin(pi/2 + kpi) = (-1)^k ).So, ( A*(-1)^k ) is the coefficient.Next, let's compute the exponential term:( 5 e^{-0.2231*1.5} )Calculating the exponent:( 0.2231 * 1.5 ≈ 0.33465 )So,( e^{-0.33465} ≈ 0.7165 )Therefore, the exponential term is:( 5 * 0.7165 ≈ 3.5825 )Putting it all together:( A*(-1)^k + 3.5825 = -2 )So,( A*(-1)^k = -2 - 3.5825 = -5.5825 )Thus,( A*(-1)^k = -5.5825 )So, ( A = -5.5825 / (-1)^k )Which simplifies to:If ( k ) is even, ( (-1)^k = 1 ), so ( A = -5.5825 )If ( k ) is odd, ( (-1)^k = -1 ), so ( A = 5.5825 )But since ( A ) is the amplitude of the sine function, it should be positive. Therefore, ( A ) must be positive, so ( (-1)^k ) must be negative. Therefore, ( k ) must be odd.So, let's take ( k = 1 ) for simplicity, which gives ( C = pi ).Therefore, ( A = 5.5825 )Wait, let me double-check:If ( k ) is odd, ( (-1)^k = -1 ), so ( A*(-1) = -5.5825 ) implies ( A = 5.5825 ). Yes, that's correct.So, ( A ≈ 5.5825 ). Let me keep more decimal places for accuracy. Let's compute ( 5.5825 ) more precisely.Wait, actually, let's backtrack a bit.We had:( A*(-1)^k = -5.5825 )If ( k ) is odd, ( (-1)^k = -1 ), so:( A*(-1) = -5.5825 ) => ( A = 5.5825 )If ( k ) is even, ( (-1)^k = 1 ), so:( A = -5.5825 ), which we don't want because amplitude should be positive.Therefore, ( A = 5.5825 ) and ( C = pi ).But let's compute ( A ) more accurately.We had:( A*(-1)^k = -5.5825 )But let's compute the exact value without approximating too early.We had:( f(1.5) = A sin(pi/2 + C) + 5 e^{-0.2231*1.5} = -2 )We know ( sin(pi/2 + C) = cos(C) ) because ( sin(pi/2 + x) = cos(x) ). Wait, is that right?Wait, no. Actually, ( sin(pi/2 + x) = cos(x) ). So, if ( C = kpi ), then ( cos(C) = cos(kpi) = (-1)^k ).Wait, so actually, ( sin(pi/2 + C) = cos(C) = (-1)^k ).Therefore, ( A*(-1)^k + 5 e^{-0.2231*1.5} = -2 )So, same as before.But perhaps I should compute ( e^{-0.2231*1.5} ) more accurately.Given that ( alpha = -ln(0.8) approx 0.22314 )So, ( alpha = 0.22314 )Thus, ( 0.22314 * 1.5 = 0.33471 )So, ( e^{-0.33471} ). Let's compute that.We know that ( e^{-0.33471} ) is approximately:Using Taylor series or calculator approximation.Alternatively, since ( e^{-0.33471} approx 1 - 0.33471 + (0.33471)^2/2 - (0.33471)^3/6 )Calculating:1 - 0.33471 = 0.66529(0.33471)^2 = 0.11203, divided by 2 is 0.056015(0.33471)^3 = 0.03753, divided by 6 is ~0.006255So, adding up:0.66529 + 0.056015 = 0.7213050.721305 - 0.006255 ≈ 0.71505So, approximately 0.71505Therefore, ( 5 * 0.71505 ≈ 3.57525 )So, back to the equation:( A*(-1)^k + 3.57525 = -2 )Thus,( A*(-1)^k = -2 - 3.57525 = -5.57525 )So, ( A = -5.57525 / (-1)^k )Again, since ( A ) must be positive, ( (-1)^k ) must be negative, so ( k ) is odd.Thus, ( A = 5.57525 )So, approximately ( A ≈ 5.575 )But let's keep more decimals for precision.Alternatively, perhaps we can express ( A ) exactly.Wait, let's see:We have:( A*(-1)^k = -5.57525 )But ( A ) is positive, so ( (-1)^k = -1 ), so ( A = 5.57525 )But let's compute ( 5.57525 ) as a fraction or exact value.Wait, perhaps we can express ( A ) in terms of exact expressions.Wait, let's see:We have:( A*(-1)^k = -5.57525 )But ( 5.57525 ) is approximately ( 5 + 0.57525 ). 0.57525 is approximately ( 0.57525 = 0.5 + 0.07525 ). Not sure if that helps.Alternatively, perhaps we can express ( A ) as ( 5 / (1 - e^{-alpha}) ) or something, but I don't think so.Alternatively, let's compute ( A ) exactly.Wait, we have:( A*(-1)^k = -5.57525 )But ( A ) is positive, so ( A = 5.57525 )But perhaps we can write it as:( A = ( -2 - 5 e^{-alpha*1.5} ) / (-1)^k )But since ( (-1)^k = -1 ), it's:( A = ( -2 - 5 e^{-alpha*1.5} ) / (-1) = 2 + 5 e^{-alpha*1.5} )Wait, that's a better way to express it.So,( A = 2 + 5 e^{-alpha*1.5} )We know ( alpha = -ln(0.8) ), so:( e^{-alpha*1.5} = e^{1.5 ln(0.8)} = (e^{ln(0.8)})^{1.5} = 0.8^{1.5} )Calculating ( 0.8^{1.5} ):( 0.8^{1} = 0.8 )( 0.8^{0.5} = sqrt{0.8} ≈ 0.8944 )So, ( 0.8^{1.5} = 0.8 * 0.8944 ≈ 0.7155 )Therefore,( A = 2 + 5 * 0.7155 ≈ 2 + 3.5775 ≈ 5.5775 )So, ( A ≈ 5.5775 )Therefore, rounding to four decimal places, ( A ≈ 5.5775 )So, summarizing the constants:- ( D = 5 )- ( alpha = -ln(0.8) ≈ 0.2231 )- ( B = pi / 3 ≈ 1.0472 )- ( C = pi ) (since ( k = 1 ))- ( A ≈ 5.5775 )Wait, let me check if ( C = pi ) is correct.Earlier, we had ( C = kpi ), and since ( k ) is odd, we took ( k = 1 ), so ( C = pi ). That seems correct.Now, let's verify if these constants satisfy the given conditions.First, at ( t = 0 ):( f(0) = 5.5775 sin(pi/3 * 0 + pi) + 5 e^{0} = 5.5775 sin(pi) + 5 = 0 + 5 = 5 ). Correct.At ( t = 1.5 ):( f(1.5) = 5.5775 sin(pi/3 * 1.5 + pi) + 5 e^{-0.2231*1.5} )Calculating the sine term:( pi/3 * 1.5 = pi/2 ), so ( pi/2 + pi = 3pi/2 )( sin(3pi/2) = -1 )So,( 5.5775 * (-1) + 5 * e^{-0.33465} ≈ -5.5775 + 5 * 0.7155 ≈ -5.5775 + 3.5775 ≈ -2 ). Correct.Good, so the constants satisfy the given conditions.Now, moving on to part 2: calculating the definite integral ( int_0^{30} f(t) dt ).Given ( f(t) = A sin(Bt + C) + D e^{-alpha t} ), the integral is:( int_0^{30} [A sin(Bt + C) + D e^{-alpha t}] dt )We can split this into two integrals:( A int_0^{30} sin(Bt + C) dt + D int_0^{30} e^{-alpha t} dt )Let's compute each integral separately.First integral: ( A int_0^{30} sin(Bt + C) dt )The integral of ( sin(Bt + C) ) with respect to t is:( -frac{1}{B} cos(Bt + C) + K )So, evaluating from 0 to 30:( A [ -frac{1}{B} cos(B*30 + C) + frac{1}{B} cos(C) ] )Simplify:( -frac{A}{B} [ cos(B*30 + C) - cos(C) ] )Second integral: ( D int_0^{30} e^{-alpha t} dt )The integral of ( e^{-alpha t} ) is ( -frac{1}{alpha} e^{-alpha t} + K )Evaluating from 0 to 30:( D [ -frac{1}{alpha} e^{-alpha*30} + frac{1}{alpha} e^{0} ] )Simplify:( frac{D}{alpha} [1 - e^{-alpha*30} ] )Putting it all together, the total integral is:( -frac{A}{B} [ cos(B*30 + C) - cos(C) ] + frac{D}{alpha} [1 - e^{-alpha*30} ] )Now, let's plug in the values we found:- ( A ≈ 5.5775 )- ( B = pi/3 ≈ 1.0472 )- ( C = pi )- ( D = 5 )- ( alpha ≈ 0.2231 )First, compute the sine integral part:( -frac{5.5775}{pi/3} [ cos(pi/3 * 30 + pi) - cos(pi) ] )Simplify:( -frac{5.5775 * 3}{pi} [ cos(10pi + pi) - cos(pi) ] )Wait, ( pi/3 * 30 = 10pi ). So,( cos(10pi + pi) = cos(11pi) )But ( cos(11pi) = cos(pi) = -1 ) because cosine has a period of ( 2pi ), so ( 11pi = 5*2pi + pi ), so it's equivalent to ( cos(pi) = -1 ).Similarly, ( cos(pi) = -1 ).Therefore,( -frac{5.5775 * 3}{pi} [ (-1) - (-1) ] = -frac{16.7325}{pi} [0] = 0 )Interesting, the sine component integral over 30 days cancels out because the cosine terms are the same.Now, the exponential integral part:( frac{5}{0.2231} [1 - e^{-0.2231*30} ] )Calculating each part:First, ( 0.2231*30 ≈ 6.693 )So, ( e^{-6.693} ) is a very small number. Let's compute it:( e^{-6.693} ≈ e^{-6} * e^{-0.693} ≈ 0.002479 * 0.5 ≈ 0.0012395 )So, approximately 0.00124.Therefore,( 1 - e^{-6.693} ≈ 1 - 0.00124 ≈ 0.99876 )Now, ( frac{5}{0.2231} ≈ 5 / 0.2231 ≈ 22.398 )So,( 22.398 * 0.99876 ≈ 22.398 * 0.99876 ≈ 22.36 )Therefore, the total integral is approximately 22.36.But let's compute it more accurately.First, ( alpha = -ln(0.8) ≈ 0.223143551 )So,( frac{D}{alpha} = 5 / 0.223143551 ≈ 22.398 )Now, ( e^{-alpha*30} = e^{-0.223143551*30} = e^{-6.69430653} )Calculating ( e^{-6.69430653} ):We know that ( e^{-6} ≈ 0.002478752 )And ( e^{-0.69430653} ≈ e^{-0.69314718} * e^{-0.00115935} ≈ 0.5 * (1 - 0.00115935) ≈ 0.49942 )Therefore,( e^{-6.69430653} ≈ 0.002478752 * 0.49942 ≈ 0.001238 )So,( 1 - e^{-6.69430653} ≈ 1 - 0.001238 ≈ 0.998762 )Thus,( frac{5}{0.223143551} * 0.998762 ≈ 22.398 * 0.998762 ≈ 22.36 )So, the total integral is approximately 22.36.But let's express it in terms of the exact constants without approximating too much.We have:( int_0^{30} f(t) dt = 0 + frac{D}{alpha} (1 - e^{-alpha*30}) )Since the sine integral part is zero.Therefore, the exact expression is:( frac{D}{alpha} (1 - e^{-30alpha}) )Given that ( D = 5 ) and ( alpha = -ln(0.8) ), we can write:( frac{5}{-ln(0.8)} (1 - e^{-30*(-ln(0.8))}) )Simplify:( frac{5}{-ln(0.8)} (1 - e^{30ln(0.8)}) )But ( e^{30ln(0.8)} = (e^{ln(0.8)})^{30} = 0.8^{30} )So,( frac{5}{-ln(0.8)} (1 - 0.8^{30}) )Since ( -ln(0.8) = ln(1/0.8) = ln(1.25) ), we can write:( frac{5}{ln(1.25)} (1 - 0.8^{30}) )But ( 0.8^{30} ) is a very small number, as we saw earlier, approximately 0.001238.So, the integral is approximately:( frac{5}{ln(1.25)} * (1 - 0.001238) ≈ frac{5}{0.22314} * 0.99876 ≈ 22.36 )Therefore, the total cumulative impact over 30 days is approximately 22.36.But since the problem asks to express the answer in terms of the constants found, we can write it as:( frac{D}{alpha} (1 - e^{-30alpha}) )Given that ( D = 5 ) and ( alpha = -ln(0.8) ), this is the exact expression.Alternatively, we can write it as:( frac{5}{-ln(0.8)} (1 - 0.8^{30}) )But since ( 0.8^{30} ) is negligible, it's approximately ( frac{5}{ln(1.25)} ), but the exact answer is better expressed with the exponential term.So, summarizing:The constants are:- ( A ≈ 5.5775 )- ( B = pi/3 )- ( C = pi )- ( D = 5 )- ( alpha ≈ 0.2231 )And the integral is:( frac{5}{0.2231} (1 - e^{-0.2231*30}) ≈ 22.36 )But to express it exactly, it's:( frac{5}{alpha} (1 - e^{-30alpha}) )Since ( alpha = -ln(0.8) ), we can write it as:( frac{5}{-ln(0.8)} (1 - 0.8^{30}) )But since ( 0.8^{30} ) is very small, it's approximately ( frac{5}{ln(1.25)} ), but the exact form is better.Therefore, the final answer for the integral is:( frac{5}{alpha} (1 - e^{-30alpha}) )Where ( alpha = -ln(0.8) ).But let me check if I can express it in terms of ( D ) and ( alpha ) only, since ( D = 5 ):Yes, ( frac{D}{alpha} (1 - e^{-30alpha}) )So, that's the expression in terms of the constants.Therefore, the cumulative impact is ( frac{D}{alpha} (1 - e^{-30alpha}) ).To recap:1. Constants: - ( A ≈ 5.5775 ) - ( B = pi/3 ) - ( C = pi ) - ( D = 5 ) - ( alpha ≈ 0.2231 )2. Integral result: - ( frac{D}{alpha} (1 - e^{-30alpha}) ) ≈ 22.36But since the problem asks to express the answer in terms of the constants found, we can leave it as ( frac{D}{alpha} (1 - e^{-30alpha}) ).Alternatively, substituting ( D = 5 ) and ( alpha = -ln(0.8) ), it becomes:( frac{5}{-ln(0.8)} (1 - e^{-30*(-ln(0.8))}) = frac{5}{ln(1.25)} (1 - 0.8^{30}) )But since ( 0.8^{30} ) is negligible, it's approximately ( frac{5}{ln(1.25)} ), but the exact form is better.So, the final answer for the integral is ( frac{5}{alpha} (1 - e^{-30alpha}) ), which is approximately 22.36.But to be precise, let's compute it more accurately.Given:( alpha = -ln(0.8) ≈ 0.223143551 )So,( frac{5}{0.223143551} ≈ 22.398 )And,( e^{-30*0.223143551} = e^{-6.69430653} ≈ 0.001238 )Thus,( 1 - 0.001238 ≈ 0.998762 )Therefore,( 22.398 * 0.998762 ≈ 22.36 )So, the integral is approximately 22.36.But since the problem asks to express the answer in terms of the constants, not numerically, we can write it as:( frac{D}{alpha} (1 - e^{-30alpha}) )Given that ( D = 5 ) and ( alpha = -ln(0.8) ), this is the exact expression.Therefore, the cumulative impact is ( frac{5}{alpha} (1 - e^{-30alpha}) ).So, to summarize:1. Constants: - ( A ≈ 5.5775 ) - ( B = pi/3 ) - ( C = pi ) - ( D = 5 ) - ( alpha ≈ 0.2231 )2. Integral result: - ( frac{5}{alpha} (1 - e^{-30alpha}) ≈ 22.36 )But since the problem asks for the answer in terms of the constants, we'll present it as ( frac{D}{alpha} (1 - e^{-30alpha}) ).Therefore, the final answer for the integral is ( frac{D}{alpha} (1 - e^{-30alpha}) ).</think>

🧠:<think>Alright, so I have this problem about sustainable gardening and optimizing water usage. It's divided into two parts. Let me try to tackle them one by one.Starting with part 1: The garden bed is a rectangle, 10 meters long and 5 meters wide. So, first, I should calculate the area of the garden bed because the absorption and evaporation rates are given per square meter. The area would be length multiplied by width, right? So that's 10 meters * 5 meters = 50 square meters. Got that down.Now, the water absorption rate is given by the function A(t) = 4t² - 3t + 5 liters per square meter per hour. They want the total amount of water absorbed during the first 3 hours. Hmm, okay, so since it's per square meter, I need to integrate A(t) over the time interval from 0 to 3 hours and then multiply by the area to get the total liters.Let me recall how integration works. The integral of A(t) from 0 to 3 will give me the total absorption per square meter over 3 hours. Then, multiplying by 50 will give the total liters absorbed by the entire garden bed.So, let's set up the integral:Total absorption per square meter = ∫₀³ (4t² - 3t + 5) dtI need to compute this integral. Let me break it down term by term.The integral of 4t² dt is (4/3)t³.The integral of -3t dt is (-3/2)t².The integral of 5 dt is 5t.So putting it all together, the integral becomes:(4/3)t³ - (3/2)t² + 5t evaluated from 0 to 3.Now, plugging in t = 3:(4/3)*(3)³ - (3/2)*(3)² + 5*(3)Calculating each term:(4/3)*27 = (4/3)*27 = 4*9 = 36(3/2)*9 = (3/2)*9 = 13.55*3 = 15So, adding them up: 36 - 13.5 + 15 = 36 + 1.5 = 37.5Wait, hold on, 36 - 13.5 is 22.5, then 22.5 + 15 is 37.5. Yeah, that's correct.Now, evaluating at t = 0, all terms become 0, so the total absorption per square meter is 37.5 liters.But wait, is that right? Let me double-check my calculations.(4/3)*(27) = 36, correct.(3/2)*(9) = 13.5, correct.5*3 = 15, correct.So 36 - 13.5 = 22.5, then 22.5 + 15 = 37.5. Yep, that's correct.So, per square meter, 37.5 liters absorbed in 3 hours.But the garden is 50 square meters, so total absorption is 37.5 * 50.Calculating that: 37.5 * 50. Let's see, 37 * 50 = 1850, and 0.5 * 50 = 25, so total is 1850 + 25 = 1875 liters.Wait, hold on, 37.5 * 50. Alternatively, 37.5 * 50 is the same as 37.5 * 5 * 10 = 187.5 * 10 = 1875 liters. Yep, that's correct.So, total water absorbed is 1875 liters over the first 3 hours.Okay, that seems solid. I think I did that right.Moving on to part 2: The evaporation rate is given by E(t) = 2e^(0.5t) liters per square meter per hour. They want the net water retained, which is total absorbed minus total evaporated over the same 3 hours.So, similar approach: integrate E(t) from 0 to 3, get the total evaporation per square meter, multiply by 50, then subtract that from the total absorption.Let me set up the integral:Total evaporation per square meter = ∫₀³ 2e^(0.5t) dtHmm, integrating exponential functions. The integral of e^(kt) dt is (1/k)e^(kt) + C. So, in this case, k is 0.5.So, the integral becomes:2 * [ (1/0.5) e^(0.5t) ] evaluated from 0 to 3.Simplify:2 * [2 e^(0.5t)] from 0 to 3.Which is 4 [e^(0.5t)] from 0 to 3.So, plugging in t = 3:4 * e^(1.5)And t = 0:4 * e^(0) = 4 * 1 = 4So, total evaporation per square meter is 4e^(1.5) - 4.Let me compute that numerically.First, e^(1.5). I know e^1 is about 2.71828, e^0.5 is about 1.64872. So, e^1.5 is e^1 * e^0.5 ≈ 2.71828 * 1.64872.Calculating that: 2.71828 * 1.64872.Let me compute 2 * 1.64872 = 3.297440.71828 * 1.64872 ≈ Let's approximate:0.7 * 1.64872 ≈ 1.15410.01828 * 1.64872 ≈ ~0.0301So, total ≈ 1.1541 + 0.0301 ≈ 1.1842So, total e^1.5 ≈ 3.29744 + 1.1842 ≈ 4.48164So, 4e^(1.5) ≈ 4 * 4.48164 ≈ 17.92656Then subtract 4: 17.92656 - 4 = 13.92656 liters per square meter.So, approximately 13.92656 liters per square meter evaporated over 3 hours.But let me check that calculation again because approximating e^1.5 might not be precise enough.Alternatively, I can use a calculator for e^1.5.But since I don't have a calculator here, maybe I can recall that e^1.5 is approximately 4.4816890703.So, 4 * 4.4816890703 ≈ 17.9267562812Subtract 4: 17.9267562812 - 4 = 13.9267562812 liters per square meter.So, approximately 13.9268 liters per square meter.Therefore, total evaporation over 50 square meters is 13.9268 * 50.Calculating that: 13.9268 * 50 = 696.34 liters.Wait, let me compute 13.9268 * 50:13 * 50 = 6500.9268 * 50 = 46.34So, total is 650 + 46.34 = 696.34 liters.So, total evaporation is approximately 696.34 liters.Earlier, total absorption was 1875 liters.Therefore, net water retained is 1875 - 696.34 = ?Calculating that: 1875 - 696.34.Let me do 1875 - 600 = 1275Then subtract 96.34: 1275 - 96.34 = 1178.66 liters.So, approximately 1178.66 liters retained.But let me check the exact value without approximating e^1.5 too early.Wait, maybe I should carry out the integral symbolically first.So, total evaporation per square meter is 4(e^(1.5) - 1). So, 4*(e^(1.5) - 1) liters.So, total evaporation is 50 * 4*(e^(1.5) - 1) = 200*(e^(1.5) - 1) liters.Similarly, total absorption is 50 * 37.5 = 1875 liters.So, net water is 1875 - 200*(e^(1.5) - 1).Compute 200*(e^(1.5) - 1):We know e^(1.5) ≈ 4.4816890703So, 4.4816890703 - 1 = 3.4816890703Multiply by 200: 3.4816890703 * 200 = 696.33781406 liters.So, net water is 1875 - 696.33781406 ≈ 1178.6621859 liters.So, approximately 1178.66 liters.So, rounding to a reasonable decimal place, maybe two decimal places: 1178.66 liters.Alternatively, if we want to express it exactly, it's 1875 - 200(e^(1.5) - 1). But since the question says to evaluate the difference, I think they want a numerical value.So, 1178.66 liters is the net water retained.Wait, let me just verify my steps again.1. Calculated area: 10*5=50 m². Correct.2. Integrated A(t) from 0 to 3:∫(4t² -3t +5) dt from 0 to 3.Antiderivative: (4/3)t³ - (3/2)t² +5t.Evaluated at 3: (4/3)*27 - (3/2)*9 +15 = 36 -13.5 +15=37.5.Multiply by 50: 37.5*50=1875 liters. Correct.3. Integrated E(t)=2e^(0.5t) from 0 to3.Antiderivative: 2*(2e^(0.5t))=4e^(0.5t).Evaluated from 0 to3: 4(e^(1.5)-1). Multiply by 50: 200(e^(1.5)-1)= approx 696.34 liters.Subtracting: 1875 -696.34=1178.66 liters.Yes, seems correct.Alternatively, maybe I can represent e^(1.5) more accurately.But since the question doesn't specify the need for an exact form, decimal approximation is fine.So, summarizing:1. Total water absorbed: 1875 liters.2. Total water evaporated: approximately 696.34 liters.Net water retained: approximately 1178.66 liters.So, I think that's the answer.Final Answer1. The total amount of water absorbed by the soil during the first 3 hours is boxed{1875} liters.2. The net water retained by the garden bed after the first 3 hours is boxed{1178.66} liters.</think>