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❓:A political science major, Alex, is analyzing voting patterns in a country with two major political parties: Party A and Party B. Alex believes that philosophical theories on voter behavior are irrelevant and instead focuses on practical, data-driven approaches. He decides to model the voting population as a continuous function based on recent polling data.1. Let the function ( V_A(x) ) represent the proportion of voters supporting Party A at a given time ( x ) (measured in months). Suppose ( V_A(x) ) is defined by the differential equation:[ frac{dV_A}{dx} = k cdot (1 - V_A(x)) cdot V_A(x) ]where ( k ) is a positive constant. If at ( x = 0 ), 40% of the voters support Party A (( V_A(0) = 0.4 )), solve the differential equation to find ( V_A(x) ).2. In addition, Alex needs to predict the point in time ( x ) where the support for Party A will surpass 60%. Based on the solution from the first sub-problem, determine the value of ( x ) when ( V_A(x) = 0.6 ).
🧠:<think>Alright, so I have this problem about modeling voter support for Party A using a differential equation. Let me try to break it down step by step.First, the problem states that the function ( V_A(x) ) represents the proportion of voters supporting Party A at time ( x ) in months. The differential equation given is:[ frac{dV_A}{dx} = k cdot (1 - V_A(x)) cdot V_A(x) ]where ( k ) is a positive constant. The initial condition is ( V_A(0) = 0.4 ).Okay, so this looks like a logistic growth model. I remember that the logistic equation is used to model population growth where there's a carrying capacity, but in this case, it's modeling the proportion of voters supporting a party. The form of the differential equation is similar to the logistic equation, which is:[ frac{dN}{dt} = rN left(1 - frac{N}{K}right) ]Here, ( r ) is the growth rate, and ( K ) is the carrying capacity. Comparing this to our equation, it seems like ( V_A(x) ) is analogous to ( N ), and ( k ) is analogous to ( r ). The term ( (1 - V_A(x)) ) is like ( left(1 - frac{N}{K}right) ), but since ( V_A(x) ) is a proportion, the carrying capacity ( K ) is 1, meaning the maximum proportion is 100%.So, the equation is a logistic differential equation with ( K = 1 ) and growth rate ( k ). The solution to the logistic equation is:[ V_A(x) = frac{1}{1 + left(frac{1 - V_A(0)}{V_A(0)}right) e^{-kx}} ]Let me verify that. Starting from the logistic equation:[ frac{dV_A}{dx} = k V_A (1 - V_A) ]This is a separable differential equation. So, I can rewrite it as:[ frac{dV_A}{V_A (1 - V_A)} = k dx ]Integrating both sides. Let me do that step by step.First, the left side integral:[ int frac{1}{V_A (1 - V_A)} dV_A ]This can be solved using partial fractions. Let me set:[ frac{1}{V_A (1 - V_A)} = frac{A}{V_A} + frac{B}{1 - V_A} ]Multiplying both sides by ( V_A (1 - V_A) ):[ 1 = A(1 - V_A) + B V_A ]Expanding:[ 1 = A - A V_A + B V_A ]Grouping terms:[ 1 = A + (B - A) V_A ]Since this must hold for all ( V_A ), the coefficients must be equal on both sides. So,For the constant term: ( A = 1 )For the ( V_A ) term: ( B - A = 0 ) => ( B = A = 1 )So, the integral becomes:[ int left( frac{1}{V_A} + frac{1}{1 - V_A} right) dV_A = int k dx ]Integrating term by term:Left side:[ int frac{1}{V_A} dV_A + int frac{1}{1 - V_A} dV_A = ln |V_A| - ln |1 - V_A| + C ]Which simplifies to:[ ln left| frac{V_A}{1 - V_A} right| + C ]Right side:[ int k dx = kx + C ]So, combining both sides:[ ln left( frac{V_A}{1 - V_A} right) = kx + C ]Exponentiating both sides to eliminate the logarithm:[ frac{V_A}{1 - V_A} = e^{kx + C} = e^{kx} cdot e^C ]Let me denote ( e^C ) as another constant, say ( C' ). So,[ frac{V_A}{1 - V_A} = C' e^{kx} ]Solving for ( V_A ):Multiply both sides by ( 1 - V_A ):[ V_A = C' e^{kx} (1 - V_A) ][ V_A = C' e^{kx} - C' e^{kx} V_A ]Bring the ( V_A ) term to the left:[ V_A + C' e^{kx} V_A = C' e^{kx} ]Factor out ( V_A ):[ V_A (1 + C' e^{kx}) = C' e^{kx} ]So,[ V_A = frac{C' e^{kx}}{1 + C' e^{kx}} ]We can rewrite this as:[ V_A = frac{1}{1 + frac{1}{C'} e^{-kx}} ]Let me denote ( frac{1}{C'} ) as another constant, say ( C'' ). So,[ V_A = frac{1}{1 + C'' e^{-kx}} ]Now, apply the initial condition ( V_A(0) = 0.4 ):At ( x = 0 ):[ 0.4 = frac{1}{1 + C'' e^{0}} = frac{1}{1 + C''} ]Solving for ( C'' ):[ 0.4 (1 + C'') = 1 ][ 0.4 + 0.4 C'' = 1 ][ 0.4 C'' = 0.6 ][ C'' = frac{0.6}{0.4} = 1.5 ]So, ( C'' = 1.5 ). Therefore, the solution is:[ V_A(x) = frac{1}{1 + 1.5 e^{-kx}} ]Alternatively, we can write this as:[ V_A(x) = frac{1}{1 + left( frac{1 - 0.4}{0.4} right) e^{-kx}} ]Because ( C'' = frac{1 - V_A(0)}{V_A(0)} = frac{0.6}{0.4} = 1.5 ). So, that's consistent with the standard logistic solution.So, that's the solution for part 1.Now, moving on to part 2. We need to find the time ( x ) when ( V_A(x) = 0.6 ).So, set ( V_A(x) = 0.6 ) in the solution:[ 0.6 = frac{1}{1 + 1.5 e^{-kx}} ]Solving for ( x ).First, take reciprocals:[ frac{1}{0.6} = 1 + 1.5 e^{-kx} ]Calculate ( frac{1}{0.6} ):[ frac{1}{0.6} = frac{10}{6} = frac{5}{3} approx 1.6667 ]So,[ frac{5}{3} = 1 + 1.5 e^{-kx} ]Subtract 1 from both sides:[ frac{5}{3} - 1 = 1.5 e^{-kx} ][ frac{2}{3} = 1.5 e^{-kx} ]Divide both sides by 1.5:[ frac{2}{3} div 1.5 = e^{-kx} ]Convert 1.5 to fraction: 1.5 = 3/2So,[ frac{2}{3} div frac{3}{2} = frac{2}{3} times frac{2}{3} = frac{4}{9} ]Thus,[ frac{4}{9} = e^{-kx} ]Take natural logarithm on both sides:[ ln left( frac{4}{9} right) = -kx ]So,[ x = -frac{1}{k} ln left( frac{4}{9} right) ]Simplify the logarithm:Note that ( ln left( frac{4}{9} right) = ln(4) - ln(9) = 2ln(2) - 2ln(3) = 2(ln(2) - ln(3)) )So,[ x = -frac{1}{k} times 2(ln(2) - ln(3)) ]Which is:[ x = frac{2}{k} (ln(3) - ln(2)) ]Alternatively, since ( ln(3/2) = ln(3) - ln(2) ), we can write:[ x = frac{2}{k} lnleft( frac{3}{2} right) ]So, that's the time when the support for Party A reaches 60%.Wait, let me double-check the algebra steps because sometimes when dealing with fractions, it's easy to make a mistake.Starting from:[ 0.6 = frac{1}{1 + 1.5 e^{-kx}} ]Multiply both sides by denominator:[ 0.6 (1 + 1.5 e^{-kx}) = 1 ][ 0.6 + 0.9 e^{-kx} = 1 ]Subtract 0.6:[ 0.9 e^{-kx} = 0.4 ]Divide both sides by 0.9:[ e^{-kx} = frac{0.4}{0.9} = frac{4}{9} ]So, same as before.Then, taking natural log:[ -kx = ln(4/9) ]Multiply both sides by -1:[ kx = -ln(4/9) = ln(9/4) ]So,[ x = frac{1}{k} lnleft( frac{9}{4} right) ]Wait, this is different from what I had earlier. Let me see.Wait, ( ln(9/4) = ln(9) - ln(4) = 2ln(3) - 2ln(2) = 2(ln(3) - ln(2)) ). So,[ x = frac{2}{k} (ln(3) - ln(2)) ]Which is the same as before. So, both ways, same result.Alternatively, ( ln(9/4) = ln( (3/2)^2 ) = 2 ln(3/2) ). So,[ x = frac{2}{k} lnleft( frac{3}{2} right) ]So, that's correct.Therefore, the time ( x ) when ( V_A(x) = 0.6 ) is ( frac{2}{k} lnleft( frac{3}{2} right) ) months.But wait, in the first approach, I had:[ x = frac{2}{k} (ln(3) - ln(2)) ]Which is the same as ( frac{2}{k} ln(3/2) ). So, both expressions are equivalent.So, that's the answer for part 2.Let me recap:1. Solved the logistic differential equation with initial condition ( V_A(0) = 0.4 ) and found the solution ( V_A(x) = frac{1}{1 + 1.5 e^{-kx}} ).2. Set ( V_A(x) = 0.6 ) and solved for ( x ), resulting in ( x = frac{2}{k} lnleft( frac{3}{2} right) ).I think that's solid. Just to make sure, let me plug the value back into the equation to verify.Let ( x = frac{2}{k} ln(3/2) ). Then,[ V_A(x) = frac{1}{1 + 1.5 e^{-k cdot frac{2}{k} ln(3/2)}} ]Simplify the exponent:[ -k cdot frac{2}{k} ln(3/2) = -2 ln(3/2) = ln( (3/2)^{-2} ) = ln( (2/3)^2 ) = ln(4/9) ]So,[ V_A(x) = frac{1}{1 + 1.5 cdot e^{ln(4/9)}} ]Since ( e^{ln(4/9)} = 4/9 ),[ V_A(x) = frac{1}{1 + 1.5 cdot (4/9)} ]Calculate ( 1.5 times 4/9 ):1.5 is 3/2, so ( (3/2) times (4/9) = (12)/18 = 2/3 )Thus,[ V_A(x) = frac{1}{1 + 2/3} = frac{1}{5/3} = 3/5 = 0.6 ]Perfect, that checks out.So, I think I did everything correctly.Final Answer1. The solution is ( boxed{V_A(x) = dfrac{1}{1 + 1.5 e^{-kx}}} ).2. The time when support reaches 60% is ( boxed{dfrac{2}{k} lnleft( dfrac{3}{2} right)} ) months.</think>
❓:Senator Johnson is analyzing the impact of a proposed tax cut on the education budget of her state. The state currently allocates 10 billion annually to education, funded entirely through tax revenue. The proposed tax cut is expected to reduce total tax revenue by 15%. Senator Johnson needs to understand the long-term economic implications of this tax cut on the education budget and overall state economy.1. Given that the elasticity of state tax revenue with respect to personal income is 0.9, estimate the percentage change in personal income necessary to offset the 15% reduction in tax revenue. Assume the relationship between tax revenue (R) and personal income (Y) can be modeled by the equation ( R = k cdot Y^{0.9} ), where ( k ) is a constant.2. If the reduction in the education budget directly correlates with a decrease in the quality of education, and this quality reduction follows a linear model where a 1% decrease in budget results in a 0.5% decrease in educational outcomes, determine the percentage decrease in educational outcomes resulting from the 15% reduction in the education budget.
🧠:<think>Alright, so Senator Johnson is looking into this tax cut proposal, and I need to help her understand the impact on the education budget and the economy. Let me try to break down the two questions step by step.Starting with the first question: We have a tax cut that's expected to reduce total tax revenue by 15%. The state currently allocates 10 billion annually to education, which is entirely funded by taxes. So, if tax revenue goes down by 15%, that would directly affect the education budget, right? But the question isn't just about the immediate impact; it's asking about the long-term economic implications. Specifically, it wants to know the percentage change in personal income needed to offset this 15% reduction in tax revenue. The elasticity of state tax revenue with respect to personal income is given as 0.9. Elasticity, if I recall correctly, measures the responsiveness of one variable to a change in another. In this case, it's how much tax revenue changes in response to a change in personal income. The formula provided is ( R = k cdot Y^{0.9} ), where ( R ) is tax revenue, ( Y ) is personal income, and ( k ) is a constant. So, elasticity is calculated as the percentage change in tax revenue divided by the percentage change in personal income. Here, the elasticity is 0.9, which means that a 1% increase in personal income leads to a 0.9% increase in tax revenue. But in this case, we have a decrease in tax revenue, so we need to find out what percentage change in personal income would offset that decrease.Let me denote the percentage change in tax revenue as ( Delta R / R ) and the percentage change in personal income as ( Delta Y / Y ). The elasticity formula is:[text{Elasticity} = frac{Delta R / R}{Delta Y / Y}]We know the elasticity is 0.9, and we need to find the percentage change in personal income (( Delta Y / Y )) that would result in a 15% decrease in tax revenue. So, plugging in the numbers:[0.9 = frac{-15%}{Delta Y / Y}]Wait, hold on. The tax revenue is decreasing by 15%, so ( Delta R / R ) is -15%. We need to find ( Delta Y / Y ) such that this equation holds. Rearranging the formula:[Delta Y / Y = frac{-15%}{0.9}]Calculating that:[Delta Y / Y = -16.666...%]So, a decrease in personal income of approximately 16.67% would lead to a 15% decrease in tax revenue. But the question is asking for the percentage change in personal income necessary to offset the 15% reduction in tax revenue. Hmm, does that mean we need to find the change in income that would bring tax revenue back to its original level?Wait, maybe I misinterpreted. If tax revenue is reduced by 15%, and we want to offset that reduction, we need to find the percentage change in personal income that would cause tax revenue to increase by 15% to offset the cut. But that might not make sense because the tax cut is reducing revenue, so to offset it, we need to increase revenue back to the original level. But how?Alternatively, perhaps the question is asking: if tax revenue is going to decrease by 15% due to the tax cut, what percentage increase in personal income is needed to keep tax revenue the same? Because if personal income increases, tax revenue would increase, potentially offsetting the tax cut.Wait, let me think again. The tax cut is reducing tax revenue by 15%, so to offset that, we need to have an increase in tax revenue elsewhere. But the model given is ( R = k cdot Y^{0.9} ). So, if Y increases, R increases. Therefore, to offset the 15% decrease in R, we need to find the percentage increase in Y such that the increase in R from Y offsets the 15% decrease.So, mathematically, we have:Original tax revenue: ( R = k cdot Y^{0.9} )After tax cut: ( R' = R - 0.15R = 0.85R )But we want to find a new Y, say ( Y' ), such that ( R' = k cdot (Y')^{0.9} ). But we also want ( R' = R ) because we want to offset the tax cut. Wait, no, actually, the tax cut is reducing R by 15%, so R' is 0.85R. To offset that, we need to find Y' such that ( k cdot (Y')^{0.9} = R ). But that would mean Y' needs to be higher than Y to bring R back to its original level.Wait, no, perhaps the question is asking: Given that the tax cut reduces R by 15%, what percentage change in Y is needed so that the new R (after both the tax cut and the change in Y) is the same as the original R. So, the net effect is zero change in R.But the model is ( R = k cdot Y^{0.9} ). If the tax cut reduces R by 15%, that's a separate factor. So, perhaps we need to model both effects.Wait, maybe I'm overcomplicating. Let's go back to the elasticity definition. Elasticity is the percentage change in R divided by the percentage change in Y. So, if we have a tax cut that reduces R by 15%, and we want to find the change in Y that would cause R to increase by 15%, thereby offsetting the tax cut.So, in that case, the percentage change in R needed is +15%, and the elasticity is 0.9, so:[0.9 = frac{15%}{Delta Y / Y}]Therefore,[Delta Y / Y = frac{15%}{0.9} = 16.666...%]So, a 16.67% increase in personal income would lead to a 15% increase in tax revenue, offsetting the 15% reduction from the tax cut.But wait, is that the correct interpretation? Because the tax cut is a policy change, not a change in Y. So, perhaps the tax cut reduces R by 15%, and we need to find the change in Y that would bring R back to its original level. So, the tax cut causes R to decrease by 15%, and we need Y to increase such that the increase in R from Y offsets the 15% decrease.So, mathematically, the tax cut reduces R by 15%, so R becomes 0.85R. We need Y to increase such that the new R is R again. So, the increase in R needed is 15% to go from 0.85R to R.So, using the elasticity formula:[text{Elasticity} = frac{Delta R / R}{Delta Y / Y}]We have ( Delta R / R = 15% ) (to offset the 15% decrease), and elasticity is 0.9, so:[0.9 = frac{15%}{Delta Y / Y}]Solving for ( Delta Y / Y ):[Delta Y / Y = frac{15%}{0.9} = 16.666...%]So, a 16.67% increase in personal income is needed to offset the 15% reduction in tax revenue.Wait, but is that the correct way to model it? Because the tax cut is a policy change, not a change in Y. So, perhaps the tax cut reduces R by 15%, and then we need to find the change in Y that would cause R to increase by 15% to offset it. So, the total change in R would be -15% (from tax cut) + 15% (from Y increase) = 0%.But in reality, the tax cut is a separate factor from Y. So, perhaps the model is that R is a function of Y and the tax rate. But in the given model, R is only a function of Y, with the elasticity given. So, perhaps the tax cut is modeled as a separate multiplier on R.Wait, maybe I need to think of it differently. The tax cut reduces R by 15%, so R becomes 0.85R. But we want R to remain the same, so we need Y to increase such that the new R (from Y) is R / 0.85, because 0.85R = k * Y'^0.9, so Y' needs to be such that k * Y'^0.9 = R / 0.85.But R = k * Y^0.9, so:k * Y'^0.9 = (k * Y^0.9) / 0.85Divide both sides by k:Y'^0.9 = Y^0.9 / 0.85Take both sides to the power of 1/0.9:Y' = (Y^0.9 / 0.85)^(1/0.9)Simplify:Y' = Y * (1 / 0.85)^(1/0.9)Calculate (1 / 0.85)^(1/0.9):First, 1 / 0.85 ≈ 1.17647Then, 1.17647^(1/0.9) ≈ 1.17647^1.1111 ≈ Let's calculate that.We can use natural logs:ln(1.17647) ≈ 0.1625Multiply by 1.1111: 0.1625 * 1.1111 ≈ 0.1806Exponentiate: e^0.1806 ≈ 1.198So, Y' ≈ Y * 1.198, which is approximately a 19.8% increase in Y.Wait, that's different from the 16.67% I calculated earlier. So, which approach is correct?I think the confusion arises from whether the tax cut is a separate factor or if it's modeled through Y. Since the model given is R = k * Y^0.9, the tax cut is likely a separate factor. So, the tax cut reduces R by 15%, so R becomes 0.85R. To offset this, we need Y to increase such that R returns to its original level. So, the new R after Y increases should be R = 0.85R * (1 + ΔY/Y)^0.9.Wait, no. Let me think again.If the tax cut reduces R by 15%, then R becomes 0.85R. But we want R to stay the same, so we need the increase in Y to cause R to increase by 15% to offset the tax cut. So, the total R after both effects should be R.So, mathematically:0.85R * (1 + ΔY/Y)^0.9 = RDivide both sides by 0.85R:(1 + ΔY/Y)^0.9 = 1 / 0.85 ≈ 1.17647Take both sides to the power of 1/0.9:1 + ΔY/Y = (1.17647)^(1/0.9) ≈ 1.17647^1.1111 ≈ 1.198So, ΔY/Y ≈ 0.198 or 19.8%So, approximately a 19.8% increase in personal income is needed to offset the 15% reduction in tax revenue.But earlier, using elasticity, I got 16.67%. So, which is correct?I think the second approach is correct because it's considering the multiplicative effect. The elasticity approach assumes a linear approximation, which is good for small changes, but here we're dealing with a 15% change, which might be large enough that the approximation isn't perfect.Alternatively, using the elasticity formula:Elasticity = %ΔR / %ΔYWe have a desired %ΔR of +15% (to offset the -15% from tax cut), so:0.9 = 15% / %ΔYThus, %ΔY = 15% / 0.9 ≈ 16.67%But this is a linear approximation. The actual calculation using the model gives 19.8%, which is higher.So, which one should I use? The question says to assume the relationship is R = k * Y^0.9, so we should use that model rather than the linear elasticity approximation.Therefore, the correct approach is to solve for Y such that 0.85R = k * Y'^0.9, but we want R to remain the same, so:R = 0.85R * (Y'/Y)^0.9Divide both sides by 0.85R:1 = 0.85 * (Y'/Y)^0.9Thus,(Y'/Y)^0.9 = 1 / 0.85 ≈ 1.17647Raise both sides to the power of 1/0.9:Y'/Y = (1.17647)^(1/0.9) ≈ 1.198So, Y' ≈ 1.198Y, which is a 19.8% increase in personal income.Therefore, the percentage change in personal income necessary to offset the 15% reduction in tax revenue is approximately 19.8%.But let me double-check the calculations:(1 / 0.85) = 1.17647Take the natural log: ln(1.17647) ≈ 0.1625Multiply by 1/0.9: 0.1625 * 1.1111 ≈ 0.1806Exponentiate: e^0.1806 ≈ 1.198Yes, that seems correct.So, the answer to the first question is approximately a 19.8% increase in personal income.Moving on to the second question: The reduction in the education budget directly correlates with a decrease in the quality of education. The model is linear, where a 1% decrease in budget results in a 0.5% decrease in educational outcomes. We need to find the percentage decrease in educational outcomes resulting from a 15% reduction in the education budget.This seems straightforward. If a 1% decrease in budget leads to a 0.5% decrease in outcomes, then a 15% decrease would lead to 15 * 0.5% = 7.5% decrease in outcomes.So, the percentage decrease in educational outcomes is 7.5%.But let me think if there's any nuance here. The model is linear, so it's a direct proportion. So, yes, 15% decrease in budget leads to 15 * 0.5% = 7.5% decrease in outcomes.Therefore, the answers are approximately 19.8% increase in personal income and a 7.5% decrease in educational outcomes.But wait, the first question was about the percentage change in personal income necessary to offset the 15% reduction in tax revenue. So, the answer is a 19.8% increase in Y.But the question might expect the answer in terms of percentage points, so 19.8% is approximately 20%.But let me see if I can express it more precisely. 19.8% is roughly 19.8%, which is 19 and 4/5 percent, or 19.8%.Alternatively, if we use more precise calculations:(1 / 0.85)^(1/0.9) = (1.176470588)^(1.111111111)Let me calculate this more accurately.First, ln(1.176470588) = 0.162518929Multiply by 1.111111111: 0.162518929 * 1.111111111 ≈ 0.180632143e^0.180632143 ≈ 1.198 (since e^0.18 ≈ 1.1972, e^0.1806 ≈ 1.198)So, yes, approximately 1.198, which is 19.8%.Therefore, the answers are:1. Approximately 19.8% increase in personal income.2. 7.5% decrease in educational outcomes.But let me check if the first question is asking for the percentage change in personal income necessary to offset the 15% reduction in tax revenue. So, if tax revenue is reduced by 15%, and we need to find the change in Y such that the new R is the same as before. So, the tax cut reduces R by 15%, so R becomes 0.85R. To offset this, we need Y to increase such that R returns to R. So, the new R after Y increases should be R = 0.85R * (Y'/Y)^0.9.Wait, no, actually, the tax cut is a separate factor. So, perhaps the model is that R = k * Y^0.9, and the tax cut reduces R by 15%, so the new R is 0.85 * k * Y^0.9. To offset this, we need Y to increase such that 0.85 * k * Y'^0.9 = k * Y^0.9.Divide both sides by k:0.85 * Y'^0.9 = Y^0.9Divide both sides by 0.85:Y'^0.9 = Y^0.9 / 0.85Take both sides to the power of 1/0.9:Y' = (Y^0.9 / 0.85)^(1/0.9) = Y * (1 / 0.85)^(1/0.9)Which is the same as before, so Y' = Y * (1.17647)^(1.1111) ≈ Y * 1.198So, 19.8% increase in Y.Yes, that's correct.Therefore, the answers are:1. Approximately 19.8% increase in personal income.2. 7.5% decrease in educational outcomes.I think that's it.</think>
❓:A Brazilian soap opera fan decides to analyze the viewing patterns of soap operas and reality music shows. He discovers that during a typical week, the viewership of his favorite soap opera can be modeled by the function ( f(t) = A sin(Bt + C) + D ), where ( t ) is the number of days since the beginning of the week, and ( A ), ( B ), ( C ), and ( D ) are constants that depend on various factors such as the storyline's intensity and its timing in the week. 1. Given that the peak viewership occurs at ( t = 3 ) days with a maximum viewership of 8 million people, and the minimum viewership of 4 million people occurs at ( t = 0 ) days, determine the values of ( A ), ( D ), and the phase shift ( C ) assuming ( B = frac{pi}{7} ).2. Meanwhile, the fan notes that the viewership of reality music shows is inversely proportional to the soap opera's viewership. If the reality music show's viewership is represented by ( g(t) = frac{k}{f(t)} ), where ( k ) is a proportional constant, and the reality show reaches a peak viewership of 10 million people, calculate ( k ) and find the day ( t ) during the week when the reality music show reaches half of its peak viewership.
🧠:<think>Alright, so I've got this problem about modeling viewership of a soap opera and a reality music show. Let me try to break it down step by step.First, part 1. The function given is ( f(t) = A sin(Bt + C) + D ). They told us that ( B = frac{pi}{7} ), so that's fixed. We need to find ( A ), ( D ), and ( C ).They gave us two key points: the maximum viewership is 8 million at ( t = 3 ), and the minimum is 4 million at ( t = 0 ).Hmm, okay. So, for a sine function of the form ( A sin(Bt + C) + D ), the maximum value is ( A + D ) and the minimum is ( -A + D ). So, if the maximum is 8 and the minimum is 4, we can set up two equations:1. ( A + D = 8 )2. ( -A + D = 4 )Let me solve these equations. If I add them together:( (A + D) + (-A + D) = 8 + 4 )( 2D = 12 )So, ( D = 6 ).Then, plugging back into the first equation: ( A + 6 = 8 ) => ( A = 2 ).Alright, so ( A = 2 ) and ( D = 6 ).Now, we need to find the phase shift ( C ). The function is ( f(t) = 2 sinleft(frac{pi}{7} t + Cright) + 6 ).We know that the maximum occurs at ( t = 3 ). For a sine function, the maximum occurs when the argument is ( frac{pi}{2} ) (since ( sin(frac{pi}{2}) = 1 )). So, let's set up the equation:( frac{pi}{7} times 3 + C = frac{pi}{2} )Solving for ( C ):( frac{3pi}{7} + C = frac{pi}{2} )( C = frac{pi}{2} - frac{3pi}{7} )Let me compute that:First, find a common denominator for the fractions. 2 and 7 have 14.( frac{pi}{2} = frac{7pi}{14} )( frac{3pi}{7} = frac{6pi}{14} )So, ( C = frac{7pi}{14} - frac{6pi}{14} = frac{pi}{14} )Therefore, ( C = frac{pi}{14} ).Wait, let me double-check. If I plug ( t = 3 ) into ( frac{pi}{7} t + C ), it should be ( frac{pi}{2} ). So:( frac{pi}{7} times 3 + frac{pi}{14} = frac{3pi}{7} + frac{pi}{14} = frac{6pi}{14} + frac{pi}{14} = frac{7pi}{14} = frac{pi}{2} ). Yep, that works.So, part 1 is done. ( A = 2 ), ( D = 6 ), ( C = frac{pi}{14} ).Moving on to part 2. The reality music show's viewership is inversely proportional to the soap opera's, so ( g(t) = frac{k}{f(t)} ). They told us that the reality show reaches a peak viewership of 10 million. So, we need to find ( k ) and then find the day ( t ) when the reality show is at half its peak, which is 5 million.First, let's find ( k ). The peak viewership of the reality show is 10 million. Since ( g(t) ) is inversely proportional to ( f(t) ), the peak of ( g(t) ) occurs when ( f(t) ) is at its minimum. Because as ( f(t) ) decreases, ( g(t) ) increases.From part 1, the minimum viewership of the soap opera is 4 million at ( t = 0 ). So, at ( t = 0 ), ( f(0) = 4 ), and ( g(0) = frac{k}{4} = 10 ) million.Solving for ( k ):( frac{k}{4} = 10 )( k = 10 times 4 = 40 )So, ( k = 40 ).Now, we need to find the day ( t ) when the reality show's viewership is half of its peak, which is 5 million.So, set ( g(t) = 5 ):( 5 = frac{40}{f(t)} )Multiply both sides by ( f(t) ):( 5 f(t) = 40 )Divide both sides by 5:( f(t) = 8 )Wait, that's interesting. So, when ( g(t) = 5 ), ( f(t) = 8 ). But from part 1, we know that the maximum viewership of the soap opera is 8 million, which occurs at ( t = 3 ).So, does that mean that the reality show is at half its peak viewership when the soap opera is at its peak? That seems to make sense because they are inversely proportional.So, ( t = 3 ) days.Wait, but let me verify. Let's compute ( f(t) ) at ( t = 3 ):( f(3) = 2 sinleft( frac{pi}{7} times 3 + frac{pi}{14} right) + 6 )Simplify the argument:( frac{3pi}{7} + frac{pi}{14} = frac{6pi}{14} + frac{pi}{14} = frac{7pi}{14} = frac{pi}{2} )So, ( f(3) = 2 sinleft( frac{pi}{2} right) + 6 = 2 times 1 + 6 = 8 ). Yep, that's correct.Therefore, when the soap opera is at its peak viewership, the reality show is at half its peak. So, the day is ( t = 3 ).But wait, let me think again. The problem says "the reality music show reaches a peak viewership of 10 million people". So, the peak of ( g(t) ) is 10 million, which occurs when ( f(t) ) is at its minimum, which is 4 million. So, that's consistent with ( k = 40 ).And when ( g(t) = 5 ), which is half of 10, that occurs when ( f(t) = 8 ), which is the maximum of the soap opera. So, that's correct.Therefore, the day ( t ) is 3.Wait, but let me make sure that this is the only solution. Because sine functions are periodic, so maybe there's another ( t ) in the week where ( f(t) = 8 ). Let's check.The function ( f(t) = 2 sinleft( frac{pi}{7} t + frac{pi}{14} right) + 6 ) has a period of ( frac{2pi}{frac{pi}{7}} = 14 ) days. But since we're only considering a week, which is 7 days, the function completes half a period in a week.So, the maximum occurs at ( t = 3 ), and the next maximum would be at ( t = 3 + 7 = 10 ), which is beyond the week. So, within the week (0 to 7 days), the maximum is only at ( t = 3 ).Therefore, the only day when ( f(t) = 8 ) is at ( t = 3 ). So, the reality show is at half its peak only on that day.Hence, ( t = 3 ).So, summarizing part 2: ( k = 40 ) and the day is ( t = 3 ).Wait, but let me just double-check my calculations for ( k ). If ( g(t) = frac{k}{f(t)} ), and the peak of ( g(t) ) is 10 million, which occurs when ( f(t) ) is at its minimum, which is 4 million. So, ( g(t) ) peaks at ( frac{k}{4} = 10 ), so ( k = 40 ). That seems correct.And when ( g(t) = 5 ), then ( f(t) = 8 ), which is the maximum of the soap opera, occurring at ( t = 3 ). So, that's correct.I think that's all. So, the answers are:1. ( A = 2 ), ( D = 6 ), ( C = frac{pi}{14} )2. ( k = 40 ), ( t = 3 )Final Answer1. ( A = boxed{2} ), ( D = boxed{6} ), and ( C = boxed{dfrac{pi}{14}} )2. ( k = boxed{40} ) and the day is ( boxed{3} )</think>
❓:Dr. Jane Doe, an experienced epidemiologist, has been analyzing the spread of a novel infectious disease. She uses a combination of statistical analysis and differential equations to model the disease dynamics. She focuses on the SIR (Susceptible-Infectious-Recovered) model, which is described by the following system of differential equations:[ begin{cases}frac{dS}{dt} = -beta frac{SI}{N} frac{dI}{dt} = beta frac{SI}{N} - gamma I frac{dR}{dt} = gamma I end{cases}]where (S(t)), (I(t)), and (R(t)) represent the number of susceptible, infectious, and recovered individuals at time (t), respectively. (beta) is the transmission rate, (gamma) is the recovery rate, and (N = S(t) + I(t) + R(t)) is the total population, which is assumed to be constant.1. Given the initial conditions (S(0) = S_0), (I(0) = I_0), and (R(0) = R_0), derive the expression for the basic reproduction number (R_0) in terms of (beta), (gamma), (S_0), and (N). Additionally, determine the condition under which the disease will lead to an epidemic (i.e., when the number of infections will increase initially).2. To verify her findings, Dr. Doe uses a dataset containing the number of new daily cases over time. She fits her model to the data using a least-squares approach. Assume she collects data ({I_{data}(t_i)}) at discrete time points ({t_i}), where (i = 1, 2, ldots, n). Formulate the least-squares problem that Dr. Doe needs to solve to estimate the parameters (beta) and (gamma).
🧠:<think>Okay, so I have this problem about the SIR model, which is a system of differential equations used to model the spread of infectious diseases. I need to tackle two parts here. Let me start with the first one.Problem 1: Derive the expression for the basic reproduction number ( R_0 ) and determine the epidemic condition.Hmm, I remember that the basic reproduction number, often denoted as ( R_0 ), is a key concept in epidemiology. It represents the average number of secondary infections produced by a single infected individual in a completely susceptible population. If ( R_0 > 1 ), an epidemic is likely to occur because each infected person is infecting more than one other person on average.Looking at the SIR model equations:[begin{cases}frac{dS}{dt} = -beta frac{SI}{N} frac{dI}{dt} = beta frac{SI}{N} - gamma I frac{dR}{dt} = gamma I end{cases}]Here, ( beta ) is the transmission rate, and ( gamma ) is the recovery rate. The total population ( N ) is constant, so ( N = S + I + R ).To find ( R_0 ), I think it's related to the initial exponential growth phase of the epidemic. When the disease first starts spreading, the number of susceptible individuals ( S ) is approximately equal to the initial susceptible population ( S_0 ), since not many people have been infected yet. So, initially, ( S approx S_0 ).In the early stages, the change in the number of infected individuals ( I ) can be approximated by the differential equation:[frac{dI}{dt} approx beta frac{S_0}{N} I - gamma I]This simplifies to:[frac{dI}{dt} approx left( beta frac{S_0}{N} - gamma right) I]This is a linear differential equation, and its solution will grow exponentially if the coefficient of ( I ) is positive. So, the condition for an epidemic (i.e., ( frac{dI}{dt} > 0 )) is:[beta frac{S_0}{N} - gamma > 0]Which simplifies to:[beta frac{S_0}{N} > gamma]Dividing both sides by ( gamma ), we get:[frac{beta}{gamma} frac{S_0}{N} > 1]So, the basic reproduction number ( R_0 ) is defined as:[R_0 = frac{beta S_0}{gamma N}]And the condition for an epidemic is when ( R_0 > 1 ).Wait, let me make sure I didn't mix up any terms. I think another way to think about ( R_0 ) is in terms of the contact rate and the duration of infectiousness. The contact rate is ( beta ), and the average infectious period is ( 1/gamma ). So, the number of contacts per infectious individual is ( beta times (1/gamma) ). But since only a fraction ( S_0/N ) of the population is susceptible, ( R_0 ) should be ( beta S_0 / (gamma N) ). Yeah, that seems consistent.So, summarizing:- ( R_0 = frac{beta S_0}{gamma N} )- Epidemic condition: ( R_0 > 1 )Problem 2: Formulate the least-squares problem to estimate ( beta ) and ( gamma ).Alright, Dr. Doe has data on the number of new daily cases over time, which I assume corresponds to the number of infected individuals ( I(t) ) at discrete time points ( t_i ). She wants to fit her SIR model to this data using a least-squares approach.In least-squares estimation, we aim to minimize the sum of the squared differences between the observed data and the model predictions. So, the goal is to find the parameters ( beta ) and ( gamma ) that make the model's ( I(t) ) as close as possible to the observed ( I_{data}(t_i) ).But here's the thing: the SIR model is a system of differential equations, which means we can't directly express ( I(t) ) in a closed-form solution easily. Instead, we need to solve the differential equations numerically for given ( beta ) and ( gamma ), and then compare the resulting ( I(t) ) to the data.So, the steps would be:1. For a given set of parameters ( beta ) and ( gamma ), solve the SIR model numerically to get the model's prediction ( hat{I}(t_i; beta, gamma) ) at each time point ( t_i ).2. Compute the residual, which is the difference between the observed data ( I_{data}(t_i) ) and the model prediction ( hat{I}(t_i; beta, gamma) ).3. Square each residual and sum them up to get the objective function, which is the sum of squared residuals (SSR).4. The least-squares problem is then to find the ( beta ) and ( gamma ) that minimize this SSR.Mathematically, the least-squares problem can be written as:[min_{beta, gamma} sum_{i=1}^{n} left( I_{data}(t_i) - hat{I}(t_i; beta, gamma) right)^2]Where ( hat{I}(t_i; beta, gamma) ) is the solution to the SIR model at time ( t_i ) given parameters ( beta ) and ( gamma ).But to write this more formally, we might need to express it in terms of the model equations. However, since the model is a system of ODEs, the exact form of ( hat{I}(t_i; beta, gamma) ) isn't straightforward. Instead, we can denote the solution as a function that depends on ( beta ) and ( gamma ).So, the least-squares problem is:Find ( beta ) and ( gamma ) that minimize:[sum_{i=1}^{n} left( I_{data}(t_i) - hat{I}(t_i; beta, gamma) right)^2]Subject to the system of differential equations:[begin{cases}frac{dS}{dt} = -beta frac{SI}{N} frac{dI}{dt} = beta frac{SI}{N} - gamma I frac{dR}{dt} = gamma I end{cases}]With initial conditions ( S(0) = S_0 ), ( I(0) = I_0 ), ( R(0) = R_0 ).Alternatively, if we want to write it in a more compact form, we can consider the vector of parameters ( theta = (beta, gamma) ) and the vector of data ( mathbf{I}_{data} = {I_{data}(t_i)} ). Then, the objective function is:[min_{theta} || mathbf{I}_{data} - mathbf{hat{I}}(theta) ||^2]Where ( mathbf{hat{I}}(theta) ) is the vector of model predictions obtained by solving the SIR model with parameters ( theta ).I think that's the gist of it. The key point is that because the model is defined by differential equations, we can't express ( hat{I}(t_i; beta, gamma) ) in a simple algebraic form, so numerical methods are required to solve the ODEs for each candidate ( beta ) and ( gamma ), and then compute the SSR.Let me double-check if I missed anything. The problem mentions that Dr. Doe uses a least-squares approach with data at discrete time points. So, yes, we need to minimize the sum of squared differences between the data and the model predictions at those specific times.I don't think there's a need to include the other variables ( S ) and ( R ) in the least-squares problem unless the data includes those as well. Since the problem states that the data is the number of new daily cases, which corresponds to ( I(t) ), we only need to fit ( I(t) ).So, to recap:- The least-squares problem is about minimizing the SSR between the observed ( I(t_i) ) and the model's ( I(t_i; beta, gamma) ).- The model is defined by the SIR ODEs, so for each ( beta ) and ( gamma ), we solve the ODEs numerically to get ( I(t_i) ).- The objective function is the sum over all data points of the squared differences.I think that's all. I don't see any other components needed for the least-squares formulation here.Final Answer1. The basic reproduction number is ( boxed{R_0 = dfrac{beta S_0}{gamma N}} ), and an epidemic occurs when ( R_0 > 1 ).2. The least-squares problem is to minimize ( sum_{i=1}^{n} left( I_{text{data}}(t_i) - hat{I}(t_i; beta, gamma) right)^2 ) with respect to ( beta ) and ( gamma ), where ( hat{I}(t_i; beta, gamma) ) is the solution of the SIR model at time ( t_i ).So, the final answers are:1. ( boxed{R_0 = dfrac{beta S_0}{gamma N}} ) and the epidemic condition is ( R_0 > 1 ).2. The least-squares problem is formulated as minimizing the sum of squared residuals between the data and the model predictions, which is:[boxed{min_{beta, gamma} sum_{i=1}^{n} left( I_{text{data}}(t_i) - hat{I}(t_i; beta, gamma) right)^2}]Where ( hat{I}(t_i; beta, gamma) ) is obtained by solving the SIR model with parameters ( beta ) and ( gamma ).Final Answer1. The basic reproduction number is ( boxed{R_0 = dfrac{beta S_0}{gamma N}} ), and an epidemic occurs when ( R_0 > 1 ).2. The least-squares problem is formulated as minimizing the sum of squared residuals between the data and the model predictions, which is:[boxed{min_{beta, gamma} sum_{i=1}^{n} left( I_{text{data}}(t_i) - hat{I}(t_i; beta, gamma) right)^2}]</think>