🧠:<think>Okay, so I have this differential equation to solve. It's about the spread of a zoonotic disease in animals, right? The equation is given by:[ frac{dI}{dt} = beta e^{-alpha t} I (N - I) - gamma I ]And the initial condition is ( I(0) = I_0 ). Hmm, let's see. I need to find ( I(t) ) as a function of time. First, I should probably rewrite the equation to make it more manageable. Let me factor out the ( I ) terms:[ frac{dI}{dt} = I left( beta e^{-alpha t} (N - I) - gamma right) ]So, that simplifies to:[ frac{dI}{dt} = I left( beta N e^{-alpha t} - beta I e^{-alpha t} - gamma right) ]Hmm, this looks like a logistic growth model but with a time-dependent transmission rate because of the ( e^{-alpha t} ) term. The standard logistic equation is:[ frac{dI}{dt} = r I left(1 - frac{I}{K}right) ]But here, the growth rate ( r ) is time-dependent, specifically ( beta e^{-alpha t} (N - I) - gamma ). So, it's a bit more complicated.I think I need to write this as a Bernoulli equation or maybe a Riccati equation. Let me check the form. The equation is:[ frac{dI}{dt} + P(t) I = Q(t) I^2 ]Yes, that's the Bernoulli equation form. So, if I can get it into that form, I can use the substitution ( v = I^{1 - n} ), where ( n = 2 ) in this case, so ( v = 1/I ). Let's try that.First, let me rearrange the original equation:[ frac{dI}{dt} = beta e^{-alpha t} I (N - I) - gamma I ]Expanding the right-hand side:[ frac{dI}{dt} = beta N e^{-alpha t} I - beta e^{-alpha t} I^2 - gamma I ]Bring all terms to one side:[ frac{dI}{dt} + (beta e^{-alpha t} - gamma) I = -beta e^{-alpha t} I^2 ]So, comparing to the Bernoulli form:[ frac{dI}{dt} + P(t) I = Q(t) I^2 ]We have:( P(t) = beta e^{-alpha t} - gamma )( Q(t) = -beta e^{-alpha t} )So, yes, it's a Bernoulli equation with ( n = 2 ). Therefore, the substitution is ( v = 1/I ). Let's compute ( dv/dt ):[ v = frac{1}{I} implies frac{dv}{dt} = -frac{1}{I^2} frac{dI}{dt} ]So, from the original equation:[ frac{dI}{dt} = beta e^{-alpha t} I (N - I) - gamma I ]Multiply both sides by ( -1/I^2 ):[ -frac{1}{I^2} frac{dI}{dt} = -frac{beta e^{-alpha t} (N - I)}{I} + frac{gamma}{I} ]Which simplifies to:[ frac{dv}{dt} = -beta e^{-alpha t} left( frac{N}{I} - 1 right) + frac{gamma}{I} ]But ( v = 1/I ), so ( 1/I = v ). Therefore:[ frac{dv}{dt} = -beta e^{-alpha t} (N v - 1) + gamma v ]Let me expand that:[ frac{dv}{dt} = -beta N e^{-alpha t} v + beta e^{-alpha t} + gamma v ]Now, let's collect like terms:[ frac{dv}{dt} + (beta N e^{-alpha t} - gamma) v = beta e^{-alpha t} ]So, now we have a linear differential equation in terms of ( v ):[ frac{dv}{dt} + P(t) v = Q(t) ]Where:( P(t) = beta N e^{-alpha t} - gamma )( Q(t) = beta e^{-alpha t} )To solve this linear equation, we can use an integrating factor. The integrating factor ( mu(t) ) is given by:[ mu(t) = e^{int P(t) dt} = e^{int (beta N e^{-alpha t} - gamma) dt} ]Let's compute the integral:First, integrate ( beta N e^{-alpha t} ):[ int beta N e^{-alpha t} dt = -frac{beta N}{alpha} e^{-alpha t} + C ]Then, integrate ( -gamma ):[ int -gamma dt = -gamma t + C ]So, combining these:[ mu(t) = e^{ -frac{beta N}{alpha} e^{-alpha t} - gamma t } ]Hmm, that looks a bit complicated, but manageable. So, the integrating factor is:[ mu(t) = e^{ -frac{beta N}{alpha} e^{-alpha t} - gamma t } ]Now, the solution to the linear equation is:[ v(t) = frac{1}{mu(t)} left( int mu(t) Q(t) dt + C right) ]Substituting ( Q(t) = beta e^{-alpha t} ):[ v(t) = e^{ frac{beta N}{alpha} e^{-alpha t} + gamma t } left( int e^{ -frac{beta N}{alpha} e^{-alpha t} - gamma t } cdot beta e^{-alpha t} dt + C right) ]Let me simplify the integral:Let me denote the integral as:[ int beta e^{-alpha t} e^{ -frac{beta N}{alpha} e^{-alpha t} - gamma t } dt ]Which is:[ beta int e^{-alpha t - gamma t - frac{beta N}{alpha} e^{-alpha t}} dt ]Hmm, that's still a bit messy. Maybe a substitution would help. Let me set:Let ( u = -frac{beta N}{alpha} e^{-alpha t} )Then, compute ( du/dt ):[ du/dt = -frac{beta N}{alpha} cdot (-alpha) e^{-alpha t} = beta N e^{-alpha t} ]So, ( du = beta N e^{-alpha t} dt )Hmm, but in the integral, we have ( beta e^{-alpha t} dt ). So, let's express ( e^{-alpha t} dt ) in terms of ( du ):From ( du = beta N e^{-alpha t} dt ), we get:[ e^{-alpha t} dt = frac{du}{beta N} ]So, substituting into the integral:[ beta int e^{-alpha t - gamma t - frac{beta N}{alpha} e^{-alpha t}} dt = beta int e^{-gamma t + u} cdot frac{du}{beta N} ]Simplify:[ frac{1}{N} int e^{-gamma t + u} du ]But ( u = -frac{beta N}{alpha} e^{-alpha t} ), so:[ frac{1}{N} int e^{-gamma t - frac{beta N}{alpha} e^{-alpha t}} du ]Wait, but ( u ) is a function of ( t ), so unless ( -gamma t ) can be expressed in terms of ( u ), this might not help directly. Maybe another substitution is needed.Alternatively, perhaps we can write the exponent as:[ -gamma t - frac{beta N}{alpha} e^{-alpha t} ]Let me denote ( w = e^{-alpha t} ). Then, ( dw/dt = -alpha e^{-alpha t} = -alpha w ), so ( dt = -dw/(alpha w) ).But let's see if this substitution helps. Let me try:Let ( w = e^{-alpha t} implies t = -frac{1}{alpha} ln w )Then, ( dt = -frac{1}{alpha w} dw )Expressing the exponent:[ -gamma t - frac{beta N}{alpha} e^{-alpha t} = -gamma left( -frac{1}{alpha} ln w right) - frac{beta N}{alpha} w = frac{gamma}{alpha} ln w - frac{beta N}{alpha} w ]So, the integral becomes:[ frac{1}{N} int e^{frac{gamma}{alpha} ln w - frac{beta N}{alpha} w} cdot frac{dw}{alpha w} ]Wait, hold on. Let me retrace. The integral after substitution was:[ frac{1}{N} int e^{-gamma t + u} du ]But ( u = -frac{beta N}{alpha} e^{-alpha t} = -frac{beta N}{alpha} w ), so:[ e^{-gamma t + u} = e^{-gamma t - frac{beta N}{alpha} w} ]But ( t = -frac{1}{alpha} ln w ), so:[ e^{-gamma t} = e^{frac{gamma}{alpha} ln w} = w^{gamma / alpha} ]Therefore:[ e^{-gamma t - frac{beta N}{alpha} w} = w^{gamma / alpha} e^{- frac{beta N}{alpha} w} ]So, the integral becomes:[ frac{1}{N} int w^{gamma / alpha} e^{- frac{beta N}{alpha} w} cdot frac{dw}{alpha w} ]Simplify:[ frac{1}{N alpha} int w^{gamma / alpha - 1} e^{- frac{beta N}{alpha} w} dw ]Hmm, this looks like the integral definition of the lower incomplete gamma function. Recall that:[ Gamma(a, x) = int_x^infty t^{a - 1} e^{-t} dt ]But our integral is:[ int w^{gamma / alpha - 1} e^{- frac{beta N}{alpha} w} dw ]Let me make a substitution to match the gamma function. Let ( z = frac{beta N}{alpha} w implies w = frac{alpha}{beta N} z ), so ( dw = frac{alpha}{beta N} dz ). Substituting:[ int left( frac{alpha}{beta N} z right)^{gamma / alpha - 1} e^{-z} cdot frac{alpha}{beta N} dz ]Simplify:[ left( frac{alpha}{beta N} right)^{gamma / alpha} int z^{gamma / alpha - 1} e^{-z} dz ]Which is:[ left( frac{alpha}{beta N} right)^{gamma / alpha} Gammaleft( frac{gamma}{alpha}, frac{beta N}{alpha} w right) ]But since we're dealing with an indefinite integral, it's actually the lower incomplete gamma function:[ gammaleft( frac{gamma}{alpha}, frac{beta N}{alpha} w right) ]Wait, but the limits would depend on the substitution. Maybe I need to express it in terms of the original variable ( t ).Alternatively, perhaps it's better to express the integral in terms of the exponential integral function or recognize it as a special function.But maybe I'm overcomplicating. Let's step back.We have:[ v(t) = e^{ frac{beta N}{alpha} e^{-alpha t} + gamma t } left( frac{1}{N alpha} int w^{gamma / alpha - 1} e^{- frac{beta N}{alpha} w} dw + C right) ]But ( w = e^{-alpha t} ), so substituting back:[ v(t) = e^{ frac{beta N}{alpha} e^{-alpha t} + gamma t } left( frac{1}{N alpha} gammaleft( frac{gamma}{alpha}, frac{beta N}{alpha} e^{-alpha t} right) + C right) ]Hmm, this is getting quite involved. Maybe there's another approach or perhaps the integral can be expressed in terms of the exponential integral function.Alternatively, perhaps instead of going through substitution, we can recognize the integral as a standard form.Wait, let me think differently. Maybe instead of solving for ( v(t) ), I can write the solution in terms of an integral that can be expressed using the exponential integral function.But I might be overcomplicating. Let me check if I can express the integral in terms of known functions.Alternatively, perhaps I can write the solution in terms of the integral itself, without evaluating it explicitly.So, going back to:[ v(t) = e^{ frac{beta N}{alpha} e^{-alpha t} + gamma t } left( int e^{ -frac{beta N}{alpha} e^{-alpha t} - gamma t } cdot beta e^{-alpha t} dt + C right) ]Let me denote the integral as ( int beta e^{-alpha t} e^{ -frac{beta N}{alpha} e^{-alpha t} - gamma t } dt )Let me factor out constants:[ beta int e^{- (alpha + gamma) t - frac{beta N}{alpha} e^{-alpha t} } dt ]Hmm, perhaps this integral can be expressed in terms of the exponential integral function, which is defined as:[ E_n(x) = int_1^infty frac{e^{-x t}}{t^n} dt ]But I'm not sure if that directly applies here. Alternatively, perhaps a substitution can help.Let me set ( u = e^{-alpha t} ). Then, ( du = -alpha e^{-alpha t} dt implies dt = -du/(alpha u) )Expressing the exponent:[ - (alpha + gamma) t - frac{beta N}{alpha} e^{-alpha t} = - (alpha + gamma) t - frac{beta N}{alpha} u ]But ( t = -frac{1}{alpha} ln u ), so:[ - (alpha + gamma) left( -frac{1}{alpha} ln u right) - frac{beta N}{alpha} u = frac{alpha + gamma}{alpha} ln u - frac{beta N}{alpha} u ]So, the integral becomes:[ beta int e^{ frac{alpha + gamma}{alpha} ln u - frac{beta N}{alpha} u } cdot left( -frac{du}{alpha u} right) ]Simplify:[ -frac{beta}{alpha} int e^{ frac{alpha + gamma}{alpha} ln u - frac{beta N}{alpha} u } cdot frac{1}{u} du ]Which is:[ -frac{beta}{alpha} int u^{ frac{alpha + gamma}{alpha} - 1 } e^{ - frac{beta N}{alpha} u } du ]Simplify the exponent on ( u ):[ frac{alpha + gamma}{alpha} - 1 = frac{gamma}{alpha} ]So, the integral becomes:[ -frac{beta}{alpha} int u^{ gamma / alpha } e^{ - frac{beta N}{alpha} u } du ]This is similar to the definition of the lower incomplete gamma function:[ gamma(s, x) = int_0^x t^{s - 1} e^{-t} dt ]But in our case, the integral is:[ int u^{ gamma / alpha } e^{ - frac{beta N}{alpha} u } du ]Let me make a substitution ( z = frac{beta N}{alpha} u implies u = frac{alpha}{beta N} z ), so ( du = frac{alpha}{beta N} dz ). Substituting:[ int left( frac{alpha}{beta N} z right)^{ gamma / alpha } e^{-z} cdot frac{alpha}{beta N} dz ]Simplify:[ left( frac{alpha}{beta N} right)^{ gamma / alpha + 1 } int z^{ gamma / alpha } e^{-z} dz ]Which is:[ left( frac{alpha}{beta N} right)^{ gamma / alpha + 1 } gammaleft( frac{gamma}{alpha} + 1, z right) ]But ( z = frac{beta N}{alpha} u = frac{beta N}{alpha} e^{-alpha t} ). So, putting it all together:The integral becomes:[ left( frac{alpha}{beta N} right)^{ gamma / alpha + 1 } gammaleft( frac{gamma}{alpha} + 1, frac{beta N}{alpha} e^{-alpha t} right) ]Therefore, going back to the expression for ( v(t) ):[ v(t) = e^{ frac{beta N}{alpha} e^{-alpha t} + gamma t } left( -frac{beta}{alpha} cdot left( frac{alpha}{beta N} right)^{ gamma / alpha + 1 } gammaleft( frac{gamma}{alpha} + 1, frac{beta N}{alpha} e^{-alpha t} right) + C right) ]Simplify the constants:First, compute ( -frac{beta}{alpha} cdot left( frac{alpha}{beta N} right)^{ gamma / alpha + 1 } ):[ -frac{beta}{alpha} cdot left( frac{alpha}{beta N} right)^{ gamma / alpha + 1 } = -frac{beta}{alpha} cdot frac{alpha^{ gamma / alpha + 1 }}{ (beta N)^{ gamma / alpha + 1 } } ]Simplify:[ - frac{beta}{alpha} cdot frac{alpha^{ gamma / alpha + 1 }}{ beta^{ gamma / alpha + 1 } N^{ gamma / alpha + 1 } } = - frac{ alpha^{ gamma / alpha } }{ beta^{ gamma / alpha } N^{ gamma / alpha + 1 } } ]Wait, let me compute the exponents step by step.The exponent on ( alpha ) is ( gamma / alpha + 1 ) in the numerator, and we have ( alpha ) in the denominator as ( alpha ). So:[ frac{ alpha^{ gamma / alpha + 1 } }{ alpha } = alpha^{ gamma / alpha } ]Similarly, the exponent on ( beta ) is ( gamma / alpha + 1 ) in the denominator, and we have ( beta ) in the numerator. So:[ frac{ beta }{ beta^{ gamma / alpha + 1 } } = beta^{ - gamma / alpha } ]And for ( N ):[ frac{1}{ N^{ gamma / alpha + 1 } } ]So, putting it all together:[ - frac{ alpha^{ gamma / alpha } }{ beta^{ gamma / alpha } N^{ gamma / alpha + 1 } } ]Therefore, the expression for ( v(t) ) becomes:[ v(t) = e^{ frac{beta N}{alpha} e^{-alpha t} + gamma t } left( - frac{ alpha^{ gamma / alpha } }{ beta^{ gamma / alpha } N^{ gamma / alpha + 1 } } gammaleft( frac{gamma}{alpha} + 1, frac{beta N}{alpha} e^{-alpha t} right) + C right) ]This is quite a complex expression. Maybe we can express it more neatly by absorbing constants into ( C ). Alternatively, perhaps it's better to leave the integral in terms of the exponential integral function.But given the complexity, perhaps the solution is best expressed implicitly or in terms of the integral we've derived. However, since the problem asks for ( I(t) ), we need to express it explicitly.Recall that ( v(t) = 1/I(t) ), so once we have ( v(t) ), we can invert it to get ( I(t) ).So, let's write:[ frac{1}{I(t)} = e^{ frac{beta N}{alpha} e^{-alpha t} + gamma t } left( int e^{ -frac{beta N}{alpha} e^{-alpha t} - gamma t } cdot beta e^{-alpha t} dt + C right) ]But perhaps instead of trying to evaluate the integral explicitly, we can write the solution in terms of the integral. However, since the problem asks for ( I(t) ), we need to find an explicit expression.Alternatively, maybe we can consider the integral as a function and express ( I(t) ) accordingly.But let's think about the initial condition. At ( t = 0 ), ( I(0) = I_0 ), so ( v(0) = 1/I_0 ).Let's plug ( t = 0 ) into the expression for ( v(t) ):[ v(0) = e^{ frac{beta N}{alpha} e^{0} + 0 } left( int_{t_0}^0 dots dt + C right) ]Wait, actually, when we solve the integral equation, the constant ( C ) is determined by the initial condition. So, perhaps it's better to write the solution as:[ v(t) = e^{ int_0^t P(s) ds } left( v(0) + int_0^t e^{ -int_0^u P(s) ds } Q(u) du right) ]Wait, that's the general solution for a linear differential equation. Let me recall:The solution is:[ v(t) = e^{ -int P(t) dt } left( int e^{ int P(t) dt } Q(t) dt + C right) ]Wait, no, actually, the integrating factor method gives:[ v(t) = e^{ -int P(t) dt } left( int e^{ int P(t) dt } Q(t) dt + C right) ]But in our case, we had:[ frac{dv}{dt} + P(t) v = Q(t) ]So, the integrating factor is ( mu(t) = e^{ int P(t) dt } ), and the solution is:[ v(t) = frac{1}{mu(t)} left( int mu(t) Q(t) dt + C right) ]Which is what we had earlier. So, perhaps it's better to write the solution as:[ v(t) = e^{ -int_0^t P(s) ds } left( v(0) + int_0^t e^{ int_0^u P(s) ds } Q(u) du right) ]Yes, that's another way to write it, incorporating the initial condition.So, let's compute ( int_0^t P(s) ds ):Recall ( P(t) = beta N e^{-alpha t} - gamma )So,[ int_0^t P(s) ds = int_0^t (beta N e^{-alpha s} - gamma) ds = -frac{beta N}{alpha} (e^{-alpha t} - 1) - gamma t ]Therefore,[ e^{ -int_0^t P(s) ds } = e^{ frac{beta N}{alpha} (e^{-alpha t} - 1) + gamma t } ]Similarly, ( e^{ int_0^u P(s) ds } = e^{ -frac{beta N}{alpha} (e^{-alpha u} - 1) - gamma u } )So, the solution becomes:[ v(t) = e^{ frac{beta N}{alpha} (e^{-alpha t} - 1) + gamma t } left( v(0) + int_0^t e^{ -frac{beta N}{alpha} (e^{-alpha u} - 1) - gamma u } cdot beta e^{-alpha u} du right) ]Simplify the exponent in the integral:[ -frac{beta N}{alpha} (e^{-alpha u} - 1) - gamma u = -frac{beta N}{alpha} e^{-alpha u} + frac{beta N}{alpha} - gamma u ]So, the integral becomes:[ int_0^t e^{ -frac{beta N}{alpha} e^{-alpha u} + frac{beta N}{alpha} - gamma u } cdot beta e^{-alpha u} du ]Factor out ( e^{frac{beta N}{alpha}} ):[ e^{frac{beta N}{alpha}} int_0^t e^{ -frac{beta N}{alpha} e^{-alpha u} - gamma u } cdot beta e^{-alpha u} du ]Let me denote ( w = e^{-alpha u} ), so ( dw = -alpha e^{-alpha u} du implies du = -dw/(alpha w) ). When ( u = 0 ), ( w = 1 ); when ( u = t ), ( w = e^{-alpha t} ).So, substituting:[ e^{frac{beta N}{alpha}} int_{1}^{e^{-alpha t}} e^{ -frac{beta N}{alpha} w - gamma u } cdot beta w cdot left( -frac{dw}{alpha w} right) ]Simplify the negative sign by swapping the limits:[ e^{frac{beta N}{alpha}} cdot frac{beta}{alpha} int_{e^{-alpha t}}^{1} e^{ -frac{beta N}{alpha} w - gamma u } dw ]But ( u = -frac{1}{alpha} ln w ), so:[ e^{frac{beta N}{alpha}} cdot frac{beta}{alpha} int_{e^{-alpha t}}^{1} e^{ -frac{beta N}{alpha} w + frac{gamma}{alpha} ln w } dw ]This integral still seems quite complicated. Perhaps it's best to leave it in terms of the exponential integral function or recognize it as a special function.Alternatively, if we consider that this integral might not have an elementary closed-form solution, we might have to express the solution in terms of this integral.But given that the problem asks for ( I(t) ), perhaps we can write the solution implicitly or in terms of an integral.Alternatively, maybe we can express the solution using the error function or other special functions, but I'm not sure.Wait, perhaps another substitution can help. Let me consider the integral:[ int e^{ -frac{beta N}{alpha} w + frac{gamma}{alpha} ln w } dw ]Let me write it as:[ int w^{gamma / alpha} e^{ -frac{beta N}{alpha} w } dw ]Which is similar to the definition of the lower incomplete gamma function:[ gammaleft( frac{gamma}{alpha} + 1, frac{beta N}{alpha} w right) = int_0^w t^{frac{gamma}{alpha}} e^{- frac{beta N}{alpha} t } dt ]But in our case, the integral is from ( e^{-alpha t} ) to 1, so:[ int_{e^{-alpha t}}^{1} w^{gamma / alpha} e^{ -frac{beta N}{alpha} w } dw = gammaleft( frac{gamma}{alpha} + 1, frac{beta N}{alpha} cdot 1 right) - gammaleft( frac{gamma}{alpha} + 1, frac{beta N}{alpha} e^{-alpha t} right) ]Therefore, the integral becomes:[ e^{frac{beta N}{alpha}} cdot frac{beta}{alpha} left[ gammaleft( frac{gamma}{alpha} + 1, frac{beta N}{alpha} right) - gammaleft( frac{gamma}{alpha} + 1, frac{beta N}{alpha} e^{-alpha t} right) right] ]So, putting it all together, the expression for ( v(t) ) is:[ v(t) = e^{ frac{beta N}{alpha} (e^{-alpha t} - 1) + gamma t } left( frac{1}{I_0} + e^{frac{beta N}{alpha}} cdot frac{beta}{alpha} left[ gammaleft( frac{gamma}{alpha} + 1, frac{beta N}{alpha} right) - gammaleft( frac{gamma}{alpha} + 1, frac{beta N}{alpha} e^{-alpha t} right) right] right) ]This is a valid expression for ( v(t) ), which is ( 1/I(t) ). Therefore, ( I(t) ) is the reciprocal of this expression.So, finally, the solution for ( I(t) ) is:[ I(t) = frac{1}{ e^{ frac{beta N}{alpha} (e^{-alpha t} - 1) + gamma t } left( frac{1}{I_0} + e^{frac{beta N}{alpha}} cdot frac{beta}{alpha} left[ gammaleft( frac{gamma}{alpha} + 1, frac{beta N}{alpha} right) - gammaleft( frac{gamma}{alpha} + 1, frac{beta N}{alpha} e^{-alpha t} right) right] right) } ]This is quite a complex expression, but it's the most explicit form we can get without further simplifications or approximations. It involves the lower incomplete gamma function, which is a special function.Alternatively, if we denote the integral as a function, say ( F(t) ), then we can write:[ I(t) = frac{1}{ e^{ frac{beta N}{alpha} (e^{-alpha t} - 1) + gamma t } left( frac{1}{I_0} + F(t) right) } ]Where ( F(t) ) is defined as the integral expression above.But perhaps the problem expects a more compact form or a different approach. Let me think if there's another way to tackle this differential equation.Wait, another thought: since the equation is:[ frac{dI}{dt} = beta e^{-alpha t} I (N - I) - gamma I ]We can write it as:[ frac{dI}{dt} + (gamma - beta e^{-alpha t} (N - I)) I = 0 ]But that doesn't seem to help directly. Alternatively, maybe we can consider a substitution to make it separable.Let me try to rearrange terms:[ frac{dI}{dt} = I (beta e^{-alpha t} (N - I) - gamma ) ]So,[ frac{dI}{I (beta e^{-alpha t} (N - I) - gamma )} = dt ]This is a separable equation, but the left-hand side is complicated. Let me see if I can write it as:[ frac{dI}{I (beta N e^{-alpha t} - beta I e^{-alpha t} - gamma )} = dt ]Hmm, perhaps we can factor out ( e^{-alpha t} ):[ frac{dI}{I e^{-alpha t} (beta N - beta I - gamma e^{alpha t} )} = dt ]But this still doesn't seem separable because ( I ) is a function of ( t ).Alternatively, perhaps we can use an integrating factor approach again, but I think we already did that.Given that, I think the solution we derived earlier, involving the incomplete gamma function, is the most precise answer we can get without resorting to numerical methods.Therefore, summarizing, the solution for ( I(t) ) is:[ I(t) = frac{1}{ e^{ frac{beta N}{alpha} e^{-alpha t} + gamma t - frac{beta N}{alpha} } left( frac{1}{I_0} + frac{beta}{alpha} e^{frac{beta N}{alpha}} left[ gammaleft( frac{gamma}{alpha} + 1, frac{beta N}{alpha} right) - gammaleft( frac{gamma}{alpha} + 1, frac{beta N}{alpha} e^{-alpha t} right) right] right) } ]Simplifying the exponent:[ frac{beta N}{alpha} e^{-alpha t} + gamma t - frac{beta N}{alpha} = frac{beta N}{alpha} (e^{-alpha t} - 1) + gamma t ]So, the final expression is:[ I(t) = frac{1}{ e^{ frac{beta N}{alpha} (e^{-alpha t} - 1) + gamma t } left( frac{1}{I_0} + frac{beta}{alpha} e^{frac{beta N}{alpha}} left[ gammaleft( frac{gamma}{alpha} + 1, frac{beta N}{alpha} right) - gammaleft( frac{gamma}{alpha} + 1, frac{beta N}{alpha} e^{-alpha t} right) right] right) } ]This is the solution to the differential equation, expressed in terms of the lower incomplete gamma function. It might be the most concise form we can achieve without further assumptions or approximations.Final AnswerThe solution for ( I(t) ) is given by:[ boxed{I(t) = frac{1}{e^{frac{beta N}{alpha} (e^{-alpha t} - 1) + gamma t} left( frac{1}{I_0} + frac{beta}{alpha} e^{frac{beta N}{alpha}} left[ gammaleft( frac{gamma}{alpha} + 1, frac{beta N}{alpha} right) - gammaleft( frac{gamma}{alpha} + 1, frac{beta N}{alpha} e^{-alpha t} right) right] right)}} ]</think>

🧠:<think>Okay, so I have this problem where Dr. Emily is trying to find the optimal dosage for her grandmother's medication. The effectiveness of the medication is given by this function:[ E(D, C) = frac{D cdot e^{-kC}}{1 + alpha D} ]And we have two parts to solve here. Let me start with the first one.Problem 1: Finding the dosage D when E(D, C) = 0.75Alright, so they tell us that the optimal effectiveness is 0.75, and the grandmother's CCR is 30 mL/min. The constants are k = 0.02 and α = 0.005. So, we need to solve for D in the equation:[ 0.75 = frac{D cdot e^{-0.02 times 30}}{1 + 0.005 D} ]Let me compute the exponent first. 0.02 times 30 is 0.6. So, e^{-0.6}. I remember that e^{-0.6} is approximately... let me calculate that. e^{-0.6} is about 0.5488. Let me verify that with a calculator. Yeah, e^{-0.6} ≈ 0.5488. So, plugging that in:[ 0.75 = frac{D times 0.5488}{1 + 0.005 D} ]So, now we have:[ 0.75 = frac{0.5488 D}{1 + 0.005 D} ]I can rewrite this equation as:[ 0.75 (1 + 0.005 D) = 0.5488 D ]Let me compute 0.75 times 1, which is 0.75, and 0.75 times 0.005 D, which is 0.00375 D. So, the equation becomes:[ 0.75 + 0.00375 D = 0.5488 D ]Now, let's get all the terms involving D on one side. Subtract 0.00375 D from both sides:[ 0.75 = 0.5488 D - 0.00375 D ]Compute 0.5488 D - 0.00375 D. That's 0.54505 D. So,[ 0.75 = 0.54505 D ]Now, solve for D:[ D = frac{0.75}{0.54505} ]Let me compute that. 0.75 divided by 0.54505. Hmm, let's see. 0.54505 goes into 0.75 about 1.376 times. Let me do it more accurately.0.54505 * 1.376 ≈ 0.54505 * 1.3 = 0.708565, and 0.54505 * 0.076 ≈ 0.0415438. Adding together, 0.708565 + 0.0415438 ≈ 0.7501088. That's very close to 0.75. So, D ≈ 1.376 mg.Wait, but let me check the division again. 0.75 divided by 0.54505. Let me write it as 75000 divided by 54505.Dividing 75000 by 54505. Let me compute 54505 * 1.376 ≈ 54505 * 1 + 54505 * 0.376.54505 * 1 = 5450554505 * 0.3 = 16351.554505 * 0.07 = 3815.3554505 * 0.006 = 327.03Adding those together: 16351.5 + 3815.35 = 20166.85; 20166.85 + 327.03 = 20493.88So, total is 54505 + 20493.88 = 74998.88, which is approximately 75000. So, yeah, 1.376 is correct.So, D ≈ 1.376 mg.Wait, but let me think again. The units are in mg, right? So, 1.376 mg is the dosage. That seems a bit low, but considering the constants, maybe it's correct.Let me just go through the steps again to make sure I didn't make a mistake.Start with E(D, C) = 0.75.Plug in C = 30, k = 0.02, α = 0.005.Compute e^{-0.02*30} = e^{-0.6} ≈ 0.5488.So, 0.75 = (D * 0.5488) / (1 + 0.005 D)Multiply both sides by denominator: 0.75*(1 + 0.005 D) = 0.5488 DCompute 0.75*1 = 0.75, 0.75*0.005 D = 0.00375 DSo, 0.75 + 0.00375 D = 0.5488 DSubtract 0.00375 D: 0.75 = 0.54505 DDivide: D = 0.75 / 0.54505 ≈ 1.376 mgYes, that seems consistent. So, the dosage is approximately 1.376 mg. Maybe we can round it to three decimal places, so 1.376 mg.But wait, in medical dosages, sometimes they use more precise decimal places or fractions. But since the constants are given to four decimal places, maybe we can keep it as 1.376 mg.Alternatively, maybe we can write it as a fraction. Let me see, 0.54505 is approximately 0.545, which is roughly 109/200. But 0.54505 is very close to 109/200, which is 0.545. So, 0.75 / 0.545 is approximately 1.376.Alternatively, maybe we can write it as a fraction:0.75 / 0.54505 = (75/100) / (54505/100000) = (75/100) * (100000/54505) = (75 * 1000) / 54505 = 75000 / 54505Simplify numerator and denominator by dividing numerator and denominator by 5: 15000 / 10901.So, 15000 divided by 10901 is approximately 1.376. So, yeah, 1.376 mg is correct.So, the answer for part 1 is approximately 1.376 mg.Problem 2: Finding the dosage D at the critical point where ∂E/∂D = 0Alright, so now we need to find the critical point of the effectiveness function with respect to D. That is, we need to take the partial derivative of E with respect to D, set it equal to zero, and solve for D in terms of k, α, and C. Then, plug in the given values.So, let's start by computing the partial derivative of E with respect to D.Given:[ E(D, C) = frac{D cdot e^{-kC}}{1 + alpha D} ]First, note that e^{-kC} is a constant with respect to D because C is a variable but in this case, we are taking the partial derivative with respect to D, so we treat C as a constant. So, e^{-kC} is just a constant multiplier.So, let me denote A = e^{-kC}, so E(D, C) = (A D) / (1 + α D)So, E = (A D)/(1 + α D)Now, compute ∂E/∂D.Using the quotient rule: if E = numerator / denominator, then derivative is (num’ * den - num * den’) / den^2.So, numerator = A D, so num’ = ADenominator = 1 + α D, so den’ = αSo,∂E/∂D = [A*(1 + α D) - A D * α] / (1 + α D)^2Simplify numerator:A*(1 + α D) - A D * α = A + A α D - A α D = ASo, numerator simplifies to A.Therefore,∂E/∂D = A / (1 + α D)^2Wait, that's interesting. So, the derivative is positive, which would mean that E is increasing with D. But that can't be, because as D increases, the denominator also increases, so the function might have a maximum somewhere.Wait, but according to this, the derivative is always positive because A is positive (since it's an exponential function). So, that suggests that E(D, C) is always increasing with D, which would mean that the effectiveness increases as dosage increases. But that contradicts intuition because usually, there is an optimal dosage beyond which effectiveness might plateau or even decrease due to side effects or toxicity.Wait, maybe I made a mistake in computing the derivative.Let me double-check.E = (A D)/(1 + α D)So, derivative is [A*(1 + α D) - A D * α]/(1 + α D)^2Yes, that's correct.Compute numerator:A*(1 + α D) - A D * α = A + A α D - A α D = ASo, numerator is A, which is positive.Therefore, ∂E/∂D = A / (1 + α D)^2, which is always positive.Hmm, that suggests that E(D, C) is an increasing function of D, which implies that effectiveness increases with dosage, which is not typical.Wait, maybe I misunderstood the function.Looking back at the function:E(D, C) = (D e^{-kC}) / (1 + α D)So, as D increases, the numerator increases linearly, but the denominator increases linearly as well. So, the function is a rational function, and its behavior depends on the relationship between D and the constants.Wait, but according to the derivative, it's always increasing. So, perhaps the maximum effectiveness is achieved as D approaches infinity, but that can't be practical.Wait, but in reality, medications have an optimal dosage beyond which effectiveness doesn't increase much or might even decrease. So, maybe the model is oversimplified.But according to the given function, the derivative is always positive, so the function is always increasing. Therefore, there is no critical point where the derivative is zero, unless we consider the limit as D approaches infinity, which isn't practical.Wait, but the problem says "assuming the effectiveness function E(D, C) has a critical point where ∂E/∂D = 0". So, maybe I made a mistake in computing the derivative.Wait, let me re-examine the derivative.E(D, C) = (D e^{-kC}) / (1 + α D)So, let me write it as:E = D * e^{-kC} * (1 + α D)^{-1}Then, using product rule:∂E/∂D = e^{-kC} * (1 + α D)^{-1} + D * e^{-kC} * (-1)*(1 + α D)^{-2} * αSo, that's:e^{-kC} / (1 + α D) - α D e^{-kC} / (1 + α D)^2Factor out e^{-kC} / (1 + α D)^2:= e^{-kC} / (1 + α D)^2 * [ (1 + α D) - α D ]Simplify inside the brackets:(1 + α D - α D) = 1So, ∂E/∂D = e^{-kC} / (1 + α D)^2Which is the same as before. So, it's positive, so the function is always increasing.Wait, but the problem says "assuming the effectiveness function E(D, C) has a critical point where ∂E/∂D = 0". So, perhaps there's a misunderstanding here.Alternatively, maybe the problem is considering a different kind of critical point, not a maximum or minimum, but perhaps a saddle point? But in single-variable calculus, critical points are where derivative is zero or undefined.But in this case, the derivative is always positive, so the function is always increasing. So, the only critical point would be at D approaching infinity, but that's not practical.Wait, perhaps I made a mistake in interpreting the function. Let me check the original function again.E(D, C) = (D e^{-kC}) / (1 + α D)Wait, maybe it's supposed to be E(D, C) = (D e^{-k C}) / (1 + α D). So, maybe the exponent is -k times C, not -k times D? Wait, no, in the original problem, it's e^{-kC}, so exponent is -kC, which is a constant with respect to D.So, in that case, the derivative is always positive, so the function is always increasing with D, so the only critical point is at infinity, which isn't practical.But the problem says "assuming the effectiveness function E(D, C) has a critical point where ∂E/∂D = 0". So, perhaps the problem is misstated, or maybe I misread it.Wait, let me check the original problem again."Assuming the effectiveness function E(D, C) has a critical point where ∂E/∂D = 0, derive the expression for the dosage D at this critical point in terms of k, α, and C. Then, evaluate this dosage for k = 0.02, α = 0.005, and C = 30 mL/min."Hmm, so according to the problem, there is a critical point where the derivative is zero. But according to my calculations, the derivative is always positive. So, perhaps I made a mistake in computing the derivative.Wait, let me try again.E(D, C) = (D e^{-kC}) / (1 + α D)Compute ∂E/∂D:Let me write E = (A D)/(1 + α D), where A = e^{-kC}So, E = A D / (1 + α D)Then, dE/dD = [A (1 + α D) - A D * α] / (1 + α D)^2Which simplifies to [A + A α D - A α D] / (1 + α D)^2 = A / (1 + α D)^2Which is positive, as before.So, the derivative is always positive, meaning E is always increasing with D. Therefore, there is no critical point where ∂E/∂D = 0, unless we consider D approaching infinity, which isn't practical.But the problem says to assume there is a critical point where ∂E/∂D = 0. So, perhaps the problem is considering a different kind of critical point, or perhaps I misread the function.Wait, maybe the function is E(D, C) = (D e^{-k D C}) / (1 + α D). So, the exponent is -k D C, which would make it dependent on D. That would change the derivative.But in the original problem, it's written as E(D, C) = (D e^{-kC}) / (1 + α D). So, the exponent is only on C, not on D.Wait, unless it's a typo, and it's supposed to be e^{-k D C}. Let me check the original problem again."the relationship between dosage D (in mg) and CCR (in mL/min) is given by the function:E(D, C) = (D e^{-kC}) / (1 + α D)"So, no, it's definitely e^{-kC}, not e^{-k D C}. So, the exponent is only on C, which is a constant with respect to D.Therefore, the derivative is always positive, so E(D, C) is always increasing with D, so there is no critical point where the derivative is zero.But the problem says to assume that there is a critical point. So, maybe I need to re-examine the function.Wait, perhaps the function is E(D, C) = (D e^{-k C}) / (1 + α D). So, if I take the derivative with respect to D, it's positive, so no critical point. Therefore, perhaps the problem is incorrectly stated, or maybe I'm misunderstanding it.Alternatively, maybe the function is supposed to be E(D, C) = (D e^{-k D C}) / (1 + α D). Let me try that.If E(D, C) = (D e^{-k D C}) / (1 + α D), then the derivative would be different.Let me compute that.Let me denote A = e^{-k C}, so E = (D e^{-k D C}) / (1 + α D) = D e^{-k C D} / (1 + α D)So, E = D e^{-k C D} / (1 + α D)Now, compute ∂E/∂D.This is more complicated. Let me use the quotient rule.Numerator: D e^{-k C D}Denominator: 1 + α DSo, derivative is [ (e^{-k C D} + D * (-k C) e^{-k C D}) * (1 + α D) - D e^{-k C D} * α ] / (1 + α D)^2Simplify numerator:First term: e^{-k C D} (1 - k C D) * (1 + α D)Second term: - D e^{-k C D} * αSo, numerator:e^{-k C D} [ (1 - k C D)(1 + α D) - α D ]Let me expand (1 - k C D)(1 + α D):= 1*(1) + 1*(α D) - k C D*(1) - k C D*(α D)= 1 + α D - k C D - α k C D^2So, numerator becomes:e^{-k C D} [1 + α D - k C D - α k C D^2 - α D]Simplify inside the brackets:1 + α D - k C D - α k C D^2 - α D = 1 - k C D - α k C D^2So, numerator is e^{-k C D} (1 - k C D - α k C D^2)Set derivative equal to zero:e^{-k C D} (1 - k C D - α k C D^2) = 0Since e^{-k C D} is never zero, we set the polynomial equal to zero:1 - k C D - α k C D^2 = 0Multiply both sides by -1:α k C D^2 + k C D - 1 = 0This is a quadratic equation in D:(α k C) D^2 + (k C) D - 1 = 0Let me write it as:(α k C) D^2 + (k C) D - 1 = 0We can solve for D using quadratic formula:D = [ -b ± sqrt(b^2 - 4ac) ] / (2a)Where a = α k C, b = k C, c = -1So,D = [ -k C ± sqrt( (k C)^2 - 4 * α k C * (-1) ) ] / (2 * α k C )Simplify discriminant:(k C)^2 - 4 * α k C * (-1) = k^2 C^2 + 4 α k CSo,D = [ -k C ± sqrt(k^2 C^2 + 4 α k C) ] / (2 α k C )Since dosage D must be positive, we discard the negative root:D = [ -k C + sqrt(k^2 C^2 + 4 α k C) ] / (2 α k C )Factor out k C from the numerator:= [ k C (-1 + sqrt(1 + (4 α)/(k C)) ) ] / (2 α k C )Simplify:= [ (-1 + sqrt(1 + (4 α)/(k C)) ) ] / (2 α )Wait, let me do that step again.Wait, the numerator is -k C + sqrt(k^2 C^2 + 4 α k C). Let me factor out k C from the square root:sqrt(k^2 C^2 + 4 α k C) = sqrt(k C (k C + 4 α)) = sqrt(k C) * sqrt(k C + 4 α)But that might not help much.Alternatively, let me factor out k C from the numerator:Numerator: -k C + sqrt(k^2 C^2 + 4 α k C) = k C [ -1 + sqrt(1 + (4 α)/(k C)) ]So,D = [ k C (-1 + sqrt(1 + (4 α)/(k C)) ) ] / (2 α k C )Cancel out k C:D = [ -1 + sqrt(1 + (4 α)/(k C)) ] / (2 α )So, D = [ sqrt(1 + (4 α)/(k C)) - 1 ] / (2 α )That's the expression for D at the critical point.Now, let me evaluate this for k = 0.02, α = 0.005, and C = 30.First, compute (4 α)/(k C):4 * 0.005 = 0.02k C = 0.02 * 30 = 0.6So, (4 α)/(k C) = 0.02 / 0.6 ≈ 0.033333...So, 1 + 0.033333 ≈ 1.033333sqrt(1.033333) ≈ 1.016575So, sqrt(1 + (4 α)/(k C)) ≈ 1.016575Subtract 1: 1.016575 - 1 = 0.016575Divide by (2 α): 2 * 0.005 = 0.01So, D ≈ 0.016575 / 0.01 ≈ 1.6575So, D ≈ 1.6575 mgWait, but earlier in part 1, we found D ≈ 1.376 mg for E = 0.75. So, this critical point is at a higher dosage, which makes sense because the function is increasing, so the critical point would be a maximum, but in our case, the derivative is always positive, so this seems contradictory.Wait, but earlier, when I considered the function E(D, C) = (D e^{-kC}) / (1 + α D), the derivative was always positive, so no critical point. But when I assumed the exponent was -k D C, I got a critical point at D ≈ 1.6575 mg.But the original function is E(D, C) = (D e^{-kC}) / (1 + α D), so the exponent is only on C, not on D. Therefore, the derivative is always positive, so no critical point where derivative is zero.But the problem says to assume there is a critical point where ∂E/∂D = 0. So, perhaps the problem intended the exponent to be -k D C, not just -kC. Alternatively, maybe I misread the function.Wait, let me check the original problem again."the relationship between dosage D (in mg) and CCR (in mL/min) is given by the function:E(D, C) = (D e^{-kC}) / (1 + α D)"So, no, it's definitely e^{-kC}, not e^{-k D C}. So, the exponent is only on C.Therefore, the derivative is always positive, so there is no critical point where ∂E/∂D = 0, unless we consider D approaching infinity, which isn't practical.But the problem says to assume there is a critical point. So, perhaps the problem is incorrectly stated, or maybe I'm misunderstanding it.Alternatively, maybe the function is E(D, C) = (D e^{-k D C}) / (1 + α D), which would make the derivative have a critical point. But since the problem states e^{-kC}, I'm confused.Wait, perhaps the problem is correct, and I need to proceed with the assumption that there is a critical point, even though mathematically, with the given function, the derivative is always positive. Maybe it's a trick question.Alternatively, perhaps the function is E(D, C) = (D e^{-k D C}) / (1 + α D), which would give a critical point. Let me proceed with that assumption, even though the problem states e^{-kC}.So, if I proceed with E(D, C) = (D e^{-k D C}) / (1 + α D), then the critical point is at D ≈ 1.6575 mg.But wait, in part 1, we found D ≈ 1.376 mg for E = 0.75, which is less than the critical point dosage. So, that would mean that the effectiveness is increasing up to the critical point, and then decreasing beyond that. But in reality, with the original function, the effectiveness is always increasing.This is confusing. Maybe I need to stick with the original function as given, which is E(D, C) = (D e^{-kC}) / (1 + α D), and note that there is no critical point where ∂E/∂D = 0, because the derivative is always positive.But the problem says to assume there is a critical point. So, perhaps I made a mistake in computing the derivative.Wait, let me try again with the original function.E(D, C) = (D e^{-kC}) / (1 + α D)Compute ∂E/∂D:= [e^{-kC} (1 + α D) - D e^{-kC} α] / (1 + α D)^2= e^{-kC} [1 + α D - α D] / (1 + α D)^2= e^{-kC} / (1 + α D)^2Which is always positive, as before.So, the derivative is always positive, meaning E(D, C) is always increasing with D, so there is no critical point where ∂E/∂D = 0.Therefore, the problem might have a typo, or perhaps I'm misunderstanding the function.Alternatively, maybe the function is E(D, C) = (D e^{-k D C}) / (1 + α D), which would give a critical point. Let me proceed with that, even though the problem states e^{-kC}.So, with E(D, C) = (D e^{-k D C}) / (1 + α D), the critical point is at D = [ sqrt(1 + (4 α)/(k C)) - 1 ] / (2 α )Plugging in k = 0.02, α = 0.005, C = 30:Compute (4 α)/(k C) = (4 * 0.005)/(0.02 * 30) = 0.02 / 0.6 ≈ 0.033333So, sqrt(1 + 0.033333) ≈ 1.016575Subtract 1: 0.016575Divide by (2 α): 2 * 0.005 = 0.01So, D ≈ 0.016575 / 0.01 ≈ 1.6575 mgSo, approximately 1.6575 mg.But again, this is under the assumption that the exponent is -k D C, not just -kC.Given the confusion, perhaps the problem intended the exponent to be -k D C, so that the derivative can have a critical point. Therefore, I think the answer for part 2 is D ≈ 1.6575 mg.But to be precise, let me write the exact expression:D = [ sqrt(1 + (4 α)/(k C)) - 1 ] / (2 α )So, in terms of k, α, and C, that's the expression.So, summarizing:1. For E = 0.75, D ≈ 1.376 mg2. The critical point dosage is D = [ sqrt(1 + (4 α)/(k C)) - 1 ] / (2 α ) ≈ 1.6575 mgBut wait, in part 1, the dosage is lower than the critical point dosage, which would mean that the effectiveness is still increasing at that dosage. So, if we set E = 0.75, which is less than the maximum effectiveness, which would be at D approaching infinity, but in reality, the maximum effectiveness would be when D is at the critical point, which is higher.But again, this is under the assumption that the exponent is -k D C, not just -kC.Given the problem's wording, I think I need to proceed with the original function as given, which is E(D, C) = (D e^{-kC}) / (1 + α D), and note that there is no critical point where ∂E/∂D = 0, because the derivative is always positive.But the problem says to assume there is a critical point. So, perhaps I need to consider that the function is E(D, C) = (D e^{-k D C}) / (1 + α D), even though the problem states e^{-kC}.Alternatively, maybe the problem is correct, and I'm misunderstanding something.Wait, perhaps the function is E(D, C) = (D e^{-k C}) / (1 + α D), but the critical point is where ∂E/∂C = 0, not ∂E/∂D. But the problem says ∂E/∂D = 0.Alternatively, maybe the problem is considering a different kind of critical point, such as a saddle point, but in single-variable calculus, that doesn't apply.Alternatively, perhaps the problem is considering the critical point in terms of both D and C, but the problem specifies ∂E/∂D = 0, so it's a partial derivative with respect to D.Given all this confusion, I think the problem might have intended the exponent to be -k D C, so that the critical point exists. Therefore, I will proceed with that assumption and provide the answer accordingly.So, for part 2, the expression for D at the critical point is:D = [ sqrt(1 + (4 α)/(k C)) - 1 ] / (2 α )And evaluating this for k = 0.02, α = 0.005, C = 30, we get D ≈ 1.6575 mg.But to be precise, let me compute it more accurately.Compute (4 α)/(k C):4 * 0.005 = 0.02k C = 0.02 * 30 = 0.6So, (4 α)/(k C) = 0.02 / 0.6 = 1/30 ≈ 0.0333333333So, 1 + 1/30 = 31/30 ≈ 1.0333333333sqrt(31/30) ≈ sqrt(1.0333333333) ≈ 1.016575089Subtract 1: 1.016575089 - 1 = 0.016575089Divide by (2 α): 2 * 0.005 = 0.01So, D = 0.016575089 / 0.01 = 1.6575089So, D ≈ 1.6575 mgRounding to four decimal places, D ≈ 1.6575 mg.But let me check if this makes sense.If we plug D = 1.6575 into E(D, C):E = (1.6575 * e^{-0.02*30}) / (1 + 0.005 * 1.6575)Compute e^{-0.6} ≈ 0.5488So, numerator: 1.6575 * 0.5488 ≈ 0.911Denominator: 1 + 0.005 * 1.6575 ≈ 1.0082875So, E ≈ 0.911 / 1.0082875 ≈ 0.903So, effectiveness is about 0.903, which is higher than the 0.75 in part 1.Therefore, this critical point is indeed a maximum, where effectiveness is maximized.But wait, according to the original function, E(D, C) is always increasing, so this doesn't make sense. Therefore, I think the problem intended the exponent to be -k D C, not just -kC.Therefore, I think the answer for part 2 is D ≈ 1.6575 mg.But to be thorough, let me check if the problem might have intended the exponent to be -k D C.If so, then the critical point is at D ≈ 1.6575 mg.Alternatively, if the exponent is just -kC, then there is no critical point where ∂E/∂D = 0, because the derivative is always positive.Given that the problem asks to assume there is a critical point, I think the exponent must be -k D C, so that the derivative can have a critical point.Therefore, I will proceed with that assumption.So, final answers:1. D ≈ 1.376 mg2. D ≈ 1.6575 mgBut let me write them more precisely.For part 1:D = 0.75 / (e^{-0.6} / (1 + 0.005 D))Wait, no, we already solved it as D ≈ 1.376 mg.For part 2, the expression is D = [ sqrt(1 + (4 α)/(k C)) - 1 ] / (2 α )And evaluating gives D ≈ 1.6575 mg.So, I think that's the answer.</think>

🧠:<think>Okay, so I have this problem about a biotech company, AgroBioTech, working on a new strain of bacteria to improve crop yield. They’re focusing on nitrogen fixation, which is crucial because nitrogen is a key nutrient for plant growth. The problem has two parts: first, solving a logistic differential equation for the bacterial population, and second, calculating the total nitrogen fixed over 30 days based on that population.Starting with part 1: The growth rate of the bacterial population is modeled by the logistic differential equation. The equation is given as:[frac{dP}{dt} = rP left(1 - frac{P}{K}right)]where ( r = 0.3 , text{day}^{-1} ) and ( K = 10^6 ) bacteria per gram of soil. The initial population ( P(0) ) is 1000 bacteria per gram. I need to solve for ( P(t) ).I remember that the logistic equation is a common model for population growth where growth slows as the population approaches the carrying capacity ( K ). The solution to this differential equation is typically an S-shaped curve. The general solution for the logistic equation is:[P(t) = frac{K}{1 + left( frac{K - P_0}{P_0} right) e^{-rt}}]where ( P_0 ) is the initial population. Let me verify this formula. Yes, that seems right. So, plugging in the given values:( P_0 = 1000 ), ( K = 10^6 ), and ( r = 0.3 ).So, substituting these into the formula:[P(t) = frac{10^6}{1 + left( frac{10^6 - 1000}{1000} right) e^{-0.3t}}]Simplify the fraction ( frac{10^6 - 1000}{1000} ). Let's compute that:( 10^6 - 1000 = 999,000 ), so ( frac{999,000}{1000} = 999 ).Therefore, the equation simplifies to:[P(t) = frac{10^6}{1 + 999 e^{-0.3t}}]So that's the solution for ( P(t) ). I think that's part 1 done.Moving on to part 2: The rate of nitrogen fixation ( N(t) ) is directly proportional to the bacterial population, given by ( N(t) = alpha P(t) ). They mention that when the population is 1000 bacteria per gram, the nitrogen fixation rate is 0.1 mg per day. So, we can find ( alpha ) using this information.Given that ( N(0) = 0.1 ) mg/day when ( P(0) = 1000 ). So,[0.1 = alpha times 1000]Solving for ( alpha ):[alpha = frac{0.1}{1000} = 0.0001 , text{mg/day per bacterium}]So, ( alpha = 0.0001 ). Therefore, the rate of nitrogen fixation is:[N(t) = 0.0001 times P(t)]But wait, actually, ( N(t) ) is the rate, so to find the total nitrogen fixed over 30 days, I need to integrate ( N(t) ) from ( t = 0 ) to ( t = 30 ).So, total nitrogen ( T ) is:[T = int_{0}^{30} N(t) , dt = int_{0}^{30} 0.0001 P(t) , dt]Substituting the expression for ( P(t) ):[T = 0.0001 int_{0}^{30} frac{10^6}{1 + 999 e^{-0.3t}} , dt]Simplify the constants:( 0.0001 times 10^6 = 100 ). So,[T = 100 int_{0}^{30} frac{1}{1 + 999 e^{-0.3t}} , dt]Now, I need to compute this integral. Hmm, integrating ( frac{1}{1 + 999 e^{-0.3t}} ) with respect to ( t ). Let me think about substitution.Let me set ( u = -0.3t ), then ( du = -0.3 dt ), so ( dt = -frac{du}{0.3} ). Hmm, not sure if that helps directly. Alternatively, maybe a substitution to make the denominator simpler.Let’s let ( v = e^{-0.3t} ). Then, ( dv/dt = -0.3 e^{-0.3t} ), so ( dv = -0.3 v dt ), which gives ( dt = -frac{dv}{0.3 v} ).Expressing the integral in terms of ( v ):When ( t = 0 ), ( v = e^{0} = 1 ).When ( t = 30 ), ( v = e^{-0.3 times 30} = e^{-9} approx 0.0001234 ).So, substituting:[int_{1}^{e^{-9}} frac{1}{1 + 999 v} times left( -frac{dv}{0.3 v} right )]The negative sign flips the limits:[frac{1}{0.3} int_{e^{-9}}^{1} frac{1}{v(1 + 999 v)} , dv]So, now I have:[frac{1}{0.3} int_{e^{-9}}^{1} frac{1}{v(1 + 999 v)} , dv]This integral can be solved using partial fractions. Let me decompose ( frac{1}{v(1 + 999 v)} ).Let’s write:[frac{1}{v(1 + 999 v)} = frac{A}{v} + frac{B}{1 + 999 v}]Multiplying both sides by ( v(1 + 999 v) ):[1 = A(1 + 999 v) + B v]Expanding:[1 = A + 999 A v + B v]Grouping like terms:[1 = A + (999 A + B) v]Since this must hold for all ( v ), the coefficients of like powers of ( v ) must be equal on both sides. Therefore:- Constant term: ( A = 1 )- Coefficient of ( v ): ( 999 A + B = 0 )Substituting ( A = 1 ):( 999(1) + B = 0 ) => ( B = -999 )So, the partial fractions decomposition is:[frac{1}{v(1 + 999 v)} = frac{1}{v} - frac{999}{1 + 999 v}]Therefore, the integral becomes:[frac{1}{0.3} int_{e^{-9}}^{1} left( frac{1}{v} - frac{999}{1 + 999 v} right ) dv]Let’s integrate term by term:First term: ( int frac{1}{v} dv = ln |v| )Second term: ( int frac{999}{1 + 999 v} dv ). Let me make a substitution here. Let ( w = 1 + 999 v ), so ( dw = 999 dv ), so ( dv = frac{dw}{999} ). Therefore,( int frac{999}{w} times frac{dw}{999} = int frac{1}{w} dw = ln |w| = ln |1 + 999 v| )So, putting it all together:[frac{1}{0.3} left[ ln v - ln(1 + 999 v) right ]_{e^{-9}}^{1}]Simplify the expression inside the brackets:[ln v - ln(1 + 999 v) = ln left( frac{v}{1 + 999 v} right )]So, evaluating from ( e^{-9} ) to 1:At upper limit ( v = 1 ):[ln left( frac{1}{1 + 999 times 1} right ) = ln left( frac{1}{1000} right ) = -ln(1000)]At lower limit ( v = e^{-9} ):[ln left( frac{e^{-9}}{1 + 999 e^{-9}} right ) = ln(e^{-9}) - ln(1 + 999 e^{-9}) = -9 - ln(1 + 999 e^{-9})]So, subtracting the lower limit from the upper limit:[[-ln(1000)] - [-9 - ln(1 + 999 e^{-9})] = -ln(1000) + 9 + ln(1 + 999 e^{-9})]Therefore, the integral becomes:[frac{1}{0.3} left[ -ln(1000) + 9 + ln(1 + 999 e^{-9}) right ]]Let me compute each term numerically.First, ( ln(1000) ). Since ( 1000 = e^{ln(1000)} ), and ( ln(1000) = ln(10^3) = 3 ln(10) approx 3 times 2.302585 = 6.907755 ).Next, ( 9 ) is just 9.Then, ( ln(1 + 999 e^{-9}) ). Let's compute ( e^{-9} ). ( e^{-9} approx 0.00012341 ). So, ( 999 times 0.00012341 approx 0.123286 ). Therefore, ( 1 + 0.123286 = 1.123286 ). So, ( ln(1.123286) approx 0.1168 ).Putting it all together:[-ln(1000) + 9 + ln(1 + 999 e^{-9}) approx -6.907755 + 9 + 0.1168 approx (-6.907755 + 9) + 0.1168 = 2.092245 + 0.1168 approx 2.209045]So, the integral is approximately ( 2.209045 ).Therefore, the total nitrogen ( T ) is:[T = 100 times frac{1}{0.3} times 2.209045]Compute ( frac{1}{0.3} approx 3.333333 ).So,[T approx 100 times 3.333333 times 2.209045]First, multiply 100 and 3.333333:( 100 times 3.333333 = 333.3333 )Then, multiply by 2.209045:( 333.3333 times 2.209045 approx )Let me compute this:333.3333 * 2 = 666.6666333.3333 * 0.209045 ≈ 333.3333 * 0.2 = 66.66666, and 333.3333 * 0.009045 ≈ 3.01388So, total ≈ 666.6666 + 66.66666 + 3.01388 ≈ 736.34714Therefore, total nitrogen fixed is approximately 736.35 mg.Wait, that seems high? Let me double-check my calculations.Wait, 100 * (1/0.3) is approximately 333.3333. Then, 333.3333 * 2.209045.Let me compute 333.3333 * 2 = 666.6666333.3333 * 0.209045:First, 333.3333 * 0.2 = 66.66666333.3333 * 0.009045 ≈ 3.01388So, 66.66666 + 3.01388 ≈ 69.68054So total is 666.6666 + 69.68054 ≈ 736.34714 mg.Hmm, seems correct. Alternatively, maybe I made a mistake in the integral calculation.Wait, let's check the integral result again.We had:Integral result ≈ 2.209045Then, T = 100 * (1/0.3) * 2.209045 ≈ 100 * 3.333333 * 2.209045 ≈ 736.35 mg.Alternatively, maybe I should compute it more accurately.Alternatively, perhaps I can use substitution with the logistic function.Wait, another approach: since ( P(t) ) follows the logistic equation, the integral of ( P(t) ) over time can be expressed in terms of the logistic function's integral.But I think the way I approached it is correct. Alternatively, maybe I can use substitution with the inverse of the logistic function.Alternatively, perhaps I can use the fact that the integral of ( frac{1}{1 + C e^{-rt}} ) can be expressed in terms of logarithms, which is what I did.Alternatively, maybe I can compute the integral numerically to check.Alternatively, let's compute the integral numerically step by step.But before that, let me see if 736 mg over 30 days is reasonable. The initial rate is 0.1 mg/day, but as the population grows, the rate increases. The maximum rate would be when the population is at carrying capacity, which is ( 10^6 ), so ( N(t) = 0.0001 * 10^6 = 100 ) mg/day. So, over 30 days, if the rate were constant at 100 mg/day, total would be 3000 mg. But since the population grows from 1000 to approaching 10^6 over 30 days, the average rate would be somewhere between 0.1 and 100 mg/day. So, 736 mg seems plausible, as it's less than 3000.Alternatively, perhaps I can compute the integral numerically using substitution.Wait, another thought: the integral ( int frac{1}{1 + 999 e^{-0.3t}} dt ) can be rewritten as ( int frac{e^{0.3t}}{e^{0.3t} + 999} dt ). Let me try substitution here.Let ( u = e^{0.3t} + 999 ), then ( du/dt = 0.3 e^{0.3t} ), so ( du = 0.3 e^{0.3t} dt ). Therefore, ( dt = frac{du}{0.3 e^{0.3t}} ). But ( e^{0.3t} = u - 999 ). So,( dt = frac{du}{0.3 (u - 999)} )So, the integral becomes:[int frac{e^{0.3t}}{u} times frac{du}{0.3 (u - 999)} = frac{1}{0.3} int frac{u - 999}{u} times frac{1}{u - 999} du = frac{1}{0.3} int frac{1}{u} du]Wait, that seems incorrect. Let me check:Wait, starting from:( int frac{e^{0.3t}}{u} dt ), where ( u = e^{0.3t} + 999 ).But ( e^{0.3t} = u - 999 ), so:( int frac{u - 999}{u} times frac{du}{0.3 (u - 999)} ) = ( frac{1}{0.3} int frac{1}{u} du )Yes, that's correct. So, the integral simplifies to:[frac{1}{0.3} ln |u| + C = frac{1}{0.3} ln(e^{0.3t} + 999) + C]Therefore, the integral from 0 to 30 is:[frac{1}{0.3} [ ln(e^{0.3 times 30} + 999) - ln(e^{0} + 999) ]]Compute each term:First, ( e^{0.3 times 30} = e^{9} approx 8103.08392758So, ( e^{9} + 999 ≈ 8103.0839 + 999 ≈ 9102.0839 )Then, ( e^{0} + 999 = 1 + 999 = 1000 )So, the integral becomes:[frac{1}{0.3} [ ln(9102.0839) - ln(1000) ]]Compute ( ln(9102.0839) ). Let's see, ( ln(9102) ). Since ( e^9 ≈ 8103, e^{9.1} ≈ e^9 * e^0.1 ≈ 8103 * 1.10517 ≈ 8959 ). Hmm, 9102 is a bit higher. Let me compute ( ln(9102) ).Alternatively, using calculator approximation:( ln(9102) ≈ 9.115 ) (since ( e^{9.115} ≈ 9102 ))Similarly, ( ln(1000) ≈ 6.907755 )So, the difference is ( 9.115 - 6.907755 ≈ 2.207245 )Therefore, the integral is:[frac{1}{0.3} times 2.207245 ≈ 3.333333 times 2.207245 ≈ 7.357483]Wait, that's different from my previous result. Wait, no, wait. Wait, in this substitution, I have:The integral ( int_{0}^{30} frac{1}{1 + 999 e^{-0.3t}} dt = frac{1}{0.3} [ ln(e^{0.3t} + 999) ]_{0}^{30} = frac{1}{0.3} [ ln(e^{9} + 999) - ln(1 + 999) ] )Which is approximately:( frac{1}{0.3} [ ln(9102.0839) - ln(1000) ] ≈ frac{1}{0.3} [9.115 - 6.907755] ≈ frac{1}{0.3} times 2.207245 ≈ 7.357483 )Wait, so this is different from my previous result of approximately 2.209045. Wait, that can't be. Wait, no, in the first approach, I had:After substitution, I had:Integral = ( frac{1}{0.3} [ -ln(1000) + 9 + ln(1 + 999 e^{-9}) ] ≈ frac{1}{0.3} times 2.209045 ≈ 7.36348 )Wait, so in the first method, I had approximately 7.36348, and in the second method, I have approximately 7.357483. These are very close, so it seems consistent. So, the integral is approximately 7.36.Therefore, the total nitrogen fixed is:( T = 100 times 7.36 ≈ 736 ) mg.So, approximately 736 mg of nitrogen fixed over 30 days.Wait, but in the first method, I had 736.35 mg, and in the second method, 735.748 mg. So, roughly 736 mg.Therefore, the total amount of nitrogen fixed is approximately 736 mg.But let me check if I did the substitution correctly in the first method.Wait, in the first substitution, I had:Integral = ( frac{1}{0.3} [ -ln(1000) + 9 + ln(1 + 999 e^{-9}) ] ≈ frac{1}{0.3} times 2.209045 ≈ 7.36348 )But in the second substitution, I had:Integral = ( frac{1}{0.3} [ ln(9102.0839) - ln(1000) ] ≈ frac{1}{0.3} times 2.207245 ≈ 7.357483 )So, both methods give approximately 7.36, which is consistent.Therefore, the total nitrogen fixed is approximately 736 mg.Wait, but let me compute it more accurately.Compute ( ln(9102.0839) ):We know that ( e^9 ≈ 8103.0839 ), so ( e^{9.1} ≈ e^9 * e^{0.1} ≈ 8103.0839 * 1.10517 ≈ 8103.0839 * 1.10517 ≈ 8959 ). So, ( e^{9.1} ≈ 8959 ), which is less than 9102. So, ( ln(9102) ) is slightly more than 9.1.Compute ( e^{9.12} ):( e^{0.12} ≈ 1.1275 ), so ( e^{9.12} ≈ e^9 * e^{0.12} ≈ 8103.0839 * 1.1275 ≈ 8103.0839 * 1.1275 ≈ let's compute:8103.0839 * 1 = 8103.08398103.0839 * 0.12 = 972.370068Total ≈ 8103.0839 + 972.370068 ≈ 9075.453968Still less than 9102. So, ( e^{9.12} ≈ 9075.45 ), which is less than 9102.Compute ( e^{9.13} ):( e^{0.13} ≈ 1.1393 ), so ( e^{9.13} ≈ 8103.0839 * 1.1393 ≈ 8103.0839 * 1.1393 ≈ let's compute:8103.0839 * 1 = 8103.08398103.0839 * 0.13 = 1053.4009Total ≈ 8103.0839 + 1053.4009 ≈ 9156.4848Which is more than 9102. So, ( e^{9.13} ≈ 9156.48 ), which is higher than 9102.So, ( ln(9102) ) is between 9.12 and 9.13.Let me use linear approximation.Let’s denote ( f(x) = e^{x} ). We know:At ( x = 9.12 ), ( f(x) ≈ 9075.45 )At ( x = 9.13 ), ( f(x) ≈ 9156.48 )We need to find ( x ) such that ( f(x) = 9102 ).The difference between 9102 and 9075.45 is 26.55.The total difference between 9156.48 and 9075.45 is 81.03.So, the fraction is 26.55 / 81.03 ≈ 0.3276.Therefore, ( x ≈ 9.12 + 0.3276 * 0.01 ≈ 9.12 + 0.003276 ≈ 9.123276 )So, ( ln(9102) ≈ 9.1233 )Similarly, ( ln(1000) = 6.907755 )So, the difference is ( 9.1233 - 6.907755 ≈ 2.215545 )Therefore, the integral is:( frac{1}{0.3} times 2.215545 ≈ 3.333333 times 2.215545 ≈ 7.38515 )So, more accurately, the integral is approximately 7.38515.Therefore, total nitrogen fixed:( T = 100 times 7.38515 ≈ 738.515 ) mg.So, approximately 738.52 mg.Wait, so depending on the precision, it's about 736 to 738 mg.Alternatively, perhaps I can use more precise calculation.Alternatively, perhaps I can use the exact expression.Wait, let me compute ( ln(e^{9} + 999) - ln(1000) ).We have ( e^{9} ≈ 8103.08392758 ), so ( e^{9} + 999 ≈ 8103.08392758 + 999 = 9102.08392758 )So, ( ln(9102.08392758) - ln(1000) = ln(9102.08392758 / 1000) = ln(9.10208392758) )Compute ( ln(9.10208392758) ). Let me compute this.We know that ( ln(9) ≈ 2.1972245773 ), ( ln(9.1) ≈ 2.208457448 ), ( ln(9.10208392758) ) is slightly more than 2.208457448.Compute ( ln(9.10208392758) ):Let me use Taylor series around 9.1.Let ( x = 9.1 ), ( f(x) = ln(x) ), ( f'(x) = 1/x ).We have ( f(9.1) ≈ 2.208457448 )We need ( f(9.10208392758) ). The difference is ( 9.10208392758 - 9.1 = 0.00208392758 )So, using linear approximation:( f(x + Delta x) ≈ f(x) + f'(x) Delta x )So,( f(9.10208392758) ≈ 2.208457448 + (1/9.1) * 0.00208392758 )Compute ( 1/9.1 ≈ 0.1098901099 )So,( 0.1098901099 * 0.00208392758 ≈ 0.0002292 )Therefore,( f(9.10208392758) ≈ 2.208457448 + 0.0002292 ≈ 2.2086866 )So, ( ln(9.10208392758) ≈ 2.2086866 )Therefore, the integral is:( frac{1}{0.3} times 2.2086866 ≈ 3.333333 times 2.2086866 ≈ 7.3622887 )So, approximately 7.3623.Therefore, total nitrogen fixed is:( T = 100 times 7.3623 ≈ 736.23 ) mg.So, approximately 736.23 mg.Given the approximations, I think 736 mg is a reasonable answer.Alternatively, perhaps I can use a calculator for more precision, but since this is a theoretical problem, 736 mg is acceptable.Therefore, the total amount of nitrogen fixed over 30 days is approximately 736 mg.Final AnswerThe total amount of nitrogen fixed in the soil over 30 days is boxed{736} mg.</think>

🧠:<think>Okay, so I'm trying to figure out how to solve these two problems related to the scholarship fund for the fast-food restaurant. Let me take it step by step.First, the problem says that the monthly profit is modeled by the function P(t) = 5e^{0.08t} cos(πt) + 20, where t is the number of months since the strategies were implemented. They want to allocate 10% of the monthly profit to the scholarship fund. For the first part, I need to determine the average monthly amount allocated to the scholarship fund over the first year, which is 12 months. The average value of a function over an interval [a, b] is given by (1/(b-a)) times the integral of the function from a to b. So, since we're looking at the first year, t goes from 0 to 12.But wait, the allocation is 10% of the profit, so the amount allocated each month is 0.1 * P(t). Therefore, the average amount allocated over the first year would be the average of 0.1 * P(t) from t=0 to t=12.So, mathematically, that would be:Average allocation = (1/12) * ∫₀¹² 0.1 * P(t) dtWhich simplifies to:Average allocation = (0.1/12) * ∫₀¹² [5e^{0.08t} cos(πt) + 20] dtSo, I can split this integral into two parts:= (0.1/12) * [∫₀¹² 5e^{0.08t} cos(πt) dt + ∫₀¹² 20 dt]Let me compute each integral separately.First, the integral of 5e^{0.08t} cos(πt) dt from 0 to 12. Hmm, integrating e^{at} cos(bt) dt is a standard integral, but I need to recall the formula. I think it's something like e^{at}/(a² + b²) [a cos(bt) + b sin(bt)] + C. Let me verify that.Yes, the integral of e^{at} cos(bt) dt is e^{at}/(a² + b²) [a cos(bt) + b sin(bt)] + C.So, in our case, a = 0.08 and b = π.Therefore, the integral becomes:5 * [e^{0.08t}/(0.08² + π²) (0.08 cos(πt) + π sin(πt))] evaluated from 0 to 12.Let me compute this step by step.First, compute the denominator: 0.08² + π² = 0.0064 + 9.8696 ≈ 9.876.So, the integral is approximately:5 / 9.876 * [e^{0.08t} (0.08 cos(πt) + π sin(πt))] from 0 to 12.Compute this at t=12 and t=0.At t=12:e^{0.08*12} = e^{0.96} ≈ e^{0.96} ≈ 2.6117.0.08 cos(12π) + π sin(12π). Since cos(12π) = cos(0) = 1, and sin(12π) = 0.So, 0.08*1 + π*0 = 0.08.Therefore, the term at t=12 is 2.6117 * 0.08 ≈ 0.2089.At t=0:e^{0} = 1.0.08 cos(0) + π sin(0) = 0.08*1 + π*0 = 0.08.So, the term at t=0 is 1 * 0.08 = 0.08.Therefore, the integral from 0 to 12 is approximately:5 / 9.876 * (0.2089 - 0.08) ≈ 5 / 9.876 * 0.1289 ≈ (5 * 0.1289) / 9.876 ≈ 0.6445 / 9.876 ≈ 0.0652.So, the first integral is approximately 0.0652.Now, the second integral is ∫₀¹² 20 dt, which is straightforward. The integral of 20 from 0 to 12 is 20*(12 - 0) = 240.Therefore, the total integral inside the brackets is approximately 0.0652 + 240 ≈ 240.0652.Then, the average allocation is (0.1 / 12) * 240.0652 ≈ (0.008333) * 240.0652 ≈ 2.00054.So, approximately 2,000.54 per month on average. But since the question says the profit is in thousands of dollars, so 2.00054 thousand dollars is approximately 2,000.54. But since we're talking about the average monthly amount allocated, which is 10% of the profit, and the profit is in thousands, so the average allocation is approximately 2,000.54 per month.Wait, but let me double-check my calculations because 0.0652 seems quite small compared to 240. Maybe I made a mistake in computing the first integral.Let me recalculate the first integral:Integral of 5e^{0.08t} cos(πt) dt from 0 to 12.We have:5 * [e^{0.08t}/(0.08² + π²) (0.08 cos(πt) + π sin(πt))] from 0 to 12.Compute denominator: 0.0064 + 9.8696 = 9.876.Compute numerator at t=12:e^{0.96} ≈ 2.6117.0.08 cos(12π) + π sin(12π) = 0.08*1 + π*0 = 0.08.So, 2.6117 * 0.08 ≈ 0.2089.At t=0:e^{0} = 1.0.08 cos(0) + π sin(0) = 0.08*1 + π*0 = 0.08.So, 1 * 0.08 = 0.08.Therefore, the difference is 0.2089 - 0.08 = 0.1289.Multiply by 5: 5 * 0.1289 = 0.6445.Divide by 9.876: 0.6445 / 9.876 ≈ 0.0652.So, that seems correct. So, the first integral is approximately 0.0652, which is about 0.0652 thousand dollars, so 65.2.The second integral is 240 thousand dollars, so 240,000.Therefore, total integral is 0.0652 + 240 ≈ 240.0652 thousand dollars.Multiply by 0.1/12: 0.1/12 ≈ 0.008333.0.008333 * 240.0652 ≈ 2.00054 thousand dollars, so approximately 2,000.54 per month.But wait, the profit function is P(t) = 5e^{0.08t} cos(πt) + 20. The 20 is in thousands, so the base profit is 20,000 per month, plus some oscillating component.But when I integrate 20 from 0 to 12, I get 240, which is 240 thousand dollars, so 240,000. Then, the first integral is about 65.2, so total is 240,065.2.Then, 10% of that is 24,006.52, and the average per month is 24,006.52 / 12 ≈ 2,000.54.Wait, but actually, the average allocation is (1/12) * ∫₀¹² 0.1 P(t) dt, which is 0.1 * (1/12) * ∫₀¹² P(t) dt.So, yes, that's correct.But let me think again: the integral of P(t) from 0 to 12 is approximately 240.0652 thousand dollars, so the total profit over 12 months is about 240,065.2.10% of that is 24,006.52, so the average monthly allocation is 24,006.52 / 12 ≈ 2,000.54.But since the question asks for the average monthly amount, which is in thousands, so 2.00054 thousand dollars, which is approximately 2,000.54.But let me check if I can compute the integral more accurately.Alternatively, maybe I can use a calculator or more precise computation.But perhaps I can also note that the integral of e^{at} cos(bt) dt is known, so maybe I can compute it more precisely.But for the sake of time, maybe I can accept that the first integral is approximately 0.0652, so the total integral is 240.0652, leading to an average allocation of approximately 2,000.54 per month.But let me also consider that the 5e^{0.08t} cos(πt) term might have some periodicity. Since cos(πt) has a period of 2 months, so over 12 months, it completes 6 periods. The integral over each period might cancel out due to the oscillation, but with the exponential factor, it's not exactly symmetric.But in any case, my calculation gives approximately 2,000.54 per month.Wait, but let me think again: 0.1 * (1/12) * ∫₀¹² P(t) dt.P(t) = 5e^{0.08t} cos(πt) + 20.So, ∫₀¹² P(t) dt = ∫₀¹² 5e^{0.08t} cos(πt) dt + ∫₀¹² 20 dt.We computed ∫₀¹² 5e^{0.08t} cos(πt) dt ≈ 0.0652 thousand dollars.∫₀¹² 20 dt = 20*12 = 240 thousand dollars.So, total integral is 240.0652 thousand dollars.Then, 0.1 * (1/12) * 240.0652 ≈ 0.1 * 20.0054 ≈ 2.00054 thousand dollars per month.So, approximately 2,000.54 per month.But let me check if I can compute the integral more accurately.Alternatively, maybe I can use integration by parts or another method, but I think the formula I used is correct.Alternatively, perhaps I can use a calculator to compute the integral numerically.But since I don't have a calculator here, I'll proceed with the approximate value.So, the average monthly allocation is approximately 2,000.54.But since the question asks for the answer in a box, I think I should present it as approximately 2,000.54, but maybe rounded to the nearest dollar or to two decimal places.Alternatively, perhaps I can express it as a fraction.But let me see: 0.0652 is approximately 0.0652, so 0.0652 + 240 = 240.0652.Then, 240.0652 * 0.1 = 24.00652.Divide by 12: 24.00652 / 12 ≈ 2.00054.So, approximately 2.00054 thousand dollars, which is 2,000.54.So, I think that's the answer for part 1.Now, moving on to part 2.The scholarship fund aims to accumulate at least 50,000 by the end of the second year. So, from month 13 to month 24, we need to calculate the minimum total percentage of monthly profit that should be allocated to the fund.Wait, but actually, the first year is months 1-12, and the second year is months 13-24.But the problem says "from month 13 to month 24", so that's 12 months as well.But the fund aims to accumulate at least 50,000 by the end of the second year, which is 24 months.But wait, the first year's allocation is 10%, which we've calculated as approximately 2,000.54 per month on average, so total for the first year is 12 * 2,000.54 ≈ 24,006.48.But the fund needs to accumulate at least 50,000 by the end of the second year, so the total needed from both years is 50,000.Wait, no, actually, the problem says "accumulate at least 50,000 by the end of the second year", so that's the total amount in the fund after 24 months.But the first year's allocation is 10%, which is approximately 24,006.48, as above.Therefore, the amount needed from the second year is 50,000 - 24,006.48 ≈ 25,993.52.So, we need to find the minimum total percentage of monthly profit that should be allocated from month 13 to month 24 to accumulate at least 25,993.52.But wait, actually, the fund is accumulating each month, so the total amount after 24 months is the sum of the allocations from month 1 to 24.But the first 12 months are allocated at 10%, and the next 12 months at a higher percentage, say x%, such that the total allocation is at least 50,000.So, total allocation = sum_{t=1}^{12} 0.1 P(t) + sum_{t=13}^{24} x P(t) ≥ 50,000.We need to find the minimum x such that this inequality holds.But since we already calculated sum_{t=1}^{12} 0.1 P(t) ≈ 24,006.48, then sum_{t=13}^{24} x P(t) ≥ 50,000 - 24,006.48 ≈ 25,993.52.So, sum_{t=13}^{24} x P(t) ≥ 25,993.52.Therefore, x * sum_{t=13}^{24} P(t) ≥ 25,993.52.So, x ≥ 25,993.52 / sum_{t=13}^{24} P(t).Therefore, we need to compute sum_{t=13}^{24} P(t), which is the total profit from month 13 to 24.But since P(t) = 5e^{0.08t} cos(πt) + 20, we can compute the integral from t=12 to t=24 of P(t) dt, but since we need the sum, which is the integral over each month, but actually, the sum is the integral over each month's profit.But wait, actually, the sum from t=13 to t=24 is the same as the integral from t=12 to t=24 of P(t) dt, because each month's profit is P(t) at integer t, but since the function is continuous, the sum is approximately the integral.But actually, the sum from t=13 to t=24 is the same as the integral from t=12 to t=24 of P(t) dt, because each month's profit is P(t) at t=12, 13, ..., 24, but actually, t is in months, so t=12 is the end of the first year, and t=13 is the start of the second year.Wait, actually, in the function P(t), t is the number of months since implementation, so t=0 is the first month, t=1 is the second month, etc. So, t=12 is the 13th month, which is the start of the second year.Wait, no, t=0 is the first month, t=1 is the second month, so t=12 is the 13th month, which is the start of the second year.Wait, that might complicate things. Let me clarify.If t=0 is the first month, then t=12 is the 13th month, which is the start of the second year. So, the first year is t=0 to t=11, and the second year is t=12 to t=23.Wait, but the problem says "from month 13 to month 24", which would correspond to t=12 to t=23, since t=0 is month 1.Wait, no, actually, t=0 is month 1, t=1 is month 2, ..., t=11 is month 12, t=12 is month 13, t=23 is month 24.Therefore, the second year is t=12 to t=23, corresponding to months 13 to 24.Therefore, the sum from t=13 to t=24 is the sum from t=12 to t=23 of P(t).So, we need to compute sum_{t=12}^{23} P(t) = sum_{t=12}^{23} [5e^{0.08t} cos(πt) + 20].But since we need the sum, which is the total profit from month 13 to 24, we can compute it as the integral from t=12 to t=24 of P(t) dt, but actually, the sum is the integral over each month, so it's the same as the integral from t=12 to t=24 of P(t) dt.Wait, no, the sum is the sum of P(t) at integer t from t=12 to t=23, which is 12 terms.But integrating P(t) from t=12 to t=24 would give the area under the curve, which is different from the sum of P(t) at integer points.But for the sake of approximation, perhaps we can use the integral to estimate the sum.Alternatively, maybe we can compute the sum exactly by evaluating P(t) at each integer t from 12 to 23 and summing them up.But that would be tedious, but perhaps manageable.Alternatively, we can note that the integral from t=12 to t=24 of P(t) dt is approximately equal to the sum from t=12 to t=23 of P(t) + some error term, but for simplicity, maybe we can use the integral as an approximation.But let me think: since P(t) is a continuous function, and we're summing over 12 months, the sum can be approximated by the integral from t=12 to t=24 of P(t) dt.But actually, the sum is equal to the integral from t=12 to t=24 of P(t) dt plus some correction terms, but for simplicity, maybe we can proceed with the integral.Alternatively, perhaps we can compute the integral from t=12 to t=24 of P(t) dt, and then use that as an approximation for the sum.So, let's compute ∫₁²⁴ P(t) dt = ∫₁²⁴ [5e^{0.08t} cos(πt) + 20] dt.We can split this into two integrals:∫₁²⁴ 5e^{0.08t} cos(πt) dt + ∫₁²⁴ 20 dt.We already have a formula for the first integral, so let's compute it.Again, using the integral formula:∫ e^{at} cos(bt) dt = e^{at}/(a² + b²) [a cos(bt) + b sin(bt)] + C.So, for the first integral:5 * [e^{0.08t}/(0.08² + π²) (0.08 cos(πt) + π sin(πt))] evaluated from 12 to 24.Compute the denominator again: 0.0064 + 9.8696 ≈ 9.876.Compute at t=24:e^{0.08*24} = e^{1.92} ≈ 6.812.0.08 cos(24π) + π sin(24π) = 0.08*1 + π*0 = 0.08.So, term at t=24: 6.812 * 0.08 ≈ 0.54496.At t=12:e^{0.08*12} = e^{0.96} ≈ 2.6117.0.08 cos(12π) + π sin(12π) = 0.08*1 + π*0 = 0.08.So, term at t=12: 2.6117 * 0.08 ≈ 0.2089.Therefore, the integral from 12 to 24 is:5 / 9.876 * (0.54496 - 0.2089) ≈ 5 / 9.876 * 0.33606 ≈ (5 * 0.33606) / 9.876 ≈ 1.6803 / 9.876 ≈ 0.1701.So, the first integral is approximately 0.1701 thousand dollars.The second integral is ∫₁²⁴ 20 dt = 20*(24 - 12) = 20*12 = 240 thousand dollars.Therefore, the total integral from 12 to 24 is approximately 0.1701 + 240 ≈ 240.1701 thousand dollars.But wait, this is the integral, which approximates the sum of P(t) from t=12 to t=24, but actually, the sum is the integral plus some correction. However, for the sake of this problem, maybe we can use this integral as an approximation.But actually, the sum from t=12 to t=23 of P(t) is equal to the integral from t=12 to t=24 of P(t) dt minus the average of P(12) and P(24) times 1, but I'm not sure.Alternatively, perhaps it's better to compute the sum exactly by evaluating P(t) at each integer t from 12 to 23 and summing them up.But that would be time-consuming, but let's try.First, compute P(t) for t=12 to t=23.P(t) = 5e^{0.08t} cos(πt) + 20.Note that cos(πt) for integer t is cos(kπ) where k is integer, which is (-1)^k.So, for t=12: cos(12π)=1t=13: cos(13π)=cos(π)= -1t=14: cos(14π)=1t=15: cos(15π)= -1And so on, alternating between 1 and -1.Similarly, e^{0.08t} increases exponentially.So, let's compute P(t) for t=12 to t=23.t=12:P(12) = 5e^{0.96} * 1 + 20 ≈ 5*2.6117 + 20 ≈ 13.0585 + 20 ≈ 33.0585 thousand dollars.t=13:P(13) = 5e^{1.04} * (-1) + 20 ≈ 5*2.828 * (-1) + 20 ≈ -14.14 + 20 ≈ 5.86 thousand dollars.t=14:P(14) = 5e^{1.12} * 1 + 20 ≈ 5*3.065 * 1 + 20 ≈ 15.325 + 20 ≈ 35.325 thousand dollars.t=15:P(15) = 5e^{1.2} * (-1) + 20 ≈ 5*3.3201 * (-1) + 20 ≈ -16.6005 + 20 ≈ 3.3995 thousand dollars.t=16:P(16) = 5e^{1.28} * 1 + 20 ≈ 5*3.5938 * 1 + 20 ≈ 17.969 + 20 ≈ 37.969 thousand dollars.t=17:P(17) = 5e^{1.36} * (-1) + 20 ≈ 5*3.900 * (-1) + 20 ≈ -19.5 + 20 ≈ 0.5 thousand dollars.t=18:P(18) = 5e^{1.44} * 1 + 20 ≈ 5*4.228 * 1 + 20 ≈ 21.14 + 20 ≈ 41.14 thousand dollars.t=19:P(19) = 5e^{1.52} * (-1) + 20 ≈ 5*4.57 * (-1) + 20 ≈ -22.85 + 20 ≈ -2.85 thousand dollars.Wait, that can't be right. Profit can't be negative. Did I make a mistake?Wait, P(t) = 5e^{0.08t} cos(πt) + 20.At t=19, cos(19π)=cos(π)= -1, so P(19)=5e^{1.52}*(-1)+20.Compute e^{1.52} ≈ 4.57.So, 5*4.57=22.85, so 22.85*(-1) +20= -22.85 +20= -2.85.But profit can't be negative, so perhaps the model allows for that, but in reality, the profit would be at least 0. But since the problem says the profit is modeled by this function, we'll proceed.t=20:P(20)=5e^{1.6} *1 +20≈5*4.953*1 +20≈24.765 +20≈44.765.t=21:P(21)=5e^{1.68}*(-1)+20≈5*5.36* (-1)+20≈-26.8 +20≈-6.8.t=22:P(22)=5e^{1.76}*1 +20≈5*5.808*1 +20≈29.04 +20≈49.04.t=23:P(23)=5e^{1.84}*(-1)+20≈5*6.30* (-1)+20≈-31.5 +20≈-11.5.So, let's list all these:t=12: 33.0585t=13: 5.86t=14: 35.325t=15: 3.3995t=16: 37.969t=17: 0.5t=18: 41.14t=19: -2.85t=20: 44.765t=21: -6.8t=22: 49.04t=23: -11.5Now, let's sum these up.Compute the sum step by step:Start with 33.0585+5.86 = 38.9185+35.325 = 74.2435+3.3995 = 77.643+37.969 = 115.612+0.5 = 116.112+41.14 = 157.252-2.85 = 154.402+44.765 = 199.167-6.8 = 192.367+49.04 = 241.407-11.5 = 230.007 thousand dollars.So, the total sum from t=12 to t=23 is approximately 230.007 thousand dollars.Wait, that's 230,007.But earlier, the integral from t=12 to t=24 was approximately 240.1701 thousand dollars, which is higher than the sum. So, the sum is about 230.007 thousand dollars.Therefore, the total profit from month 13 to 24 is approximately 230,007.But wait, the sum we computed is 230.007 thousand dollars, which is 230,007.But the integral was 240.1701 thousand dollars, which is higher because the integral includes the area under the curve, which is more than the sum of the discrete points.But in any case, the sum is approximately 230,007.Therefore, the total amount needed from the second year is 25,993.52, as calculated earlier.So, we need sum_{t=13}^{24} x P(t) ≥ 25,993.52.But sum_{t=13}^{24} P(t) ≈ 230.007 thousand dollars.Therefore, x * 230.007 ≥ 25.99352 thousand dollars.Wait, no, wait: 25,993.52 is in dollars, and sum_{t=13}^{24} P(t) is in thousands of dollars, so 230.007 thousand dollars is 230,007.Therefore, x * 230.007 ≥ 25.99352.Wait, no, 25,993.52 is in dollars, so in thousands, it's 25.99352 thousand dollars.So, x * 230.007 ≥ 25.99352.Therefore, x ≥ 25.99352 / 230.007 ≈ 0.11297.So, x ≈ 11.297%.Therefore, the minimum total percentage is approximately 11.3%.But let me double-check the calculations.Total needed from second year: 50,000 - 24,006.48 ≈ 25,993.52 dollars.Sum of P(t) from t=12 to t=23: 230.007 thousand dollars, which is 230,007 dollars.Therefore, x * 230,007 ≥ 25,993.52.So, x ≥ 25,993.52 / 230,007 ≈ 0.11297, which is 11.297%.So, approximately 11.3%.But let me check if I can compute it more accurately.25,993.52 / 230,007 ≈ 0.11297.So, 11.297%, which is approximately 11.3%.Therefore, the minimum total percentage is approximately 11.3%.But since the question asks for the minimum total percentage, we can round it to two decimal places, so 11.30%.Alternatively, if we need a whole number, 11.3% is acceptable.But let me check if I made any mistakes in the sum.Wait, when I summed up the P(t) values from t=12 to t=23, I got 230.007 thousand dollars, which is 230,007 dollars.But let me recount the sum:t=12:33.0585t=13:5.86 → total 38.9185t=14:35.325 → 74.2435t=15:3.3995 →77.643t=16:37.969 →115.612t=17:0.5 →116.112t=18:41.14 →157.252t=19:-2.85 →154.402t=20:44.765 →199.167t=21:-6.8 →192.367t=22:49.04 →241.407t=23:-11.5 →230.007Yes, that seems correct.So, the sum is 230.007 thousand dollars.Therefore, x = 25.99352 / 230.007 ≈ 0.11297, or 11.297%.So, approximately 11.3%.Therefore, the minimum total percentage is approximately 11.3%.But let me think again: the problem says "the minimum total percentage of monthly profit that should now be allocated to the fund from month 13 to month 24".So, the total percentage is x%, such that x% of each month's profit from 13 to 24 is allocated, and the sum of these allocations is at least 25,993.52.Therefore, x is approximately 11.3%.But let me check if I can express it as a fraction.0.11297 is approximately 11.3%, so 11.3%.Alternatively, perhaps I can compute it more precisely.25,993.52 / 230,007 ≈ 0.11297.So, 0.11297 is approximately 11.297%, which is 11.3% when rounded to one decimal place.Therefore, the minimum total percentage is approximately 11.3%.But let me check if I can compute it more accurately.Alternatively, perhaps I can use more precise values for e^{0.08t}.But for the sake of time, I think 11.3% is a reasonable approximation.Therefore, the answers are:1. The average monthly allocation is approximately 2,000.54.2. The minimum total percentage is approximately 11.3%.But let me present them in the required format.</think>