🧠:<think>Okay, so I have these two math problems to solve related to skeleton racing. Let me take them one at a time.Starting with the first problem: Emily is looking at the speed of a sled, which is modeled by the quadratic function ( v(t) = -4t^2 + 16t + 20 ). I need to find the time ( t ) when the sled reaches its maximum speed and then calculate that maximum speed.Hmm, quadratic functions. I remember that a quadratic function in the form ( at^2 + bt + c ) has its vertex at ( t = -frac{b}{2a} ). Since the coefficient of ( t^2 ) is negative (-4), the parabola opens downward, meaning the vertex is the maximum point. So, that should give me the time when the maximum speed occurs.Let me write that down:The time ( t ) at which the maximum speed occurs is ( t = -frac{b}{2a} ).In this equation, ( a = -4 ) and ( b = 16 ).So plugging in:( t = -frac{16}{2*(-4)} = -frac{16}{-8} = 2 ) seconds.Okay, so the maximum speed happens at 2 seconds. Now, to find the maximum speed, I need to plug ( t = 2 ) back into the original equation ( v(t) ).Calculating ( v(2) ):( v(2) = -4*(2)^2 + 16*(2) + 20 ).First, ( (2)^2 = 4 ), so:( -4*4 = -16 ).Then, ( 16*2 = 32 ).So, adding them up:( -16 + 32 + 20 ).( -16 + 32 = 16 ).( 16 + 20 = 36 ).So, the maximum speed is 36 meters per second at 2 seconds.Wait, that seems pretty fast for a sled. Maybe I did the calculations right. Let me double-check.Original equation: ( v(t) = -4t^2 + 16t + 20 ).At ( t = 2 ):( -4*(4) + 16*(2) + 20 = -16 + 32 + 20 = 36 ). Yep, that's correct.Alright, so part 1 is done. The sled reaches maximum speed at 2 seconds, and that speed is 36 m/s.Moving on to problem 2: The track's height ( h(x) ) is given by ( h(x) = 0.1x^3 - 0.5x^2 + 2sin(x) ). I need to find the height at ( x = 5 ) meters and the slope of the track at that point.First, let's compute ( h(5) ).So, plugging ( x = 5 ) into the equation:( h(5) = 0.1*(5)^3 - 0.5*(5)^2 + 2sin(5) ).Calculating each term step by step.First term: ( 0.1*(5)^3 ).( 5^3 = 125 ).( 0.1*125 = 12.5 ).Second term: ( -0.5*(5)^2 ).( 5^2 = 25 ).( -0.5*25 = -12.5 ).Third term: ( 2sin(5) ).Hmm, I need to calculate ( sin(5) ). But wait, is the angle in radians or degrees? In calculus, unless specified, it's usually radians. So, I'll assume radians.Calculating ( sin(5) ). Let me recall that ( pi ) is approximately 3.1416, so 5 radians is a bit more than ( pi ) (which is about 3.14), so it's in the second quadrant where sine is positive.But to get the exact value, I might need a calculator. Since I don't have one here, I can approximate it or remember that ( sin(5) ) is approximately... let me think.Wait, 5 radians is about 286 degrees (since 5*(180/π) ≈ 286 degrees). So, sine of 286 degrees is sine of (360 - 74) degrees, which is -sin(74). But wait, in radians, 5 radians is more than π (3.14), so it's in the third quadrant? Wait, no, π is about 3.14, 3π/2 is about 4.712, so 5 radians is just a bit more than 3π/2, which is 4.712. So, 5 radians is in the fourth quadrant, where sine is negative.Wait, hold on. Let me clarify:π radians = 180 degreesSo, 5 radians = 5*(180/π) ≈ 5*57.3 ≈ 286.5 degrees.286.5 degrees is in the fourth quadrant (270 to 360 degrees), where sine is negative.So, ( sin(5) ) is negative. Let me recall that ( sin(5) ) is approximately -0.9589. Wait, is that right?Wait, actually, let me think of the unit circle. At 5 radians, which is about 286.5 degrees, the sine is negative. The reference angle would be 360 - 286.5 = 73.5 degrees. So, ( sin(5) = -sin(73.5) ). ( sin(73.5) ) is approximately 0.9589, so ( sin(5) ≈ -0.9589 ).Therefore, ( 2sin(5) ≈ 2*(-0.9589) ≈ -1.9178 ).So, putting it all together:First term: 12.5Second term: -12.5Third term: -1.9178Adding them up:12.5 - 12.5 = 00 - 1.9178 ≈ -1.9178So, ( h(5) ≈ -1.9178 ) meters.Wait, that's negative? The height is negative? That would mean it's below the starting point or something. Maybe the track goes below ground level at that point? Or perhaps the coordinate system is such that positive is upwards, so negative is downwards.But let me double-check my calculations because getting a negative height seems a bit odd, but maybe it's correct.First term: 0.1*(125) = 12.5Second term: -0.5*(25) = -12.5Third term: 2*sin(5) ≈ 2*(-0.9589) ≈ -1.9178So, 12.5 - 12.5 - 1.9178 = -1.9178. So yes, that's correct.So, the height at x=5 meters is approximately -1.9178 meters.Hmm, okay. Maybe the track has a dip there.Now, moving on to the slope of the track at x=5 meters. The slope is the derivative of the height function with respect to x, which is ( h'(x) ).So, let's compute ( h'(x) ).Given ( h(x) = 0.1x^3 - 0.5x^2 + 2sin(x) ).The derivative term by term:- The derivative of ( 0.1x^3 ) is ( 0.3x^2 ).- The derivative of ( -0.5x^2 ) is ( -1x ).- The derivative of ( 2sin(x) ) is ( 2cos(x) ).So, putting it all together:( h'(x) = 0.3x^2 - x + 2cos(x) ).Now, evaluate this at x=5.So, ( h'(5) = 0.3*(5)^2 - 5 + 2cos(5) ).Calculating each term:First term: 0.3*(25) = 7.5Second term: -5Third term: 2cos(5). Again, angle in radians.Calculating ( cos(5) ). 5 radians is approximately 286.5 degrees, which is in the fourth quadrant where cosine is positive. The reference angle is 73.5 degrees, so ( cos(5) = cos(73.5) ≈ 0.2837 ).Therefore, ( 2cos(5) ≈ 2*0.2837 ≈ 0.5674 ).Adding them up:7.5 - 5 + 0.5674 = (7.5 - 5) + 0.5674 = 2.5 + 0.5674 ≈ 3.0674.So, the slope at x=5 meters is approximately 3.0674.Wait, the slope is positive, meaning the track is rising at that point? But the height was negative, so it's going up from a lower point.But let me double-check the calculations.First term: 0.3*(25) = 7.5Second term: -5Third term: 2cos(5). Cos(5 radians) is approximately 0.2837, so 2*0.2837 ≈ 0.5674.So, 7.5 - 5 = 2.5; 2.5 + 0.5674 ≈ 3.0674.Yes, that seems correct.So, the slope is approximately 3.0674.Hmm, that's a pretty steep slope. Let me convert that to a percentage or something to get a sense.Slope as a derivative is rise over run, so 3.0674 means for every meter horizontally, the track rises about 3.0674 meters. That's a very steep slope, almost 307% grade. That seems extremely steep for a skeleton track. Maybe I made a mistake.Wait, let me think. The derivative is the instantaneous rate of change, which is the slope of the tangent line at that point. So, if the slope is 3.0674, that means the track is rising quite steeply at x=5 meters.But in reality, skeleton tracks have varying slopes, but 3 meters up for every 1 meter forward is very steep. Maybe that's correct for a particularly steep section.Alternatively, perhaps I made a mistake in the derivative.Wait, let me check the derivative again.Given ( h(x) = 0.1x^3 - 0.5x^2 + 2sin(x) ).Derivative:- ( d/dx [0.1x^3] = 0.3x^2 )- ( d/dx [-0.5x^2] = -1x )- ( d/dx [2sin(x)] = 2cos(x) )So, ( h'(x) = 0.3x^2 - x + 2cos(x) ). That seems correct.So, plugging in x=5:0.3*(25) = 7.5-52cos(5) ≈ 0.5674So, 7.5 - 5 + 0.5674 ≈ 3.0674. So, that's correct.Hmm, okay. So, the slope is positive and quite steep at that point.Now, using these findings, I need to comment on the potential challenges a racer might face at this point of the track.So, at x=5 meters, the height is approximately -1.92 meters, which is below the starting point, and the slope is approximately 3.07, which is a steep upward slope.So, a racer approaching this point would be going downhill before this point, reaching a certain speed, and then encountering a steep uphill. That could be challenging because the sled would need to maintain enough speed to get over the hill without losing too much momentum. If the slope is too steep, the sled might slow down significantly, making it harder to maintain control or even stop if necessary.Additionally, the negative height might indicate that the track is below ground level, which could affect the racer's visibility or the aerodynamics of the sled, though I'm not sure about that.Moreover, the combination of a dip (negative height) followed by a steep slope could create a situation where the sled might experience a sudden change in direction or force, potentially leading to loss of control or difficulty in maintaining the optimal racing line.Therefore, racers need to be cautious at such points, ensuring they have enough speed to navigate the steep slope and maintain control through the dip and incline.Wait, but the height is negative, so it's a dip, and then the slope is positive, meaning it's rising. So, the track goes down to a dip at x=5 meters and then starts rising. So, the racer would be going downhill, reaching the dip, and then immediately starting to climb a steep slope.That could be tricky because the racer might be going fast downhill, then hit a point where the track starts to rise, which could cause the sled to slow down if not handled properly. The transition from a dip to a steep uphill could also create a moment where the racer needs to adjust their body position quickly to maintain momentum and control.Additionally, the negative height might mean that the track is below the surrounding area, which could affect lighting or visibility, though that's more of an environmental factor rather than a track design challenge.In terms of physics, the steep slope would require the racer to have enough kinetic energy to overcome the potential energy gain as they ascend the slope. If the slope is too steep, it might be difficult to maintain speed, which could lead to a slower overall time.So, in summary, the challenges at this point would include managing speed through a dip and then a steep uphill, maintaining control during the transition, and ensuring enough momentum to ascend without losing too much speed.I think that covers the potential challenges.Final Answer1. The sled reaches its maximum speed at boxed{2} seconds, with a maximum speed of boxed{36} meters per second.2. The height of the track at ( x = 5 ) meters is approximately boxed{-1.92} meters, and the slope of the track at this point is approximately boxed{3.07}.Racers might face challenges at this point due to the steep slope and the dip in the track, requiring careful control and sufficient speed to navigate effectively.</think>

🧠:<think>Okay, so I have this problem about two types of lunar rocks, A and B, and their formation rates are given by these differential equations. I need to solve for their explicit solutions and then find the time when their combined amount is half of their total carrying capacity. Hmm, let me break this down step by step.First, looking at the differential equations for both rocks. They both look similar, which is good because maybe I can solve one and then apply the same method to the other.For rock A, the equation is:[ frac{dA}{dt} = k_1 A left(1 - frac{A}{C}right) ]And for rock B:[ frac{dB}{dt} = k_2 B left(1 - frac{B}{D}right) ]These look like logistic growth equations. I remember that the logistic equation has the form:[ frac{dx}{dt} = r x left(1 - frac{x}{K}right) ]where ( r ) is the growth rate and ( K ) is the carrying capacity. So, yeah, these are logistic models for the formation of rocks A and B.I need to solve these differential equations to find ( A(t) ) and ( B(t) ). The standard solution for the logistic equation is:[ x(t) = frac{K}{1 + left(frac{K - x_0}{x_0}right) e^{-rt}} ]Where ( x_0 ) is the initial amount. So, applying this to rock A, substituting ( K = C ), ( r = k_1 ), and ( x_0 = A_0 ), the solution should be:[ A(t) = frac{C}{1 + left(frac{C - A_0}{A_0}right) e^{-k_1 t}} ]Similarly, for rock B, substituting ( K = D ), ( r = k_2 ), and ( x_0 = B_0 ), the solution is:[ B(t) = frac{D}{1 + left(frac{D - B_0}{B_0}right) e^{-k_2 t}} ]Wait, let me make sure I got that right. The standard solution is:[ x(t) = frac{K x_0}{x_0 + (K - x_0) e^{-rt}} ]Which can also be written as:[ x(t) = frac{K}{1 + left(frac{K - x_0}{x_0}right) e^{-rt}} ]Yes, that's correct. So, my expressions for ( A(t) ) and ( B(t) ) are right.So, that solves the first part. Now, moving on to the second sub-problem: finding the time ( t^* ) when the combined amount ( A(t^*) + B(t^*) = frac{C + D}{2} ).Hmm, okay. So, I need to set up the equation:[ A(t^*) + B(t^*) = frac{C + D}{2} ]And solve for ( t^* ). Let me write down the expressions for ( A(t) ) and ( B(t) ) again:[ A(t) = frac{C}{1 + left(frac{C - A_0}{A_0}right) e^{-k_1 t}} ][ B(t) = frac{D}{1 + left(frac{D - B_0}{B_0}right) e^{-k_2 t}} ]So, adding these together:[ frac{C}{1 + left(frac{C - A_0}{A_0}right) e^{-k_1 t^*}} + frac{D}{1 + left(frac{D - B_0}{B_0}right) e^{-k_2 t^*}} = frac{C + D}{2} ]Hmm, this looks complicated. It's a transcendental equation because ( t^* ) appears in the exponents. I don't think there's an analytical solution for ( t^* ) here. Maybe I need to solve this numerically?But wait, the problem says "determine the time ( t^* )", so perhaps there's a way to express it in terms of logarithms or something? Let me see.Alternatively, maybe if I make some assumptions or simplify the problem? But the problem doesn't specify any particular values for ( A_0, B_0, C, D, k_1, k_2 ), so I think the solution has to be in terms of these parameters.Wait, maybe I can set ( u = e^{-k_1 t^*} ) and ( v = e^{-k_2 t^*} ), but that might not help much because we still have two variables and one equation.Alternatively, if I consider the case where ( k_1 = k_2 ), but the problem doesn't specify that. So, perhaps it's not possible to find an explicit formula for ( t^* ) without numerical methods.But since the problem is asking to determine ( t^* ), maybe I can express it in terms of inverse functions or logarithms? Let me try to manipulate the equation.Let me denote:[ A(t) = frac{C}{1 + M e^{-k_1 t}} ][ B(t) = frac{D}{1 + N e^{-k_2 t}} ]Where ( M = frac{C - A_0}{A_0} ) and ( N = frac{D - B_0}{B_0} ).So, the equation becomes:[ frac{C}{1 + M e^{-k_1 t^*}} + frac{D}{1 + N e^{-k_2 t^*}} = frac{C + D}{2} ]Let me denote ( x = e^{-k_1 t^*} ) and ( y = e^{-k_2 t^*} ). Then, the equation becomes:[ frac{C}{1 + M x} + frac{D}{1 + N y} = frac{C + D}{2} ]But we also have the relationships:[ x = e^{-k_1 t^*} ][ y = e^{-k_2 t^*} ]So, ( y = e^{-k_2 t^*} = left(e^{-k_1 t^*}right)^{k_2 / k_1} = x^{k_2 / k_1} )So, ( y = x^{k_2 / k_1} )Therefore, we can write the equation as:[ frac{C}{1 + M x} + frac{D}{1 + N x^{k_2 / k_1}} = frac{C + D}{2} ]This still looks complicated, but maybe I can consider specific cases or see if there's a symmetry.Alternatively, perhaps if I let ( t^* ) be such that both ( A(t^*) ) and ( B(t^*) ) are half their carrying capacities? But that's not necessarily the case because the combined amount is half the total, not each individually.Wait, if ( A(t^*) + B(t^*) = frac{C + D}{2} ), it doesn't mean each is half. It could be that one is more than half and the other is less.Hmm, maybe I can consider the case where ( A(t^*) = frac{C}{2} ) and ( B(t^*) = frac{D}{2} ). Then, their sum would be ( frac{C + D}{2} ). But is that the only solution? Probably not, because depending on the growth rates and initial conditions, the sum could reach half the total carrying capacity at different points where neither is exactly half.But maybe in some cases, it's when both are half. Let me check.If ( A(t^*) = frac{C}{2} ), then:[ frac{C}{2} = frac{C}{1 + M e^{-k_1 t^*}} ][ frac{1}{2} = frac{1}{1 + M e^{-k_1 t^*}} ][ 1 + M e^{-k_1 t^*} = 2 ][ M e^{-k_1 t^*} = 1 ][ e^{-k_1 t^*} = frac{1}{M} ][ -k_1 t^* = lnleft(frac{1}{M}right) ][ t^* = -frac{1}{k_1} lnleft(frac{1}{M}right) ][ t^* = frac{1}{k_1} ln M ]Similarly, for ( B(t^*) = frac{D}{2} ):[ t^* = frac{1}{k_2} ln N ]So, unless ( frac{1}{k_1} ln M = frac{1}{k_2} ln N ), which would require specific relationships between ( k_1, k_2, M, N ), the times when each reaches half their carrying capacity would be different.Therefore, unless the problem specifies that ( t^* ) is when both are half, which it doesn't, we can't assume that. So, the general case is more complicated.Given that, perhaps the only way to solve for ( t^* ) is numerically. But since the problem is asking for an explicit solution, maybe I need to find a way to express ( t^* ) in terms of logarithms or something.Wait, perhaps if I consider the case where ( k_1 = k_2 ), then ( y = x ), and the equation becomes:[ frac{C}{1 + M x} + frac{D}{1 + N x} = frac{C + D}{2} ]But even then, it's still a nonlinear equation in ( x ), which might not have an analytical solution.Alternatively, maybe I can rearrange the equation:[ frac{C}{1 + M e^{-k_1 t^*}} + frac{D}{1 + N e^{-k_2 t^*}} = frac{C + D}{2} ]Let me subtract ( frac{C + D}{2} ) from both sides:[ frac{C}{1 + M e^{-k_1 t^*}} + frac{D}{1 + N e^{-k_2 t^*}} - frac{C + D}{2} = 0 ]Let me denote ( u = e^{-k_1 t^*} ) and ( v = e^{-k_2 t^*} ), so the equation becomes:[ frac{C}{1 + M u} + frac{D}{1 + N v} - frac{C + D}{2} = 0 ]But we also have ( u = e^{-k_1 t^*} ) and ( v = e^{-k_2 t^*} ), so ( v = u^{k_2 / k_1} ) as before.So, substituting ( v = u^{k_2 / k_1} ), we get:[ frac{C}{1 + M u} + frac{D}{1 + N u^{k_2 / k_1}} - frac{C + D}{2} = 0 ]This is still a complicated equation in ( u ). I don't think there's a way to solve this analytically unless specific values are given.Wait, maybe if I consider the case where ( k_1 = k_2 ), then ( k_2 / k_1 = 1 ), so ( v = u ), and the equation becomes:[ frac{C}{1 + M u} + frac{D}{1 + N u} - frac{C + D}{2} = 0 ]Let me combine the fractions:Let me find a common denominator, which would be ( (1 + M u)(1 + N u) ):[ frac{C(1 + N u) + D(1 + M u) - frac{C + D}{2}(1 + M u)(1 + N u)}{(1 + M u)(1 + N u)} = 0 ]So, the numerator must be zero:[ C(1 + N u) + D(1 + M u) - frac{C + D}{2}(1 + M u)(1 + N u) = 0 ]Expanding each term:First term: ( C + C N u )Second term: ( D + D M u )Third term: ( frac{C + D}{2} (1 + M u + N u + M N u^2) )So, putting it all together:[ C + C N u + D + D M u - frac{C + D}{2} (1 + (M + N)u + M N u^2) = 0 ]Let me compute each part:Left side: ( (C + D) + (C N + D M) u )Right side: ( frac{C + D}{2} + frac{C + D}{2} (M + N) u + frac{C + D}{2} M N u^2 )So, moving everything to the left side:[ (C + D) + (C N + D M) u - frac{C + D}{2} - frac{C + D}{2} (M + N) u - frac{C + D}{2} M N u^2 = 0 ]Simplify term by term:1. ( (C + D) - frac{C + D}{2} = frac{C + D}{2} )2. ( (C N + D M) u - frac{C + D}{2} (M + N) u )Let me factor out ( u ):( u left[ C N + D M - frac{(C + D)(M + N)}{2} right] )3. ( - frac{C + D}{2} M N u^2 )So, the equation becomes:[ frac{C + D}{2} + u left[ C N + D M - frac{(C + D)(M + N)}{2} right] - frac{C + D}{2} M N u^2 = 0 ]Let me compute the coefficient of ( u ):[ C N + D M - frac{(C + D)(M + N)}{2} ]Expanding ( (C + D)(M + N) ):[ C M + C N + D M + D N ]So, half of that is:[ frac{C M + C N + D M + D N}{2} ]Therefore, the coefficient of ( u ) is:[ C N + D M - frac{C M + C N + D M + D N}{2} ][ = frac{2 C N + 2 D M - C M - C N - D M - D N}{2} ][ = frac{(2 C N - C N) + (2 D M - D M) - C M - D N}{2} ][ = frac{C N + D M - C M - D N}{2} ][ = frac{C(N - M) + D(M - N)}{2} ][ = frac{(C - D)(N - M)}{2} ]So, putting it all back together, the equation is:[ frac{C + D}{2} + u cdot frac{(C - D)(N - M)}{2} - frac{C + D}{2} M N u^2 = 0 ]Multiply both sides by 2 to eliminate denominators:[ (C + D) + u (C - D)(N - M) - (C + D) M N u^2 = 0 ]Rearranged:[ - (C + D) M N u^2 + (C - D)(N - M) u + (C + D) = 0 ]This is a quadratic equation in ( u ):[ [ - (C + D) M N ] u^2 + [ (C - D)(N - M) ] u + (C + D) = 0 ]Let me write it as:[ (C + D) M N u^2 - (C - D)(N - M) u - (C + D) = 0 ]Wait, I multiplied both sides by -1 to make the coefficient of ( u^2 ) positive:[ (C + D) M N u^2 - (C - D)(N - M) u - (C + D) = 0 ]So, quadratic in ( u ):[ A u^2 + B u + C = 0 ]Where:( A = (C + D) M N )( B = - (C - D)(N - M) )( C = - (C + D) )Wait, but the constant term is ( - (C + D) ), which is negative. Hmm.Using the quadratic formula:[ u = frac{ -B pm sqrt{B^2 - 4AC} }{2A} ]Plugging in:[ u = frac{ (C - D)(N - M) pm sqrt{ [ - (C - D)(N - M) ]^2 - 4 (C + D) M N (- (C + D)) } }{ 2 (C + D) M N } ]Simplify the discriminant:First, ( [ - (C - D)(N - M) ]^2 = (C - D)^2 (N - M)^2 )Second term: ( -4 (C + D) M N (- (C + D)) = 4 (C + D)^2 M N )So, discriminant:[ (C - D)^2 (N - M)^2 + 4 (C + D)^2 M N ]So, putting it all together:[ u = frac{ (C - D)(N - M) pm sqrt{ (C - D)^2 (N - M)^2 + 4 (C + D)^2 M N } }{ 2 (C + D) M N } ]This is quite a complicated expression, but it's a solution for ( u ). Remember, ( u = e^{-k_1 t^*} ), so once we have ( u ), we can solve for ( t^* ):[ t^* = -frac{1}{k_1} ln u ]But this is under the assumption that ( k_1 = k_2 ). If ( k_1 neq k_2 ), then this approach doesn't work because ( v ) isn't equal to ( u ), and we can't reduce it to a single variable equation.So, in the case where ( k_1 = k_2 ), we can find ( t^* ) using this quadratic solution. But in the general case, it's more complicated.Given that the problem doesn't specify ( k_1 = k_2 ), I think the answer might be that ( t^* ) cannot be expressed in a closed-form solution and must be found numerically. However, since the problem asks to "determine" the time, maybe it expects an expression in terms of logarithms, assuming some simplification.Alternatively, perhaps the problem expects me to consider that the combined carrying capacity is ( C + D ), and the halfway point is ( (C + D)/2 ). Maybe if I assume that both rocks reach their halfway points at the same time, but as I saw earlier, that requires specific conditions.Wait, maybe I can consider the case where ( A(t) ) and ( B(t) ) are both growing logistically, and their sum is also logistic? But I don't think that's necessarily the case because the sum of two logistic functions isn't itself logistic.Alternatively, maybe I can use the fact that the logistic function approaches its carrying capacity asymptotically, so the sum would approach ( C + D ), and we need the time when it's halfway there.But again, without knowing the specific parameters, it's hard to say.Wait, perhaps I can make a substitution. Let me denote ( S(t) = A(t) + B(t) ). Then, ( S(t) = frac{C}{1 + M e^{-k_1 t}} + frac{D}{1 + N e^{-k_2 t}} ). We need ( S(t^*) = frac{C + D}{2} ).This is a transcendental equation in ( t^* ), which generally doesn't have an analytical solution. So, unless there's some symmetry or specific parameter values, we can't solve for ( t^* ) explicitly.Therefore, the answer is that ( t^* ) must be found numerically, given the parameters ( A_0, B_0, C, D, k_1, k_2 ).But wait, the problem says "determine the time ( t^* )", so maybe it expects an expression in terms of logarithms, assuming that both ( A(t) ) and ( B(t) ) are growing such that their sum reaches half the total carrying capacity at a time that can be expressed in terms of their individual times to reach half capacity.But as I saw earlier, unless ( k_1 = k_2 ) and some other conditions, that's not straightforward.Alternatively, maybe if I consider the time when each rock has formed half of their respective carrying capacities, but as I saw, that would require ( t^* ) to satisfy two different equations, which is only possible if the parameters are set such that both are satisfied.Alternatively, perhaps I can consider the time when the sum is half the total, regardless of individual contributions. But without more information, I can't find an explicit formula.Wait, maybe I can use the fact that for each rock, the time to reach half the carrying capacity is ( t_{1/2} = frac{1}{k} lnleft(frac{K - x_0}{x_0}right) ). So, for rock A, ( t_{1/2,A} = frac{1}{k_1} lnleft(frac{C - A_0}{A_0}right) ), and similarly for rock B.But how does that help me with the combined sum?Alternatively, maybe I can set ( t^* ) such that ( A(t^*) = frac{C}{2} ) and ( B(t^*) = frac{D}{2} ), but as I saw earlier, this requires ( t^* ) to satisfy both:[ t^* = frac{1}{k_1} lnleft(frac{C - A_0}{A_0}right) ][ t^* = frac{1}{k_2} lnleft(frac{D - B_0}{B_0}right) ]Which is only possible if:[ frac{1}{k_1} lnleft(frac{C - A_0}{A_0}right) = frac{1}{k_2} lnleft(frac{D - B_0}{B_0}right) ]Which is a specific condition that may not hold in general.Therefore, unless given specific values or additional constraints, I think the time ( t^* ) cannot be expressed in a closed-form solution and must be determined numerically.But the problem didn't specify any particular constraints or values, so perhaps the answer is that ( t^* ) is the solution to the equation:[ frac{C}{1 + left(frac{C - A_0}{A_0}right) e^{-k_1 t}} + frac{D}{1 + left(frac{D - B_0}{B_0}right) e^{-k_2 t}} = frac{C + D}{2} ]Which can be solved numerically for ( t ).Alternatively, maybe I can express it in terms of the inverse function, but I don't think that's helpful.Wait, perhaps if I let ( t^* ) be such that ( e^{-k_1 t^*} = x ) and ( e^{-k_2 t^*} = y ), then I have two equations:1. ( frac{C}{1 + M x} + frac{D}{1 + N y} = frac{C + D}{2} )2. ( y = x^{k_2 / k_1} )So, substituting equation 2 into equation 1, I get:[ frac{C}{1 + M x} + frac{D}{1 + N x^{k_2 / k_1}} = frac{C + D}{2} ]This is a single equation in ( x ), but it's still transcendental. So, unless ( k_2 / k_1 ) is 1, which would make it quadratic as before, otherwise, it's not solvable analytically.Therefore, I think the conclusion is that ( t^* ) cannot be expressed in a closed-form solution and must be found numerically.But wait, the problem says "determine the time ( t^* )", so maybe it expects an expression in terms of logarithms, assuming that the sum reaches half the total carrying capacity at a time that can be expressed as a combination of the individual half-times.But I don't see a straightforward way to combine them.Alternatively, maybe I can consider the case where ( A(t) ) and ( B(t) ) are growing exponentially initially, before the carrying capacity kicks in. But that might not help either.Wait, perhaps if I consider the initial exponential growth phase, where ( A(t) approx A_0 e^{k_1 t} ) and ( B(t) approx B_0 e^{k_2 t} ), then the sum would be ( A_0 e^{k_1 t} + B_0 e^{k_2 t} = frac{C + D}{2} ). But this is only valid for early times when ( A(t) ll C ) and ( B(t) ll D ). So, this might give an approximate solution for ( t^* ), but it's not exact.Alternatively, maybe I can use the fact that the logistic function can be approximated by exponential growth for small ( t ), but again, this is an approximation.Given all this, I think the answer is that ( t^* ) must be found numerically, as there's no closed-form solution for the given equation.But wait, let me check if the problem specifies any particular values or if I missed something. The problem states:"Given the initial conditions ( A(0) = A_0 ) and ( B(0) = B_0 ), where ( A_0 ) and ( B_0 ) are the initial amounts of rocks A and B respectively, solve the following sub-problems:"So, no specific values are given, just general expressions. Therefore, I think the answer is that ( t^* ) is the solution to the equation:[ frac{C}{1 + left(frac{C - A_0}{A_0}right) e^{-k_1 t}} + frac{D}{1 + left(frac{D - B_0}{B_0}right) e^{-k_2 t}} = frac{C + D}{2} ]Which must be solved numerically.But the problem says "determine the time ( t^* )", so maybe it's expecting an expression in terms of logarithms, assuming that the sum reaches half the total carrying capacity at a time that can be expressed as a combination of the individual times.Alternatively, perhaps if I consider that the sum ( A(t) + B(t) ) can be approximated as a logistic function with carrying capacity ( C + D ), but I don't think that's valid because the sum of two logistic functions isn't itself logistic.Alternatively, maybe I can consider the time when the sum is half the total carrying capacity, which is ( (C + D)/2 ), and use the fact that each term is a logistic function. But without knowing the relationship between the parameters, it's hard to say.Wait, perhaps if I consider the case where ( A_0 = B_0 = 0 ), but that's not given. If ( A_0 = 0 ) and ( B_0 = 0 ), then the solutions would be ( A(t) = frac{C}{1 + (C/A_0) e^{-k_1 t}} ), but if ( A_0 = 0 ), it's undefined. So, that's not helpful.Alternatively, perhaps if ( A_0 = C ) and ( B_0 = D ), but then ( A(t) = C ) and ( B(t) = D ) for all ( t ), so the sum is always ( C + D ), which is more than half, so that's not helpful.Wait, maybe I can consider the case where ( A_0 = C/2 ) and ( B_0 = D/2 ), then ( A(t) = C/(1 + (C/(C/2)) e^{-k_1 t}) = C/(1 + 2 e^{-k_1 t}) ), similarly for ( B(t) ). Then, the sum would be ( C/(1 + 2 e^{-k_1 t}) + D/(1 + 2 e^{-k_2 t}) ). Setting this equal to ( (C + D)/2 ), we get:[ frac{C}{1 + 2 e^{-k_1 t}} + frac{D}{1 + 2 e^{-k_2 t}} = frac{C + D}{2} ]But again, this is a transcendental equation and doesn't have an analytical solution unless ( k_1 = k_2 ).So, in conclusion, I think that without specific values or additional constraints, the time ( t^* ) cannot be expressed in a closed-form solution and must be determined numerically.Therefore, the answer to part 1 is the explicit solutions for ( A(t) ) and ( B(t) ) as logistic functions, and the answer to part 2 is that ( t^* ) must be found numerically by solving the equation ( A(t) + B(t) = (C + D)/2 ).But wait, the problem says "determine the time ( t^* )", so maybe it expects an expression in terms of logarithms, assuming that the sum reaches half the total carrying capacity at a time that can be expressed as a combination of the individual times.Alternatively, perhaps if I consider that the sum ( A(t) + B(t) ) can be expressed in terms of the individual solutions, but I don't see a way to combine them into a single logarithmic expression.Alternatively, maybe I can use the fact that the logistic function can be inverted. For each rock, the time to reach a certain amount is given by:For rock A:[ t = frac{1}{k_1} lnleft( frac{C - A(t)}{A(t)} cdot frac{A_0}{C - A_0} right) ]Similarly for rock B:[ t = frac{1}{k_2} lnleft( frac{D - B(t)}{B(t)} cdot frac{B_0}{D - B_0} right) ]But since ( A(t^*) + B(t^*) = (C + D)/2 ), we have two equations:1. ( A(t^*) = frac{C}{1 + M e^{-k_1 t^*}} )2. ( B(t^*) = frac{D}{1 + N e^{-k_2 t^*}} )3. ( A(t^*) + B(t^*) = frac{C + D}{2} )But this system of equations doesn't have an analytical solution, so I think the answer is that ( t^* ) must be found numerically.Therefore, summarizing:1. The explicit solutions are the logistic functions for ( A(t) ) and ( B(t) ) as derived earlier.2. The time ( t^* ) must be determined numerically by solving the equation ( A(t) + B(t) = frac{C + D}{2} ).But since the problem is asking to "determine" ( t^* ), maybe it expects an expression in terms of the inverse functions, but I don't think that's feasible.Alternatively, perhaps if I consider that the sum ( A(t) + B(t) ) can be expressed in terms of the individual solutions, but I don't see a way to combine them into a single logarithmic expression.Therefore, I think the answer is that ( t^* ) is the solution to the equation:[ frac{C}{1 + left(frac{C - A_0}{A_0}right) e^{-k_1 t}} + frac{D}{1 + left(frac{D - B_0}{B_0}right) e^{-k_2 t}} = frac{C + D}{2} ]Which must be solved numerically.So, to wrap up, the explicit solutions for ( A(t) ) and ( B(t) ) are the logistic functions, and ( t^* ) is found by solving the above equation numerically.</think>

🧠:<think>Alright, so I have this problem about a director reimagining classic TV shows, and they're using a mathematical model to optimize viewership. The problem has two parts. Let me tackle them one by one.Starting with part 1: They give me a function P(t) = A cos(Bt) + C sin(Dt) + E. The constants A, C, and E are given as 5, 3, and 10 respectively. The director wants to reimagine 12 episodes, and the total viewership V is the integral of P(t) over [0, 12π]. They want P(t) to have exactly 6 complete cycles within that interval. I need to find B and D.Hmm, okay. So, first, I remember that the period of a cosine function cos(Bt) is 2π/B, and similarly, the period of a sine function sin(Dt) is 2π/D. Since the interval is [0, 12π], and they want exactly 6 cycles, that means each function should complete 6 cycles over 12π.Wait, hold on. Is that for each function individually, or combined? The problem says P(t) has exactly 6 complete cycles. So, the overall function P(t) should have 6 cycles over 12π. But P(t) is a combination of two periodic functions with potentially different periods. That could complicate things because the overall function might not be strictly periodic unless the periods are commensurate.But maybe the problem is assuming that each component has 6 cycles over 12π. Let me think.If each component has 6 cycles over 12π, then for the cosine term, the period would be 12π / 6 = 2π. So, 2π/B = 2π => B = 1. Similarly, for the sine term, 2π/D = 2π => D = 1.But wait, if both B and D are 1, then both functions have the same period, 2π, so over 12π, each completes 6 cycles. That makes sense. But is that the only possibility?Alternatively, maybe the overall function P(t) has 6 cycles. But since P(t) is a combination of two periodic functions, unless they have the same frequency, the overall function might not be periodic. So, perhaps the director wants each component to have 6 cycles, so that the overall function's behavior is predictable.Alternatively, maybe the director wants the sum to have 6 cycles, but that might require the frequencies to be such that their combination results in 6 beats or something. But I think it's more straightforward.Given that the problem says P(t) has exactly 6 complete cycles, and since it's a sum of two periodic functions, it's likely that each component must individually have 6 cycles over the interval. Otherwise, the sum might not have a clear number of cycles.So, let's go with that. So, for the cosine term, period is 2π/B, and over 12π, the number of cycles is (12π)/(2π/B) = (12π * B)/(2π) = 6B. They want this to be 6, so 6B = 6 => B = 1.Similarly, for the sine term, period is 2π/D, so number of cycles over 12π is (12π)/(2π/D) = (12π * D)/(2π) = 6D. They want this to be 6, so 6D = 6 => D = 1.So, both B and D are 1.Wait, but let me double-check. If B and D are both 1, then P(t) = 5 cos(t) + 3 sin(t) + 10. The integral over [0, 12π] would be the area under this curve. But the number of cycles is 6 for each component, so the overall function would have a combination of these cycles. But does the sum have exactly 6 cycles? Hmm, not necessarily. Because when you add two sine/cosine functions with the same frequency, you get another sinusoidal function with the same frequency. So, in this case, 5 cos(t) + 3 sin(t) can be written as R cos(t - φ), where R is the amplitude and φ is the phase shift. So, the combined function would have the same frequency, so the number of cycles would still be 6 over 12π. So, that makes sense.So, I think B = 1 and D = 1 is correct.Moving on to part 2: After determining B and D, which are both 1, the director wants to find the critical points of P(t) within [0, 12π] and determine their nature.So, critical points occur where the derivative P’(t) is zero or undefined. Since P(t) is a combination of sine and cosine functions, its derivative will also be smooth, so we just need to find where P’(t) = 0.First, let's write P(t):P(t) = 5 cos(t) + 3 sin(t) + 10Compute the derivative:P’(t) = -5 sin(t) + 3 cos(t)Set derivative equal to zero:-5 sin(t) + 3 cos(t) = 0Let me solve for t:-5 sin(t) + 3 cos(t) = 0Let's rearrange:3 cos(t) = 5 sin(t)Divide both sides by cos(t):3 = 5 tan(t)So, tan(t) = 3/5Therefore, t = arctan(3/5) + kπ, where k is an integer.Now, we need to find all t in [0, 12π] such that t = arctan(3/5) + kπ.First, compute arctan(3/5). Let me calculate that.arctan(3/5) is approximately, since tan(30 degrees) is about 0.577, and 3/5 is 0.6, which is slightly larger. So, arctan(0.6) is approximately 0.5404 radians, which is roughly 30.96 degrees.So, the solutions are t ≈ 0.5404 + kπ, for k = 0, 1, 2, ..., up to k where t <= 12π.Since 12π is approximately 37.699 radians, and each k adds π (≈3.1416), so the number of solutions is 12π / π = 12, but since we start at k=0, we have 12 + 1 = 13 solutions? Wait, no, because when k=12, t ≈0.5404 +12π ≈0.5404 +37.699≈38.24, which is beyond 12π≈37.699. So, k can go from 0 to 11, giving 12 solutions.Wait, let me check:For k=0: t≈0.5404k=1: t≈0.5404 +3.1416≈3.682k=2:≈3.682 +3.1416≈6.8236...k=11:≈0.5404 +11π≈0.5404 +34.5575≈35.0979, which is less than 12π≈37.699k=12:≈0.5404 +12π≈38.24, which is beyond 12π.So, total of 12 critical points.But wait, let me think again. Since tan(t) has a period of π, so in each interval of π, there is one solution. So, over 12π, there are 12 intervals of π, hence 12 solutions.So, 12 critical points.Now, to determine their nature, we need to check whether each critical point is a local maximum, minimum, or saddle point.Since P(t) is a combination of sine and cosine functions, which are smooth and periodic, the critical points will alternate between maxima and minima.But let's confirm by using the second derivative test.Compute the second derivative:P’’(t) = -5 cos(t) - 3 sin(t)At each critical point t = arctan(3/5) + kπ, let's evaluate P’’(t).First, let's note that at t = arctan(3/5) + kπ, sin(t) and cos(t) can be determined.Let me denote θ = arctan(3/5). So, tan(θ) = 3/5, which implies that in a right triangle, the opposite side is 3, adjacent is 5, hypotenuse is sqrt(3² +5²)=sqrt(34).So, sin(θ)=3/sqrt(34), cos(θ)=5/sqrt(34).Now, for t = θ + kπ.Case 1: k even, say k=2m. Then t = θ + 2mπ. So, sin(t)=sin(θ)=3/sqrt(34), cos(t)=cos(θ)=5/sqrt(34).Case 2: k odd, say k=2m+1. Then t = θ + (2m+1)π. So, sin(t)=sin(θ + π)= -sin(θ)= -3/sqrt(34), cos(t)=cos(θ + π)= -cos(θ)= -5/sqrt(34).So, let's compute P’’(t) for both cases.Case 1: k even.P’’(t) = -5 cos(t) - 3 sin(t) = -5*(5/sqrt(34)) - 3*(3/sqrt(34)) = (-25 -9)/sqrt(34) = (-34)/sqrt(34) = -sqrt(34)Which is negative, so the function is concave down, hence local maximum.Case 2: k odd.P’’(t) = -5 cos(t) - 3 sin(t) = -5*(-5/sqrt(34)) - 3*(-3/sqrt(34)) = (25 +9)/sqrt(34) = 34/sqrt(34) = sqrt(34)Which is positive, so the function is concave up, hence local minimum.Therefore, the critical points alternate between local maxima and minima.So, in the interval [0,12π], we have 12 critical points, alternating between maxima and minima.Since we start at t≈0.5404, which is a local maximum (since P’’(t) is negative there), then the next critical point at t≈3.682 is a local minimum, then t≈6.8236 is a maximum, and so on, alternating.So, in total, 6 local maxima and 6 local minima.Wait, but 12 critical points, alternating, so 6 of each.Yes, that makes sense.So, to summarize:For part 1, B=1 and D=1.For part 2, there are 12 critical points in [0,12π], alternating between local maxima and minima, starting with a maximum at t≈0.5404, then a minimum at t≈3.682, and so on, ending with a maximum at t≈35.0979 (since 12 is even, the last one would be a maximum? Wait, let's check.Wait, k goes from 0 to 11, which is 12 terms. Starting with k=0: maximum, k=1: minimum, k=2: maximum,..., k=11: minimum or maximum?Wait, k=0: maximumk=1: minimumk=2: maximum...k=11: since 11 is odd, it's a minimum.So, total of 6 maxima (k=0,2,4,6,8,10) and 6 minima (k=1,3,5,7,9,11).So, 6 local maxima and 6 local minima.Therefore, the critical points are at t = arctan(3/5) + kπ for k=0,1,...,11, with alternating nature starting with maximum.So, I think that's the solution.Final Answer1. The values of ( B ) and ( D ) are both boxed{1}.2. The critical points occur at ( t = arctanleft(frac{3}{5}right) + kpi ) for ( k = 0, 1, 2, ldots, 11 ). These points alternate between local maxima and minima, starting with a local maximum at ( t = arctanleft(frac{3}{5}right) ).The final answers are:1. ( B = boxed{1} ) and ( D = boxed{1} ).2. The critical points are ( t = arctanleft(frac{3}{5}right) + kpi ) for ( k = 0, 1, ldots, 11 ), with alternating local maxima and minima.</think>

🧠:<think>Okay, so I need to figure out the natural frequency of this compound pendulum that Dr. Clara Evans is studying. The system is a rod of length ( L ) and mass ( m ), with a point mass ( M ) at the end. It's modeled as a compound pendulum, so I remember that compound pendulums have a certain formula for their natural frequency.First, I think I need to find the moment of inertia of the entire system about the pivot point, which is the knee joint. The rod is uniform, so its moment of inertia about one end is given as ( frac{1}{3}mL^2 ). Then, the point mass ( M ) is at the end of the rod, so its moment of inertia about the pivot is ( ML^2 ). So, the total moment of inertia ( I ) should be the sum of these two, right?So, ( I = frac{1}{3}mL^2 + ML^2 ). That simplifies to ( I = L^2left( frac{m}{3} + M right) ). Okay, that makes sense.Next, I need to consider the torque acting on the system due to gravity. For a pendulum, the torque is given by ( tau = -mgd sintheta ), where ( d ) is the distance from the pivot to the center of mass. But in this case, it's a compound pendulum with two masses: the rod and the point mass.Wait, actually, for a compound pendulum, the formula for natural frequency is ( omega_n = sqrt{frac{mgd}{I}} ), where ( d ) is the distance from the pivot to the center of mass of the entire system. So, I need to find the center of mass of the system.The center of mass ( d ) can be calculated by taking the weighted average of the centers of mass of each component. The rod's center of mass is at ( frac{L}{2} ) from the pivot, and the point mass ( M ) is at ( L ) from the pivot.So, the total mass is ( m + M ). The center of mass ( d ) is:( d = frac{m cdot frac{L}{2} + M cdot L}{m + M} )Simplifying that:( d = frac{frac{mL}{2} + ML}{m + M} = frac{L(m/2 + M)}{m + M} )Okay, so now I have ( d ). Then, the torque is ( tau = - (m + M) g d sintheta ), but wait, actually, no. The torque is the sum of the torques from each mass. The torque from the rod is ( -mg cdot frac{L}{2} sintheta ) and the torque from the point mass is ( -Mg cdot L sintheta ). So, the total torque is:( tau = - left( frac{mL}{2} + ML right) g sintheta )But since ( d ) is the center of mass, the total torque can also be expressed as ( - (m + M) g d sintheta ). Either way, when we write the equation of motion, we can use torque equals moment of inertia times angular acceleration.So, the equation of motion for small angles is ( I ddot{theta} = - (m + M) g d sintheta approx - (m + M) g d theta ), since ( sintheta approx theta ) for small angles.Therefore, the equation becomes:( I ddot{theta} + (m + M) g d theta = 0 )Which is a simple harmonic oscillator equation, so the natural frequency ( omega_n ) is:( omega_n = sqrt{frac{(m + M) g d}{I}} )Now, plugging in the expressions for ( d ) and ( I ):First, ( d = frac{L(m/2 + M)}{m + M} ), so:( (m + M) g d = (m + M) g cdot frac{L(m/2 + M)}{m + M} = g L (m/2 + M) )And ( I = L^2 (m/3 + M) )So, ( omega_n = sqrt{frac{g L (m/2 + M)}{L^2 (m/3 + M)}} = sqrt{frac{g (m/2 + M)}{L (m/3 + M)}} )Simplify the expression inside the square root:Let me write it as:( sqrt{frac{g}{L} cdot frac{m/2 + M}{m/3 + M}} )To make it cleaner, maybe factor out ( m ) and ( M ):( sqrt{frac{g}{L} cdot frac{frac{m}{2} + M}{frac{m}{3} + M}} )Alternatively, we can write the numerator and denominator with a common denominator:Numerator: ( frac{m}{2} + M = frac{m + 2M}{2} )Denominator: ( frac{m}{3} + M = frac{m + 3M}{3} )So, the fraction becomes:( frac{frac{m + 2M}{2}}{frac{m + 3M}{3}} = frac{3(m + 2M)}{2(m + 3M)} )Therefore, ( omega_n = sqrt{frac{g}{L} cdot frac{3(m + 2M)}{2(m + 3M)}} )Simplify further:( omega_n = sqrt{frac{3g(m + 2M)}{2L(m + 3M)}} )So, that's the natural frequency.Wait, let me double-check the steps. I calculated the center of mass correctly, right? Yes, ( d = frac{m cdot frac{L}{2} + M cdot L}{m + M} ). Then, the torque is ( (m + M) g d ), and the moment of inertia is ( I = frac{1}{3}mL^2 + ML^2 ). Plugging into the formula, I get ( omega_n = sqrt{frac{(m + M) g d}{I}} ), which simplifies to the expression above.Alternatively, another approach is to consider the potential energy and kinetic energy, then use Lagrangian mechanics. But since it's a simple pendulum, the torque method should suffice.Okay, moving on to part 2. Now, damping is introduced with a linear viscous damping coefficient ( c ). So, the damping torque is ( tau_d = -c dot{theta} ). So, the equation of motion becomes:( I ddot{theta} + c dot{theta} + (m + M) g d theta = 0 )Which is a second-order linear differential equation with damping. The standard form is:( ddot{theta} + frac{c}{I} dot{theta} + frac{(m + M) g d}{I} theta = 0 )The damped natural frequency ( omega_d ) is given by:( omega_d = omega_n sqrt{1 - left( frac{zeta}{sqrt{1 - zeta^2}} right)^2} )Wait, no, actually, the damped natural frequency is ( omega_d = omega_n sqrt{1 - zeta^2} ), where ( zeta ) is the damping ratio, defined as ( zeta = frac{c}{2 sqrt{I k}} ), where ( k ) is the stiffness, which in this case is ( (m + M) g d ).So, let's compute ( zeta ):( zeta = frac{c}{2 sqrt{I cdot (m + M) g d}} )But ( omega_n = sqrt{frac{(m + M) g d}{I}} ), so ( sqrt{frac{(m + M) g d}{I}} = omega_n ), which implies ( sqrt{I cdot (m + M) g d} = sqrt{I} cdot sqrt{(m + M) g d} = sqrt{I} cdot omega_n sqrt{I} ) Wait, no, let me think.Actually, ( sqrt{I cdot (m + M) g d} = sqrt{I} cdot sqrt{(m + M) g d} ). But ( sqrt{(m + M) g d} = sqrt{I} cdot omega_n ), because ( omega_n^2 = frac{(m + M) g d}{I} ), so ( (m + M) g d = I omega_n^2 ). Therefore, ( sqrt{(m + M) g d} = sqrt{I} omega_n ).So, ( sqrt{I cdot (m + M) g d} = sqrt{I} cdot sqrt{I} omega_n = I omega_n ).Therefore, ( zeta = frac{c}{2 I omega_n} )Then, the damped natural frequency is:( omega_d = omega_n sqrt{1 - zeta^2} = omega_n sqrt{1 - left( frac{c}{2 I omega_n} right)^2 } )Simplify:( omega_d = omega_n sqrt{1 - frac{c^2}{4 I^2 omega_n^2}} )But ( omega_n^2 = frac{(m + M) g d}{I} ), so ( I omega_n^2 = (m + M) g d ). Therefore, ( I^2 omega_n^2 = I (m + M) g d ).So, substituting back:( omega_d = omega_n sqrt{1 - frac{c^2}{4 I (m + M) g d}} )Alternatively, we can express ( omega_d ) directly in terms of the given parameters.But perhaps it's better to express ( omega_d ) in terms of ( omega_n ) and ( c ). Alternatively, express it without ( omega_n ).Given that ( omega_n = sqrt{frac{3g(m + 2M)}{2L(m + 3M)}} ), as found earlier, and ( I = L^2 (m/3 + M) ), we can write:( omega_d = sqrt{frac{3g(m + 2M)}{2L(m + 3M)}} cdot sqrt{1 - left( frac{c}{2 I sqrt{frac{3g(m + 2M)}{2L(m + 3M)}}} right)^2 } )Wait, that seems complicated. Maybe it's better to express ( omega_d ) as:( omega_d = sqrt{frac{(m + M) g d}{I} - left( frac{c}{2I} right)^2} )Because the general form for damped systems is ( omega_d = sqrt{omega_n^2 - left( frac{c}{2m} right)^2} ), but here it's rotational, so instead of mass, it's moment of inertia.Wait, let me recall. For a translational system, ( omega_d = sqrt{omega_n^2 - left( frac{c}{2m} right)^2} ). For rotational systems, it's similar but with moment of inertia. So, yes, ( omega_d = sqrt{omega_n^2 - left( frac{c}{2I} right)^2} ).Therefore, substituting ( omega_n^2 = frac{(m + M) g d}{I} ), we have:( omega_d = sqrt{frac{(m + M) g d}{I} - left( frac{c}{2I} right)^2} )But we can express ( d ) and ( I ) in terms of ( L, m, M ):( d = frac{L(m/2 + M)}{m + M} )( I = L^2 (m/3 + M) )So, plugging these into ( omega_d ):First, compute ( frac{(m + M) g d}{I} ):( frac{(m + M) g cdot frac{L(m/2 + M)}{m + M}}{L^2 (m/3 + M)} = frac{g L (m/2 + M)}{L^2 (m/3 + M)} = frac{g (m/2 + M)}{L (m/3 + M)} )Which is ( omega_n^2 ), as before.Then, ( left( frac{c}{2I} right)^2 = frac{c^2}{4 I^2} = frac{c^2}{4 L^4 (m/3 + M)^2} )So, putting it all together:( omega_d = sqrt{ frac{g (m/2 + M)}{L (m/3 + M)} - frac{c^2}{4 L^4 (m/3 + M)^2} } )Hmm, that seems a bit messy. Maybe factor out ( frac{1}{L (m/3 + M)} ) from both terms inside the square root:( omega_d = sqrt{ frac{1}{L (m/3 + M)} left( g (m/2 + M) - frac{c^2}{4 L^3 (m/3 + M)} right) } )But I'm not sure if that's helpful. Alternatively, perhaps express it as:( omega_d = sqrt{ frac{3g(m + 2M)}{2L(m + 3M)} - frac{c^2}{4 L^4 (m/3 + M)^2} } )But this is getting complicated. Maybe it's better to leave it in terms of ( omega_n ) and ( c ), as ( omega_d = sqrt{omega_n^2 - left( frac{c}{2I} right)^2} ), but since ( omega_n ) is already expressed in terms of ( L, m, M, g ), and ( I ) is known, perhaps that's acceptable.Alternatively, express ( omega_d ) as:( omega_d = sqrt{ frac{3g(m + 2M)}{2L(m + 3M)} - frac{c^2}{4 L^4 (m/3 + M)^2} } )But this seems too unwieldy. Maybe it's better to write it in terms of ( omega_n ) and ( c ):( omega_d = sqrt{ omega_n^2 - left( frac{c}{2I} right)^2 } )Given that ( I = L^2 (m/3 + M) ), so ( frac{c}{2I} = frac{c}{2 L^2 (m/3 + M)} ), so:( omega_d = sqrt{ omega_n^2 - left( frac{c}{2 L^2 (m/3 + M)} right)^2 } )But perhaps the problem expects the expression in terms of the given parameters without substituting ( omega_n ). Let me check the problem statement again.It says: "Derive the equation of motion for the damped oscillation and express the damped natural frequency ( omega_d ) in terms of the given parameters ( L, m, M, g, c )."So, perhaps I should express ( omega_d ) directly in terms of ( L, m, M, g, c ) without substituting ( omega_n ).Given that, let's recall that for a damped rotational system, the damped natural frequency is:( omega_d = sqrt{ frac{k}{I} - left( frac{c}{2I} right)^2 } )Where ( k ) is the torsional spring constant, which in this case is ( (m + M) g d ).So, ( k = (m + M) g d ), and ( d = frac{L(m/2 + M)}{m + M} ), so ( k = (m + M) g cdot frac{L(m/2 + M)}{m + M} = g L (m/2 + M) )Therefore, ( omega_d = sqrt{ frac{g L (m/2 + M)}{I} - left( frac{c}{2I} right)^2 } )But ( I = L^2 (m/3 + M) ), so substituting:( omega_d = sqrt{ frac{g L (m/2 + M)}{L^2 (m/3 + M)} - frac{c^2}{4 L^4 (m/3 + M)^2} } )Simplify the first term:( frac{g L (m/2 + M)}{L^2 (m/3 + M)} = frac{g (m/2 + M)}{L (m/3 + M)} )So, ( omega_d = sqrt{ frac{g (m/2 + M)}{L (m/3 + M)} - frac{c^2}{4 L^4 (m/3 + M)^2} } )Alternatively, factor out ( frac{1}{L (m/3 + M)} ):( omega_d = sqrt{ frac{1}{L (m/3 + M)} left( g (m/2 + M) - frac{c^2}{4 L^3 (m/3 + M)} right) } )But this still seems complicated. Maybe it's better to leave it as:( omega_d = sqrt{ frac{g (m/2 + M)}{L (m/3 + M)} - left( frac{c}{2 L^2 (m/3 + M)} right)^2 } )But perhaps the problem expects a simplified expression, so let me see if I can combine the terms under a common denominator.Let me denote ( A = frac{g (m/2 + M)}{L (m/3 + M)} ) and ( B = frac{c^2}{4 L^4 (m/3 + M)^2} ), so ( omega_d = sqrt{A - B} ).To combine them, express ( A ) with denominator ( L^4 (m/3 + M)^2 ):( A = frac{g (m/2 + M) L^3 (m/3 + M)}{L^4 (m/3 + M)^2} = frac{g (m/2 + M) L^3 (m/3 + M)}{L^4 (m/3 + M)^2} = frac{g (m/2 + M) (m/3 + M)}{L (m/3 + M)^2} )Wait, that might not help. Alternatively, let's compute ( A - B ):( A - B = frac{g (m/2 + M)}{L (m/3 + M)} - frac{c^2}{4 L^4 (m/3 + M)^2} )To combine these, find a common denominator, which would be ( 4 L^4 (m/3 + M)^2 ):So,( A = frac{g (m/2 + M) cdot 4 L^3 (m/3 + M)}{4 L^4 (m/3 + M)^2} )( B = frac{c^2}{4 L^4 (m/3 + M)^2} )Therefore,( A - B = frac{4 g L^3 (m/2 + M)(m/3 + M) - c^2}{4 L^4 (m/3 + M)^2} )So,( omega_d = sqrt{ frac{4 g L^3 (m/2 + M)(m/3 + M) - c^2}{4 L^4 (m/3 + M)^2} } = frac{ sqrt{4 g L^3 (m/2 + M)(m/3 + M) - c^2} }{2 L^2 (m/3 + M)} } )Simplify numerator inside the square root:( 4 g L^3 (m/2 + M)(m/3 + M) = 4 g L^3 left( frac{m}{2} + M right) left( frac{m}{3} + M right) )Let me expand ( left( frac{m}{2} + M right) left( frac{m}{3} + M right) ):( frac{m^2}{6} + frac{mM}{2} + frac{mM}{3} + M^2 = frac{m^2}{6} + frac{3mM + 2mM}{6} + M^2 = frac{m^2}{6} + frac{5mM}{6} + M^2 )So, ( 4 g L^3 left( frac{m^2}{6} + frac{5mM}{6} + M^2 right) = 4 g L^3 left( frac{m^2 + 5mM + 6M^2}{6} right) = frac{4 g L^3 (m^2 + 5mM + 6M^2)}{6} = frac{2 g L^3 (m^2 + 5mM + 6M^2)}{3} )Therefore, the numerator inside the square root becomes:( frac{2 g L^3 (m^2 + 5mM + 6M^2)}{3} - c^2 )So, putting it all together:( omega_d = frac{ sqrt{ frac{2 g L^3 (m^2 + 5mM + 6M^2)}{3} - c^2 } }{2 L^2 (m/3 + M)} } )Simplify the denominator:( 2 L^2 (m/3 + M) = 2 L^2 left( frac{m + 3M}{3} right) = frac{2 L^2 (m + 3M)}{3} )So,( omega_d = frac{ sqrt{ frac{2 g L^3 (m^2 + 5mM + 6M^2)}{3} - c^2 } }{ frac{2 L^2 (m + 3M)}{3} } } = frac{3}{2 L^2 (m + 3M)} cdot sqrt{ frac{2 g L^3 (m^2 + 5mM + 6M^2)}{3} - c^2 } )This is getting quite involved. Maybe it's better to leave the expression as:( omega_d = sqrt{ frac{g (m/2 + M)}{L (m/3 + M)} - frac{c^2}{4 L^4 (m/3 + M)^2} } )Alternatively, factor out ( frac{1}{L} ) from the first term and ( frac{1}{L^4} ) from the second term, but I don't think that helps much.Perhaps the problem expects the expression in terms of ( omega_n ) and ( c ), so:( omega_d = sqrt{ omega_n^2 - left( frac{c}{2I} right)^2 } )Given that ( I = L^2 (m/3 + M) ), so:( omega_d = sqrt{ omega_n^2 - frac{c^2}{4 L^4 (m/3 + M)^2} } )But since ( omega_n^2 = frac{g (m/2 + M)}{L (m/3 + M)} ), we can write:( omega_d = sqrt{ frac{g (m/2 + M)}{L (m/3 + M)} - frac{c^2}{4 L^4 (m/3 + M)^2} } )I think this is as simplified as it gets without making it too convoluted. So, that's the expression for ( omega_d ).To recap:1. The natural frequency ( omega_n ) is ( sqrt{frac{3g(m + 2M)}{2L(m + 3M)}} ).2. The damped natural frequency ( omega_d ) is ( sqrt{ frac{g (m/2 + M)}{L (m/3 + M)} - frac{c^2}{4 L^4 (m/3 + M)^2} } ).I think that's the answer they're looking for.</think>