🧠:<think>Okay, so I have this problem about modeling the tension in a romance novel using a sinusoidal function. The function is given by ( T(t) = A sin(Bt + C) + D ). I need to find the constants ( A, B, C, ) and ( D ). Then, in part 2, I have to figure out when the tension falls below 20. Let me start with part 1.First, the problem says the tension follows a sinusoidal pattern. I remember that sinusoidal functions have a certain amplitude, period, phase shift, and vertical shift. The general form is ( A sin(Bt + C) + D ), where:- ( A ) is the amplitude, which is half the difference between the maximum and minimum values.- ( B ) affects the period; the period is ( frac{2pi}{B} ).- ( C ) is the phase shift, which shifts the graph left or right.- ( D ) is the vertical shift, which moves the graph up or down.Given that the average tension over one cycle is 50, that should be the vertical shift ( D ), right? Because the average of a sinusoidal function is its vertical shift. So, ( D = 50 ).Next, the amplitude is given as 30. The amplitude ( A ) is the maximum deviation from the average. So, ( A = 30 ). That means the tension ranges from ( 50 - 30 = 20 ) to ( 50 + 30 = 80 ). Wait, but the problem mentions the tension never falls below 20, which matches the minimum here. So, that makes sense.Now, the tension reaches its peak at chapter 5 and chapter 15. So, the peaks are at ( t = 5 ) and ( t = 15 ). The minimum tension occurs at chapter 10. So, the trough is at ( t = 10 ).Let me visualize this. The function peaks at 5 and 15, with a trough at 10. So, the distance between two peaks is 10 chapters, which would be the period. So, the period ( P ) is 10. Since the period of a sine function is ( frac{2pi}{B} ), we can solve for ( B ):( P = frac{2pi}{B} Rightarrow 10 = frac{2pi}{B} Rightarrow B = frac{2pi}{10} = frac{pi}{5} ).So, ( B = frac{pi}{5} ).Now, I need to find ( C ), the phase shift. Let's think about the standard sine function ( sin(Bt + C) ). The standard sine function peaks at ( frac{pi}{2} ), troughs at ( frac{3pi}{2} ), and so on. So, in our case, the first peak is at ( t = 5 ). Let's set up the equation for the peak.The function ( T(t) = A sin(Bt + C) + D ) reaches its maximum when ( sin(Bt + C) = 1 ). So, at ( t = 5 ):( sinleft( frac{pi}{5} times 5 + C right) = 1 Rightarrow sin(pi + C) = 1 ).Wait, ( sin(pi + C) = 1 ). Hmm, but ( sin(pi + C) = -sin(C) ). So, ( -sin(C) = 1 Rightarrow sin(C) = -1 ). So, ( C = frac{3pi}{2} + 2pi k ), where ( k ) is an integer. But since we can adjust the phase shift by adding multiples of the period, let's choose the simplest one. Let me take ( C = frac{3pi}{2} ).Wait, but let me check if this works for the trough at ( t = 10 ). The trough is when ( sin(Bt + C) = -1 ). So, at ( t = 10 ):( sinleft( frac{pi}{5} times 10 + C right) = -1 Rightarrow sin(2pi + C) = -1 ).But ( sin(2pi + C) = sin(C) ). So, ( sin(C) = -1 ). Which is consistent with our previous result ( C = frac{3pi}{2} ).Wait, but if I plug ( C = frac{3pi}{2} ) into the function, let's see:( T(t) = 30 sinleft( frac{pi}{5} t + frac{3pi}{2} right) + 50 ).Let me check the peak at ( t = 5 ):( frac{pi}{5} times 5 = pi ), so ( pi + frac{3pi}{2} = frac{5pi}{2} ). ( sinleft( frac{5pi}{2} right) = 1 ). Correct.At ( t = 10 ):( frac{pi}{5} times 10 = 2pi ), so ( 2pi + frac{3pi}{2} = frac{7pi}{2} ). ( sinleft( frac{7pi}{2} right) = -1 ). Correct.At ( t = 15 ):( frac{pi}{5} times 15 = 3pi ), so ( 3pi + frac{3pi}{2} = frac{9pi}{2} ). ( sinleft( frac{9pi}{2} right) = 1 ). Correct.So, that seems to work. Therefore, ( C = frac{3pi}{2} ).Wait, but sometimes phase shifts are represented differently. Let me think if there's another way to represent this. Alternatively, we can write the function as ( sin(Bt + C) ), but sometimes it's written as ( sin(B(t - h)) ) where ( h ) is the phase shift. So, ( C = -Bh ). So, in our case, ( C = frac{3pi}{2} ), so ( h = -C/B = -frac{3pi}{2} / frac{pi}{5} = -frac{15}{2} ). So, the phase shift is ( -7.5 ), meaning the graph is shifted 7.5 chapters to the left. But since we're dealing with chapters, which are discrete, but the function is continuous, so it's okay.But maybe I can represent ( C ) in a different form. Let me see. Alternatively, since the function peaks at ( t = 5 ), which is the first peak, maybe we can write the function as ( sin(Bt + C) ) such that at ( t = 5 ), the argument is ( frac{pi}{2} ), which is where sine peaks.So, ( B times 5 + C = frac{pi}{2} ). We already have ( B = frac{pi}{5} ), so:( frac{pi}{5} times 5 + C = frac{pi}{2} Rightarrow pi + C = frac{pi}{2} Rightarrow C = -frac{pi}{2} ).Wait, that's different from what I had before. So, which one is correct?Wait, if I set ( C = -frac{pi}{2} ), then the function becomes ( T(t) = 30 sinleft( frac{pi}{5} t - frac{pi}{2} right) + 50 ).Let me check the peak at ( t = 5 ):( frac{pi}{5} times 5 - frac{pi}{2} = pi - frac{pi}{2} = frac{pi}{2} ). ( sinleft( frac{pi}{2} right) = 1 ). Correct.At ( t = 10 ):( frac{pi}{5} times 10 - frac{pi}{2} = 2pi - frac{pi}{2} = frac{3pi}{2} ). ( sinleft( frac{3pi}{2} right) = -1 ). Correct.At ( t = 15 ):( frac{pi}{5} times 15 - frac{pi}{2} = 3pi - frac{pi}{2} = frac{5pi}{2} ). ( sinleft( frac{5pi}{2} right) = 1 ). Correct.So, both representations are correct, but with different phase shifts. So, which one should I choose? It depends on how the function is defined. Since the problem doesn't specify any particular form, either is acceptable, but perhaps the second one is simpler because it results in a phase shift that's a multiple of ( pi/2 ), which is often preferred.Wait, but let me think again. When I initially set ( C = frac{3pi}{2} ), the function was ( 30 sinleft( frac{pi}{5} t + frac{3pi}{2} right) + 50 ). But when I set ( C = -frac{pi}{2} ), it's ( 30 sinleft( frac{pi}{5} t - frac{pi}{2} right) + 50 ). These are actually the same function because ( sin(theta + frac{3pi}{2}) = sin(theta - frac{pi}{2} + 2pi) = sin(theta - frac{pi}{2}) ). So, they are equivalent because sine has a period of ( 2pi ). So, both are correct, but ( C = -frac{pi}{2} ) is simpler.Therefore, I think it's better to take ( C = -frac{pi}{2} ).So, summarizing:- ( A = 30 )- ( B = frac{pi}{5} )- ( C = -frac{pi}{2} )- ( D = 50 )Let me write the function:( T(t) = 30 sinleft( frac{pi}{5} t - frac{pi}{2} right) + 50 ).Alternatively, using co-function identity, ( sin(theta - frac{pi}{2}) = -cos(theta) ). So, the function can also be written as:( T(t) = -30 cosleft( frac{pi}{5} t right) + 50 ).But maybe that's complicating things. The original form is fine.Now, moving on to part 2. The editor wants to ensure that the tension level never falls below 20. So, we need to find all intervals where ( T(t) < 20 ).Given ( T(t) = 30 sinleft( frac{pi}{5} t - frac{pi}{2} right) + 50 ), we set up the inequality:( 30 sinleft( frac{pi}{5} t - frac{pi}{2} right) + 50 < 20 ).Subtract 50 from both sides:( 30 sinleft( frac{pi}{5} t - frac{pi}{2} right) < -30 ).Divide both sides by 30:( sinleft( frac{pi}{5} t - frac{pi}{2} right) < -1 ).Wait, but the sine function only ranges between -1 and 1. So, ( sin(theta) < -1 ) is never true. That can't be right. Wait, did I make a mistake?Wait, let's go back. The minimum tension is 20, which is exactly the lower bound of the function. So, the tension reaches 20 at the troughs, but never goes below. So, the tension is always ( geq 20 ). Therefore, there are no intervals where ( T(t) < 20 ). So, the answer is that there are no such intervals.But wait, let me double-check my calculations.Starting from ( T(t) = 30 sinleft( frac{pi}{5} t - frac{pi}{2} right) + 50 ).Set ( T(t) < 20 ):( 30 sinleft( frac{pi}{5} t - frac{pi}{2} right) + 50 < 20 )Subtract 50:( 30 sinleft( frac{pi}{5} t - frac{pi}{2} right) < -30 )Divide by 30:( sinleft( frac{pi}{5} t - frac{pi}{2} right) < -1 )But since the sine function can't be less than -1, this inequality has no solution. Therefore, the tension never falls below 20. So, the answer is that there are no chapters where ( T(t) < 20 ).Wait, but the problem says "using the constants derived in part 1, determine all intervals of chapters ( t ) for which the tension ( T(t) ) is below this threshold." So, perhaps I made a mistake in part 1.Wait, in part 1, I concluded that the minimum tension is 20, which is exactly the threshold. So, the tension reaches 20 but doesn't go below. So, the tension is always ( geq 20 ). Therefore, there are no intervals where ( T(t) < 20 ).Alternatively, if the problem had a different threshold, say 25, then we would have intervals where ( T(t) < 25 ). But since 20 is the minimum, it's never below.So, the answer is that there are no such intervals.Wait, but let me think again. Maybe I made a mistake in the function. Let me plot the function or think about its behavior.The function is ( T(t) = 30 sinleft( frac{pi}{5} t - frac{pi}{2} right) + 50 ). Let's see, when ( t = 0 ):( T(0) = 30 sinleft( -frac{pi}{2} right) + 50 = 30(-1) + 50 = 20 ).At ( t = 5 ):( T(5) = 30 sinleft( pi - frac{pi}{2} right) + 50 = 30 sinleft( frac{pi}{2} right) + 50 = 30(1) + 50 = 80 ).At ( t = 10 ):( T(10) = 30 sinleft( 2pi - frac{pi}{2} right) + 50 = 30 sinleft( frac{3pi}{2} right) + 50 = 30(-1) + 50 = 20 ).At ( t = 15 ):( T(15) = 30 sinleft( 3pi - frac{pi}{2} right) + 50 = 30 sinleft( frac{5pi}{2} right) + 50 = 30(1) + 50 = 80 ).So, the function oscillates between 20 and 80, peaking at 80 at chapters 5, 15, etc., and troughing at 20 at chapters 10, 20, etc. So, the tension never goes below 20. Therefore, the answer is correct.So, in part 2, there are no intervals where ( T(t) < 20 ).Wait, but the problem says "using the constants derived in part 1, determine all intervals of chapters ( t ) for which the tension ( T(t) ) is below this threshold." So, maybe I should express it as "no solution" or "there are no such intervals."Alternatively, perhaps I made a mistake in part 1. Let me check again.In part 1, I found ( A = 30 ), ( D = 50 ), so the function ranges from 20 to 80. The peaks are at 80, troughs at 20. So, the tension never goes below 20. So, part 2 is correct.Therefore, the answer is that there are no chapters where the tension is below 20.Wait, but let me think again. Maybe the function is written differently. If I had written it as ( T(t) = -30 cosleft( frac{pi}{5} t right) + 50 ), then the minimum is 20, maximum 80, same as before. So, same conclusion.Alternatively, perhaps the function is written as ( T(t) = 30 sinleft( frac{pi}{5} t + frac{3pi}{2} right) + 50 ), which is equivalent. So, same result.Therefore, I think my conclusion is correct.So, to summarize:1. The constants are ( A = 30 ), ( B = frac{pi}{5} ), ( C = -frac{pi}{2} ), ( D = 50 ).2. There are no intervals where ( T(t) < 20 ).But wait, let me check if the function is indeed correct. Let me plug in ( t = 0 ):( T(0) = 30 sinleft( -frac{pi}{2} right) + 50 = -30 + 50 = 20 ). Correct.At ( t = 2.5 ):( T(2.5) = 30 sinleft( frac{pi}{5} times 2.5 - frac{pi}{2} right) + 50 = 30 sinleft( frac{pi}{2} - frac{pi}{2} right) + 50 = 30 sin(0) + 50 = 0 + 50 = 50 ). So, halfway between 0 and 5, the tension is average.At ( t = 5 ):( T(5) = 30 sinleft( pi - frac{pi}{2} right) + 50 = 30 sinleft( frac{pi}{2} right) + 50 = 30 + 50 = 80 ). Correct.At ( t = 7.5 ):( T(7.5) = 30 sinleft( frac{3pi}{2} - frac{pi}{2} right) + 50 = 30 sin(pi) + 50 = 0 + 50 = 50 ). Correct.At ( t = 10 ):( T(10) = 30 sinleft( 2pi - frac{pi}{2} right) + 50 = 30 sinleft( frac{3pi}{2} right) + 50 = -30 + 50 = 20 ). Correct.So, the function behaves as expected. Therefore, part 1 is correct, and part 2 is correct in that the tension never goes below 20.Therefore, the final answers are:1. ( A = 30 ), ( B = frac{pi}{5} ), ( C = -frac{pi}{2} ), ( D = 50 ).2. There are no intervals where ( T(t) < 20 ).</think>

🧠:<think>Alright, so I have two problems to solve here. Let me tackle them one by one. Starting with the first problem: Dr. Thorne's student needs to find the coefficients ( a ), ( b ), and ( c ) of the quadratic function ( G(t) = at^2 + bt + c ). The data points given are on day 1, 3, and 5 with grain amounts 200, 180, and 150 units respectively. Okay, so I know that a quadratic function has three coefficients, and we have three data points, which should be enough to set up a system of equations. Let me write those equations out.For day 1: ( G(1) = a(1)^2 + b(1) + c = a + b + c = 200 ).For day 3: ( G(3) = a(3)^2 + b(3) + c = 9a + 3b + c = 180 ).For day 5: ( G(5) = a(5)^2 + b(5) + c = 25a + 5b + c = 150 ).So now I have three equations:1. ( a + b + c = 200 ) 2. ( 9a + 3b + c = 180 ) 3. ( 25a + 5b + c = 150 )I need to solve this system for ( a ), ( b ), and ( c ). Let me subtract the first equation from the second to eliminate ( c ):Equation 2 - Equation 1: ( (9a + 3b + c) - (a + b + c) = 180 - 200 ) Simplify: ( 8a + 2b = -20 ) Divide both sides by 2: ( 4a + b = -10 ) Let me call this Equation 4.Similarly, subtract Equation 2 from Equation 3: ( (25a + 5b + c) - (9a + 3b + c) = 150 - 180 ) Simplify: ( 16a + 2b = -30 ) Divide both sides by 2: ( 8a + b = -15 ) Let me call this Equation 5.Now, I have two equations: 4. ( 4a + b = -10 ) 5. ( 8a + b = -15 )Subtract Equation 4 from Equation 5 to eliminate ( b ): ( (8a + b) - (4a + b) = -15 - (-10) ) Simplify: ( 4a = -5 ) So, ( a = -5/4 ) or ( -1.25 ).Now plug ( a = -1.25 ) into Equation 4: ( 4*(-1.25) + b = -10 ) Calculate: ( -5 + b = -10 ) So, ( b = -10 + 5 = -5 ).Now, go back to Equation 1 to find ( c ): ( a + b + c = 200 ) Substitute ( a = -1.25 ) and ( b = -5 ): ( -1.25 - 5 + c = 200 ) Simplify: ( -6.25 + c = 200 ) So, ( c = 200 + 6.25 = 206.25 ).Let me double-check these values with the original equations.For Equation 1: ( -1.25 + (-5) + 206.25 = 200 ). Yes, that's correct.For Equation 2: ( 9*(-1.25) + 3*(-5) + 206.25 = -11.25 -15 + 206.25 = 180 ). Correct.For Equation 3: ( 25*(-1.25) + 5*(-5) + 206.25 = -31.25 -25 + 206.25 = 150 ). Correct.Alright, so the coefficients are ( a = -1.25 ), ( b = -5 ), and ( c = 206.25 ).Moving on to the second problem: The grain distribution function is given by ( D(t) = k e^{-lambda t} ). We have two data points: on day 1, 100 units were distributed, and on day 4, 50 units were distributed.So, let's write the equations based on these points.For day 1: ( D(1) = k e^{-lambda *1} = k e^{-lambda} = 100 ).For day 4: ( D(4) = k e^{-lambda *4} = k e^{-4lambda} = 50 ).So, we have two equations:1. ( k e^{-lambda} = 100 ) 2. ( k e^{-4lambda} = 50 )I need to solve for ( k ) and ( lambda ). Let me divide Equation 2 by Equation 1 to eliminate ( k ):( frac{k e^{-4lambda}}{k e^{-lambda}} = frac{50}{100} ) Simplify: ( e^{-4lambda + lambda} = 0.5 ) Which is: ( e^{-3lambda} = 0.5 )Take the natural logarithm of both sides: ( -3lambda = ln(0.5) ) So, ( lambda = frac{ln(0.5)}{-3} )Calculate ( ln(0.5) ): ( ln(0.5) approx -0.6931 ) Thus, ( lambda = frac{-0.6931}{-3} approx 0.2310 )So, ( lambda approx 0.2310 ).Now, substitute ( lambda ) back into Equation 1 to find ( k ): ( k e^{-0.2310} = 100 ) Calculate ( e^{-0.2310} ): ( e^{-0.2310} approx 0.794 ) So, ( k * 0.794 = 100 ) Thus, ( k = 100 / 0.794 approx 125.93 )Let me verify this with Equation 2: ( D(4) = 125.93 e^{-0.2310*4} ) Calculate exponent: ( -0.2310 *4 = -0.924 ) ( e^{-0.924} approx 0.396 ) Multiply by 125.93: ( 125.93 * 0.396 ≈ 50 ). Perfect, that's correct.So, the constants are ( k approx 125.93 ) and ( lambda approx 0.2310 ).Wait, but maybe I should express ( k ) more precisely. Let me recalculate ( k ).From Equation 1: ( k = 100 / e^{-lambda} ). Since ( lambda = frac{ln(0.5)}{-3} ), then ( e^{-lambda} = e^{ln(0.5)/3} = (e^{ln(0.5)})^{1/3} = (0.5)^{1/3} approx 0.7937 ).So, ( k = 100 / 0.7937 ≈ 125.99 ). So, approximately 126.But let me see if I can express ( k ) and ( lambda ) in exact terms.From ( e^{-3lambda} = 0.5 ), so ( -3lambda = ln(0.5) ), so ( lambda = -ln(0.5)/3 = ln(2)/3 ), since ( ln(0.5) = -ln(2) ).So, ( lambda = ln(2)/3 ).Similarly, ( k = 100 / e^{-lambda} = 100 e^{lambda} = 100 e^{ln(2)/3} = 100 * 2^{1/3} ).Since ( e^{ln(2)/3} = 2^{1/3} ).So, ( k = 100 * 2^{1/3} ).Calculating ( 2^{1/3} ) is approximately 1.26, so ( k approx 126 ), which matches my earlier approximation.So, exact expressions are ( k = 100 cdot 2^{1/3} ) and ( lambda = frac{ln 2}{3} ).Alternatively, ( k ) can be written as ( 100 sqrt[3]{2} ).I think that's a better way to present it rather than the decimal approximation.So, summarizing:1. The quadratic function coefficients are ( a = -1.25 ), ( b = -5 ), and ( c = 206.25 ).2. The distribution function constants are ( k = 100 sqrt[3]{2} ) and ( lambda = frac{ln 2}{3} ).I should probably write these in LaTeX format as per the instructions.Final Answer1. The coefficients are ( a = boxed{-frac{5}{4}} ), ( b = boxed{-5} ), and ( c = boxed{frac{825}{4}} ).2. The constants are ( k = boxed{100 sqrt[3]{2}} ) and ( lambda = boxed{dfrac{ln 2}{3}} ).</think>

🧠:<think>Okay, so I have this problem about learning efficiency, E(t), and it's modeled by a differential equation. The equation is dE/dt = kE(M - E) - hE². Hmm, let me try to parse this.First, part 1 is asking for the equilibrium points and their stability. I remember that equilibrium points are where dE/dt = 0. So I need to set the right-hand side equal to zero and solve for E.So, let's write that equation:kE(M - E) - hE² = 0.Let me expand that:kE(M - E) = kME - kE².So the equation becomes:kME - kE² - hE² = 0.Combine like terms:kME - (k + h)E² = 0.Factor out E:E(kM - (k + h)E) = 0.So, the solutions are when E = 0 or when kM - (k + h)E = 0.Solving the second equation:kM = (k + h)ESo, E = (kM)/(k + h).Therefore, the equilibrium points are E = 0 and E = (kM)/(k + h).Now, to determine their stability, I need to look at the derivative of dE/dt with respect to E, evaluated at each equilibrium point. If the derivative is negative, it's a stable equilibrium; if positive, it's unstable.So, let's compute d/dE [dE/dt] = d/dE [kE(M - E) - hE²].First, expand the original equation:dE/dt = kME - kE² - hE² = kME - (k + h)E².Taking derivative with respect to E:d²E/dt² = kM - 2(k + h)E.Wait, actually, no. Wait, the derivative of dE/dt with respect to E is the Jacobian, which is the slope of the function at E. So, let me compute it correctly.Given f(E) = kE(M - E) - hE² = kME - (k + h)E².Then, f'(E) = kM - 2(k + h)E.So, at E = 0:f'(0) = kM - 0 = kM. Since k and M are positive constants, this is positive. Therefore, E = 0 is an unstable equilibrium.At E = (kM)/(k + h):f'(E) = kM - 2(k + h)*(kM)/(k + h) = kM - 2kM = -kM.Which is negative because k and M are positive. So, E = (kM)/(k + h) is a stable equilibrium.So, that's part 1 done.Moving on to part 2. We have specific values: k = 0.1, h = 0.01, M = 100. Initial condition E(0) = 20.We need to solve the differential equation and describe the long-term behavior.First, let me write down the differential equation with these values:dE/dt = 0.1*E*(100 - E) - 0.01*E².Simplify:0.1*E*(100 - E) = 10E - 0.1E².So, dE/dt = 10E - 0.1E² - 0.01E² = 10E - 0.11E².So, the equation is dE/dt = 10E - 0.11E².This is a logistic-type equation, but let me write it in standard form.dE/dt = E*(10 - 0.11E).So, it's a Bernoulli equation, and we can solve it using separation of variables.Let me write it as:dE / [E*(10 - 0.11E)] = dt.We can use partial fractions to integrate the left side.Let me set:1 / [E*(10 - 0.11E)] = A/E + B/(10 - 0.11E).Multiply both sides by E*(10 - 0.11E):1 = A*(10 - 0.11E) + B*E.Let me solve for A and B.Expanding:1 = 10A - 0.11A E + B E.Grouping terms:1 = 10A + (B - 0.11A) E.Since this must hold for all E, the coefficients of like terms must be equal.So,10A = 1 => A = 1/10 = 0.1.And,B - 0.11A = 0 => B = 0.11A = 0.11*0.1 = 0.011.Therefore, the partial fractions decomposition is:1 / [E*(10 - 0.11E)] = 0.1/E + 0.011/(10 - 0.11E).So, integrating both sides:∫ [0.1/E + 0.011/(10 - 0.11E)] dE = ∫ dt.Compute the integrals.First integral:0.1 ∫ (1/E) dE = 0.1 ln|E| + C.Second integral:0.011 ∫ [1/(10 - 0.11E)] dE.Let me make substitution u = 10 - 0.11E, then du/dE = -0.11 => dE = du / (-0.11).So, integral becomes:0.011 ∫ [1/u] * (du / (-0.11)) = 0.011 / (-0.11) ∫ (1/u) du = -0.1 ∫ (1/u) du = -0.1 ln|u| + C = -0.1 ln|10 - 0.11E| + C.Putting it all together:0.1 ln|E| - 0.1 ln|10 - 0.11E| = t + C.We can combine the logs:0.1 [ln|E| - ln|10 - 0.11E|] = t + C.Which is:0.1 ln|E / (10 - 0.11E)| = t + C.Multiply both sides by 10:ln|E / (10 - 0.11E)| = 10t + C'.Exponentiate both sides:E / (10 - 0.11E) = C'' e^{10t}.Where C'' = e^{C'}, which is just another constant.Let me write this as:E = (10 - 0.11E) C'' e^{10t}.Bring all E terms to one side:E + 0.11E C'' e^{10t} = 10 C'' e^{10t}.Factor E:E [1 + 0.11 C'' e^{10t}] = 10 C'' e^{10t}.Therefore,E = [10 C'' e^{10t}] / [1 + 0.11 C'' e^{10t}].Let me write this as:E(t) = [10 C'' e^{10t}] / [1 + 0.11 C'' e^{10t}].We can simplify this expression by letting K = C''. So,E(t) = (10 K e^{10t}) / (1 + 0.11 K e^{10t}).To find K, we use the initial condition E(0) = 20.So, plug t = 0:20 = (10 K e^{0}) / (1 + 0.11 K e^{0}) => 20 = (10 K) / (1 + 0.11 K).Multiply both sides by denominator:20 (1 + 0.11 K) = 10 K.Expand:20 + 2.2 K = 10 K.Subtract 2.2 K:20 = 7.8 K.Thus, K = 20 / 7.8 ≈ 2.5641.But let me compute it exactly:20 / 7.8 = 200 / 78 = 100 / 39 ≈ 2.5641.So, K = 100/39.Therefore, the solution is:E(t) = (10*(100/39) e^{10t}) / (1 + 0.11*(100/39) e^{10t}).Simplify numerator and denominator:Numerator: (1000/39) e^{10t}.Denominator: 1 + (11/100)*(100/39) e^{10t} = 1 + (11/39) e^{10t}.So,E(t) = (1000/39 e^{10t}) / (1 + (11/39) e^{10t}).We can factor out 1/39 in the denominator:E(t) = (1000/39 e^{10t}) / [ (39 + 11 e^{10t}) / 39 ] = (1000 e^{10t}) / (39 + 11 e^{10t}).Simplify numerator and denominator:Factor numerator and denominator:E(t) = (1000 e^{10t}) / (11 e^{10t} + 39).We can write this as:E(t) = (1000 / 11) * [ e^{10t} / (e^{10t} + 39/11) ].But maybe it's clearer to leave it as:E(t) = 1000 e^{10t} / (11 e^{10t} + 39).Now, to describe the long-term behavior as t approaches infinity.As t → ∞, e^{10t} dominates, so the denominator is approximately 11 e^{10t}, and numerator is 1000 e^{10t}.Thus, E(t) ≈ 1000 e^{10t} / (11 e^{10t}) = 1000 / 11 ≈ 90.909.But wait, let's compute 1000 / 11: 11*90 = 990, so 1000 - 990 = 10, so 90 + 10/11 ≈ 90.909.But wait, in part 1, we found the equilibrium points at E = 0 and E = (kM)/(k + h). Let me compute that with given k, h, M.Given k = 0.1, h = 0.01, M = 100.So, (kM)/(k + h) = (0.1*100)/(0.1 + 0.01) = 10 / 0.11 ≈ 90.909.Which matches our long-term behavior. So, as t → ∞, E(t) approaches approximately 90.909.Therefore, the learning efficiency asymptotically approaches 1000/11 ≈ 90.909.So, summarizing:1. The equilibrium points are E = 0 (unstable) and E = (kM)/(k + h) (stable).2. With the given constants and initial condition, the solution is E(t) = 1000 e^{10t} / (11 e^{10t} + 39), and as t approaches infinity, E(t) approaches 1000/11 ≈ 90.909.I think that's it. Let me just double-check my calculations.In part 1, solving for equilibrium points: correct, E=0 and E=(kM)/(k+h). Stability: correct, E=0 is unstable, the other is stable.In part 2, solving the differential equation: I used partial fractions, integrated, applied initial condition, solved for K, and simplified. The algebra seems correct. The limit as t→infty: yes, the exponential terms dominate, so it approaches 1000/11.Yes, I think that's solid.</think>

🧠:<think>Okay, so I have this problem about a pharmaceutical company looking into new packaging materials. They want to increase the shelf life and stability of their medications by using materials with certain properties, specifically focusing on permeability to oxygen. There are two sub-problems here, and I need to solve both. Let me take them one by one.Starting with Sub-problem 1: The permeability P is given by the function P(x, y, z) = k * e^(-αx) * y² * sin(βz). The constants are k = 0.8, α = 0.1, and β = π/180. I need to find P when x = 2 mm, y = 25°C, and z = 60%. Alright, let's break this down. First, I should plug in all the given values into the function. Let me write that out step by step.So, P = 0.8 * e^(-0.1 * 2) * (25)^2 * sin( (π/180) * 60 )Let me compute each part separately.First, compute the exponent part: -0.1 * 2 = -0.2. So, e^(-0.2). I remember that e^(-0.2) is approximately... hmm, e^0.2 is about 1.2214, so e^(-0.2) would be 1/1.2214 ≈ 0.8187. Let me double-check that with a calculator if I can. Wait, actually, I can recall that e^(-0.2) is approximately 0.8187. Yeah, that seems right.Next, compute y squared: 25 squared is 625. That's straightforward.Then, compute the argument inside the sine function: (π/180) * 60. Since π/180 converts degrees to radians, so 60 degrees in radians is π/3. So, sin(π/3) is equal to sqrt(3)/2, which is approximately 0.8660.Now, putting it all together:P = 0.8 * 0.8187 * 625 * 0.8660Let me compute this step by step.First, multiply 0.8 and 0.8187: 0.8 * 0.8187 ≈ 0.65496.Next, multiply that result by 625: 0.65496 * 625. Let's see, 0.65496 * 600 = 392.976, and 0.65496 * 25 = 16.374. Adding those together: 392.976 + 16.374 = 409.35.Then, multiply that by 0.8660: 409.35 * 0.8660. Hmm, let's compute that. 409.35 * 0.8 = 327.48, and 409.35 * 0.066 ≈ 27.0171. Adding those together: 327.48 + 27.0171 ≈ 354.4971.So, approximately 354.4971. Let me check if that makes sense. The permeability is quite high, but given that y is 25°C, which is squared, and z is 60%, which is a significant sine value, maybe that's correct. Let me just verify each step again.Wait, hold on, 0.8 * 0.8187 is indeed approximately 0.65496. Then, 0.65496 * 625: 625 is 5^4, so 625 * 0.65496. Let me compute 625 * 0.6 = 375, 625 * 0.05496 ≈ 625 * 0.05 = 31.25, and 625 * 0.00496 ≈ 3.1. So, 375 + 31.25 + 3.1 ≈ 409.35. That seems correct.Then, 409.35 * 0.8660: Let's compute 409.35 * 0.8 = 327.48, and 409.35 * 0.066 ≈ 27.0171. Adding them gives 354.4971. So, approximately 354.5. Hmm, that seems quite high for permeability, but maybe that's the case. Alternatively, perhaps I made a mistake in the order of operations or in interpreting the formula.Wait, let me check the original function again: P(x, y, z) = k * e^(-αx) * y² * sin(βz). So, it's all multiplied together. So, my computation seems correct. So, unless there's a unit issue or something, but the units are all given in mm, degrees Celsius, and percentage, so I think that's okay.So, I think the value is approximately 354.5. But let me compute it more accurately using a calculator step by step.Compute e^(-0.2): Let me use a calculator for better precision. e^(-0.2) is approximately 0.818730753.Then, 25 squared is 625.sin(60 degrees): sin(60°) is sqrt(3)/2 ≈ 0.8660254038.Now, compute 0.8 * 0.818730753: 0.8 * 0.818730753 = 0.6549846024.Multiply that by 625: 0.6549846024 * 625. Let's compute 0.6549846024 * 600 = 392.99076144, and 0.6549846024 * 25 = 16.37461506. Adding them together: 392.99076144 + 16.37461506 ≈ 409.3653765.Now, multiply by 0.8660254038: 409.3653765 * 0.8660254038. Let's compute this.First, 409.3653765 * 0.8 = 327.4923012.Then, 409.3653765 * 0.0660254038 ≈ Let's compute 409.3653765 * 0.06 = 24.56192259, and 409.3653765 * 0.0060254038 ≈ 2.4665.Adding those together: 24.56192259 + 2.4665 ≈ 27.02842259.Now, add that to 327.4923012: 327.4923012 + 27.02842259 ≈ 354.5207238.So, approximately 354.52. So, rounding to two decimal places, that's 354.52. But since the question didn't specify the number of decimal places, maybe we can leave it as is or round to a reasonable number.Alternatively, perhaps the question expects a certain number of decimal places. Let me check the given values: k is 0.8, which is one decimal place, α is 0.1, one decimal, β is π/180, which is exact. The inputs are x=2, y=25, z=60, all integers. So, maybe the answer should be given to two decimal places or something. Alternatively, perhaps it's better to write it as 354.52 or 354.5.But wait, let me think again. The function is P(x, y, z) = k * e^(-αx) * y² * sin(βz). So, all the operations are multiplicative, so the result is a product of these terms. So, 0.8 * e^(-0.2) * 625 * sin(60°). So, yeah, the calculation seems correct.Wait, but 354.5 seems quite high. Maybe the units are in some specific units? The problem didn't specify the units for permeability, just that it's a value. So, perhaps that's okay.Alternatively, maybe I made a mistake in interpreting the formula. Let me check again: P(x, y, z) = k * e^(-αx) * y² * sin(βz). So, yes, all multiplied together. So, 0.8 * e^(-0.2) * 625 * sin(60°). So, yeah, that's correct.Alternatively, maybe the formula is P(x, y, z) = k * e^(-αx) * y² * sin(βz). So, perhaps the order is correct. So, I think my computation is correct.So, for Sub-problem 1, the value of P is approximately 354.52.Wait, but let me check if I did the multiplication correctly. 0.8 * e^(-0.2) is 0.8 * 0.8187 ≈ 0.65496. Then, 0.65496 * 625 is 409.35, and then 409.35 * 0.8660 is approximately 354.52. Yeah, that seems correct.Okay, moving on to Sub-problem 2: The company needs to maintain an average permeability of 0.05 for optimal drug stability. Using the same function P(x, y, z), determine the necessary thickness x of the material if the average temperature y is 30°C and the average humidity z is 50%. The constants k, α, and β are the same as in Sub-problem 1.So, we need to find x such that P(x, 30, 50) = 0.05.Given P(x, y, z) = 0.8 * e^(-0.1x) * y² * sin(βz). So, plugging in y = 30, z = 50, and P = 0.05.So, 0.05 = 0.8 * e^(-0.1x) * (30)^2 * sin( (π/180)*50 )Let me compute each part step by step.First, compute y squared: 30 squared is 900.Next, compute the argument inside the sine function: (π/180)*50. That's 50 degrees in radians. So, sin(50°). Let me compute sin(50°). Using a calculator, sin(50°) is approximately 0.7660444431.So, now, plug these into the equation:0.05 = 0.8 * e^(-0.1x) * 900 * 0.7660444431Let me compute the constants first: 0.8 * 900 * 0.7660444431.Compute 0.8 * 900 first: 0.8 * 900 = 720.Then, 720 * 0.7660444431 ≈ Let's compute 720 * 0.7 = 504, 720 * 0.0660444431 ≈ 720 * 0.06 = 43.2, and 720 * 0.0060444431 ≈ 4.35. So, adding those together: 504 + 43.2 + 4.35 ≈ 551.55.Wait, let me compute it more accurately: 720 * 0.7660444431.Compute 720 * 0.7 = 504.720 * 0.06 = 43.2.720 * 0.0060444431 ≈ 720 * 0.006 = 4.32, and 720 * 0.0000444431 ≈ 0.032. So, total ≈ 4.32 + 0.032 ≈ 4.352.So, total is 504 + 43.2 + 4.352 ≈ 551.552.So, approximately 551.552.So, the equation becomes:0.05 = 551.552 * e^(-0.1x)Now, we need to solve for x.First, divide both sides by 551.552:e^(-0.1x) = 0.05 / 551.552 ≈ 0.00009067.So, e^(-0.1x) ≈ 0.00009067.Now, take the natural logarithm of both sides:ln(e^(-0.1x)) = ln(0.00009067)Simplify left side: -0.1x = ln(0.00009067)Compute ln(0.00009067). Let me calculate that.ln(0.0001) is approximately -9.2103, since e^(-9.2103) ≈ 0.0001.But 0.00009067 is slightly less than 0.0001, so the ln should be slightly less than -9.2103, meaning more negative.Let me compute ln(0.00009067):Using a calculator, ln(0.00009067) ≈ -9.396.Wait, let me check: e^(-9.396) ≈ e^(-9) * e^(-0.396) ≈ 0.0001234 * 0.673 ≈ 0.000083. Hmm, that's less than 0.00009067. So, maybe my approximation is off.Alternatively, perhaps I can use the exact value.Let me compute ln(0.00009067):We can write 0.00009067 as 9.067 * 10^(-5).So, ln(9.067 * 10^(-5)) = ln(9.067) + ln(10^(-5)).ln(9.067) ≈ 2.205, and ln(10^(-5)) = -5 * ln(10) ≈ -5 * 2.302585 ≈ -11.5129.So, total ≈ 2.205 - 11.5129 ≈ -9.3079.Wait, but earlier, I thought e^(-9.396) ≈ 0.000083, which is less than 0.00009067, so perhaps the exact value is around -9.3079.Wait, let me compute e^(-9.3079):e^(-9) ≈ 0.00012341, e^(-0.3079) ≈ 0.733. So, e^(-9.3079) ≈ 0.00012341 * 0.733 ≈ 0.0000900.Which is very close to 0.00009067. So, ln(0.00009067) ≈ -9.3079.So, approximately -9.3079.So, we have:-0.1x ≈ -9.3079Divide both sides by -0.1:x ≈ (-9.3079)/(-0.1) ≈ 93.079.So, x ≈ 93.079 mm.Wait, that seems quite thick. Is that reasonable? Let me check my steps again.We have P = 0.05 = 0.8 * e^(-0.1x) * 900 * sin(50°).Compute 0.8 * 900 = 720.720 * sin(50°) ≈ 720 * 0.7660 ≈ 551.52.So, 0.05 = 551.52 * e^(-0.1x).So, e^(-0.1x) = 0.05 / 551.52 ≈ 0.00009067.Taking natural log: ln(0.00009067) ≈ -9.3079.So, -0.1x = -9.3079 => x = 93.079 mm.Hmm, 93 mm is about 9.3 cm, which seems quite thick for packaging material. Maybe that's correct, given the exponential decay in the permeability function. Let me think: the permeability decreases exponentially with thickness, so to get a very low permeability, you need a significant thickness.Alternatively, perhaps I made a mistake in the calculation. Let me verify the steps again.Compute P = 0.8 * e^(-0.1x) * y² * sin(βz).Given y = 30, z = 50, P = 0.05.So, 0.05 = 0.8 * e^(-0.1x) * 30² * sin(50°).Compute 30² = 900.Compute sin(50°) ≈ 0.7660.So, 0.8 * 900 = 720.720 * 0.7660 ≈ 551.52.So, 0.05 = 551.52 * e^(-0.1x).So, e^(-0.1x) = 0.05 / 551.52 ≈ 0.00009067.Taking natural log: ln(0.00009067) ≈ -9.3079.So, -0.1x = -9.3079 => x = 93.079 mm.Yes, that seems correct. So, the necessary thickness is approximately 93.08 mm.Wait, but let me check if I can write it more accurately. Let me compute ln(0.00009067) more precisely.Using a calculator, ln(0.00009067) ≈ -9.3079.So, x ≈ 93.079 mm.Rounding to a reasonable number of decimal places, maybe two: 93.08 mm.Alternatively, if the question expects an integer, maybe 93 mm.But let me see: 93.079 is approximately 93.08, so I think 93.08 mm is acceptable.Alternatively, perhaps I can write it as 93.1 mm.But let me check the exact calculation again.Compute 0.05 / 551.52:0.05 / 551.52 ≈ 0.00009067.Compute ln(0.00009067):Using a calculator, ln(0.00009067) ≈ -9.3079.So, x ≈ 93.079 mm.So, I think that's correct.Therefore, the necessary thickness x is approximately 93.08 mm.Wait, but let me think about the function again. The permeability decreases exponentially with x, so to get a very low P, you need a large x. So, 93 mm is correct in that sense.Alternatively, maybe I can express the answer in terms of exact expressions, but since the question asks for the value, I think the numerical value is expected.So, summarizing:Sub-problem 1: P ≈ 354.52.Sub-problem 2: x ≈ 93.08 mm.Wait, but let me check if I can write the exact expression for x.We had:0.05 = 0.8 * e^(-0.1x) * 900 * sin(50°)So, solving for x:e^(-0.1x) = 0.05 / (0.8 * 900 * sin(50°))Take natural log:-0.1x = ln(0.05 / (0.8 * 900 * sin(50°)))So,x = -10 * ln(0.05 / (0.8 * 900 * sin(50°)))But perhaps that's not necessary, since we already computed the numerical value.Alternatively, if I want to write it in terms of exact expressions, but I think the numerical value is sufficient.So, I think I've solved both sub-problems.</think>