🧠:<think>Alright, so I'm trying to figure out this problem about two interconnected water tanks. Let me start by understanding the setup. There are two tanks, Tank A and Tank B. Tank A is elevated above Tank B, which probably means that water can flow from A to B due to gravity. The cross-sectional areas are given: 3 square meters for Tank A and 5 square meters for Tank B. The initial volumes are 20 cubic meters in Tank A and 10 cubic meters in Tank B. The flow rate between them is given by Q = k(h_A - h_B), where k is 0.5. Okay, so the first part asks to formulate the differential equations governing the heights h_A(t) and h_B(t). Hmm. I remember that the volume of water in a tank is related to the cross-sectional area and the height of the water. So, Volume = Area × Height. Therefore, the height can be expressed as Volume divided by Area. Since the flow rate Q is the rate at which water is moving from Tank A to Tank B, I think this will affect the volumes (and hence the heights) of both tanks over time. Specifically, the volume in Tank A will decrease at a rate equal to Q, and the volume in Tank B will increase at the same rate. Let me denote the volume in Tank A as V_A(t) and in Tank B as V_B(t). Then, the heights h_A(t) and h_B(t) can be written as:h_A(t) = V_A(t) / 3h_B(t) = V_B(t) / 5Since the flow rate Q is the rate of change of volume, we can write:dV_A/dt = -QdV_B/dt = QBecause water is leaving Tank A and entering Tank B. Given that Q = k(h_A - h_B), and k = 0.5, so Q = 0.5(h_A - h_B). Therefore, substituting Q into the differential equations:dV_A/dt = -0.5(h_A - h_B)dV_B/dt = 0.5(h_A - h_B)But since h_A = V_A / 3 and h_B = V_B / 5, we can substitute these into the equations:dV_A/dt = -0.5((V_A / 3) - (V_B / 5))dV_B/dt = 0.5((V_A / 3) - (V_B / 5))Alternatively, since we need differential equations in terms of h_A and h_B, maybe it's better to express everything in terms of h_A and h_B directly.Let me think. Since V_A = 3 h_A and V_B = 5 h_B, then dV_A/dt = 3 dh_A/dt and dV_B/dt = 5 dh_B/dt.So substituting into the previous equations:3 dh_A/dt = -0.5(h_A - h_B)5 dh_B/dt = 0.5(h_A - h_B)So, simplifying these:dh_A/dt = (-0.5 / 3)(h_A - h_B) = (-1/6)(h_A - h_B)dh_B/dt = (0.5 / 5)(h_A - h_B) = (1/10)(h_A - h_B)So, the differential equations are:dh_A/dt = (-1/6)(h_A - h_B)dh_B/dt = (1/10)(h_A - h_B)That seems right. Let me double-check. The flow rate is proportional to the height difference, so the rate at which Tank A loses water is proportional to (h_A - h_B), and Tank B gains water at the same rate. Since the cross-sectional areas are different, the rates of change of the heights will be different. Yes, so the differential equations are correct. Now, moving on to part 2: determining the equilibrium heights. At equilibrium, the heights should stabilize, meaning that dh_A/dt = 0 and dh_B/dt = 0. So, setting the derivatives equal to zero:0 = (-1/6)(h_A - h_B)0 = (1/10)(h_A - h_B)From both equations, we get that h_A - h_B = 0, so h_A = h_B at equilibrium.So, in equilibrium, both tanks have the same water level. But wait, the tanks have different cross-sectional areas. So, does that mean the volumes will be different? Let me think. Yes, because Volume = Area × Height. So, if h_A = h_B at equilibrium, then V_A = 3 h and V_B = 5 h, where h is the equilibrium height. But initially, Tank A has 20 m³ and Tank B has 10 m³. So, the total volume is 30 m³. At equilibrium, the total volume should still be 30 m³, right? Because there's no inflow or outflow other than the connecting pipe, so the total volume is conserved.So, V_A + V_B = 30 m³. But V_A = 3 h and V_B = 5 h, so 3 h + 5 h = 8 h = 30. Therefore, h = 30 / 8 = 3.75 meters. So, the equilibrium height is 3.75 meters for both tanks. Wait, but Tank A is elevated above Tank B. Does that affect the equilibrium? Hmm, in the problem statement, it's mentioned that Tank A is elevated above Tank B, but the flow is from A to B, so the elevation difference is already accounted for in the flow rate equation Q = k(h_A - h_B). But at equilibrium, h_A = h_B, so the elevation difference doesn't cause any net flow. That makes sense because if they are at the same height, there's no driving force for the water to flow. But wait, actually, Tank A is elevated, so if h_A = h_B, the actual water levels would have Tank A's water level lower than its elevation? Or is the elevation difference already considered in the height measurements? Hmm, the problem says "the heights of the water levels in the two tanks." So, h_A is the height of the water in Tank A, and h_B is the height in Tank B. Since Tank A is elevated, if h_A = h_B, the actual vertical distance between the water surfaces would be the elevation difference of the tanks plus h_A - h_B. But since h_A = h_B, the net height difference is just the elevation difference. Wait, but the flow rate depends on the height difference between the water levels, which is h_A - h_B. If Tank A is elevated, then h_A is measured from the base of Tank A, and h_B is measured from the base of Tank B. So, if Tank A is elevated, say, by some height H, then the actual vertical distance between the water surfaces would be H + h_A - h_B. But in the problem statement, it just says the connecting pipe allows water to flow from A to B at a rate proportional to the height difference between the water levels in the two tanks. So, I think in this context, h_A and h_B are the heights of the water relative to their respective bases, regardless of the elevation of the tanks. So, when h_A = h_B, the water levels are equal relative to their own bases, but Tank A is still elevated above Tank B. But for the flow rate, it's just the difference in water levels, not considering the elevation. So, if h_A = h_B, then Q = 0, which is correct for equilibrium. So, going back, the equilibrium height is 3.75 meters for both tanks. But let me verify the total volume. At equilibrium, V_A = 3 * 3.75 = 11.25 m³V_B = 5 * 3.75 = 18.75 m³Total volume = 11.25 + 18.75 = 30 m³, which matches the initial total volume. So that's correct.Therefore, the equilibrium heights are both 3.75 meters.Wait, but Tank A is elevated, so does that mean that the water level in Tank A is 3.75 meters above its base, and Tank B is 3.75 meters above its base, which is lower than Tank A's base? So, the actual vertical distance between the water surfaces would be the elevation difference of Tank A plus 3.75 - 3.75 = elevation difference. But since the flow rate depends only on the water level difference, which is zero, so no flow. Yes, that makes sense. So, the equilibrium is when the water levels are equal relative to their own bases, regardless of the elevation of the tanks themselves.So, summarizing:1. The differential equations are:dh_A/dt = (-1/6)(h_A - h_B)dh_B/dt = (1/10)(h_A - h_B)2. The equilibrium heights are both 3.75 meters.I think that's it. Let me just make sure I didn't make any calculation errors.For the differential equations:We started with dV_A/dt = -Q and dV_B/dt = Q.Q = 0.5(h_A - h_B)Expressed in terms of h_A and h_B:dV_A/dt = -0.5(h_A - h_B) => 3 dh_A/dt = -0.5(h_A - h_B) => dh_A/dt = (-0.5/3)(h_A - h_B) = (-1/6)(h_A - h_B)Similarly, dV_B/dt = 0.5(h_A - h_B) => 5 dh_B/dt = 0.5(h_A - h_B) => dh_B/dt = (0.5/5)(h_A - h_B) = (1/10)(h_A - h_B)Yes, that's correct.For the equilibrium:Total volume is conserved, so V_A + V_B = 30 m³.At equilibrium, h_A = h_B = h.So, V_A = 3h, V_B = 5h.3h + 5h = 8h = 30 => h = 3.75 m.Yes, that's correct.So, I think I've got it.</think>

🧠:<think>Okay, so I have this differential equation problem to solve. It's about modeling the recovery of stroke survivors using a second-order differential equation. Let me take it step by step.First, the equation given is:[ frac{d^2R}{dt^2} + 5frac{dR}{dt} + 6R = 0 ]I need to find the general solution for ( R(t) ). Since this is a linear homogeneous differential equation with constant coefficients, I remember that the standard approach is to find the characteristic equation and solve for its roots.So, the characteristic equation for this DE would be:[ r^2 + 5r + 6 = 0 ]Let me solve this quadratic equation. The discriminant is ( b^2 - 4ac = 25 - 24 = 1 ), which is positive, so there are two real roots.Using the quadratic formula:[ r = frac{-b pm sqrt{b^2 - 4ac}}{2a} = frac{-5 pm 1}{2} ]So, the roots are:[ r_1 = frac{-5 + 1}{2} = frac{-4}{2} = -2 ][ r_2 = frac{-5 - 1}{2} = frac{-6}{2} = -3 ]Therefore, the general solution of the homogeneous equation is:[ R_h(t) = C_1 e^{-2t} + C_2 e^{-3t} ]Where ( C_1 ) and ( C_2 ) are constants determined by initial conditions.So that's part 1 done. The general solution is ( R(t) = C_1 e^{-2t} + C_2 e^{-3t} ).Moving on to part 2. Dr. Smith introduces ( M(t) = e^{-2t} sin(3t) ) as the impact of medical care. The total recovery function ( R(t) ) is a combination of the homogeneous solution and a particular solution that includes ( M(t) ).Wait, actually, the original equation is homogeneous, but now we need to include ( M(t) ) as a nonhomogeneous term. So, the equation becomes:[ frac{d^2R}{dt^2} + 5frac{dR}{dt} + 6R = M(t) ][ frac{d^2R}{dt^2} + 5frac{dR}{dt} + 6R = e^{-2t} sin(3t) ]So, now it's a nonhomogeneous differential equation, and we need to find a particular solution ( R_p(t) ) such that the general solution is:[ R(t) = R_h(t) + R_p(t) ]Given that ( R_h(t) = C_1 e^{-2t} + C_2 e^{-3t} ), we need to find ( R_p(t) ).To find the particular solution, I can use the method of undetermined coefficients. The nonhomogeneous term is ( e^{-2t} sin(3t) ). First, I note that the homogeneous solution already includes ( e^{-2t} ) as a term (since ( r = -2 ) is a root). Therefore, when the nonhomogeneous term is of the form ( e^{alpha t} sin(beta t) ) and ( alpha + ibeta ) is a root of the characteristic equation, we need to multiply by ( t ) to find a suitable particular solution.In this case, ( alpha = -2 ) and ( beta = 3 ). Since ( -2 ) is a root, but ( -2 + 3i ) is not a root (the roots are -2 and -3, which are real), so actually, ( e^{-2t} sin(3t) ) is not part of the homogeneous solution. Wait, let me think again.Wait, the homogeneous solution has ( e^{-2t} ) and ( e^{-3t} ). The nonhomogeneous term is ( e^{-2t} sin(3t) ). So, the exponential part ( e^{-2t} ) is already present in the homogeneous solution, but the sine part is not. So, does that mean we need to multiply by ( t ) to account for the exponential part?Yes, because the exponential term is already in the homogeneous solution, but the sine term is not. So, to find the particular solution, we can assume a form:[ R_p(t) = t e^{-2t} (A cos(3t) + B sin(3t)) ]Where ( A ) and ( B ) are constants to be determined.Alternatively, sometimes people write it as ( e^{-2t} (A cos(3t) + B sin(3t)) ), but since ( e^{-2t} ) is part of the homogeneous solution, we need to multiply by ( t ) to ensure linear independence.So, let's proceed with:[ R_p(t) = t e^{-2t} (A cos(3t) + B sin(3t)) ]Now, I need to compute the first and second derivatives of ( R_p(t) ) and substitute them into the differential equation.First, let's compute ( R_p'(t) ):Let me denote ( u = t e^{-2t} ) and ( v = A cos(3t) + B sin(3t) ). Then, ( R_p(t) = u v ).Using the product rule:[ R_p'(t) = u' v + u v' ]Compute ( u' ):( u = t e^{-2t} ), so:[ u' = e^{-2t} + t (-2) e^{-2t} = e^{-2t} (1 - 2t) ]Compute ( v' ):( v = A cos(3t) + B sin(3t) ), so:[ v' = -3A sin(3t) + 3B cos(3t) ]Therefore,[ R_p'(t) = e^{-2t} (1 - 2t) (A cos(3t) + B sin(3t)) + t e^{-2t} (-3A sin(3t) + 3B cos(3t)) ]Let me factor out ( e^{-2t} ):[ R_p'(t) = e^{-2t} [ (1 - 2t)(A cos(3t) + B sin(3t)) + t (-3A sin(3t) + 3B cos(3t)) ] ]Now, let's compute the second derivative ( R_p''(t) ). This will get a bit messy, but let's take it step by step.Again, using the product rule on ( R_p'(t) = e^{-2t} [ ... ] ), let me denote ( w = e^{-2t} ) and ( z = [ (1 - 2t)(A cos(3t) + B sin(3t)) + t (-3A sin(3t) + 3B cos(3t)) ] ). So, ( R_p'(t) = w z ).Then,[ R_p''(t) = w' z + w z' ]Compute ( w' ):( w = e^{-2t} ), so ( w' = -2 e^{-2t} )Compute ( z' ):( z = (1 - 2t)(A cos(3t) + B sin(3t)) + t (-3A sin(3t) + 3B cos(3t)) )Let me differentiate term by term.First term: ( (1 - 2t)(A cos(3t) + B sin(3t)) )Derivative:-2*(A cos(3t) + B sin(3t)) + (1 - 2t)*(-3A sin(3t) + 3B cos(3t))Second term: ( t (-3A sin(3t) + 3B cos(3t)) )Derivative:(-3A sin(3t) + 3B cos(3t)) + t*(-9A cos(3t) - 9B sin(3t))So, putting it all together:z' = [ -2(A cos(3t) + B sin(3t)) + (1 - 2t)(-3A sin(3t) + 3B cos(3t)) ] + [ (-3A sin(3t) + 3B cos(3t)) + t*(-9A cos(3t) - 9B sin(3t)) ]This is getting quite complicated. Maybe I should try a different approach, like using complex exponentials or another method, but perhaps it's manageable.Alternatively, maybe I can substitute ( R_p(t) ) into the differential equation and equate coefficients. Let me try that.So, the differential equation is:[ R'' + 5 R' + 6 R = e^{-2t} sin(3t) ]We have ( R_p(t) = t e^{-2t} (A cos(3t) + B sin(3t)) )Let me compute ( R_p'' + 5 R_p' + 6 R_p ) and set it equal to ( e^{-2t} sin(3t) ), then solve for A and B.First, compute ( R_p(t) ):[ R_p = t e^{-2t} (A cos(3t) + B sin(3t)) ]Compute ( R_p' ):As before, ( R_p' = e^{-2t} [ (1 - 2t)(A cos(3t) + B sin(3t)) + t (-3A sin(3t) + 3B cos(3t)) ] )Compute ( R_p'' ):This is going to be a bit involved, but let's proceed.Let me denote ( R_p' = e^{-2t} [ (1 - 2t)(A cos(3t) + B sin(3t)) + t (-3A sin(3t) + 3B cos(3t)) ] )Let me denote the expression inside the brackets as ( S(t) ). So, ( R_p' = e^{-2t} S(t) )Then, ( R_p'' = (-2) e^{-2t} S(t) + e^{-2t} S'(t) )So, ( R_p'' = e^{-2t} [ -2 S(t) + S'(t) ] )Therefore, ( R_p'' + 5 R_p' + 6 R_p = e^{-2t} [ -2 S(t) + S'(t) ] + 5 e^{-2t} S(t) + 6 t e^{-2t} (A cos(3t) + B sin(3t)) )Factor out ( e^{-2t} ):[ e^{-2t} [ (-2 S + S') + 5 S + 6 t (A cos(3t) + B sin(3t)) ] ]Simplify inside the brackets:[ (-2 S + S' + 5 S) + 6 t (A cos(3t) + B sin(3t)) ][ (3 S + S') + 6 t (A cos(3t) + B sin(3t)) ]Now, substitute ( S(t) ):Recall ( S(t) = (1 - 2t)(A cos(3t) + B sin(3t)) + t (-3A sin(3t) + 3B cos(3t)) )So, let's compute ( 3 S(t) + S'(t) + 6 t (A cos(3t) + B sin(3t)) )First, compute ( 3 S(t) ):[ 3 (1 - 2t)(A cos(3t) + B sin(3t)) + 3 t (-3A sin(3t) + 3B cos(3t)) ]Then, compute ( S'(t) ):Earlier, we had:z' = [ -2(A cos(3t) + B sin(3t)) + (1 - 2t)(-3A sin(3t) + 3B cos(3t)) ] + [ (-3A sin(3t) + 3B cos(3t)) + t*(-9A cos(3t) - 9B sin(3t)) ]Wait, actually, ( S(t) ) is the same as z(t), so S'(t) is z'(t), which we computed earlier.But perhaps it's better to compute S'(t) directly.Wait, S(t) is:[ S(t) = (1 - 2t)(A cos(3t) + B sin(3t)) + t (-3A sin(3t) + 3B cos(3t)) ]So, to find S'(t), differentiate term by term:First term: ( (1 - 2t)(A cos(3t) + B sin(3t)) )Derivative:-2*(A cos(3t) + B sin(3t)) + (1 - 2t)*(-3A sin(3t) + 3B cos(3t))Second term: ( t (-3A sin(3t) + 3B cos(3t)) )Derivative:(-3A sin(3t) + 3B cos(3t)) + t*(-9A cos(3t) - 9B sin(3t))So, combining these:S'(t) = -2(A cos(3t) + B sin(3t)) + (1 - 2t)(-3A sin(3t) + 3B cos(3t)) + (-3A sin(3t) + 3B cos(3t)) + t*(-9A cos(3t) - 9B sin(3t))Now, let's expand and collect like terms.First, expand the terms:1. -2A cos(3t) - 2B sin(3t)2. (1 - 2t)(-3A sin(3t) + 3B cos(3t)) = -3A sin(3t) + 3B cos(3t) + 6A t sin(3t) - 6B t cos(3t)3. -3A sin(3t) + 3B cos(3t)4. -9A t cos(3t) - 9B t sin(3t)Now, combine all these:-2A cos(3t) - 2B sin(3t)-3A sin(3t) + 3B cos(3t) + 6A t sin(3t) - 6B t cos(3t)-3A sin(3t) + 3B cos(3t)-9A t cos(3t) - 9B t sin(3t)Now, let's collect like terms:For cos(3t):-2A cos(3t) + 3B cos(3t) + 3B cos(3t) -6B t cos(3t) -9A t cos(3t)For sin(3t):-2B sin(3t) -3A sin(3t) -3A sin(3t) +6A t sin(3t) -9B t sin(3t)So, let's compute each:cos(3t) terms:-2A + 3B + 3B = (-2A + 6B) cos(3t)Then, the terms with t:-6B t -9A t = (-6B -9A) t cos(3t)sin(3t) terms:-2B -3A -3A = (-2B -6A) sin(3t)Then, the terms with t:6A t -9B t = (6A -9B) t sin(3t)So, overall:S'(t) = [ (-2A + 6B) + (-6B -9A) t ] cos(3t) + [ (-2B -6A) + (6A -9B) t ] sin(3t)Now, going back to the expression:3 S(t) + S'(t) + 6 t (A cos(3t) + B sin(3t))We already have S(t):S(t) = (1 - 2t)(A cos(3t) + B sin(3t)) + t (-3A sin(3t) + 3B cos(3t))Let me expand S(t):= A cos(3t) + B sin(3t) - 2t A cos(3t) - 2t B sin(3t) -3A t sin(3t) + 3B t cos(3t)So, S(t) = [A cos(3t) + B sin(3t)] + t [ -2A cos(3t) - 2B sin(3t) -3A sin(3t) + 3B cos(3t) ]= [A cos(3t) + B sin(3t)] + t [ (-2A + 3B) cos(3t) + (-2B -3A) sin(3t) ]Therefore, 3 S(t) = 3 [A cos(3t) + B sin(3t)] + 3 t [ (-2A + 3B) cos(3t) + (-2B -3A) sin(3t) ]So, 3 S(t) = 3A cos(3t) + 3B sin(3t) + 3t (-2A + 3B) cos(3t) + 3t (-2B -3A) sin(3t)Now, let's write down 3 S(t) + S'(t) + 6 t (A cos(3t) + B sin(3t)):First, 3 S(t):3A cos(3t) + 3B sin(3t) + 3t (-2A + 3B) cos(3t) + 3t (-2B -3A) sin(3t)Plus S'(t):[ (-2A + 6B) + (-6B -9A) t ] cos(3t) + [ (-2B -6A) + (6A -9B) t ] sin(3t)Plus 6 t (A cos(3t) + B sin(3t)):6A t cos(3t) + 6B t sin(3t)Now, let's combine all these terms.First, collect the cos(3t) terms:From 3 S(t):3A cos(3t) + 3t (-2A + 3B) cos(3t)From S'(t):[ (-2A + 6B) + (-6B -9A) t ] cos(3t)From 6 t term:6A t cos(3t)So, combining:3A + (-2A + 6B) + [ 3*(-2A + 3B) + (-6B -9A) + 6A ] tSimilarly for sin(3t):From 3 S(t):3B sin(3t) + 3t (-2B -3A) sin(3t)From S'(t):[ (-2B -6A) + (6A -9B) t ] sin(3t)From 6 t term:6B t sin(3t)So, combining:3B + (-2B -6A) + [ 3*(-2B -3A) + (6A -9B) + 6B ] tLet me compute the coefficients step by step.First, for cos(3t):Constant terms:3A + (-2A + 6B) = (3A - 2A) + 6B = A + 6BCoefficient of t:3*(-2A + 3B) + (-6B -9A) + 6A= (-6A + 9B) + (-6B -9A) + 6ACombine like terms:-6A -9A +6A = (-6 -9 +6)A = (-9)A9B -6B = 3BSo, coefficient of t: -9A + 3BSimilarly, for sin(3t):Constant terms:3B + (-2B -6A) = (3B -2B) -6A = B -6ACoefficient of t:3*(-2B -3A) + (6A -9B) + 6B= (-6B -9A) + (6A -9B) + 6BCombine like terms:-6B -9B +6B = (-6 -9 +6)B = (-9)B-9A +6A = (-3A)So, coefficient of t: -9B -3ATherefore, putting it all together:3 S(t) + S'(t) + 6 t (A cos(3t) + B sin(3t)) =[ (A + 6B) + (-9A + 3B) t ] cos(3t) + [ (B -6A) + (-9B -3A) t ] sin(3t)But this expression is equal to the left-hand side of the differential equation, which is ( e^{-2t} sin(3t) ). Therefore, we have:[ e^{-2t} [ (A + 6B) + (-9A + 3B) t ] cos(3t) + [ (B -6A) + (-9B -3A) t ] sin(3t) ] = e^{-2t} sin(3t) ]Since ( e^{-2t} ) is never zero, we can divide both sides by ( e^{-2t} ), resulting in:[ [ (A + 6B) + (-9A + 3B) t ] cos(3t) + [ (B -6A) + (-9B -3A) t ] sin(3t) = sin(3t) ]Now, for this equation to hold for all t, the coefficients of ( cos(3t) ) and ( sin(3t) ) must match on both sides. On the right-hand side, the coefficient of ( cos(3t) ) is 0, and the coefficient of ( sin(3t) ) is 1.Therefore, we can set up the following equations:For ( cos(3t) ):1. Coefficient of ( cos(3t) ): ( (A + 6B) + (-9A + 3B) t = 0 ) for all t.But this is a polynomial in t, so the coefficients of t and the constant term must both be zero.So,- Coefficient of t: ( -9A + 3B = 0 )- Constant term: ( A + 6B = 0 )Similarly, for ( sin(3t) ):2. Coefficient of ( sin(3t) ): ( (B -6A) + (-9B -3A) t = 1 ) for all t.Again, this is a polynomial in t, so:- Coefficient of t: ( -9B -3A = 0 )- Constant term: ( B -6A = 1 )So, now we have a system of four equations:From ( cos(3t) ):1. ( -9A + 3B = 0 ) --> Let's call this equation (1)2. ( A + 6B = 0 ) --> Equation (2)From ( sin(3t) ):3. ( -9B -3A = 0 ) --> Equation (3)4. ( B -6A = 1 ) --> Equation (4)Wait, actually, equations (1) and (3) are similar. Let me check:Equation (1): -9A + 3B = 0Equation (3): -9B -3A = 0Let me write them:Equation (1): -9A + 3B = 0 --> Divide by 3: -3A + B = 0 --> B = 3AEquation (3): -9B -3A = 0 --> Divide by -3: 3B + A = 0 --> A = -3BBut from equation (1), B = 3A, substitute into equation (3):A = -3B = -3*(3A) = -9ASo, A = -9A --> 10A = 0 --> A = 0Then, from equation (1): B = 3A = 0But then, from equation (4): B -6A = 1 --> 0 -0 = 1 --> 0 =1, which is a contradiction.Hmm, that's a problem. It seems inconsistent. That suggests that perhaps my assumption for the particular solution was incorrect, or perhaps I made a mistake in the differentiation.Wait, let me double-check the differentiation steps because this inconsistency suggests an error.Alternatively, maybe I should have multiplied by t^2 instead of t because the exponential term is already in the homogeneous solution. Wait, no, the exponential term is e^{-2t}, which is a solution, but the nonhomogeneous term is e^{-2t} sin(3t). So, since the exponential part is already in the homogeneous solution, but the sine part is not, so we only need to multiply by t once. So, my initial assumption was correct.Alternatively, perhaps I made an error in computing S'(t). Let me check.Wait, in the expression for S'(t), I had:S'(t) = [ (-2A + 6B) + (-6B -9A) t ] cos(3t) + [ (-2B -6A) + (6A -9B) t ] sin(3t)But when I combined 3 S(t) + S'(t) + 6 t (A cos(3t) + B sin(3t)), I might have made a mistake in the coefficients.Alternatively, perhaps it's better to use the method of undetermined coefficients with complex functions.Let me try that approach.Let me write the nonhomogeneous term as the imaginary part of ( e^{-2t} e^{i3t} ). So, ( e^{-2t} sin(3t) = text{Im}(e^{(-2 + 3i)t}) )So, let me assume a particular solution of the form ( R_p(t) = t e^{(-2 + 3i)t} ) (since -2 + 3i is not a root of the characteristic equation, but wait, actually, the homogeneous solution has roots -2 and -3, so -2 + 3i is not a root. Therefore, we don't need to multiply by t. Wait, but in the homogeneous solution, we have e^{-2t}, but not e^{(-2 + 3i)t}. So, actually, perhaps the particular solution can be assumed as ( R_p(t) = e^{-2t} (A cos(3t) + B sin(3t)) ) without multiplying by t.Wait, but earlier, I thought that since e^{-2t} is part of the homogeneous solution, we need to multiply by t. But in the complex approach, since -2 + 3i is not a root, we don't need to multiply by t. Hmm, this is conflicting.Wait, perhaps the confusion arises because in the complex approach, the particular solution is assumed as ( e^{rt} ), where r is the root of the characteristic equation of the nonhomogeneous term. Since r = -2 + 3i is not a root of the homogeneous equation, we don't need to multiply by t. Therefore, the particular solution can be written as ( e^{-2t} (A cos(3t) + B sin(3t)) ).But earlier, I thought that since e^{-2t} is part of the homogeneous solution, we need to multiply by t. Maybe that's incorrect because the homogeneous solution has e^{-2t}, but the nonhomogeneous term is e^{-2t} sin(3t), which is a product of e^{-2t} and sin(3t). So, perhaps the correct approach is to use the method of undetermined coefficients with a particular solution of the form ( e^{-2t} (A cos(3t) + B sin(3t)) ), without multiplying by t, because the exponential term is already in the homogeneous solution, but the sine term is not. Wait, no, actually, the product is considered as a single term, so if the exponential part is already in the homogeneous solution, we need to multiply by t.Wait, let me clarify. The method says that if the nonhomogeneous term is of the form ( e^{alpha t} cos(beta t) ) or ( e^{alpha t} sin(beta t) ), and ( alpha + ibeta ) is a root of the characteristic equation, then we need to multiply by t^k, where k is the multiplicity of the root.In our case, the nonhomogeneous term is ( e^{-2t} sin(3t) ). The characteristic equation for the nonhomogeneous term would have roots -2 ± 3i. However, our homogeneous equation has roots -2 and -3, which are real. Therefore, -2 + 3i is not a root of the homogeneous equation, so we don't need to multiply by t. Therefore, the particular solution can be assumed as ( e^{-2t} (A cos(3t) + B sin(3t)) ).Wait, that contradicts my earlier thought. Let me check.The rule is: if the nonhomogeneous term is of the form ( e^{alpha t} cos(beta t) ) or ( e^{alpha t} sin(beta t) ), and ( alpha + ibeta ) is a root of the characteristic equation, then multiply by t^k, where k is the multiplicity.In our case, the nonhomogeneous term is ( e^{-2t} sin(3t) ), so ( alpha = -2 ), ( beta = 3 ). The characteristic equation for the nonhomogeneous term would have roots ( -2 pm 3i ). However, our homogeneous equation has roots -2 and -3, which are real. Therefore, ( -2 + 3i ) is not a root of the homogeneous equation, so we don't need to multiply by t. Therefore, the particular solution can be assumed as ( e^{-2t} (A cos(3t) + B sin(3t)) ).Wait, but earlier, I thought that since e^{-2t} is part of the homogeneous solution, we need to multiply by t. But perhaps that's incorrect because the rule is about the complex roots, not just the exponential part. So, since -2 + 3i is not a root, we don't need to multiply by t.Therefore, let me try assuming ( R_p(t) = e^{-2t} (A cos(3t) + B sin(3t)) ).Let me compute ( R_p'' + 5 R_p' + 6 R_p ) and set it equal to ( e^{-2t} sin(3t) ).Compute ( R_p(t) = e^{-2t} (A cos(3t) + B sin(3t)) )Compute ( R_p'(t) ):Using product rule:( R_p' = (-2) e^{-2t} (A cos(3t) + B sin(3t)) + e^{-2t} (-3A sin(3t) + 3B cos(3t)) )Factor out ( e^{-2t} ):( R_p' = e^{-2t} [ -2(A cos(3t) + B sin(3t)) + (-3A sin(3t) + 3B cos(3t)) ] )Simplify inside:= ( e^{-2t} [ (-2A + 3B) cos(3t) + (-2B -3A) sin(3t) ] )Compute ( R_p''(t) ):Differentiate ( R_p'(t) ):Again, using product rule:( R_p'' = (-2) e^{-2t} [ (-2A + 3B) cos(3t) + (-2B -3A) sin(3t) ] + e^{-2t} [ 3(-2A + 3B) sin(3t) - 3(-2B -3A) cos(3t) ] )Simplify:= ( e^{-2t} [ 2(2A - 3B) cos(3t) + 2(2B + 3A) sin(3t) + (-6A + 9B) sin(3t) + (6B + 9A) cos(3t) ] )Combine like terms:cos(3t) terms: ( 2(2A - 3B) + (6B + 9A) = (4A -6B +6B +9A) = 13A )sin(3t) terms: ( 2(2B + 3A) + (-6A +9B) = (4B +6A -6A +9B) = 13B )So, ( R_p'' = e^{-2t} (13A cos(3t) + 13B sin(3t)) )Now, compute ( R_p'' + 5 R_p' + 6 R_p ):= ( e^{-2t} (13A cos(3t) + 13B sin(3t)) + 5 e^{-2t} [ (-2A + 3B) cos(3t) + (-2B -3A) sin(3t) ] + 6 e^{-2t} (A cos(3t) + B sin(3t)) )Factor out ( e^{-2t} ):= ( e^{-2t} [ 13A cos(3t) + 13B sin(3t) + 5(-2A + 3B) cos(3t) + 5(-2B -3A) sin(3t) + 6A cos(3t) + 6B sin(3t) ] )Now, combine like terms:cos(3t) terms:13A + 5*(-2A +3B) +6A = 13A -10A +15B +6A = (13 -10 +6)A +15B = 9A +15Bsin(3t) terms:13B +5*(-2B -3A) +6B =13B -10B -15A +6B = (13 -10 +6)B -15A =9B -15ASo, the entire expression becomes:( e^{-2t} [ (9A +15B) cos(3t) + (9B -15A) sin(3t) ] )Set this equal to ( e^{-2t} sin(3t) ):So,( (9A +15B) cos(3t) + (9B -15A) sin(3t) = sin(3t) )Therefore, we can set up the equations:1. ( 9A +15B = 0 ) (coefficient of cos(3t))2. ( 9B -15A = 1 ) (coefficient of sin(3t))Now, solve this system.From equation 1:9A +15B =0 --> Divide by 3: 3A +5B=0 --> 3A = -5B --> A = (-5/3)BSubstitute into equation 2:9B -15A =1= 9B -15*(-5/3 B) =1= 9B +25B =1=34B=1 --> B=1/34Then, A= (-5/3)*(1/34)= -5/(102)= -5/102So, A= -5/102, B=1/34Therefore, the particular solution is:( R_p(t) = e^{-2t} ( (-5/102) cos(3t) + (1/34) sin(3t) ) )Simplify:We can write this as:( R_p(t) = e^{-2t} left( -frac{5}{102} cos(3t) + frac{1}{34} sin(3t) right) )Alternatively, factor out 1/34:= ( e^{-2t} cdot frac{1}{34} left( -frac{5}{3} cos(3t) + sin(3t) right) )But perhaps it's fine as is.Therefore, the general solution is:( R(t) = R_h(t) + R_p(t) = C_1 e^{-2t} + C_2 e^{-3t} + e^{-2t} left( -frac{5}{102} cos(3t) + frac{1}{34} sin(3t) right) )Now, apply the initial conditions to find ( C_1 ) and ( C_2 ).Given:1. ( R(0) = 2 )2. ( R'(0) = -1 )First, compute ( R(0) ):( R(0) = C_1 e^{0} + C_2 e^{0} + e^{0} left( -frac{5}{102} cos(0) + frac{1}{34} sin(0) right) )Simplify:= ( C_1 + C_2 + [ -frac{5}{102} *1 + 0 ] )= ( C_1 + C_2 - frac{5}{102} = 2 )So, equation (1): ( C_1 + C_2 = 2 + frac{5}{102} = frac{204}{102} + frac{5}{102} = frac{209}{102} )Next, compute ( R'(t) ):First, let's find the derivative of the general solution.( R(t) = C_1 e^{-2t} + C_2 e^{-3t} + e^{-2t} left( -frac{5}{102} cos(3t) + frac{1}{34} sin(3t) right) )Compute ( R'(t) ):= ( -2 C_1 e^{-2t} -3 C_2 e^{-3t} + frac{d}{dt} [ e^{-2t} ( -frac{5}{102} cos(3t) + frac{1}{34} sin(3t) ) ] )Compute the derivative of the particular solution term:Let me denote ( f(t) = e^{-2t} ( -frac{5}{102} cos(3t) + frac{1}{34} sin(3t) ) )Then, ( f'(t) = -2 e^{-2t} ( -frac{5}{102} cos(3t) + frac{1}{34} sin(3t) ) + e^{-2t} ( frac{5}{102} *3 sin(3t) + frac{1}{34} *3 cos(3t) ) )Simplify:= ( e^{-2t} [ 2( frac{5}{102} cos(3t) - frac{1}{34} sin(3t) ) + ( frac{15}{102} sin(3t) + frac{3}{34} cos(3t) ) ] )Simplify each term:First term inside the brackets:2*(5/102 cos(3t) -1/34 sin(3t)) = (10/102) cos(3t) - (2/34) sin(3t) = (5/51) cos(3t) - (1/17) sin(3t)Second term:15/102 sin(3t) + 3/34 cos(3t) = (5/34) sin(3t) + (3/34) cos(3t)Combine both terms:cos(3t): 5/51 + 3/34 = (10/102 + 9/102) = 19/102sin(3t): -1/17 +5/34 = (-2/34 +5/34)=3/34Therefore, ( f'(t) = e^{-2t} ( frac{19}{102} cos(3t) + frac{3}{34} sin(3t) ) )So, putting it all together, ( R'(t) = -2 C_1 e^{-2t} -3 C_2 e^{-3t} + e^{-2t} ( frac{19}{102} cos(3t) + frac{3}{34} sin(3t) ) )Now, evaluate ( R'(0) ):= ( -2 C_1 e^{0} -3 C_2 e^{0} + e^{0} ( frac{19}{102} cos(0) + frac{3}{34} sin(0) ) )= ( -2 C_1 -3 C_2 + ( frac{19}{102} *1 + 0 ) )= ( -2 C_1 -3 C_2 + frac{19}{102} = -1 )So, equation (2): ( -2 C_1 -3 C_2 = -1 - frac{19}{102} = -frac{102}{102} - frac{19}{102} = -frac{121}{102} )Now, we have the system:1. ( C_1 + C_2 = frac{209}{102} )2. ( -2 C_1 -3 C_2 = -frac{121}{102} )Let me write this as:Equation (1): ( C_1 + C_2 = frac{209}{102} )Equation (2): ( -2 C_1 -3 C_2 = -frac{121}{102} )Let me solve this system.Multiply equation (1) by 2:2 C_1 + 2 C_2 = 418/102Now, add to equation (2):(2 C_1 + 2 C_2) + (-2 C_1 -3 C_2) = 418/102 + (-121)/102Simplify:(0 C_1 - C_2) = (418 -121)/102 = 297/102So, -C_2 = 297/102 --> C_2 = -297/102Simplify:297 ÷ 3 = 99, 102 ÷3=34 --> -99/34So, C_2 = -99/34Now, from equation (1):C_1 + (-99/34) = 209/102Convert 209/102 to 34 denominator:209/102 = (209 ÷ 3)/(102 ÷3)= 69.666.../34, but better to find a common denominator.Alternatively, express 209/102 as (209*1)/(102*1). Let me compute 209/102:209 ÷ 102 = 2 with remainder 5 (since 102*2=204, 209-204=5). So, 209/102 = 2 +5/102Similarly, C_1 = 209/102 - C_2 = 209/102 - (-99/34) = 209/102 +99/34Convert 99/34 to denominator 102:99/34 = (99*3)/(34*3)=297/102So, C_1 = 209/102 +297/102 = (209 +297)/102 =506/102Simplify:506 ÷ 2=253, 102 ÷2=51 --> 253/51So, C_1=253/51Therefore, the specific solution is:( R(t) = frac{253}{51} e^{-2t} - frac{99}{34} e^{-3t} + e^{-2t} left( -frac{5}{102} cos(3t) + frac{1}{34} sin(3t) right) )We can factor out ( e^{-2t} ) from the last two terms:= ( frac{253}{51} e^{-2t} - frac{99}{34} e^{-3t} + e^{-2t} left( -frac{5}{102} cos(3t) + frac{1}{34} sin(3t) right) )Alternatively, combine the ( e^{-2t} ) terms:= ( left( frac{253}{51} - frac{5}{102} cos(3t) + frac{1}{34} sin(3t) right) e^{-2t} - frac{99}{34} e^{-3t} )But perhaps it's clearer to leave it as is.So, the specific solution is:( R(t) = frac{253}{51} e^{-2t} - frac{99}{34} e^{-3t} + e^{-2t} left( -frac{5}{102} cos(3t) + frac{1}{34} sin(3t) right) )Alternatively, we can write the constants as decimals for simplicity, but since the problem doesn't specify, fractions are acceptable.So, to summarize:1. The general solution is ( R(t) = C_1 e^{-2t} + C_2 e^{-3t} )2. The particular solution is ( R_p(t) = e^{-2t} left( -frac{5}{102} cos(3t) + frac{1}{34} sin(3t) right) )3. The specific solution with initial conditions is:( R(t) = frac{253}{51} e^{-2t} - frac{99}{34} e^{-3t} + e^{-2t} left( -frac{5}{102} cos(3t) + frac{1}{34} sin(3t) right) )I think this is the final answer.</think>

🧠:<think>Alright, so I've got this problem about a film producer who wants to adapt comic book properties into movies. She has a budget and wants to maximize her expected success score. Let me try to break this down step by step.Starting with part 1. She has n comic book properties, each with a success score s_i and a cost c_i. She needs to choose a subset of these properties such that the total cost doesn't exceed her budget B, and the sum of their success scores is maximized. Hmm, this sounds familiar. It's like the classic knapsack problem where you have items with weights and values, and you want to maximize the value without exceeding the weight limit.So, in the knapsack analogy, each comic book property is an item. The "weight" of each item is the cost c_i, and the "value" is the success score s_i. The budget B is the knapsack's capacity. The goal is to select items (properties) such that the total weight (cost) is within B and the total value (success score) is as high as possible.To formulate this as an optimization problem, I can define a binary variable x_i for each property i. x_i will be 1 if the property is selected and 0 otherwise. Then, the problem becomes:Maximize Σ (s_i * x_i) for all i from 1 to nSubject to Σ (c_i * x_i) ≤ BAnd x_i ∈ {0,1} for all i.This is a 0-1 knapsack problem. The solution will give the maximum success score achievable within the budget.Now, for part 2, there's an added twist: synergy effects. If properties belong to the same comic book universe, the total success score is amplified by a factor α_k, where α_k = 1 + 0.1 * |U_k|. So, if a universe has more properties selected, the amplification factor increases.This complicates things because now the success score isn't just the sum of individual s_i, but it's scaled based on the number of properties chosen from each universe. So, we need to model this in our optimization.First, let's think about how to represent this. Suppose there are m different comic book universes. Each universe k has a set U_k of properties. If we select t properties from universe k, then the success score for that universe becomes (Σ s_i for i in selected U_k) * α_k, where α_k = 1 + 0.1 * t.Wait, but α_k depends on how many properties we select from universe k. So, it's not a fixed multiplier; it changes based on the selection. This makes the problem non-linear because the multiplier depends on the number of selected items in each universe.Hmm, how can we model this? Maybe we can introduce another variable for each universe to track how many properties are selected from it. Let me denote y_k as the number of properties selected from universe k. Then, α_k = 1 + 0.1 * y_k.But then, the total success score would be Σ (Σ s_i * x_i for i in U_k) * α_k for all k. Wait, no, that might not be accurate because each property is in exactly one universe, right? So, the total success score is the sum over all universes of (sum of s_i for selected properties in U_k) multiplied by α_k.But since each property is in only one universe, the total success score can be written as Σ (s_i * x_i) * α_k for each i. Wait, no, that's not quite right. Because α_k is specific to universe k, and each property is in one universe, so for each property i in universe k, its contribution to the total success is s_i * x_i multiplied by α_k.But α_k depends on how many properties are selected from universe k, which is y_k = Σ x_i for i in U_k. So, the total success score is Σ (s_i * x_i) * (1 + 0.1 * y_k) for each i.But this seems a bit tangled because y_k is dependent on x_i. Maybe we can express it differently.Alternatively, for each universe k, the contribution to the total success is (Σ s_i * x_i for i in U_k) * (1 + 0.1 * y_k). So, the total success is the sum over all k of [ (Σ s_i * x_i for i in U_k) * (1 + 0.1 * y_k) ].This is a non-linear term because y_k is a sum of x_i's, which are binary variables. So, the problem becomes a non-linear integer programming problem.To handle this, maybe we can linearize the expression. Let's consider that for each universe k, the contribution is (Σ s_i * x_i) + 0.1 * (Σ s_i * x_i) * y_k.But y_k is Σ x_i for i in U_k, so 0.1 * (Σ s_i * x_i) * y_k is 0.1 * Σ s_i * x_i * Σ x_j for j in U_k.This is a quadratic term in x_i and x_j, which complicates things. It might not be easy to linearize this without some clever manipulation.Alternatively, perhaps we can model this by considering that selecting more properties from a universe increases the multiplier for all properties in that universe. So, for each universe, the more properties you select, the higher the multiplier, which could encourage selecting more properties from the same universe to get a higher amplification.But how does this affect the selection? It might be beneficial to cluster selections within a universe to get a higher multiplier, even if some properties in other universes have higher individual success scores.This could lead to a situation where even if a property in another universe has a higher s_i, the amplification from selecting multiple properties in a single universe might make it more attractive.So, the problem now is to select a subset of properties, possibly multiple from the same universe, such that the total cost is within budget, and the total success score, which is the sum of individual s_i multiplied by their universe's amplification factor, is maximized.This seems like a variation of the knapsack problem with a non-linear reward structure. It might not have a straightforward greedy solution, and exact solutions might require more complex algorithms or heuristics.Perhaps a dynamic programming approach could be adapted, but the non-linearity complicates things. Alternatively, we might need to use integer programming techniques with the non-linear terms handled through piecewise linear approximations or other methods.In summary, part 1 is a standard 0-1 knapsack problem, while part 2 introduces a non-linear amplification factor based on the number of properties selected from each universe, making it a more complex optimization problem that likely requires a different approach or more advanced techniques to solve.</think>

🧠:<think>Okay, so I have this problem about optimizing the drag coefficient for racing suits. It's divided into two parts. Let me start with the first one.Problem 1: Finding Critical PointsThe drag coefficient is given by:[ C_d(f, theta) = a cdot left(frac{1}{1 + b cdot e^{-c cdot f}}right) cdot cos^2(theta) ]With constants ( a = 1.5 ), ( b = 2.0 ), and ( c = 0.5 ).I need to find the critical points of this function with respect to both ( f ) and ( theta ). Critical points occur where the partial derivatives are zero or undefined. Since the function is smooth, I just need to set the partial derivatives to zero.Let me first compute the partial derivative with respect to ( f ).Let me denote the function as:[ C_d(f, theta) = 1.5 cdot left(frac{1}{1 + 2e^{-0.5f}}right) cdot cos^2(theta) ]Let me simplify the expression inside the function:Let ( g(f) = frac{1}{1 + 2e^{-0.5f}} ). So, ( C_d = 1.5 cdot g(f) cdot cos^2(theta) ).First, find ( frac{partial C_d}{partial f} ):The derivative of ( g(f) ) with respect to ( f ):[ g'(f) = frac{d}{df} left( frac{1}{1 + 2e^{-0.5f}} right) ]Using the chain rule:Let ( u = 1 + 2e^{-0.5f} ), so ( g(f) = 1/u ), so ( g'(f) = -1/u^2 cdot du/df ).Compute ( du/df ):[ du/df = 2 cdot (-0.5) e^{-0.5f} = -e^{-0.5f} ]Thus,[ g'(f) = -1/(1 + 2e^{-0.5f})^2 cdot (-e^{-0.5f}) = frac{e^{-0.5f}}{(1 + 2e^{-0.5f})^2} ]Therefore,[ frac{partial C_d}{partial f} = 1.5 cdot frac{e^{-0.5f}}{(1 + 2e^{-0.5f})^2} cdot cos^2(theta) ]To find critical points, set this equal to zero:[ 1.5 cdot frac{e^{-0.5f}}{(1 + 2e^{-0.5f})^2} cdot cos^2(theta) = 0 ]Since ( e^{-0.5f} ) is always positive, and ( 1.5 ) and ( (1 + 2e^{-0.5f})^2 ) are positive, the only way this product is zero is if ( cos^2(theta) = 0 ).So, ( cos(theta) = 0 ) which implies ( theta = pi/2 + kpi ) for integer ( k ). But since ( theta ) is in radians, and probably within a range, say ( [0, 2pi) ), the critical points with respect to ( f ) occur when ( theta = pi/2 ) or ( 3pi/2 ).But wait, actually, when taking partial derivatives, we consider each variable separately. So, for critical points, both partial derivatives must be zero simultaneously.Wait, actually, critical points in multivariable functions are points where both partial derivatives are zero. So, I need to set both ( partial C_d / partial f = 0 ) and ( partial C_d / partial theta = 0 ).So, let's compute the partial derivative with respect to ( theta ):[ frac{partial C_d}{partial theta} = 1.5 cdot frac{1}{1 + 2e^{-0.5f}} cdot 2cos(theta)(-sin(theta)) ]Simplify:[ frac{partial C_d}{partial theta} = -3 cdot frac{cos(theta)sin(theta)}{1 + 2e^{-0.5f}} ]Set this equal to zero:[ -3 cdot frac{cos(theta)sin(theta)}{1 + 2e^{-0.5f}} = 0 ]Again, since the denominator is always positive, and the constant is non-zero, we have:[ cos(theta)sin(theta) = 0 ]Which implies either ( cos(theta) = 0 ) or ( sin(theta) = 0 ).So, ( theta = 0, pi/2, pi, 3pi/2, 2pi ), etc.Now, to find critical points, both partial derivatives must be zero. So, let's see:From ( partial C_d / partial f = 0 ), we have ( cos^2(theta) = 0 ), so ( theta = pi/2, 3pi/2 ).From ( partial C_d / partial theta = 0 ), we have ( cos(theta)sin(theta) = 0 ), so ( theta = 0, pi/2, pi, 3pi/2, 2pi ).So, the overlapping solutions are ( theta = pi/2, 3pi/2 ).Now, for these ( theta ), we need to check if ( partial C_d / partial f = 0 ). But as we saw earlier, ( partial C_d / partial f ) is zero only when ( cos^2(theta) = 0 ), which is exactly at ( theta = pi/2, 3pi/2 ). So, for these ( theta ), the partial derivative with respect to ( f ) is zero regardless of ( f ).Wait, that's interesting. So, for ( theta = pi/2, 3pi/2 ), the partial derivative with respect to ( f ) is zero for any ( f ). But does that mean that along these ( theta ), the function is constant with respect to ( f )?Wait, let me check the function ( C_d(f, theta) ). At ( theta = pi/2 ), ( cos(pi/2) = 0 ), so ( C_d(f, pi/2) = 0 ) regardless of ( f ). Similarly, at ( theta = 3pi/2 ), ( cos(3pi/2) = 0 ), so again ( C_d = 0 ).Therefore, along these ( theta ), the function is zero for any ( f ). So, in terms of critical points, these are points where the function is minimized (since drag coefficient is being minimized). So, these are minima.But wait, in terms of critical points, they are points where the gradient is zero. So, yes, these are critical points.But are there any other critical points? Let's see.Suppose ( theta ) is not ( pi/2 ) or ( 3pi/2 ). Then, ( partial C_d / partial f ) is not zero, so the only critical points are at ( theta = pi/2, 3pi/2 ), and any ( f ). But since at these ( theta ), the function is zero regardless of ( f ), so all points along these ( theta ) lines are critical points.But in terms of optimization, the minimum drag coefficient is zero, achieved when ( theta = pi/2 ) or ( 3pi/2 ), regardless of ( f ). So, the critical points are all points where ( theta = pi/2 ) or ( 3pi/2 ), and ( f ) can be any value.But wait, that seems a bit strange. Let me think again.Wait, the function ( C_d(f, theta) ) is a product of two terms: one depending on ( f ) and another on ( theta ). The term depending on ( f ) is always positive because ( a ), ( b ), ( c ) are positive, and ( e^{-0.5f} ) is positive, so ( 1/(1 + 2e^{-0.5f}) ) is between 0 and 1. The term ( cos^2(theta) ) is also always non-negative.Therefore, the function ( C_d(f, theta) ) is non-negative. The minimum value is zero, achieved when ( cos^2(theta) = 0 ), i.e., ( theta = pi/2, 3pi/2 ), regardless of ( f ).So, in terms of critical points, all points where ( theta = pi/2 ) or ( 3pi/2 ) are critical points, and they are minima.But wait, what about the partial derivative with respect to ( f )? For ( theta ) not equal to ( pi/2 ) or ( 3pi/2 ), the partial derivative with respect to ( f ) is not zero, so those points are not critical points. Therefore, the only critical points are along ( theta = pi/2 ) and ( 3pi/2 ), for any ( f ).But let me check if the function has any maxima or other critical points.Looking at the function ( C_d(f, theta) ), it's a product of a sigmoid-like function in ( f ) and ( cos^2(theta) ). The sigmoid function ( 1/(1 + 2e^{-0.5f}) ) increases with ( f ), approaching 1 as ( f ) approaches infinity, and approaching 1/3 as ( f ) approaches negative infinity.So, for fixed ( theta ), ( C_d ) increases with ( f ). Therefore, for ( theta ) not equal to ( pi/2 ) or ( 3pi/2 ), the function is increasing in ( f ), so the minimum occurs at the smallest possible ( f ), but since ( f ) can be any positive value (as it's oscillations per meter), the minimum is achieved as ( f ) approaches zero.Wait, but in terms of critical points, we are looking for points where the gradient is zero, not necessarily global minima or maxima.So, for ( theta ) not equal to ( pi/2 ) or ( 3pi/2 ), the partial derivative with respect to ( f ) is positive, meaning the function is increasing in ( f ), so there's no critical point in ( f ) unless ( f ) is at a boundary. But since ( f ) can be any positive real number, there's no critical point in ( f ) except where the derivative is zero, which only happens when ( cos^2(theta) = 0 ).Therefore, the only critical points are along ( theta = pi/2 ) and ( 3pi/2 ), for any ( f ), and these are minima.So, to summarize, the critical points are all points where ( theta = pi/2 ) or ( 3pi/2 ), and ( f ) can be any value. These are minima because the function is zero there, which is the lowest possible value.Problem 2: Finding ( d ) to Minimize Average Drag CoefficientNow, the drag coefficient is modified to:[ C_d'(f, theta, d) = a cdot left(frac{1}{1 + b cdot e^{-c cdot f}}right) cdot cos^2(theta) + d cdot sin(f cdot theta) ]We need to find the value of ( d ) such that the average drag coefficient over a full cycle of ( theta ) (from ( 0 ) to ( 2pi )) is minimized for a fixed ( f = 10 ).First, let's note that ( f = 10 ), so the function becomes:[ C_d'(10, theta, d) = 1.5 cdot left(frac{1}{1 + 2e^{-0.5 cdot 10}}right) cdot cos^2(theta) + d cdot sin(10 cdot theta) ]Compute the constant term:First, compute ( e^{-0.5 cdot 10} = e^{-5} approx 0.006737947 ).So,[ frac{1}{1 + 2e^{-5}} approx frac{1}{1 + 2 cdot 0.006737947} = frac{1}{1 + 0.013475894} approx frac{1}{1.013475894} approx 0.9868 ]Therefore,[ C_d'(10, theta, d) approx 1.5 cdot 0.9868 cdot cos^2(theta) + d cdot sin(10theta) ]Compute ( 1.5 cdot 0.9868 approx 1.4802 )So,[ C_d'(10, theta, d) approx 1.4802 cos^2(theta) + d sin(10theta) ]We need to find the average of this function over ( theta ) from ( 0 ) to ( 2pi ), and then find ( d ) that minimizes this average.The average value ( overline{C_d'} ) is:[ overline{C_d'} = frac{1}{2pi} int_{0}^{2pi} C_d'(10, theta, d) dtheta ]Substitute the expression:[ overline{C_d'} = frac{1}{2pi} int_{0}^{2pi} left(1.4802 cos^2(theta) + d sin(10theta)right) dtheta ]We can split the integral:[ overline{C_d'} = frac{1.4802}{2pi} int_{0}^{2pi} cos^2(theta) dtheta + frac{d}{2pi} int_{0}^{2pi} sin(10theta) dtheta ]Compute each integral separately.First integral:[ int_{0}^{2pi} cos^2(theta) dtheta ]We know that ( cos^2(theta) = frac{1 + cos(2theta)}{2} ), so:[ int_{0}^{2pi} cos^2(theta) dtheta = int_{0}^{2pi} frac{1 + cos(2theta)}{2} dtheta = frac{1}{2} int_{0}^{2pi} 1 dtheta + frac{1}{2} int_{0}^{2pi} cos(2theta) dtheta ]Compute each part:[ frac{1}{2} int_{0}^{2pi} 1 dtheta = frac{1}{2} [ theta ]_{0}^{2pi} = frac{1}{2} (2pi - 0) = pi ][ frac{1}{2} int_{0}^{2pi} cos(2theta) dtheta = frac{1}{2} left[ frac{sin(2theta)}{2} right]_{0}^{2pi} = frac{1}{4} [ sin(4pi) - sin(0) ] = 0 ]So, the first integral is ( pi ).Second integral:[ int_{0}^{2pi} sin(10theta) dtheta ]The integral of ( sin(ktheta) ) over ( 0 ) to ( 2pi ) is zero for any integer ( k ), because it's a full period.Therefore, the second integral is zero.So, putting it all together:[ overline{C_d'} = frac{1.4802}{2pi} cdot pi + frac{d}{2pi} cdot 0 = frac{1.4802}{2} approx 0.7401 ]Wait, that's interesting. The average drag coefficient is approximately 0.7401, regardless of ( d ), because the integral of ( sin(10theta) ) over ( 0 ) to ( 2pi ) is zero.But the problem says to find ( d ) such that the average drag coefficient is minimized. However, the average does not depend on ( d ) because the integral of the sine term is zero. Therefore, the average is constant for any ( d ).But that can't be right. Maybe I made a mistake.Wait, let me double-check. The average is:[ overline{C_d'} = frac{1.4802}{2pi} cdot pi + frac{d}{2pi} cdot 0 = 0.7401 ]Yes, that's correct. So, the average is constant, independent of ( d ). Therefore, ( d ) does not affect the average drag coefficient over a full cycle of ( theta ).But the problem says to find ( d ) such that the average is minimized. Since the average is constant, any ( d ) would give the same average. Therefore, there is no unique ( d ) that minimizes the average; it's already minimized for any ( d ).But that seems counterintuitive. Maybe I misinterpreted the problem.Wait, perhaps the problem is to minimize the average drag coefficient, but the average is already minimized because the sine term averages out to zero. So, regardless of ( d ), the average is the same. Therefore, ( d ) can be any value, but to minimize the average, we can choose ( d = 0 ) to eliminate the sine term, but since the average is already the same, maybe the answer is ( d = 0 ).Wait, but the average is the same for any ( d ). So, perhaps the question is to find ( d ) such that the average is minimized, but since the average is fixed, ( d ) can be any value. However, maybe the problem is to minimize the average over ( theta ), but considering that the sine term can have both positive and negative contributions, but over a full cycle, it cancels out.Alternatively, perhaps the problem is to minimize the average, but the average is fixed, so ( d ) doesn't affect it. Therefore, the minimal average is achieved for any ( d ), but to make the function as low as possible, perhaps set ( d = 0 ).Wait, but the problem says "the average drag coefficient over a full cycle of ( theta ) (from ( 0 ) to ( 2pi )) is minimized for a fixed ( f = 10 )."Since the average is fixed, regardless of ( d ), the minimal average is already achieved, so ( d ) can be any value. But maybe the problem expects us to set ( d ) such that the average is minimized, but since the average is fixed, perhaps the answer is ( d = 0 ).Alternatively, perhaps I made a mistake in computing the average. Let me check again.Wait, the function is:[ C_d'(10, theta, d) = 1.4802 cos^2(theta) + d sin(10theta) ]The average is:[ frac{1}{2pi} int_{0}^{2pi} [1.4802 cos^2(theta) + d sin(10theta)] dtheta ]Which is:[ frac{1.4802}{2pi} int_{0}^{2pi} cos^2(theta) dtheta + frac{d}{2pi} int_{0}^{2pi} sin(10theta) dtheta ]As before, the first integral is ( pi ), so:[ frac{1.4802}{2pi} cdot pi = 0.7401 ]The second integral is zero, so the average is 0.7401, regardless of ( d ).Therefore, the average is constant, so ( d ) does not affect it. Therefore, the minimal average is 0.7401, achieved for any ( d ). So, perhaps the answer is that any ( d ) minimizes the average, but since the problem asks for the value of ( d ), maybe ( d = 0 ) is the answer, as it removes the sine term.But I'm not sure. Alternatively, perhaps the problem is to minimize the average, but since the average is fixed, the minimal average is 0.7401, and ( d ) can be any value. But the problem says "find the value of ( d )", so maybe it's expecting ( d = 0 ).Alternatively, perhaps I misapplied the integral. Let me think again.Wait, perhaps the problem is to minimize the average, but the average is fixed, so ( d ) can be any value. But maybe the problem is to minimize the average, considering that the sine term can have both positive and negative contributions, but over a full cycle, it cancels out. Therefore, the average is fixed, so ( d ) does not affect it. Therefore, the minimal average is achieved for any ( d ), but to make the function as low as possible, perhaps set ( d = 0 ).But I'm not sure. Alternatively, perhaps the problem is to minimize the average, but the average is fixed, so ( d ) can be any value. Therefore, the answer is that any ( d ) minimizes the average, but since the problem asks for a specific value, perhaps ( d = 0 ).Wait, but in the expression, ( d ) is multiplied by ( sin(10theta) ). If we set ( d = 0 ), the term disappears, and the average is just the average of the first term, which is 0.7401. If ( d ) is non-zero, the average remains the same, but the function itself has oscillations. So, perhaps the minimal average is achieved regardless of ( d ), but to minimize the maximum drag coefficient, we might set ( d = 0 ). But the problem specifically asks to minimize the average, which is fixed.Therefore, the answer is that any ( d ) minimizes the average, but since the problem asks for a specific value, perhaps ( d = 0 ).But I'm not entirely confident. Alternatively, perhaps I made a mistake in computing the average. Let me check again.Wait, the average is:[ overline{C_d'} = frac{1}{2pi} int_{0}^{2pi} [1.4802 cos^2(theta) + d sin(10theta)] dtheta ]Which is:[ frac{1.4802}{2pi} int_{0}^{2pi} cos^2(theta) dtheta + frac{d}{2pi} int_{0}^{2pi} sin(10theta) dtheta ]As before, the first integral is ( pi ), so:[ frac{1.4802}{2pi} cdot pi = 0.7401 ]The second integral is zero, so the average is 0.7401, regardless of ( d ).Therefore, the average is constant, so ( d ) does not affect it. Therefore, the minimal average is 0.7401, achieved for any ( d ). So, perhaps the answer is that any ( d ) minimizes the average, but since the problem asks for the value of ( d ), maybe ( d = 0 ).Alternatively, perhaps the problem expects us to set ( d ) such that the average is minimized, but since the average is fixed, the minimal average is achieved for any ( d ). Therefore, the answer is that ( d ) can be any real number, but to minimize the average, it's already minimized regardless of ( d ).But the problem says "find the value of ( d )", so perhaps the answer is ( d = 0 ).Wait, but if ( d ) is zero, the average is 0.7401. If ( d ) is non-zero, the average remains the same. So, perhaps the answer is ( d = 0 ) because it removes the oscillating term, making the drag coefficient smoother, but the average is the same.Alternatively, perhaps the problem is to minimize the average, but since the average is fixed, the answer is that any ( d ) minimizes the average, but since the problem asks for a specific value, perhaps ( d = 0 ).I think I'll go with ( d = 0 ) as the answer, because it removes the oscillating term, which might cause higher drag in some regions, even though the average is the same.But I'm not entirely sure. Alternatively, perhaps the problem is to minimize the average, but since the average is fixed, the answer is that any ( d ) minimizes the average, but since the problem asks for a specific value, perhaps ( d = 0 ).Wait, but in the expression, ( d ) is multiplied by ( sin(10theta) ). If we set ( d = 0 ), the term disappears, and the average is just the average of the first term, which is 0.7401. If ( d ) is non-zero, the average remains the same, but the function itself has oscillations. So, perhaps the minimal average is achieved regardless of ( d ), but to minimize the maximum drag coefficient, we might set ( d = 0 ). But the problem specifically asks to minimize the average, which is fixed.Therefore, the answer is that any ( d ) minimizes the average, but since the problem asks for a specific value, perhaps ( d = 0 ).But I'm still not entirely confident. Alternatively, perhaps the problem is to minimize the average, but the average is fixed, so the answer is that ( d ) can be any value, but since the problem asks for a specific value, perhaps ( d = 0 ).I think I'll conclude that ( d = 0 ) is the answer.</think>