🧠:<think>Okay, so I have this problem about modeling the rise of an artist's popularity using a logistic growth function. The function given is ( P(t) = frac{K}{1 + Ae^{-Bt}} ). I need to find the constants ( A ) and ( B ) given some initial conditions, and then use those to find the time when the popularity reaches 90. Let me try to break this down step by step.First, the problem states that at ( t = 0 ), the popularity index ( P(0) ) was 5. So, plugging ( t = 0 ) into the logistic function should give me 5. Let me write that out:( P(0) = frac{K}{1 + Ae^{-B cdot 0}} )Since ( e^{-B cdot 0} = e^0 = 1 ), this simplifies to:( 5 = frac{K}{1 + A} )We are also told that the carrying capacity ( K ) is 100. So substituting ( K = 100 ) into the equation:( 5 = frac{100}{1 + A} )Now, I can solve for ( A ). Let's do that. Multiply both sides by ( 1 + A ):( 5(1 + A) = 100 )Divide both sides by 5:( 1 + A = 20 )Subtract 1 from both sides:( A = 19 )Okay, so I found ( A = 19 ). That wasn't too bad. Now, moving on to finding ( B ). The problem gives another condition: after 12 months, the popularity index ( P(12) ) was 60. Let's plug that into the logistic function.( P(12) = frac{100}{1 + 19e^{-B cdot 12}} = 60 )So, we have:( 60 = frac{100}{1 + 19e^{-12B}} )Let me solve for ( B ). First, multiply both sides by ( 1 + 19e^{-12B} ):( 60(1 + 19e^{-12B}) = 100 )Divide both sides by 60:( 1 + 19e^{-12B} = frac{100}{60} )Simplify ( frac{100}{60} ) to ( frac{5}{3} ):( 1 + 19e^{-12B} = frac{5}{3} )Subtract 1 from both sides:( 19e^{-12B} = frac{5}{3} - 1 )Calculate ( frac{5}{3} - 1 = frac{2}{3} ):( 19e^{-12B} = frac{2}{3} )Now, divide both sides by 19:( e^{-12B} = frac{2}{3 times 19} )Calculate ( 3 times 19 = 57 ):( e^{-12B} = frac{2}{57} )To solve for ( B ), take the natural logarithm of both sides:( ln(e^{-12B}) = lnleft(frac{2}{57}right) )Simplify the left side:( -12B = lnleft(frac{2}{57}right) )Now, solve for ( B ):( B = -frac{1}{12} lnleft(frac{2}{57}right) )Hmm, let me compute that. First, compute ( lnleft(frac{2}{57}right) ). Since ( frac{2}{57} ) is a small number, the natural log will be negative. Let me calculate it:( lnleft(frac{2}{57}right) = ln(2) - ln(57) approx 0.6931 - 4.0432 = -3.3501 )So, ( B = -frac{1}{12} times (-3.3501) = frac{3.3501}{12} approx 0.2792 )So, ( B ) is approximately 0.2792. Let me double-check my calculations to make sure I didn't make a mistake.Starting from ( P(12) = 60 ):( 60 = frac{100}{1 + 19e^{-12B}} )Multiply both sides by denominator:( 60(1 + 19e^{-12B}) = 100 )Divide by 60:( 1 + 19e^{-12B} = frac{5}{3} )Subtract 1:( 19e^{-12B} = frac{2}{3} )Divide by 19:( e^{-12B} = frac{2}{57} )Take natural log:( -12B = ln(2/57) )So, ( B = -frac{1}{12} ln(2/57) ). Calculating ( ln(2/57) approx -3.3501 ), so ( B approx 0.2792 ). That seems correct.So, summarizing part 1, ( A = 19 ) and ( B approx 0.2792 ).Moving on to part 2, we need to calculate the time ( t ) when the popularity index ( P(t) ) reaches 90. Using the logistic function with ( K = 100 ), ( A = 19 ), and ( B approx 0.2792 ):( 90 = frac{100}{1 + 19e^{-0.2792t}} )Let me solve for ( t ). First, multiply both sides by the denominator:( 90(1 + 19e^{-0.2792t}) = 100 )Divide both sides by 90:( 1 + 19e^{-0.2792t} = frac{100}{90} )Simplify ( frac{100}{90} ) to ( frac{10}{9} approx 1.1111 ):( 1 + 19e^{-0.2792t} = 1.1111 )Subtract 1 from both sides:( 19e^{-0.2792t} = 0.1111 )Divide both sides by 19:( e^{-0.2792t} = frac{0.1111}{19} approx 0.005847 )Take the natural logarithm of both sides:( ln(e^{-0.2792t}) = ln(0.005847) )Simplify the left side:( -0.2792t = ln(0.005847) )Calculate ( ln(0.005847) ). Let me compute that:( ln(0.005847) approx -5.135 )So,( -0.2792t = -5.135 )Divide both sides by -0.2792:( t = frac{-5.135}{-0.2792} approx 18.39 ) months.So, approximately 18.39 months. Let me check my steps again.Starting from ( P(t) = 90 ):( 90 = frac{100}{1 + 19e^{-0.2792t}} )Multiply both sides:( 90(1 + 19e^{-0.2792t}) = 100 )Divide by 90:( 1 + 19e^{-0.2792t} = frac{10}{9} approx 1.1111 )Subtract 1:( 19e^{-0.2792t} = 0.1111 )Divide by 19:( e^{-0.2792t} approx 0.005847 )Take natural log:( -0.2792t = ln(0.005847) approx -5.135 )Divide:( t approx 18.39 ) months.That seems consistent. So, approximately 18.39 months, which is about 1 year and 6.39 months, or roughly 1 year and 6 months. But since the question asks for the time ( t ), I can leave it as approximately 18.39 months.Wait, let me check my calculation for ( ln(0.005847) ). Maybe I approximated too much earlier.Calculating ( ln(0.005847) ):We know that ( ln(0.01) approx -4.6052 ), and ( 0.005847 ) is about half of 0.011694, which is roughly ( e^{-5} approx 0.0067379 ). So, ( ln(0.0067379) = -5 ). Since 0.005847 is slightly less than 0.0067379, the natural log should be slightly less than -5, maybe around -5.135 as I had before. So, that seems correct.Therefore, ( t approx 18.39 ) months. Let me see if I can express this as a fraction or something more precise, but since the question doesn't specify, decimal is probably fine.Alternatively, if I use more precise values for ( B ), maybe the result would be slightly different. Let me recalculate ( B ) with more precision.Earlier, I approximated ( ln(2/57) approx -3.3501 ). Let me compute it more accurately.Compute ( ln(2) approx 0.69314718056 )Compute ( ln(57) ). Let's see, ( ln(50) approx 3.91202 ), ( ln(57) ) is a bit higher. Let me compute it:( ln(57) = ln(50 times 1.14) = ln(50) + ln(1.14) approx 3.91202 + 0.13353 = 4.04555 )So, ( ln(2/57) = ln(2) - ln(57) approx 0.693147 - 4.04555 approx -3.3524 )So, ( B = -frac{1}{12} times (-3.3524) approx 0.279367 )So, ( B approx 0.279367 ). Let's use this more precise value.Now, going back to part 2, when ( P(t) = 90 ):( 90 = frac{100}{1 + 19e^{-0.279367t}} )Multiply both sides:( 90(1 + 19e^{-0.279367t}) = 100 )Divide by 90:( 1 + 19e^{-0.279367t} = frac{10}{9} approx 1.111111 )Subtract 1:( 19e^{-0.279367t} = 0.111111 )Divide by 19:( e^{-0.279367t} = frac{0.111111}{19} approx 0.0058474 )Take natural log:( -0.279367t = ln(0.0058474) )Compute ( ln(0.0058474) ). Let me use a calculator for more precision.( ln(0.0058474) approx -5.135 ). Let me confirm:We know that ( e^{-5} approx 0.0067379 ), which is larger than 0.0058474. So, ( ln(0.0058474) ) is less than -5.Compute ( e^{-5.135} ):First, compute ( e^{-5} approx 0.0067379 )Compute ( e^{-0.135} approx 1 - 0.135 + 0.135^2/2 - 0.135^3/6 approx 1 - 0.135 + 0.0091125 - 0.000256 approx 0.8748565 )So, ( e^{-5.135} = e^{-5} times e^{-0.135} approx 0.0067379 times 0.8748565 approx 0.00589 )But we have ( e^{-0.279367t} = 0.0058474 ), which is slightly less than 0.00589. So, ( -0.279367t ) is slightly less than -5.135, meaning ( t ) is slightly more than 18.39 months.Wait, let me compute ( ln(0.0058474) ) more accurately.Using a calculator, ( ln(0.0058474) approx -5.135 ). Let me check with a calculator:Compute ( ln(0.0058474) ):Since ( ln(0.005) approx -5.2983 ), and ( 0.0058474 ) is larger than 0.005, so ( ln(0.0058474) ) is greater than -5.2983, which is consistent with -5.135.Wait, actually, no. Wait, 0.0058474 is greater than 0.005, so the natural log should be less negative, meaning closer to zero. So, ( ln(0.0058474) ) is greater than ( ln(0.005) approx -5.2983 ). So, if I have ( ln(0.0058474) approx -5.135 ), that is correct because it's between -5.2983 and -5.So, ( -0.279367t = -5.135 )Thus, ( t = frac{5.135}{0.279367} approx 18.39 ) months.Wait, let me compute that division more accurately.Compute ( 5.135 / 0.279367 ):First, 0.279367 * 18 = 5.028606Subtract that from 5.135: 5.135 - 5.028606 = 0.106394Now, 0.279367 * 0.38 ≈ 0.106159So, 18 + 0.38 ≈ 18.38 months.So, approximately 18.38 months, which is about 18.38 months. So, rounding to two decimal places, 18.38 months.But let me see if I can express this as a fraction or something. 0.38 months is roughly 0.38 * 30 ≈ 11.4 days. So, approximately 18 months and 11 days. But since the question asks for the time ( t ), I think 18.38 months is acceptable.Alternatively, if I use more precise calculations, maybe it's 18.38 months.Wait, let me check the exact value:( t = frac{ln(100/90 - 1) / 19}{-B} ). Wait, no, let me re-express it.Wait, actually, let me go back to the equation:( 90 = frac{100}{1 + 19e^{-Bt}} )Multiply both sides by denominator:( 90(1 + 19e^{-Bt}) = 100 )Divide by 90:( 1 + 19e^{-Bt} = frac{10}{9} )Subtract 1:( 19e^{-Bt} = frac{1}{9} )Divide by 19:( e^{-Bt} = frac{1}{171} )Take natural log:( -Bt = ln(1/171) = -ln(171) )So,( t = frac{ln(171)}{B} )Compute ( ln(171) ):( ln(171) = ln(100 times 1.71) = ln(100) + ln(1.71) approx 4.60517 + 0.5378 approx 5.14297 )So, ( t = frac{5.14297}{B} )We have ( B = frac{-1}{12} ln(2/57) approx 0.279367 )So,( t approx frac{5.14297}{0.279367} approx 18.39 ) months.So, that's consistent with my earlier calculation.Therefore, the time when the popularity index reaches 90 is approximately 18.39 months.Let me just recap:1. Found ( A = 19 ) using ( P(0) = 5 ).2. Found ( B approx 0.279367 ) using ( P(12) = 60 ).3. Used these values to solve for ( t ) when ( P(t) = 90 ), resulting in ( t approx 18.39 ) months.I think that's thorough. I don't see any mistakes in the calculations now.</think>

🧠:<think>Alright, so I have this problem about two starships moving in 3D space, and I need to figure out when they're closest to each other and then ensure their minimum distance is at least D. Hmm, okay, let's break this down step by step.First, the positions of the starships are given by parametric equations. Starship A is moving with position vector r_A(t) = [a cos(ωt + φ), b sin(ωt + φ), c t], and Starship B is at r_B(t) = [d cos(αt + β), e sin(αt + β), f t]. So, both are moving in some elliptical paths in the x-y plane and linearly in the z-direction.To find the closest approach, I think I need to find the time t where the distance between r_A(t) and r_B(t) is minimized. The distance squared between them would be the square of the distance, which is easier to work with because the square root can complicate things when taking derivatives.So, let's denote the distance squared between the two starships as D(t)^2. That would be:D(t)^2 = [a cos(ωt + φ) - d cos(αt + β)]² + [b sin(ωt + φ) - e sin(αt + β)]² + [c t - f t]²Simplify that:D(t)^2 = [a cos(ωt + φ) - d cos(αt + β)]² + [b sin(ωt + φ) - e sin(αt + β)]² + [(c - f) t]²Okay, so to find the minimum distance, I need to find the t that minimizes D(t)^2. To do that, I can take the derivative of D(t)^2 with respect to t, set it equal to zero, and solve for t.Let me compute the derivative d/dt [D(t)^2]. Let's denote each component as x, y, z for simplicity.Let x(t) = a cos(ωt + φ) - d cos(αt + β)y(t) = b sin(ωt + φ) - e sin(αt + β)z(t) = (c - f) tThen, D(t)^2 = x(t)^2 + y(t)^2 + z(t)^2So, the derivative is:d/dt [D(t)^2] = 2x(t) * x’(t) + 2y(t) * y’(t) + 2z(t) * z’(t)Compute each derivative:x’(t) = -a ω sin(ωt + φ) + d α sin(αt + β)y’(t) = b ω cos(ωt + φ) - e α cos(αt + β)z’(t) = (c - f)So, putting it all together:d/dt [D(t)^2] = 2 [x(t) * (-a ω sin(ωt + φ) + d α sin(αt + β)) + y(t) * (b ω cos(ωt + φ) - e α cos(αt + β)) + z(t) * (c - f)]Set this derivative equal to zero for minima:x(t) * (-a ω sin(ωt + φ) + d α sin(αt + β)) + y(t) * (b ω cos(ωt + φ) - e α cos(αt + β)) + z(t) * (c - f) = 0So, that's the equation we need to solve for t. Hmm, this looks pretty complicated. It's a transcendental equation because of the sine and cosine terms with different frequencies ω and α. Solving this analytically might be tough unless ω = α, which would make it a bit simpler.Wait, if ω ≠ α, this equation might not have a closed-form solution. So, maybe the animator has to choose ω and α such that they are equal? Or perhaps the problem expects a general expression in terms of the constants, even if it's not solvable explicitly.Alternatively, maybe we can express t in terms of the other constants, but it's going to be messy. Let me think if there's another approach.Alternatively, maybe we can write the relative position vector as r_A(t) - r_B(t) and then find when its derivative is zero. But that's essentially what I did above.Alternatively, perhaps we can write the relative motion in terms of a single trigonometric function, but given the different frequencies, that might not be straightforward.Wait, maybe if we consider the relative velocity vector. The minimum distance occurs when the relative velocity is perpendicular to the relative position vector. So, the dot product of (r_A - r_B) and (r_A’ - r_B’) should be zero.Which is exactly the condition we have above.So, in that case, the equation is:(r_A - r_B) · (r_A’ - r_B’) = 0Which is exactly the derivative of D(t)^2 set to zero.So, that's the condition we have.But solving this equation for t is non-trivial because it's a combination of sine and cosine terms with different frequencies.So, unless ω = α, which would make the frequencies the same, and perhaps we can combine terms, but even then, it might not be straightforward.Alternatively, if ω and α are commensurate, meaning their ratio is a rational number, we might be able to find a solution, but in general, it's difficult.So, perhaps the answer is that the time t is given implicitly by the equation:[a cos(ωt + φ) - d cos(αt + β)] * [-a ω sin(ωt + φ) + d α sin(αt + β)] + [b sin(ωt + φ) - e sin(αt + β)] * [b ω cos(ωt + φ) - e α cos(αt + β)] + [(c - f) t] * (c - f) = 0Which is the equation we derived above.But that seems too complicated. Maybe we can write it more neatly.Let me denote θ_A = ωt + φ and θ_B = αt + β.Then, the equation becomes:[a cos θ_A - d cos θ_B] * [-a ω sin θ_A + d α sin θ_B] + [b sin θ_A - e sin θ_B] * [b ω cos θ_A - e α cos θ_B] + (c - f)^2 t = 0Hmm, that might not help much, but perhaps we can expand the terms.Let's compute each part:First term: [a cos θ_A - d cos θ_B] * [-a ω sin θ_A + d α sin θ_B]= -a^2 ω cos θ_A sin θ_A + a d ω cos θ_A sin θ_B + a d α cos θ_B sin θ_A - d^2 α cos θ_B sin θ_BSecond term: [b sin θ_A - e sin θ_B] * [b ω cos θ_A - e α cos θ_B]= b^2 ω sin θ_A cos θ_A - b e ω sin θ_A cos θ_B - b e α sin θ_B cos θ_A + e^2 α sin θ_B cos θ_BThird term: (c - f)^2 tSo, putting all together:- a^2 ω cos θ_A sin θ_A + a d ω cos θ_A sin θ_B + a d α cos θ_B sin θ_A - d^2 α cos θ_B sin θ_B + b^2 ω sin θ_A cos θ_A - b e ω sin θ_A cos θ_B - b e α sin θ_B cos θ_A + e^2 α sin θ_B cos θ_B + (c - f)^2 t = 0Now, notice that -a^2 ω cos θ_A sin θ_A + b^2 ω sin θ_A cos θ_A = ω (b^2 - a^2) sin θ_A cos θ_ASimilarly, -d^2 α cos θ_B sin θ_B + e^2 α sin θ_B cos θ_B = α (e^2 - d^2) sin θ_B cos θ_BThen, the cross terms:a d ω cos θ_A sin θ_B + a d α cos θ_B sin θ_A - b e ω sin θ_A cos θ_B - b e α sin θ_B cos θ_ALet me factor these:= a d ω cos θ_A sin θ_B + a d α cos θ_B sin θ_A - b e ω sin θ_A cos θ_B - b e α sin θ_B cos θ_A= [a d ω cos θ_A sin θ_B - b e ω sin θ_A cos θ_B] + [a d α cos θ_B sin θ_A - b e α sin θ_B cos θ_A]Factor out ω and α:= ω [a d cos θ_A sin θ_B - b e sin θ_A cos θ_B] + α [a d cos θ_B sin θ_A - b e sin θ_B cos θ_A]Hmm, notice that cos θ_A sin θ_B - sin θ_A cos θ_B = sin(θ_B - θ_A)Similarly, cos θ_B sin θ_A - sin θ_B cos θ_A = sin(θ_A - θ_B) = -sin(θ_B - θ_A)So, substituting that:= ω [a d sin(θ_B - θ_A) ] + α [ -a d sin(θ_B - θ_A) ]= [ω a d - α a d] sin(θ_B - θ_A)= a d (ω - α) sin(θ_B - θ_A)So, putting it all together, the equation becomes:ω (b^2 - a^2) sin θ_A cos θ_A + α (e^2 - d^2) sin θ_B cos θ_B + a d (ω - α) sin(θ_B - θ_A) + (c - f)^2 t = 0Hmm, that's a bit simpler, but still quite complex.Let me recall that sin θ cos θ = (1/2) sin(2θ), so we can write:= (ω (b^2 - a^2)/2) sin(2θ_A) + (α (e^2 - d^2)/2) sin(2θ_B) + a d (ω - α) sin(θ_B - θ_A) + (c - f)^2 t = 0But I don't know if that helps much.Alternatively, maybe we can write θ_B - θ_A = (α t + β) - (ω t + φ) = (α - ω) t + (β - φ)Let me denote Δθ = θ_B - θ_A = (α - ω) t + (β - φ)So, sin(θ_B - θ_A) = sin(Δθ)Similarly, sin(2θ_A) and sin(2θ_B) can be written in terms of θ_A and θ_B, but I don't see a straightforward way to combine them.So, perhaps this is as far as we can go analytically. Therefore, the equation we need to solve is:(ω (b^2 - a^2)/2) sin(2θ_A) + (α (e^2 - d^2)/2) sin(2θ_B) + a d (ω - α) sin(Δθ) + (c - f)^2 t = 0Where θ_A = ω t + φ, θ_B = α t + β, and Δθ = (α - ω) t + (β - φ)This is a transcendental equation in t, which likely doesn't have an analytical solution unless specific conditions on the constants are met.Therefore, the answer to part 1 is that the time t at which the two starships are closest is given implicitly by the equation above, which cannot be solved explicitly for t without additional constraints on the constants.But wait, the problem says "Express your answer in terms of the given constants." So, perhaps they expect an expression, even if it's implicit.Alternatively, maybe we can write it as:[ a cos(ωt + φ) - d cos(αt + β) ] * [ -a ω sin(ωt + φ) + d α sin(αt + β) ] + [ b sin(ωt + φ) - e sin(αt + β) ] * [ b ω cos(ωt + φ) - e α cos(αt + β) ] + (c - f)^2 t = 0Which is the same as what I had earlier.So, perhaps that's the answer they're looking for.Moving on to part 2: Given that the animator wants the starships to have a minimum distance of at least D units at all times, derive the conditions on the constants.Hmm, so we need to ensure that the minimum distance between the two starships is always ≥ D. So, the minimum of D(t) is ≥ D.But since D(t) is the distance, and we're dealing with parametric equations, perhaps we can find the minimum distance and set it ≥ D.But finding the minimum distance is equivalent to finding the minimum of D(t)^2, which we already considered.But since solving for t is difficult, perhaps we can analyze the problem differently.Alternatively, maybe we can consider the relative motion in the x-y plane and the z-direction separately.In the x-y plane, the relative position is:x_rel(t) = a cos(ωt + φ) - d cos(αt + β)y_rel(t) = b sin(ωt + φ) - e sin(αt + β)And in the z-direction:z_rel(t) = (c - f) tSo, the distance squared is x_rel(t)^2 + y_rel(t)^2 + z_rel(t)^2To ensure that this is always ≥ D^2, we need:x_rel(t)^2 + y_rel(t)^2 + z_rel(t)^2 ≥ D^2 for all t.But z_rel(t) is linear in t, so as t increases, z_rel(t) will dominate if c ≠ f. So, unless c = f, the distance will eventually go to infinity. Therefore, to have a minimum distance, we must have c = f, otherwise, the distance can be made arbitrarily small by choosing t where x_rel(t) and y_rel(t) are small, but z_rel(t) is also small. Wait, no, if c ≠ f, then as t increases, z_rel(t) grows without bound, but the x and y components are oscillatory. So, the distance will oscillate but with an increasing trend. Therefore, the minimum distance would occur at some finite t.But if c = f, then z_rel(t) = 0 for all t, so the distance is determined solely by the x and y components.Therefore, if c ≠ f, the minimum distance is achieved at some finite t, and we need to ensure that this minimum is ≥ D.If c = f, then the distance is always the distance in the x-y plane, which is periodic, so we need to ensure that the minimum distance in the x-y plane is ≥ D.So, perhaps we can split into two cases:Case 1: c ≠ fIn this case, the minimum distance occurs at some t where the derivative is zero, as we considered earlier. So, we need to ensure that the minimum value of D(t) is ≥ D.But since solving for t is difficult, perhaps we can bound the distance.Alternatively, maybe we can use the fact that the distance squared is x_rel^2 + y_rel^2 + z_rel^2.If we can find the minimum of x_rel^2 + y_rel^2, and then add z_rel^2, which is a quadratic function, we can find the overall minimum.But since z_rel(t) is linear, the overall minimum distance might be a combination of the x-y minimum and the z-component.Alternatively, perhaps we can consider the problem as minimizing D(t)^2 = x_rel(t)^2 + y_rel(t)^2 + (c - f)^2 t^2So, to ensure that this is always ≥ D^2, we need:x_rel(t)^2 + y_rel(t)^2 + (c - f)^2 t^2 ≥ D^2 for all t.But x_rel(t)^2 + y_rel(t)^2 is the distance squared in the x-y plane, which is periodic if ω and α are commensurate, or quasi-periodic otherwise.But regardless, it's oscillatory. So, the minimum of x_rel(t)^2 + y_rel(t)^2 occurs at some t, say t0, and the minimum distance squared is x0^2 + y0^2.Then, the overall distance squared is x0^2 + y0^2 + (c - f)^2 t0^2.But since t0 is the time where x_rel and y_rel are minimized, which might not correspond to the same t where z_rel is minimized.Wait, actually, the minimum of D(t)^2 occurs where the derivative is zero, which is the equation we had earlier. So, it's not necessarily the case that the x-y minimum and the z minimum occur at the same t.Therefore, perhaps we can find the minimum of D(t)^2 by considering the minimum of x_rel^2 + y_rel^2 + z_rel^2, which is a combination of oscillatory and linear terms.Alternatively, maybe we can use the Cauchy-Schwarz inequality or some other inequality to bound the distance.Alternatively, perhaps we can consider the worst-case scenario where x_rel and y_rel are at their maximum, but that might not help.Wait, actually, to ensure that D(t) ≥ D for all t, we need that for all t,x_rel(t)^2 + y_rel(t)^2 + z_rel(t)^2 ≥ D^2Which can be rewritten as:x_rel(t)^2 + y_rel(t)^2 ≥ D^2 - z_rel(t)^2But z_rel(t)^2 is non-negative, so the right-hand side is D^2 minus something non-negative. Therefore, to have x_rel(t)^2 + y_rel(t)^2 ≥ D^2 - z_rel(t)^2 for all t, we need that x_rel(t)^2 + y_rel(t)^2 is always greater than or equal to D^2 minus the maximum of z_rel(t)^2.But z_rel(t)^2 is (c - f)^2 t^2, which goes to infinity as t increases unless c = f.Wait, if c ≠ f, then as t increases, z_rel(t)^2 becomes very large, so D(t)^2 becomes very large, so the condition D(t) ≥ D is automatically satisfied for large t. The problem is for small t, near t=0, where z_rel(t) is small, so the distance is dominated by x_rel(t)^2 + y_rel(t)^2.Therefore, to ensure that D(t) ≥ D for all t, it's sufficient to ensure that the minimum of x_rel(t)^2 + y_rel(t)^2 is ≥ D^2, because for t where z_rel(t) is non-zero, the distance will be larger.Wait, is that correct? Let's think.If c ≠ f, then as t increases, z_rel(t) increases, so D(t) increases. Therefore, the minimum distance occurs at some finite t, possibly near t=0. So, if the minimum of x_rel(t)^2 + y_rel(t)^2 is ≥ D^2, then for all t, D(t)^2 ≥ D^2.But if c = f, then z_rel(t) = 0 for all t, so the distance is purely x_rel(t)^2 + y_rel(t)^2, which is periodic. Therefore, in this case, we need the minimum of x_rel(t)^2 + y_rel(t)^2 ≥ D^2.So, in both cases, whether c = f or not, we need the minimum of x_rel(t)^2 + y_rel(t)^2 ≥ D^2.But wait, if c ≠ f, then even if the x-y distance is less than D, the z-component might make the total distance larger. However, near t=0, z_rel(t) is small, so the distance is approximately x_rel(0)^2 + y_rel(0)^2. So, to ensure that D(t) ≥ D for all t, we need that x_rel(t)^2 + y_rel(t)^2 ≥ D^2 for all t, because if c ≠ f, the z-component can only increase the distance beyond that.Wait, no, actually, if c ≠ f, then as t increases, z_rel(t) increases, so D(t) increases. Therefore, the minimum distance occurs at the minimal t, which is t=0 or somewhere near t=0.Wait, but t can be negative as well, depending on the context. If t is allowed to be negative, then z_rel(t) can be negative, but squared, it's still positive. So, the minimal distance might occur at t where x_rel(t) and y_rel(t) are minimized, regardless of z.But if c ≠ f, then z_rel(t) is a linear function, so it can be positive or negative, but squared, it's always positive. So, the minimal distance occurs when x_rel(t)^2 + y_rel(t)^2 is minimized, regardless of z.Wait, no, because D(t)^2 = x_rel(t)^2 + y_rel(t)^2 + z_rel(t)^2. So, if x_rel(t)^2 + y_rel(t)^2 is minimized at some t, but z_rel(t) at that t is non-zero, then D(t)^2 is larger than the minimum of x_rel(t)^2 + y_rel(t)^2.Therefore, the minimal D(t)^2 is the minimum of x_rel(t)^2 + y_rel(t)^2 + z_rel(t)^2, which is not necessarily the same as the minimum of x_rel(t)^2 + y_rel(t)^2.So, perhaps we need to ensure that the minimal value of D(t)^2 is ≥ D^2.But since solving for t is difficult, perhaps we can find a lower bound for D(t)^2.Alternatively, maybe we can use the fact that x_rel(t)^2 + y_rel(t)^2 ≥ (a - d)^2 + (b - e)^2 - something, but I'm not sure.Alternatively, perhaps we can use the triangle inequality.Wait, x_rel(t) = a cos(θ_A) - d cos(θ_B)Similarly, y_rel(t) = b sin(θ_A) - e sin(θ_B)So, x_rel(t)^2 + y_rel(t)^2 = [a cos θ_A - d cos θ_B]^2 + [b sin θ_A - e sin θ_B]^2Expanding this:= a² cos² θ_A - 2 a d cos θ_A cos θ_B + d² cos² θ_B + b² sin² θ_A - 2 b e sin θ_A sin θ_B + e² sin² θ_BCombine terms:= (a² cos² θ_A + b² sin² θ_A) + (d² cos² θ_B + e² sin² θ_B) - 2 [a d cos θ_A cos θ_B + b e sin θ_A sin θ_B]Hmm, notice that a² cos² θ_A + b² sin² θ_A is the squared distance of Starship A from the origin in x-y plane, and similarly for Starship B.But I don't know if that helps.Alternatively, perhaps we can write this as:= (a cos θ_A - d cos θ_B)^2 + (b sin θ_A - e sin θ_B)^2Which is the squared distance between two points on two ellipses.Alternatively, perhaps we can consider the maximum and minimum possible values of this expression.But it's complicated because θ_A and θ_B are functions of t with different frequencies.Alternatively, perhaps we can use the fact that for any t, the expression is bounded below by something.Wait, using the Cauchy-Schwarz inequality, perhaps.But I'm not sure.Alternatively, maybe we can consider the worst-case scenario where x_rel(t) and y_rel(t) are as small as possible, which would correspond to the starships being as close as possible in the x-y plane.But again, without knowing the exact relationship between θ_A and θ_B, it's hard to find the minimum.Alternatively, perhaps we can consider the maximum of x_rel(t)^2 + y_rel(t)^2 over t, but we need the minimum.Wait, maybe we can use the fact that for any t,x_rel(t)^2 + y_rel(t)^2 ≥ (a - d)^2 + (b - e)^2 - something.But I don't think that's directly applicable.Alternatively, perhaps we can consider the maximum possible overlap between the two ellipses.Wait, if we think of the x-y motion of each starship as an ellipse, then the minimal distance between the two ellipses would be the minimal distance between any two points on the ellipses.But since the ellipses are parameterized with different frequencies, the minimal distance might not be straightforward.Alternatively, if the ellipses are fixed, the minimal distance between them is fixed, but in our case, the ellipses are functions of t, so they might be rotating or something.Wait, actually, no. The ellipses are fixed in shape, but their positions are functions of t.Wait, no, actually, the parametric equations are:Starship A: x = a cos(ωt + φ), y = b sin(ωt + φ)Starship B: x = d cos(αt + β), y = e sin(αt + β)So, in the x-y plane, each starship is moving along an ellipse with semi-axes a and b for A, and d and e for B.The relative position is the difference between these two points on the ellipses.So, the minimal distance between the two ellipses is a known problem, but in our case, the ellipses are moving with different angular frequencies, so the minimal distance can vary.But perhaps, to ensure that the minimal distance is always ≥ D, we can ensure that the minimal distance between the two ellipses is ≥ D, regardless of their relative phases.But that might be too restrictive, because even if the ellipses are close at some phase difference, the linear motion in z might make the overall distance larger.But as we discussed earlier, if c ≠ f, the z-component will eventually dominate, so the minimal distance is determined near t=0.Wait, but if c ≠ f, then as t increases, z_rel(t) increases, so the distance becomes larger. Therefore, the minimal distance occurs at t=0 or somewhere near t=0.Wait, let's compute D(0):At t=0,x_rel(0) = a cos(φ) - d cos(β)y_rel(0) = b sin(φ) - e sin(β)z_rel(0) = 0So, D(0)^2 = [a cos φ - d cos β]^2 + [b sin φ - e sin β]^2Similarly, as t approaches infinity, D(t) approaches infinity if c ≠ f.Therefore, the minimal distance occurs either at t=0 or somewhere else where the derivative is zero.But to ensure that D(t) ≥ D for all t, we need that the minimal value of D(t) is ≥ D.But since solving for t is difficult, perhaps we can ensure that D(0) ≥ D, and that the minimal distance in the x-y plane is ≥ D.But that might not be sufficient, because the minimal distance could occur at some t ≠ 0.Alternatively, perhaps we can consider that the minimal distance in the x-y plane is the minimal value of x_rel(t)^2 + y_rel(t)^2, and set that to be ≥ D^2.But again, finding the minimal value is difficult.Alternatively, perhaps we can use the fact that the minimal distance between two points on the ellipses is at least | |A| - |B| |, where |A| and |B| are the distances from the origin.But that's not necessarily true because the ellipses can be oriented in any way.Alternatively, perhaps we can consider the maximum of the minimal distances.Wait, maybe it's better to consider the problem in terms of the relative velocity and position.But I'm not sure.Alternatively, perhaps we can use the concept of the distance between two parametric curves.But that's a complex problem.Alternatively, maybe we can consider the worst-case scenario where the x-y distance is minimized, and the z-component is also minimized.But if c ≠ f, the z-component is minimized at t=0, so the minimal distance is D(0).Therefore, perhaps to ensure that D(t) ≥ D for all t, it's sufficient to ensure that D(0) ≥ D.But wait, that might not be the case because the minimal distance could occur at some t ≠ 0 where x_rel(t) and y_rel(t) are smaller than at t=0.But if c ≠ f, then as t increases, z_rel(t) increases, so the distance becomes larger. Therefore, the minimal distance occurs at the minimal t, which is t=0.Wait, but t can be negative as well. If t is allowed to be negative, then z_rel(t) can be negative, but squared, it's still positive. So, the minimal distance could occur at t where z_rel(t) is zero, but that's only if c = f.Wait, no, if c ≠ f, then z_rel(t) is linear, so it's zero at t = 0 only if c = f.Wait, no, z_rel(t) = (c - f) t. So, it's zero only at t=0 if c ≠ f.Therefore, if c ≠ f, the minimal distance occurs at t=0, because for t > 0 or t < 0, z_rel(t) increases, making D(t) larger.Therefore, in that case, to ensure D(t) ≥ D for all t, it's sufficient to ensure that D(0) ≥ D.But wait, is that correct? Let me think.Suppose c ≠ f, so z_rel(t) = (c - f) t.At t=0, z_rel(t)=0, so D(0) is the distance in x-y plane.As t increases, z_rel(t) increases, so D(t) increases.Similarly, as t decreases (negative t), z_rel(t) becomes negative, but squared, it's positive, so D(t) increases as |t| increases.Therefore, the minimal distance occurs at t=0.Therefore, in this case, to ensure D(t) ≥ D for all t, it's sufficient to ensure that D(0) ≥ D.So, the condition is:[a cos φ - d cos β]^2 + [b sin φ - e sin β]^2 ≥ D^2But wait, what if c = f?If c = f, then z_rel(t) = 0 for all t, so the distance is purely in the x-y plane, which is periodic. Therefore, the minimal distance is the minimal value of x_rel(t)^2 + y_rel(t)^2 over t.So, in this case, we need to ensure that the minimal distance in the x-y plane is ≥ D.But finding the minimal distance between two parametric ellipses is non-trivial.Alternatively, perhaps we can consider the minimal distance between the two ellipses, which is a known problem, but it's complicated.Alternatively, perhaps we can use the fact that the minimal distance between two ellipses is at least the difference between their semi-major axes or something, but that's not necessarily true.Alternatively, perhaps we can use the fact that the minimal distance between two points on the ellipses is at least the distance between their centers minus the sum of their radii, but again, that's for circles, not ellipses.Alternatively, perhaps we can consider the maximum and minimum possible values of x_rel(t)^2 + y_rel(t)^2.But given the parametric forms, it's difficult.Alternatively, perhaps we can use the fact that for any t,x_rel(t)^2 + y_rel(t)^2 ≥ (a - d)^2 + (b - e)^2 - something.But I don't think that's directly applicable.Alternatively, perhaps we can consider the worst-case scenario where the two ellipses are as close as possible.But without knowing the exact relationship between the parameters, it's hard to derive a general condition.Alternatively, perhaps we can consider that the minimal distance between the two ellipses is zero if they intersect, so to ensure a minimal distance D, we need that the ellipses do not come closer than D.But determining the minimal distance between two ellipses is a complex problem, and I don't think there's a simple formula for it.Therefore, perhaps the answer is that if c ≠ f, then the minimal distance occurs at t=0, so we need:[a cos φ - d cos β]^2 + [b sin φ - e sin β]^2 ≥ D^2And if c = f, then we need the minimal distance in the x-y plane to be ≥ D, which requires that the minimal value of x_rel(t)^2 + y_rel(t)^2 is ≥ D^2. But since finding that minimal value is difficult, perhaps we can impose that the distance between the ellipses is always ≥ D, which would require certain conditions on a, b, d, e, ω, α, φ, β.But without a specific method to compute the minimal distance, it's hard to derive explicit conditions.Alternatively, perhaps we can use the fact that the minimal distance between two ellipses is at least the distance between their centers minus the sum of their "radii" in some sense, but I'm not sure.Alternatively, perhaps we can consider the maximum of the minimal distances.Wait, maybe it's better to consider that if the two ellipses are far enough apart, their minimal distance is ≥ D.But without specific information, it's hard to define.Alternatively, perhaps we can consider that the minimal distance between the two ellipses is given by the minimal value of x_rel(t)^2 + y_rel(t)^2, which is a function of t, and set that to be ≥ D^2.But since we can't solve for t, perhaps we can use some inequality.Alternatively, perhaps we can use the fact that for any t,x_rel(t)^2 + y_rel(t)^2 ≥ (a - d)^2 + (b - e)^2 - 2 sqrt{(a d)^2 + (b e)^2}But I'm not sure if that's a valid inequality.Alternatively, perhaps we can use the triangle inequality in reverse.Wait, |A - B| ≥ ||A| - |B||So, applying that to vectors in x-y plane:||A(t) - B(t)|| ≥ ||A(t)|| - ||B(t)||Where ||A(t)|| = sqrt{(a cos θ_A)^2 + (b sin θ_A)^2}Similarly, ||B(t)|| = sqrt{(d cos θ_B)^2 + (e sin θ_B)^2}Therefore,x_rel(t)^2 + y_rel(t)^2 ≥ (||A(t)|| - ||B(t)||)^2So, to ensure that x_rel(t)^2 + y_rel(t)^2 ≥ D^2, it's sufficient that (||A(t)|| - ||B(t)||)^2 ≥ D^2, which implies that ||A(t)|| - ||B(t)|| ≥ D or ||A(t)|| - ||B(t)|| ≤ -D.But since ||A(t)|| and ||B(t)|| are non-negative, the second inequality would imply that ||A(t)|| ≤ ||B(t)|| - D, which is more restrictive.But this approach might not be useful because it requires that the difference in magnitudes is always ≥ D, which is a stronger condition than needed.Alternatively, perhaps we can consider that the minimal distance between the two ellipses is at least D, which would require that the ellipses do not intersect and are separated by at least D.But determining the minimal distance between two ellipses is a complex problem, and I don't think there's a simple formula for it.Therefore, perhaps the answer is that if c ≠ f, then the minimal distance is at t=0, so we need:[a cos φ - d cos β]^2 + [b sin φ - e sin β]^2 ≥ D^2And if c = f, then we need the minimal distance in the x-y plane to be ≥ D, which requires that the minimal value of x_rel(t)^2 + y_rel(t)^2 is ≥ D^2. But since solving for t is difficult, we might need to impose additional conditions on the parameters to ensure this.Alternatively, perhaps we can consider that the minimal distance in the x-y plane is at least D, which would require that the two ellipses are separated by at least D. But without a specific method to compute this, it's hard to derive explicit conditions.Therefore, perhaps the answer is:If c ≠ f, then [a cos φ - d cos β]^2 + [b sin φ - e sin β]^2 ≥ D^2If c = f, then the minimal distance in the x-y plane must be ≥ D, which requires that the minimal value of x_rel(t)^2 + y_rel(t)^2 is ≥ D^2. However, since solving for t is difficult, additional constraints on a, b, d, e, ω, α, φ, β are needed to ensure this.But the problem says "derive the conditions on the constants", so perhaps they expect a general condition.Alternatively, perhaps we can consider that the minimal distance is achieved when the relative velocity is zero, which is the condition we derived earlier. Therefore, to ensure that the minimal distance is ≥ D, we need that the minimal value of D(t) is ≥ D, which can be expressed as:The minimal value of sqrt{[a cos(ωt + φ) - d cos(αt + β)]² + [b sin(ωt + φ) - e sin(αt + β)]² + [(c - f) t]²} is ≥ DBut since we can't solve for t explicitly, perhaps we can't write a more specific condition.Alternatively, perhaps we can use the fact that the minimal distance occurs at t=0 if c ≠ f, so the condition is [a cos φ - d cos β]^2 + [b sin φ - e sin β]^2 ≥ D^2And if c = f, then we need the minimal distance in the x-y plane to be ≥ D, which would require that the minimal value of [a cos(ωt + φ) - d cos(αt + β)]² + [b sin(ωt + φ) - e sin(αt + β)]² is ≥ D^2But again, without solving for t, we can't write an explicit condition.Therefore, perhaps the answer is:To ensure that the minimum distance between the two starships is at least D units at all times, the following conditions must be satisfied:1. If c ≠ f, then [a cos φ - d cos β]^2 + [b sin φ - e sin β]^2 ≥ D^22. If c = f, then the minimal value of [a cos(ωt + φ) - d cos(αt + β)]² + [b sin(ωt + φ) - e sin(αt + β)]² over t must be ≥ D^2But since the second condition is difficult to express without solving for t, perhaps we can leave it as that.Alternatively, perhaps we can consider that the minimal distance in the x-y plane is given by the minimal value of the expression, which can be found by solving the equation we derived earlier, but since it's transcendental, we can't write it explicitly.Therefore, the conditions are:- If c ≠ f, then [a cos φ - d cos β]^2 + [b sin φ - e sin β]^2 ≥ D^2- If c = f, then the minimal distance in the x-y plane, which is the minimal value of [a cos(ωt + φ) - d cos(αt + β)]² + [b sin(ωt + φ) - e sin(αt + β)]² over t, must be ≥ D^2But since the problem asks to derive the conditions, perhaps we can write them as:For all t, [a cos(ωt + φ) - d cos(αt + β)]² + [b sin(ωt + φ) - e sin(αt + β)]² + [(c - f) t]² ≥ D²But that's just restating the requirement.Alternatively, perhaps we can consider that the minimal distance is achieved when the derivative is zero, so we can write that the minimal distance squared is equal to the expression evaluated at the t that satisfies the derivative condition, and set that to be ≥ D².But since we can't solve for t, perhaps we can't write it explicitly.Therefore, the answer is:The time t at which the two starships are closest is given implicitly by the equation:[a cos(ωt + φ) - d cos(αt + β)] * [-a ω sin(ωt + φ) + d α sin(αt + β)] + [b sin(ωt + φ) - e sin(αt + β)] * [b ω cos(ωt + φ) - e α cos(αt + β)] + (c - f)^2 t = 0And to ensure the minimum distance is at least D, the following must hold:If c ≠ f, then [a cos φ - d cos β]^2 + [b sin φ - e sin β]^2 ≥ D²If c = f, then the minimal value of [a cos(ωt + φ) - d cos(αt + β)]² + [b sin(ωt + φ) - e sin(αt + β)]² over t must be ≥ D²But since the minimal value when c = f is difficult to express without solving for t, perhaps we can't write a more specific condition.Therefore, the conditions are:1. If c ≠ f, then [a cos φ - d cos β]^2 + [b sin φ - e sin β]^2 ≥ D²2. If c = f, then the minimal distance between the two ellipses in the x-y plane must be ≥ D, which requires that the minimal value of [a cos(ωt + φ) - d cos(αt + β)]² + [b sin(ωt + φ) - e sin(αt + β)]² is ≥ D²But without solving for t, we can't write an explicit condition for the second case.Therefore, perhaps the answer is:To ensure that the minimum distance between the two starships is at least D units at all times, the following conditions must be satisfied:1. If c ≠ f, then [a cos φ - d cos β]^2 + [b sin φ - e sin β]^2 ≥ D²2. If c = f, then the minimal distance between the two ellipses in the x-y plane must be ≥ D, which can be ensured by appropriate selection of a, b, d, e, ω, α, φ, β such that the ellipses do not come closer than D.But since the problem asks to derive the conditions on the constants, perhaps we can write it as:For all t, [a cos(ωt + φ) - d cos(αt + β)]² + [b sin(ωt + φ) - e sin(αt + β)]² + [(c - f) t]² ≥ D²But that's just restating the requirement.Alternatively, perhaps we can write that the minimal distance is given by the expression evaluated at the t that satisfies the derivative condition, and set that to be ≥ D.But since we can't solve for t, perhaps we can't write it explicitly.Therefore, the answer is:1. The time t at which the two starships are closest is given implicitly by the equation:[a cos(ωt + φ) - d cos(αt + β)] * [-a ω sin(ωt + φ) + d α sin(αt + β)] + [b sin(ωt + φ) - e sin(αt + β)] * [b ω cos(ωt + φ) - e α cos(αt + β)] + (c - f)^2 t = 02. To ensure the minimum distance is at least D, the following must hold:If c ≠ f, then [a cos φ - d cos β]^2 + [b sin φ - e sin β]^2 ≥ D²If c = f, then the minimal value of [a cos(ωt + φ) - d cos(αt + β)]² + [b sin(ωt + φ) - e sin(αt + β)]² over t must be ≥ D²But since the minimal value when c = f is difficult to express without solving for t, perhaps we can't write a more specific condition.Therefore, the conditions are as above.</think>

🧠:<think>Alright, so I have this differential equation problem to solve. It's about modeling the number of views on a French learning video over time. The equation given is:[ frac{dV}{dt} = kVleft(1 - frac{V}{L}right) ]Hmm, okay. This looks familiar. I think it's a logistic growth model. Yeah, logistic equation is used to model population growth with a carrying capacity, which in this case is the maximum number of views, L. So, the growth rate is proportional to both the current number of views and the remaining capacity.The first part asks for the general solution V(t) given the initial condition V(0) = V₀. So, I need to solve this differential equation. Let me recall how to solve logistic equations. It's a separable equation, right? So, I can rewrite it as:[ frac{dV}{dt} = kVleft(1 - frac{V}{L}right) ]Which can be rewritten as:[ frac{dV}{Vleft(1 - frac{V}{L}right)} = k dt ]Now, I need to integrate both sides. The left side looks a bit tricky, but I remember partial fractions can help here. Let me set up the integral:[ int frac{1}{Vleft(1 - frac{V}{L}right)} dV = int k dt ]Let me simplify the integrand on the left. Let me rewrite the denominator:[ Vleft(1 - frac{V}{L}right) = V cdot left(frac{L - V}{L}right) = frac{V(L - V)}{L} ]So, the integral becomes:[ int frac{L}{V(L - V)} dV = int k dt ]That simplifies to:[ L int left( frac{1}{V} + frac{1}{L - V} right) dV = int k dt ]Wait, how did I get that? Let me check. Partial fractions decomposition for 1/(V(L - V)) is A/V + B/(L - V). Let me solve for A and B.So, 1 = A(L - V) + B VLet me plug in V = 0: 1 = A(L - 0) + B(0) => A = 1/LSimilarly, plug in V = L: 1 = A(0) + B L => B = 1/LSo, indeed, 1/(V(L - V)) = (1/L)(1/V + 1/(L - V))Therefore, the integral becomes:[ L cdot frac{1}{L} int left( frac{1}{V} + frac{1}{L - V} right) dV = int k dt ]Simplify L*(1/L) to 1:[ int left( frac{1}{V} + frac{1}{L - V} right) dV = int k dt ]Now, integrate term by term:Integral of 1/V dV is ln|V|, and integral of 1/(L - V) dV is -ln|L - V|.So, putting it together:[ ln|V| - ln|L - V| = kt + C ]Combine the logs:[ lnleft|frac{V}{L - V}right| = kt + C ]Exponentiate both sides to eliminate the natural log:[ frac{V}{L - V} = e^{kt + C} = e^{kt} cdot e^C ]Let me denote e^C as another constant, say, C₁.So,[ frac{V}{L - V} = C₁ e^{kt} ]Now, solve for V. Multiply both sides by (L - V):[ V = C₁ e^{kt} (L - V) ]Expand the right side:[ V = C₁ L e^{kt} - C₁ V e^{kt} ]Bring all terms with V to the left:[ V + C₁ V e^{kt} = C₁ L e^{kt} ]Factor V:[ V (1 + C₁ e^{kt}) = C₁ L e^{kt} ]Solve for V:[ V = frac{C₁ L e^{kt}}{1 + C₁ e^{kt}} ]Now, apply the initial condition V(0) = V₀. Let's plug t = 0:[ V₀ = frac{C₁ L e^{0}}{1 + C₁ e^{0}} = frac{C₁ L}{1 + C₁} ]Solve for C₁:Multiply both sides by (1 + C₁):[ V₀ (1 + C₁) = C₁ L ]Expand:[ V₀ + V₀ C₁ = C₁ L ]Bring terms with C₁ to one side:[ V₀ = C₁ L - V₀ C₁ ]Factor C₁:[ V₀ = C₁ (L - V₀) ]Therefore,[ C₁ = frac{V₀}{L - V₀} ]So, substitute back into the expression for V(t):[ V(t) = frac{left( frac{V₀}{L - V₀} right) L e^{kt}}{1 + left( frac{V₀}{L - V₀} right) e^{kt}} ]Simplify numerator and denominator:Numerator: (V₀ L / (L - V₀)) e^{kt}Denominator: 1 + (V₀ / (L - V₀)) e^{kt} = (L - V₀ + V₀ e^{kt}) / (L - V₀)So, V(t) becomes:[ V(t) = frac{V₀ L e^{kt} / (L - V₀)}{(L - V₀ + V₀ e^{kt}) / (L - V₀)} ]The (L - V₀) cancels out:[ V(t) = frac{V₀ L e^{kt}}{L - V₀ + V₀ e^{kt}} ]We can factor out V₀ e^{kt} in the denominator:Wait, actually, let me write it as:[ V(t) = frac{V₀ L e^{kt}}{L - V₀ + V₀ e^{kt}} ]Alternatively, factor L in the denominator:Wait, maybe factor e^{kt} in the denominator:[ V(t) = frac{V₀ L e^{kt}}{L - V₀ + V₀ e^{kt}} = frac{V₀ L e^{kt}}{L - V₀ + V₀ e^{kt}} ]Alternatively, factor out V₀:Wait, perhaps it's better to leave it as is. So, that's the general solution.Alternatively, we can write it as:[ V(t) = frac{L}{1 + left( frac{L - V₀}{V₀} right) e^{-kt}} ]Let me check that. Let's see:Starting from:[ V(t) = frac{V₀ L e^{kt}}{L - V₀ + V₀ e^{kt}} ]Divide numerator and denominator by V₀ e^{kt}:Numerator: (V₀ L e^{kt}) / (V₀ e^{kt}) = LDenominator: (L - V₀ + V₀ e^{kt}) / (V₀ e^{kt}) = (L - V₀)/(V₀ e^{kt}) + 1So,[ V(t) = frac{L}{1 + frac{L - V₀}{V₀} e^{-kt}} ]Yes, that's another way to write it, which is the standard form of the logistic function.So, either form is acceptable, but perhaps the second form is more elegant.So, the general solution is:[ V(t) = frac{L}{1 + left( frac{L - V₀}{V₀} right) e^{-kt}} ]Alright, that's part 1 done.Now, part 2: The creator wants to reach 80% of L, which is 0.8 L, as quickly as possible. So, we need to find the time t when V(t) = 0.8 L.So, set V(t) = 0.8 L and solve for t.Using the general solution:[ 0.8 L = frac{L}{1 + left( frac{L - V₀}{V₀} right) e^{-kt}} ]Divide both sides by L:[ 0.8 = frac{1}{1 + left( frac{L - V₀}{V₀} right) e^{-kt}} ]Take reciprocal of both sides:[ frac{1}{0.8} = 1 + left( frac{L - V₀}{V₀} right) e^{-kt} ]Simplify 1/0.8 = 1.25:[ 1.25 = 1 + left( frac{L - V₀}{V₀} right) e^{-kt} ]Subtract 1 from both sides:[ 0.25 = left( frac{L - V₀}{V₀} right) e^{-kt} ]Solve for e^{-kt}:[ e^{-kt} = frac{0.25 V₀}{L - V₀} ]Take natural logarithm of both sides:[ -kt = lnleft( frac{0.25 V₀}{L - V₀} right) ]Multiply both sides by -1:[ kt = -lnleft( frac{0.25 V₀}{L - V₀} right) = lnleft( frac{L - V₀}{0.25 V₀} right) ]Therefore,[ t = frac{1}{k} lnleft( frac{L - V₀}{0.25 V₀} right) ]Simplify the fraction inside the log:[ frac{L - V₀}{0.25 V₀} = frac{4(L - V₀)}{V₀} ]So,[ t = frac{1}{k} lnleft( frac{4(L - V₀)}{V₀} right) ]Alternatively, we can write it as:[ t = frac{1}{k} lnleft(4 cdot frac{L - V₀}{V₀}right) ]Which is the same as:[ t = frac{1}{k} left[ ln 4 + lnleft( frac{L - V₀}{V₀} right) right] ]But perhaps the first expression is simpler.So, that's the time it takes to reach 80% of the maximum views.Let me recap:1. Solved the logistic differential equation, found the general solution.2. Plugged in V(t) = 0.8 L, solved for t, got t = (1/k) ln(4*(L - V₀)/V₀)So, that's the answer.Final Answer1. The general solution is (boxed{V(t) = dfrac{L}{1 + left( dfrac{L - V_0}{V_0} right) e^{-kt}}}).2. The time to reach 80% of the maximum views is (boxed{t = dfrac{1}{k} lnleft( dfrac{4(L - V_0)}{V_0} right)}).</think>

🧠:<think>Alright, so I've got these two problems about Brigg Britannia F.C.'s season. Let me try to figure them out step by step. Starting with the first problem: They played 30 home matches with a total attendance of 600,000. The attendance per match is normally distributed with a mean of 20,000 and a variance of 64,000,000. I need to find the probability that a randomly chosen match had attendance between 18,000 and 22,000.Hmm, okay. So, since the attendance follows a normal distribution, I can use the properties of the normal curve to find this probability. The mean is 20,000, which is the center of the distribution, and the variance is 64,000,000. First, I should find the standard deviation because that's needed for the z-score calculations. The variance is the square of the standard deviation, so taking the square root of 64,000,000 should give me the standard deviation. Let me calculate that:Standard deviation (σ) = sqrt(64,000,000) = 8,000.Okay, so σ is 8,000. Now, I need to find the probability that attendance is between 18,000 and 22,000. Since the mean is 20,000, 18,000 is 2,000 below the mean, and 22,000 is 2,000 above. To find this probability, I can convert these attendance numbers into z-scores. The z-score formula is:z = (X - μ) / σWhere X is the attendance, μ is the mean, and σ is the standard deviation.Calculating the z-scores for 18,000 and 22,000:For 18,000:z1 = (18,000 - 20,000) / 8,000 = (-2,000) / 8,000 = -0.25For 22,000:z2 = (22,000 - 20,000) / 8,000 = 2,000 / 8,000 = 0.25So, the z-scores are -0.25 and 0.25. Now, I need to find the area under the standard normal curve between these two z-scores. This will give me the probability that attendance falls between 18,000 and 22,000.I remember that the total area under the normal curve is 1, and the curve is symmetric around the mean. The area between -0.25 and 0.25 can be found by looking up the cumulative probabilities for these z-scores and subtracting them.Looking up z = 0.25 in the standard normal distribution table, the cumulative probability is approximately 0.5987. Similarly, for z = -0.25, the cumulative probability is approximately 0.3015.So, the area between -0.25 and 0.25 is 0.5987 - 0.3015 = 0.2972.Therefore, the probability that a randomly chosen match had attendance between 18,000 and 22,000 is approximately 29.72%.Wait, let me double-check that. The z-scores are correct, right? 18,000 is 2,000 below the mean, which is 20,000, and 2,000 divided by 8,000 is indeed 0.25. So, the z-scores are correct. And the cumulative probabilities for 0.25 and -0.25 are 0.5987 and 0.3015, so subtracting gives 0.2972. Yeah, that seems right.Okay, moving on to the second problem. The team won 60% of their home matches, so that's 18 out of 30. They want the probability that they won exactly 18 matches out of 30, assuming each match is independent.This sounds like a binomial probability problem. The binomial distribution gives the probability of having exactly k successes in n independent trials, with the probability of success on a single trial being p.The formula for the binomial probability is:P(k) = C(n, k) * p^k * (1-p)^(n-k)Where C(n, k) is the combination of n things taken k at a time.In this case, n = 30, k = 18, and p = 0.6.So, I need to calculate C(30, 18) * (0.6)^18 * (0.4)^12.First, let me compute the combination C(30, 18). That's the number of ways to choose 18 successes out of 30 trials.C(30, 18) = 30! / (18! * (30 - 18)!) = 30! / (18! * 12!)Calculating factorials can be cumbersome, but I remember that C(n, k) is equal to C(n, n - k), so C(30, 18) is the same as C(30, 12). Maybe that's easier to compute, but not sure.Alternatively, I can use the formula:C(n, k) = n! / (k! (n - k)!) But computing 30! directly is impractical. Instead, I can use the multiplicative formula for combinations:C(n, k) = (n * (n - 1) * ... * (n - k + 1)) / (k * (k - 1) * ... * 1)So, for C(30, 18):C(30, 18) = (30 * 29 * 28 * ... * 13) / (18 * 17 * ... * 1)But that's still a lot of multiplication. Maybe I can compute it step by step or use logarithms to simplify, but perhaps it's easier to use a calculator or look up the value.Wait, I might remember that C(30, 18) is 86,493,225. Let me verify that.Alternatively, I can compute it as:C(30, 18) = C(30, 12) = 86,493,225. Yes, that's correct.So, C(30, 18) = 86,493,225.Now, compute (0.6)^18 and (0.4)^12.Calculating (0.6)^18:0.6^1 = 0.60.6^2 = 0.360.6^3 = 0.2160.6^4 = 0.12960.6^5 = 0.077760.6^6 = 0.0466560.6^7 = 0.02799360.6^8 = 0.016796160.6^9 = 0.0100776960.6^10 = 0.00604661760.6^11 = 0.003627970560.6^12 = 0.0021767823360.6^13 = 0.00130606940160.6^14 = 0.000783641640960.6^15 = 0.0004701849845760.6^16 = 0.0002821109907460.6^17 = 0.0001692665944480.6^18 = 0.000101559956669So, approximately 0.00010156.Similarly, (0.4)^12:0.4^1 = 0.40.4^2 = 0.160.4^3 = 0.0640.4^4 = 0.02560.4^5 = 0.010240.4^6 = 0.0040960.4^7 = 0.00163840.4^8 = 0.000655360.4^9 = 0.0002621440.4^10 = 0.00010485760.4^11 = 0.000041943040.4^12 = 0.000016777216So, approximately 0.000016777.Now, multiply all these together:P(18) = C(30, 18) * (0.6)^18 * (0.4)^12 ≈ 86,493,225 * 0.00010156 * 0.000016777.Let me compute this step by step.First, multiply 86,493,225 by 0.00010156:86,493,225 * 0.00010156 ≈ Let's compute 86,493,225 * 0.0001 = 8,649.3225Then, 86,493,225 * 0.00000156 ≈ 86,493,225 * 0.000001 = 86.493225, and 86,493,225 * 0.00000056 ≈ 48.4362056So, total is approximately 8,649.3225 + 86.493225 + 48.4362056 ≈ 8,784.2519.Wait, that seems a bit off. Maybe a better way is to compute 86,493,225 * 0.00010156.Let me write it as 86,493,225 * 1.0156e-4.Compute 86,493,225 * 1.0156e-4:First, 86,493,225 * 1e-4 = 8,649.3225Then, 86,493,225 * 0.0156e-4 = 86,493,225 * 1.56e-6 ≈ 134.956So, total is approximately 8,649.3225 + 134.956 ≈ 8,784.2785.So, approximately 8,784.28.Now, multiply this by 0.000016777:8,784.28 * 0.000016777 ≈ Let's compute 8,784.28 * 1.6777e-5.First, 8,784.28 * 1e-5 = 0.0878428Then, 8,784.28 * 0.6777e-5 ≈ 8,784.28 * 6.777e-6 ≈ 0.05936So, total is approximately 0.0878428 + 0.05936 ≈ 0.1472.Therefore, the probability is approximately 0.1472, or 14.72%.Wait, that seems a bit high. Let me check the calculations again.Alternatively, maybe I can use logarithms or exponentials to compute this more accurately, but that might be time-consuming.Alternatively, I can use the fact that the binomial distribution can be approximated by the normal distribution when n is large, but since n=30 and p=0.6, it's not extremely large, but maybe it's still a decent approximation.But since the question asks for the exact probability, I should stick with the binomial formula.Alternatively, maybe I made an error in the multiplication steps.Let me try a different approach. Let's compute (0.6)^18 * (0.4)^12 first.(0.6)^18 ≈ 0.00010156(0.4)^12 ≈ 0.000016777Multiplying these together: 0.00010156 * 0.000016777 ≈ 1.703e-9Now, multiply by C(30,18) = 86,493,225:86,493,225 * 1.703e-9 ≈ Let's compute 86,493,225 * 1.703e-9.First, 86,493,225 * 1e-9 = 0.086493225Then, 86,493,225 * 0.703e-9 ≈ 86,493,225 * 7.03e-10 ≈ 0.06088So, total is approximately 0.086493225 + 0.06088 ≈ 0.14737.So, approximately 0.1474, or 14.74%.That's consistent with my previous calculation. So, the probability is approximately 14.74%.Wait, but let me check if C(30,18) is indeed 86,493,225. Let me compute it step by step.C(30,18) = 30! / (18! * 12!) But computing factorials is tedious, but I can use the multiplicative formula:C(30,18) = (30 * 29 * 28 * 27 * 26 * 25 * 24 * 23 * 22 * 21 * 20 * 19) / (12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1)Let me compute numerator and denominator separately.Numerator:30 * 29 = 870870 * 28 = 24,36024,360 * 27 = 657,720657,720 * 26 = 17,100,72017,100,720 * 25 = 427,518,000427,518,000 * 24 = 10,260,432,00010,260,432,000 * 23 = 235,990, (Wait, let me compute step by step)Wait, maybe it's better to compute in smaller steps:30 * 29 = 870870 * 28 = 24,36024,360 * 27 = 657,720657,720 * 26 = 17,100,72017,100,720 * 25 = 427,518,000427,518,000 * 24 = 10,260,432,00010,260,432,000 * 23 = 235,990, (Wait, 10,260,432,000 * 20 = 205,208,640,000, plus 10,260,432,000 * 3 = 30,781,296,000, total 235,989,936,000)235,989,936,000 * 22 = 5,191,778,592,0005,191,778,592,000 * 21 = 109,027,350,432,000109,027,350,432,000 * 20 = 2,180,547,008,640,0002,180,547,008,640,000 * 19 = 41,430,393,164,160,000So, numerator is 41,430,393,164,160,000.Denominator:12 * 11 = 132132 * 10 = 1,3201,320 * 9 = 11,88011,880 * 8 = 95,04095,040 * 7 = 665,280665,280 * 6 = 3,991,6803,991,680 * 5 = 19,958,40019,958,400 * 4 = 79,833,60079,833,600 * 3 = 239,500,800239,500,800 * 2 = 479,001,600479,001,600 * 1 = 479,001,600So, denominator is 479,001,600.Now, C(30,18) = numerator / denominator = 41,430,393,164,160,000 / 479,001,600.Let me compute this division.First, simplify the numbers:41,430,393,164,160,000 ÷ 479,001,600.Let me write both numbers in scientific notation:Numerator: 4.143039316416e16Denominator: 4.790016e8So, dividing them: (4.143039316416e16) / (4.790016e8) ≈ (4.143039316416 / 4.790016) * 10^(16-8) ≈ (0.865) * 10^8 ≈ 8.65e7 ≈ 86,500,000.Wait, but earlier I thought it was 86,493,225. So, that's consistent. So, C(30,18) ≈ 86,493,225.So, the calculation seems correct.Therefore, the probability is approximately 0.1474, or 14.74%.Wait, but let me check if that makes sense. The expected number of wins is 30 * 0.6 = 18. So, the probability of exactly 18 wins should be the highest, but it's still around 14.7%, which seems reasonable.Alternatively, I can use the normal approximation to the binomial distribution to estimate this probability.The mean μ = n*p = 30*0.6 = 18.The variance σ² = n*p*(1-p) = 30*0.6*0.4 = 7.2So, σ = sqrt(7.2) ≈ 2.6833Using the continuity correction, since we're approximating a discrete distribution with a continuous one, we can consider the interval from 17.5 to 18.5.Compute z-scores:z1 = (17.5 - 18) / 2.6833 ≈ -0.186z2 = (18.5 - 18) / 2.6833 ≈ 0.186Looking up these z-scores in the standard normal table:For z = 0.186, the cumulative probability is approximately 0.5735.For z = -0.186, the cumulative probability is approximately 0.4265.So, the area between them is 0.5735 - 0.4265 = 0.147.So, the normal approximation gives approximately 14.7%, which matches our exact calculation of approximately 14.74%. So, that seems consistent.Therefore, the probability that they won exactly 18 home matches is approximately 14.74%.So, summarizing:1. The probability of attendance between 18,000 and 22,000 is approximately 29.72%.2. The probability of winning exactly 18 matches is approximately 14.74%.I think that's it. I double-checked the calculations, especially the combination part, and it seems correct. The normal approximation also supports the binomial result, so I'm confident in these answers.</think>