🧠:<think>Alright, so I've got this problem about two brothers, Alex and Ben, who are disputing an inheritance over a piece of land. The land is described by these parametric equations:[ x(t) = a cos^3(t) + b sin(t) ][ y(t) = a sin^3(t) + b cos(t) ]where (a) and (b) are constants, and (t) ranges from 0 to (2pi). The first part asks me to find the area enclosed by this curve using Green's Theorem. The second part is about dividing this area into two equal parts with a new boundary curve.Starting with part 1: Finding the area using Green's Theorem. I remember that Green's Theorem relates a line integral around a simple closed curve to a double integral over the region it encloses. Specifically, the area can be calculated using the formula:[ text{Area} = frac{1}{2} oint_{C} (x , dy - y , dx) ]So, I need to compute this line integral. Since the curve is given parametrically, I can express (dx) and (dy) in terms of (t). Let me compute (dx/dt) and (dy/dt) first.Calculating (dx/dt):[ frac{dx}{dt} = frac{d}{dt} [a cos^3(t) + b sin(t)] ][ = a cdot 3 cos^2(t) (-sin(t)) + b cos(t) ][ = -3a cos^2(t) sin(t) + b cos(t) ]Similarly, calculating (dy/dt):[ frac{dy}{dt} = frac{d}{dt} [a sin^3(t) + b cos(t)] ][ = a cdot 3 sin^2(t) cos(t) - b sin(t) ][ = 3a sin^2(t) cos(t) - b sin(t) ]Now, plugging these into the area formula:[ text{Area} = frac{1}{2} int_{0}^{2pi} [x(t) cdot frac{dy}{dt} - y(t) cdot frac{dx}{dt}] , dt ]Let me write out the integrand:[ x(t) cdot frac{dy}{dt} - y(t) cdot frac{dx}{dt} ]Substituting the expressions:First, compute (x(t) cdot frac{dy}{dt}):[ [a cos^3(t) + b sin(t)] cdot [3a sin^2(t) cos(t) - b sin(t)] ]Expanding this:[ a cos^3(t) cdot 3a sin^2(t) cos(t) + a cos^3(t) cdot (-b sin(t)) + b sin(t) cdot 3a sin^2(t) cos(t) + b sin(t) cdot (-b sin(t)) ]Simplify term by term:1. ( 3a^2 cos^4(t) sin^2(t) )2. ( -ab cos^3(t) sin(t) )3. ( 3ab sin^3(t) cos(t) )4. ( -b^2 sin^2(t) )Now, compute (y(t) cdot frac{dx}{dt}):[ [a sin^3(t) + b cos(t)] cdot [-3a cos^2(t) sin(t) + b cos(t)] ]Expanding this:[ a sin^3(t) cdot (-3a cos^2(t) sin(t)) + a sin^3(t) cdot b cos(t) + b cos(t) cdot (-3a cos^2(t) sin(t)) + b cos(t) cdot b cos(t) ]Simplify term by term:1. ( -3a^2 sin^4(t) cos^2(t) )2. ( ab sin^3(t) cos(t) )3. ( -3ab cos^3(t) sin(t) )4. ( b^2 cos^2(t) )Now, subtracting (y(t) cdot frac{dx}{dt}) from (x(t) cdot frac{dy}{dt}):So, the integrand becomes:[ [3a^2 cos^4(t) sin^2(t) - ab cos^3(t) sin(t) + 3ab sin^3(t) cos(t) - b^2 sin^2(t)] - [-3a^2 sin^4(t) cos^2(t) + ab sin^3(t) cos(t) - 3ab cos^3(t) sin(t) + b^2 cos^2(t)] ]Let me distribute the negative sign:[ 3a^2 cos^4(t) sin^2(t) - ab cos^3(t) sin(t) + 3ab sin^3(t) cos(t) - b^2 sin^2(t) + 3a^2 sin^4(t) cos^2(t) - ab sin^3(t) cos(t) + 3ab cos^3(t) sin(t) - b^2 cos^2(t) ]Now, let's combine like terms:1. Terms with (a^2): - (3a^2 cos^4(t) sin^2(t)) - (3a^2 sin^4(t) cos^2(t)) Combine these: (3a^2 cos^2(t) sin^2(t) [cos^2(t) + sin^2(t)]) Since (cos^2(t) + sin^2(t) = 1), this simplifies to (3a^2 cos^2(t) sin^2(t))2. Terms with (ab): - (-ab cos^3(t) sin(t)) - (3ab sin^3(t) cos(t)) - (-ab sin^3(t) cos(t)) - (3ab cos^3(t) sin(t)) Let's see: The first and last terms: (-ab cos^3(t) sin(t) + 3ab cos^3(t) sin(t) = 2ab cos^3(t) sin(t)) The middle terms: (3ab sin^3(t) cos(t) - ab sin^3(t) cos(t) = 2ab sin^3(t) cos(t)) So total (ab) terms: (2ab cos^3(t) sin(t) + 2ab sin^3(t) cos(t)) Factor out 2ab sin(t) cos(t): (2ab sin(t) cos(t) [cos^2(t) + sin^2(t)] = 2ab sin(t) cos(t))3. Terms with (b^2): - (-b^2 sin^2(t)) - (-b^2 cos^2(t)) Combine: (-b^2 (sin^2(t) + cos^2(t)) = -b^2)So, putting it all together, the integrand simplifies to:[ 3a^2 cos^2(t) sin^2(t) + 2ab sin(t) cos(t) - b^2 ]Therefore, the area is:[ text{Area} = frac{1}{2} int_{0}^{2pi} [3a^2 cos^2(t) sin^2(t) + 2ab sin(t) cos(t) - b^2] , dt ]Now, let's compute each integral separately.First, let's compute ( int_{0}^{2pi} 3a^2 cos^2(t) sin^2(t) , dt ).I recall that ( cos^2(t) sin^2(t) = frac{1}{4} sin^2(2t) ), since ( sin(2t) = 2 sin(t) cos(t) ), so ( sin^2(2t) = 4 sin^2(t) cos^2(t) ), hence ( sin^2(t) cos^2(t) = frac{1}{4} sin^2(2t) ).Therefore, the integral becomes:[ 3a^2 cdot frac{1}{4} int_{0}^{2pi} sin^2(2t) , dt = frac{3a^2}{4} int_{0}^{2pi} sin^2(2t) , dt ]The integral of ( sin^2(k t) ) over a full period is ( pi/k ). Here, (k = 2), so the integral over (0) to (2pi) is ( pi/2 cdot 2 = pi ). Wait, let me think.Wait, more accurately, ( int_{0}^{2pi} sin^2(2t) , dt ). Let me compute it.Using the identity ( sin^2(x) = frac{1 - cos(2x)}{2} ):[ int_{0}^{2pi} sin^2(2t) , dt = int_{0}^{2pi} frac{1 - cos(4t)}{2} , dt ][ = frac{1}{2} int_{0}^{2pi} 1 , dt - frac{1}{2} int_{0}^{2pi} cos(4t) , dt ][ = frac{1}{2} [2pi] - frac{1}{2} [0] ][ = pi ]So, the first integral is ( frac{3a^2}{4} cdot pi = frac{3pi a^2}{4} ).Next, compute ( int_{0}^{2pi} 2ab sin(t) cos(t) , dt ).Again, using the identity ( sin(t) cos(t) = frac{1}{2} sin(2t) ):[ 2ab cdot frac{1}{2} int_{0}^{2pi} sin(2t) , dt = ab int_{0}^{2pi} sin(2t) , dt ]The integral of ( sin(2t) ) over (0) to (2pi) is:[ left[ -frac{1}{2} cos(2t) right]_0^{2pi} = -frac{1}{2} [cos(4pi) - cos(0)] = -frac{1}{2} [1 - 1] = 0 ]So, the second integral is 0.Third, compute ( int_{0}^{2pi} (-b^2) , dt ):[ -b^2 int_{0}^{2pi} dt = -b^2 [2pi - 0] = -2pi b^2 ]Putting all together, the area is:[ frac{1}{2} left( frac{3pi a^2}{4} + 0 - 2pi b^2 right) ][ = frac{1}{2} left( frac{3pi a^2}{4} - 2pi b^2 right) ][ = frac{3pi a^2}{8} - pi b^2 ]Wait, let me check the arithmetic:Wait, no. Wait, the integrand was:[ 3a^2 cos^2(t) sin^2(t) + 2ab sin(t) cos(t) - b^2 ]So, integrating each term:First term: ( frac{3pi a^2}{4} )Second term: 0Third term: ( -2pi b^2 )So, the entire integral is ( frac{3pi a^2}{4} - 2pi b^2 )Then, multiplying by 1/2:[ frac{1}{2} left( frac{3pi a^2}{4} - 2pi b^2 right) = frac{3pi a^2}{8} - pi b^2 ]Wait, that seems correct.But let me think again. Is the area positive? Since (a) and (b) are constants, but depending on their values, the area could be positive or negative. But since it's an area, it should be positive. So, perhaps I made a mistake in the sign somewhere.Wait, let's go back. The integrand after simplification was:[ 3a^2 cos^2(t) sin^2(t) + 2ab sin(t) cos(t) - b^2 ]So, integrating term by term:1. (3a^2 cos^2(t) sin^2(t)): integrated to ( frac{3pi a^2}{4} )2. (2ab sin(t) cos(t)): integrated to 03. (-b^2): integrated to (-2pi b^2)So, the total integral is ( frac{3pi a^2}{4} - 2pi b^2 )Then, multiplying by 1/2:[ frac{1}{2} times left( frac{3pi a^2}{4} - 2pi b^2 right) = frac{3pi a^2}{8} - pi b^2 ]Hmm, so the area is ( frac{3pi a^2}{8} - pi b^2 ). But this could be negative if ( frac{3a^2}{8} < b^2 ). Since area should be positive, perhaps I made an error in the sign when computing the integrand.Wait, let's double-check the earlier steps when subtracting ( y(t) dx/dt ) from ( x(t) dy/dt ).Wait, the integrand is ( x(t) dy/dt - y(t) dx/dt ). Let me check if I did that correctly.Yes, I think so. Let me verify the expansion:When I expanded (x(t) dy/dt), I had:1. ( 3a^2 cos^4(t) sin^2(t) )2. ( -ab cos^3(t) sin(t) )3. ( 3ab sin^3(t) cos(t) )4. ( -b^2 sin^2(t) )And for ( y(t) dx/dt ):1. ( -3a^2 sin^4(t) cos^2(t) )2. ( ab sin^3(t) cos(t) )3. ( -3ab cos^3(t) sin(t) )4. ( b^2 cos^2(t) )Then, subtracting ( y(t) dx/dt ) from ( x(t) dy/dt ):So, the integrand is:[ x(t) dy/dt - y(t) dx/dt = [3a^2 cos^4(t) sin^2(t) - ab cos^3(t) sin(t) + 3ab sin^3(t) cos(t) - b^2 sin^2(t)] - [-3a^2 sin^4(t) cos^2(t) + ab sin^3(t) cos(t) - 3ab cos^3(t) sin(t) + b^2 cos^2(t)] ]Which becomes:[ 3a^2 cos^4(t) sin^2(t) - ab cos^3(t) sin(t) + 3ab sin^3(t) cos(t) - b^2 sin^2(t) + 3a^2 sin^4(t) cos^2(t) - ab sin^3(t) cos(t) + 3ab cos^3(t) sin(t) - b^2 cos^2(t) ]Then, combining like terms:- ( 3a^2 cos^4(t) sin^2(t) + 3a^2 sin^4(t) cos^2(t) = 3a^2 cos^2(t) sin^2(t) (cos^2(t) + sin^2(t)) = 3a^2 cos^2(t) sin^2(t) )- ( -ab cos^3(t) sin(t) + 3ab cos^3(t) sin(t) = 2ab cos^3(t) sin(t) )- ( 3ab sin^3(t) cos(t) - ab sin^3(t) cos(t) = 2ab sin^3(t) cos(t) )- ( -b^2 sin^2(t) - b^2 cos^2(t) = -b^2 (sin^2(t) + cos^2(t)) = -b^2 )So, the integrand is:[ 3a^2 cos^2(t) sin^2(t) + 2ab sin(t) cos(t) - b^2 ]Which is what I had earlier. So, integrating that gives:[ frac{3pi a^2}{4} + 0 - 2pi b^2 ]Then, multiplying by 1/2:[ frac{3pi a^2}{8} - pi b^2 ]So, unless ( frac{3a^2}{8} > b^2 ), the area would be positive. If ( frac{3a^2}{8} < b^2 ), the area would be negative, which doesn't make sense. So, perhaps I made a mistake in the sign when applying Green's Theorem.Wait, Green's Theorem requires the curve to be oriented counterclockwise. Is the given parametric curve oriented counterclockwise? Let me check.Looking at the parametric equations:[ x(t) = a cos^3(t) + b sin(t) ][ y(t) = a sin^3(t) + b cos(t) ]As ( t ) increases from 0 to ( 2pi ), let's see the direction. At ( t = 0 ), ( x = a ), ( y = b ). At ( t = pi/2 ), ( x = b ), ( y = a ). At ( t = pi ), ( x = -a ), ( y = -b ). At ( t = 3pi/2 ), ( x = -b ), ( y = -a ). So, it seems to be moving counterclockwise, as the points are moving in a counterclockwise manner around the origin.Therefore, the orientation is correct, so the area should be positive. Therefore, perhaps I made a mistake in the sign when computing the integrand.Wait, let me check the integrand again. The integrand is ( x dy/dt - y dx/dt ). Let me verify the signs when expanding.Wait, when I expanded ( x(t) dy/dt ), I had:[ [a cos^3(t) + b sin(t)] cdot [3a sin^2(t) cos(t) - b sin(t)] ]Which is correct.Similarly, ( y(t) dx/dt ) was:[ [a sin^3(t) + b cos(t)] cdot [-3a cos^2(t) sin(t) + b cos(t)] ]Which is also correct.Then, when subtracting ( y(t) dx/dt ), I distributed the negative sign correctly.Wait, perhaps I made a mistake in the sign when computing ( x(t) dy/dt - y(t) dx/dt ). Let me re-express the integrand as:[ x frac{dy}{dt} - y frac{dx}{dt} ]Which is:[ [a cos^3(t) + b sin(t)][3a sin^2(t) cos(t) - b sin(t)] - [a sin^3(t) + b cos(t)][-3a cos^2(t) sin(t) + b cos(t)] ]Expanding both products:First product:1. ( a cos^3(t) cdot 3a sin^2(t) cos(t) = 3a^2 cos^4(t) sin^2(t) )2. ( a cos^3(t) cdot (-b sin(t)) = -ab cos^3(t) sin(t) )3. ( b sin(t) cdot 3a sin^2(t) cos(t) = 3ab sin^3(t) cos(t) )4. ( b sin(t) cdot (-b sin(t)) = -b^2 sin^2(t) )Second product:1. ( a sin^3(t) cdot (-3a cos^2(t) sin(t)) = -3a^2 sin^4(t) cos^2(t) )2. ( a sin^3(t) cdot b cos(t) = ab sin^3(t) cos(t) )3. ( b cos(t) cdot (-3a cos^2(t) sin(t)) = -3ab cos^3(t) sin(t) )4. ( b cos(t) cdot b cos(t) = b^2 cos^2(t) )So, subtracting the second product from the first:1. ( 3a^2 cos^4(t) sin^2(t) - (-3a^2 sin^4(t) cos^2(t)) = 3a^2 cos^4(t) sin^2(t) + 3a^2 sin^4(t) cos^2(t) )2. ( -ab cos^3(t) sin(t) - ab sin^3(t) cos(t) = -ab cos^3(t) sin(t) - ab sin^3(t) cos(t) )3. ( 3ab sin^3(t) cos(t) - (-3ab cos^3(t) sin(t)) = 3ab sin^3(t) cos(t) + 3ab cos^3(t) sin(t) )4. ( -b^2 sin^2(t) - b^2 cos^2(t) = -b^2 (sin^2(t) + cos^2(t)) = -b^2 )Wait, hold on. I think I made a mistake in the subtraction step. Let me clarify:The integrand is:[ [3a^2 cos^4(t) sin^2(t) - ab cos^3(t) sin(t) + 3ab sin^3(t) cos(t) - b^2 sin^2(t)] - [-3a^2 sin^4(t) cos^2(t) + ab sin^3(t) cos(t) - 3ab cos^3(t) sin(t) + b^2 cos^2(t)] ]So, distributing the negative sign:[ 3a^2 cos^4(t) sin^2(t) - ab cos^3(t) sin(t) + 3ab sin^3(t) cos(t) - b^2 sin^2(t) + 3a^2 sin^4(t) cos^2(t) - ab sin^3(t) cos(t) + 3ab cos^3(t) sin(t) - b^2 cos^2(t) ]Now, combining like terms:- ( 3a^2 cos^4(t) sin^2(t) + 3a^2 sin^4(t) cos^2(t) = 3a^2 cos^2(t) sin^2(t) (cos^2(t) + sin^2(t)) = 3a^2 cos^2(t) sin^2(t) )- ( -ab cos^3(t) sin(t) + 3ab cos^3(t) sin(t) = 2ab cos^3(t) sin(t) )- ( 3ab sin^3(t) cos(t) - ab sin^3(t) cos(t) = 2ab sin^3(t) cos(t) )- ( -b^2 sin^2(t) - b^2 cos^2(t) = -b^2 (sin^2(t) + cos^2(t)) = -b^2 )So, the integrand is:[ 3a^2 cos^2(t) sin^2(t) + 2ab sin(t) cos(t) - b^2 ]Which is the same as before. So, integrating this gives:[ frac{3pi a^2}{4} + 0 - 2pi b^2 ]Thus, the area is:[ frac{1}{2} left( frac{3pi a^2}{4} - 2pi b^2 right) = frac{3pi a^2}{8} - pi b^2 ]Wait, but this result can be negative if ( b^2 > frac{3a^2}{8} ). Since area can't be negative, perhaps I made a mistake in the orientation or the parametrization.Alternatively, maybe I should have taken the absolute value. But in Green's Theorem, the sign depends on the orientation. Since we assumed the curve is counterclockwise, the area should be positive. So, perhaps the correct expression is:[ text{Area} = left| frac{3pi a^2}{8} - pi b^2 right| ]But I'm not sure. Alternatively, maybe I made a mistake in the parametrization. Let me check the parametrization again.Wait, looking at the parametric equations:[ x(t) = a cos^3(t) + b sin(t) ][ y(t) = a sin^3(t) + b cos(t) ]This seems similar to a Lissajous figure but with cubic terms. Alternatively, perhaps it's a type of hypocycloid or something else.Wait, another approach: Maybe I can rewrite the parametric equations in terms of multiple angles or use trigonometric identities to simplify.Alternatively, perhaps I can consider specific cases. For example, if ( b = 0 ), then the parametric equations become:[ x(t) = a cos^3(t) ][ y(t) = a sin^3(t) ]Which is a standard astroid, and its area is known to be ( frac{3pi a^2}{8} ). So, in that case, our formula gives ( frac{3pi a^2}{8} - 0 = frac{3pi a^2}{8} ), which is correct.If ( a = 0 ), then the parametric equations become:[ x(t) = b sin(t) ][ y(t) = b cos(t) ]Which is a circle of radius ( b ), but traversed clockwise because as ( t ) increases, ( x ) starts at 0, goes to ( b ), then back to 0, etc., while ( y ) starts at ( b ), goes to 0, then to ( -b ), etc. So, it's a circle but oriented clockwise. Therefore, the area computed by Green's Theorem would be negative, which is ( -pi b^2 ). But since we take the absolute value, the area is ( pi b^2 ).But in our formula, when ( a = 0 ), the area is ( 0 - pi b^2 = -pi b^2 ), which is negative, as expected because the orientation is clockwise. So, the formula is correct in terms of sign depending on orientation.But in the problem statement, it says the curve does not intersect itself, so it's a simple closed curve. Therefore, the area should be positive if oriented counterclockwise. So, perhaps in our case, the curve is oriented counterclockwise, so the area should be positive. Therefore, maybe ( frac{3pi a^2}{8} - pi b^2 ) must be positive, so ( frac{3a^2}{8} > b^2 ).Alternatively, perhaps I made a mistake in the parametrization direction. Let me check at ( t = 0 ), ( x = a ), ( y = b ). At ( t = pi/2 ), ( x = b ), ( y = a ). So, moving from ( (a, b) ) to ( (b, a) ), which is a counterclockwise movement if ( a > b ), but if ( b > a ), it might be different.Wait, perhaps the orientation is not strictly counterclockwise. Maybe the curve can have different orientations depending on ( a ) and ( b ). But the problem states that the curve does not intersect itself, so it's a simple closed curve, but the orientation might affect the sign.But regardless, the magnitude of the area is ( left| frac{3pi a^2}{8} - pi b^2 right| ). However, since the problem says to assume the curve does not intersect itself, perhaps ( a ) and ( b ) are such that the area is positive. Therefore, the area is ( frac{3pi a^2}{8} - pi b^2 ).Wait, but when ( a = 0 ), we get a negative area, which is correct because the curve is oriented clockwise. So, perhaps the correct area is ( frac{3pi a^2}{8} - pi b^2 ), and depending on the orientation, it could be positive or negative. But since the problem says to find the area, we can take the absolute value, but perhaps the answer is simply ( frac{3pi a^2}{8} - pi b^2 ), assuming the orientation is correct.Alternatively, maybe I made a mistake in the integrand. Let me try a different approach. Maybe using a substitution or recognizing the curve.Wait, another idea: Maybe the parametric equations can be rewritten in terms of multiple angles.Let me consider ( x(t) = a cos^3(t) + b sin(t) ) and ( y(t) = a sin^3(t) + b cos(t) ).I recall that ( cos^3(t) ) and ( sin^3(t) ) can be expressed using multiple angle identities.For example:[ cos^3(t) = frac{3 cos(t) + cos(3t)}{4} ][ sin^3(t) = frac{3 sin(t) - sin(3t)}{4} ]So, substituting these into ( x(t) ) and ( y(t) ):[ x(t) = a left( frac{3 cos(t) + cos(3t)}{4} right) + b sin(t) ][ = frac{3a}{4} cos(t) + frac{a}{4} cos(3t) + b sin(t) ]Similarly,[ y(t) = a left( frac{3 sin(t) - sin(3t)}{4} right) + b cos(t) ][ = frac{3a}{4} sin(t) - frac{a}{4} sin(3t) + b cos(t) ]So, now, ( x(t) ) and ( y(t) ) are expressed in terms of multiple angles. Maybe this can help in computing the area.But I'm not sure if this helps directly. Alternatively, perhaps using complex analysis or Fourier series, but that might be overcomplicating.Alternatively, perhaps using polar coordinates, but the parametric equations are in terms of Cartesian coordinates.Alternatively, perhaps I can use the shoelace formula, but that applies to polygons, not parametric curves.Alternatively, perhaps using the formula for the area of a parametric curve, which is indeed what I did earlier.Wait, another thought: Maybe I can use the fact that the parametric equations are a combination of a circle and a hypocycloid or something similar.Wait, if ( a = b ), what happens? Let me see:If ( a = b ), then:[ x(t) = a cos^3(t) + a sin(t) = a (cos^3(t) + sin(t)) ][ y(t) = a sin^3(t) + a cos(t) = a (sin^3(t) + cos(t)) ]Not sure if that helps.Alternatively, perhaps I can consider the parametric equations as a sum of two vectors: one from ( a cos^3(t), a sin^3(t) ) and another from ( b sin(t), b cos(t) ).The first part, ( a cos^3(t), a sin^3(t) ), is an astroid scaled by ( a ), and the second part, ( b sin(t), b cos(t) ), is a circle of radius ( b ) but shifted in phase.So, the curve is a combination of an astroid and a circle. Therefore, the area would be the sum of the areas of the astroid and the circle, but since they are combined parametrically, it's not simply additive.Wait, but in our earlier calculation, when ( b = 0 ), the area is ( frac{3pi a^2}{8} ), which is correct for the astroid. When ( a = 0 ), the area is ( -pi b^2 ), which is correct for the circle oriented clockwise. So, perhaps the formula is correct.Therefore, the area is ( frac{3pi a^2}{8} - pi b^2 ). But since area can't be negative, perhaps the correct answer is the absolute value, but the problem doesn't specify, so I think we can proceed with the formula as is.So, the area is ( frac{3pi a^2}{8} - pi b^2 ).Now, moving on to part 2: Given that Alex and Ben want to split the land into two regions of equal area, find the parametric equations for a new curve that serves as a boundary line dividing the land into these two equal regions. The boundary line should be a function of ( t ) and should lie entirely within the original curve.Hmm, this is more complex. So, we need to find a curve that splits the original area into two equal parts. Since the original curve is parametrized by ( t ), perhaps the dividing curve can also be parametrized by ( t ), but scaled or shifted in some way.One approach is to find a curve that is a scaled version of the original curve, such that the area enclosed by the new curve is half of the original area.Alternatively, perhaps the dividing curve can be found by taking a certain fraction of the original curve's parametric equations.But let me think more carefully.Suppose the original curve has area ( A = frac{3pi a^2}{8} - pi b^2 ). We need a new curve that encloses an area of ( A/2 ).Assuming that the new curve is similar to the original curve, perhaps scaled by a factor ( k ), such that the area scales by ( k^2 ). Therefore, if the original area is ( A ), the new area would be ( k^2 A ). Setting ( k^2 A = A/2 ), we get ( k = frac{1}{sqrt{2}} ).But wait, scaling the parametric equations by ( k ) would scale the area by ( k^2 ). So, if we scale ( a ) and ( b ) by ( k ), the new area would be ( frac{3pi (k a)^2}{8} - pi (k b)^2 = k^2 left( frac{3pi a^2}{8} - pi b^2 right) = k^2 A ).So, setting ( k^2 A = A/2 ), we get ( k = frac{1}{sqrt{2}} ).Therefore, the new parametric equations would be:[ x_{text{new}}(t) = frac{a}{sqrt{2}} cos^3(t) + frac{b}{sqrt{2}} sin(t) ][ y_{text{new}}(t) = frac{a}{sqrt{2}} sin^3(t) + frac{b}{sqrt{2}} cos(t) ]But wait, this assumes that scaling both ( a ) and ( b ) by ( 1/sqrt{2} ) would result in a curve that encloses half the area. However, this might not necessarily be the case because the original curve is a combination of an astroid and a circle, and scaling both parameters might not linearly affect the area in a straightforward way.Wait, but in our earlier calculation, the area is linear in ( a^2 ) and ( b^2 ), so scaling ( a ) and ( b ) by ( k ) would scale the area by ( k^2 ). Therefore, scaling both ( a ) and ( b ) by ( 1/sqrt{2} ) would indeed scale the area by ( 1/2 ).But wait, let me verify this. Suppose we scale ( a ) and ( b ) by ( k ). Then, the new area would be:[ frac{3pi (k a)^2}{8} - pi (k b)^2 = k^2 left( frac{3pi a^2}{8} - pi b^2 right) = k^2 A ]So, yes, scaling both ( a ) and ( b ) by ( k ) scales the area by ( k^2 ). Therefore, to get half the area, set ( k = 1/sqrt{2} ).Therefore, the new parametric equations would be:[ x_{text{new}}(t) = frac{a}{sqrt{2}} cos^3(t) + frac{b}{sqrt{2}} sin(t) ][ y_{text{new}}(t) = frac{a}{sqrt{2}} sin^3(t) + frac{b}{sqrt{2}} cos(t) ]But wait, is this the only way? Or is there another way to split the area?Alternatively, perhaps the dividing curve is a radial line from the origin, but given the parametric form, it's not clear.Alternatively, perhaps the dividing curve is another parametric curve that is a function of ( t ), but not necessarily a scaled version.Wait, another approach: Since the original curve is parametrized by ( t ), perhaps the dividing curve can be found by taking a certain function of ( t ), such as ( t/2 ) or something else, but I'm not sure.Alternatively, perhaps the dividing curve is a function that, for each ( t ), lies halfway between the origin and the original curve. But that might not necessarily split the area equally.Wait, perhaps using the concept of a parallel curve or offset curve. But that might be more complex.Alternatively, perhaps the dividing curve is another parametric curve that is a function of ( t ), such as ( x(t) = k x_{text{original}}(t) ), ( y(t) = k y_{text{original}}(t) ), where ( k ) is a scaling factor. As we discussed earlier, scaling ( a ) and ( b ) by ( 1/sqrt{2} ) would give a curve with half the area.But wait, in the original parametric equations, both ( a ) and ( b ) are constants. So, scaling them would change the shape of the curve, not just scale it uniformly. Therefore, the new curve might not be similar to the original curve.Wait, but in our earlier calculation, the area scales as ( k^2 ) when both ( a ) and ( b ) are scaled by ( k ). So, perhaps this is a valid approach.Alternatively, perhaps the dividing curve is a function of ( t ) that is a linear combination of the original parametric equations. For example, ( x(t) = c x_{text{original}}(t) ), ( y(t) = c y_{text{original}}(t) ), where ( c ) is a constant. Then, the area enclosed by this new curve would be ( c^2 ) times the original area. Therefore, setting ( c = 1/sqrt{2} ) would give a curve with half the area.But wait, in this case, the new curve would be a scaled version of the original curve, which might not necessarily lie entirely within the original curve. Because scaling down the curve would bring it closer to the origin, but depending on the shape, parts of it might not lie within the original curve.Wait, but the problem states that the boundary line should lie entirely within the original curve. So, scaling down the curve by ( 1/sqrt{2} ) might achieve this, as it would be entirely inside the original curve.But I'm not entirely sure. Let me think.Alternatively, perhaps the dividing curve is a function that, for each ( t ), is a point along the line from the origin to the original curve at ( t ), but only going halfway. So, ( x(t) = frac{1}{2} x_{text{original}}(t) ), ( y(t) = frac{1}{2} y_{text{original}}(t) ). But this would scale the area by ( 1/4 ), which is not half.Wait, no. The area scales with the square of the scaling factor. So, scaling by ( 1/sqrt{2} ) would scale the area by ( 1/2 ), as desired.Therefore, the parametric equations for the dividing curve would be:[ x(t) = frac{a}{sqrt{2}} cos^3(t) + frac{b}{sqrt{2}} sin(t) ][ y(t) = frac{a}{sqrt{2}} sin^3(t) + frac{b}{sqrt{2}} cos(t) ]But let me verify this. If I scale both ( a ) and ( b ) by ( 1/sqrt{2} ), then the new area would be:[ frac{3pi (a/sqrt{2})^2}{8} - pi (b/sqrt{2})^2 = frac{3pi a^2}{8 cdot 2} - frac{pi b^2}{2} = frac{3pi a^2}{16} - frac{pi b^2}{2} ]Wait, that's not half of the original area. The original area was ( frac{3pi a^2}{8} - pi b^2 ). Half of that would be ( frac{3pi a^2}{16} - frac{pi b^2}{2} ), which is exactly what we get. So, yes, scaling both ( a ) and ( b ) by ( 1/sqrt{2} ) gives a curve with half the area.But wait, in our earlier calculation, when we scaled ( a ) and ( b ) by ( k ), the area scaled by ( k^2 ). So, if we set ( k = 1/sqrt{2} ), the area becomes ( (1/2) A ), which is correct.Therefore, the parametric equations for the dividing curve are:[ x(t) = frac{a}{sqrt{2}} cos^3(t) + frac{b}{sqrt{2}} sin(t) ][ y(t) = frac{a}{sqrt{2}} sin^3(t) + frac{b}{sqrt{2}} cos(t) ]But let me think again. Is this the only way? Or is there another curve that can divide the area equally?Alternatively, perhaps the dividing curve is a function that, for each ( t ), is a point along the line from the origin to the original curve at ( t ), but only going halfway. So, ( x(t) = frac{1}{2} x_{text{original}}(t) ), ( y(t) = frac{1}{2} y_{text{original}}(t) ). But as I thought earlier, this would scale the area by ( 1/4 ), which is not half.Wait, no, scaling by ( 1/2 ) would scale the area by ( 1/4 ), but scaling by ( 1/sqrt{2} ) scales the area by ( 1/2 ), which is what we need.Therefore, the parametric equations for the dividing curve are as above.But let me check if this curve lies entirely within the original curve. Since we are scaling both ( a ) and ( b ) by ( 1/sqrt{2} ), which is approximately 0.707, the new curve should be entirely inside the original curve, as each point is closer to the origin.Therefore, the parametric equations for the boundary line dividing the land into two equal regions are:[ x(t) = frac{a}{sqrt{2}} cos^3(t) + frac{b}{sqrt{2}} sin(t) ][ y(t) = frac{a}{sqrt{2}} sin^3(t) + frac{b}{sqrt{2}} cos(t) ]Alternatively, we can factor out ( 1/sqrt{2} ):[ x(t) = frac{1}{sqrt{2}} left( a cos^3(t) + b sin(t) right) ][ y(t) = frac{1}{sqrt{2}} left( a sin^3(t) + b cos(t) right) ]Which is a more compact way to write it.Therefore, the parametric equations for the boundary line are:[ x(t) = frac{a}{sqrt{2}} cos^3(t) + frac{b}{sqrt{2}} sin(t) ][ y(t) = frac{a}{sqrt{2}} sin^3(t) + frac{b}{sqrt{2}} cos(t) ]for ( t in [0, 2pi] ).But wait, let me think again. Is this the only possible dividing curve? Or are there other curves that can divide the area equally?For example, perhaps a straight line could divide the area, but given the parametric nature of the curve, it's not obvious. Alternatively, perhaps a curve that is a function of ( t ) but not necessarily a scaled version.But given the complexity of the original curve, scaling seems like a reasonable approach, and it satisfies the condition of lying entirely within the original curve.Therefore, I think this is a valid solution.</think>

🧠:<think>Alright, so I've got these two math problems related to a film director's plans, which is pretty interesting. Let me try to tackle them one by one. Starting with the first problem: the director is using linear and nonlinear storytelling techniques, modeled by the functions f(x) = 3x + 2 and g(x) = x² - 4x + 6. They want to find the points where these two narratives intersect, meaning where f(x) equals g(x). That makes sense because those points would be where the storytelling transitions from linear to nonlinear or vice versa.So, to find the points of intersection, I need to set f(x) equal to g(x) and solve for x. That means:3x + 2 = x² - 4x + 6Hmm, okay, let's rearrange this equation to bring all terms to one side so I can solve for x. Subtract 3x and 2 from both sides:0 = x² - 4x + 6 - 3x - 2Simplify the right side:x² - 7x + 4 = 0Wait, let me check that again. 6 - 2 is 4, and -4x - 3x is -7x. Yeah, that seems right.So now I have a quadratic equation: x² - 7x + 4 = 0. To solve this, I can use the quadratic formula, which is x = [7 ± sqrt(49 - 16)] / 2. Because the quadratic formula is x = [-b ± sqrt(b² - 4ac)] / 2a, right?Here, a = 1, b = -7, c = 4. So plugging in:x = [7 ± sqrt(49 - 16)] / 2Calculate the discriminant: 49 - 16 is 33. So sqrt(33) is approximately 5.7446, but I'll keep it exact for now.So, x = [7 + sqrt(33)] / 2 and x = [7 - sqrt(33)] / 2.Let me compute these to decimal to get a sense of where these points are. sqrt(33) is about 5.7446, so:First solution: (7 + 5.7446)/2 ≈ 12.7446/2 ≈ 6.3723Second solution: (7 - 5.7446)/2 ≈ 1.2554/2 ≈ 0.6277So, the points of intersection are at approximately x ≈ 0.6277 and x ≈ 6.3723.But wait, the problem just says to find the points of intersection, so I should probably present them as exact values rather than decimal approximations. So, x = [7 ± sqrt(33)] / 2.But to be thorough, I should also find the corresponding y-values for these x's to get the full coordinates. Let's do that using f(x) = 3x + 2.First, for x = [7 + sqrt(33)] / 2:y = 3*([7 + sqrt(33)] / 2) + 2 = (21 + 3sqrt(33))/2 + 2 = (21 + 3sqrt(33) + 4)/2 = (25 + 3sqrt(33))/2Similarly, for x = [7 - sqrt(33)] / 2:y = 3*([7 - sqrt(33)] / 2) + 2 = (21 - 3sqrt(33))/2 + 2 = (21 - 3sqrt(33) + 4)/2 = (25 - 3sqrt(33))/2So, the points of intersection are ([7 + sqrt(33)] / 2, (25 + 3sqrt(33))/2) and ([7 - sqrt(33)] / 2, (25 - 3sqrt(33))/2).Let me just double-check my calculations to make sure I didn't make a mistake. Starting from f(x) = g(x):3x + 2 = x² - 4x + 6Bring all terms to left: x² - 7x + 4 = 0Quadratic formula: x = [7 ± sqrt(49 - 16)] / 2 = [7 ± sqrt(33)] / 2. Yep, that's correct.Then plugging back into f(x):3x + 2. For x = [7 + sqrt(33)] / 2:3*(7 + sqrt(33))/2 + 2 = (21 + 3sqrt(33))/2 + 4/2 = (25 + 3sqrt(33))/2. Correct.Same for the other x. Looks good.So, that's the first problem done. Now, moving on to the second problem.The director wants to represent justice as an area bounded by two curves: h(x) = sin(x) and k(x) = e^{-x} over the interval [0, π]. So, I need to calculate the area between these two curves from x = 0 to x = π.First, I should visualize these two functions to understand which one is on top and which is on the bottom in this interval. Because the area between two curves is the integral of the top function minus the bottom function.So, let's think about h(x) = sin(x) over [0, π]. Sin(x) starts at 0, goes up to 1 at π/2, then back down to 0 at π.k(x) = e^{-x} is an exponential decay function. At x = 0, it's e^{0} = 1. As x increases, it decreases towards 0. So, at x = π, it's e^{-π} ≈ 0.0432.So, at x = 0, both functions are equal to 1. Then, as x increases, sin(x) goes up to 1 at π/2, while e^{-x} is decreasing. So, somewhere between 0 and π, these two functions must intersect again.Therefore, the area between them will be split into two regions: from 0 to the intersection point, one function is on top, and from the intersection point to π, the other is on top.Wait, actually, let's check if they intersect more than once in [0, π]. At x=0, both are 1. Then, sin(x) increases to 1 at π/2, while e^{-x} decreases. So, sin(x) is above e^{-x} from 0 to some point where they cross again.Wait, actually, let's set sin(x) = e^{-x} and solve for x in [0, π]. That will give the points of intersection.So, sin(x) = e^{-x}This is a transcendental equation, meaning it can't be solved algebraically, so we'll have to approximate the solution numerically.But since we're calculating the area, we need to know where they intersect so we can split the integral accordingly.Let me denote the intersection point as c, where c is in (0, π). Let's approximate c.At x = 0: sin(0) = 0, e^{-0} = 1. Wait, no, wait: at x=0, sin(0)=0, but e^{-0}=1, so they aren't equal at x=0. Wait, hold on, earlier I thought both were 1 at x=0, but no: sin(0)=0, e^{-0}=1. So, actually, at x=0, sin(x) is 0 and e^{-x} is 1. So, e^{-x} is above sin(x) at x=0.Wait, that contradicts my earlier thought. Let me recast:Wait, h(x) = sin(x) and k(x) = e^{-x}. At x=0, sin(0)=0, e^{-0}=1. So, k(x) is above h(x) at x=0.Then, as x increases, sin(x) increases to 1 at π/2, while k(x) decreases from 1 to e^{-π/2} ≈ 0.2079.So, sin(x) starts below k(x), then crosses it somewhere between 0 and π/2, then continues to rise above k(x) until π/2, then starts to fall back down, while k(x) continues to decrease.So, perhaps sin(x) crosses k(x) once between 0 and π/2, and then again between π/2 and π? Or maybe only once?Wait, let's evaluate at x=π/2:sin(π/2)=1, e^{-π/2}≈0.2079. So, sin(x) is above k(x) at π/2.At x=π: sin(π)=0, e^{-π}≈0.0432. So, sin(x) is below k(x) at π.Therefore, the curves must cross somewhere between π/2 and π as well.Wait, so actually, there are two points of intersection: one between 0 and π/2, and another between π/2 and π.Wait, but at x=0, sin(x)=0 < k(x)=1.At x=π/2, sin(x)=1 > k(x)=~0.2079.So, by the Intermediate Value Theorem, since sin(x) goes from 0 to 1 and k(x) goes from 1 to ~0.2079, they must cross once between 0 and π/2.Similarly, at x=π/2, sin(x)=1 > k(x)=~0.2079.At x=π, sin(x)=0 < k(x)=~0.0432.So, sin(x) goes from 1 to 0, while k(x) goes from ~0.2079 to ~0.0432. So, sin(x) starts above k(x) at π/2 and ends below it at π, so they must cross once between π/2 and π.Therefore, there are two points of intersection: c1 in (0, π/2) and c2 in (π/2, π). Therefore, the area between the curves will be split into three regions:1. From 0 to c1: k(x) is above sin(x).2. From c1 to c2: sin(x) is above k(x).3. From c2 to π: k(x) is above sin(x).Wait, but actually, since we're integrating from 0 to π, and the curves cross twice, the area will be the integral from 0 to c1 of (k(x) - sin(x)) dx plus the integral from c1 to c2 of (sin(x) - k(x)) dx plus the integral from c2 to π of (k(x) - sin(x)) dx.But that seems a bit complicated. Alternatively, perhaps I can compute the total area as the integral from 0 to π of |sin(x) - e^{-x}| dx. But since we have two crossing points, it's better to split the integral at those points.But since we don't know c1 and c2 exactly, we might need to approximate them numerically.Alternatively, maybe I can compute the area without finding the exact points, but I think it's necessary to find c1 and c2 to compute the definite integrals correctly.So, first, let's find c1 and c2 where sin(x) = e^{-x}.Let me denote c1 as the solution in (0, π/2) and c2 as the solution in (π/2, π).To approximate c1:We can use numerical methods like the Newton-Raphson method.Let me define the function F(x) = sin(x) - e^{-x}. We need to find x where F(x) = 0.First, let's find c1 in (0, π/2):Compute F(0) = sin(0) - e^{0} = 0 - 1 = -1F(π/2) = sin(π/2) - e^{-π/2} ≈ 1 - 0.2079 ≈ 0.7921So, F(x) goes from -1 to ~0.7921, so by Intermediate Value Theorem, there is a root between 0 and π/2.Let's pick an initial guess. Let's try x=1 (since π/2 ≈1.5708). Let's compute F(1):sin(1) ≈0.8415, e^{-1}≈0.3679, so F(1)=0.8415 - 0.3679≈0.4736>0So, F(1)=0.4736>0, while F(0)=-1<0. So, the root is between 0 and 1.Let's try x=0.5:sin(0.5)≈0.4794, e^{-0.5}≈0.6065, so F(0.5)=0.4794 - 0.6065≈-0.1271<0So, F(0.5)≈-0.1271, F(1)=0.4736. So, the root is between 0.5 and 1.Let's try x=0.75:sin(0.75)≈0.6816, e^{-0.75}≈0.4724, so F(0.75)=0.6816 - 0.4724≈0.2092>0So, F(0.75)=0.2092>0, F(0.5)= -0.1271<0. So, root between 0.5 and 0.75.Let's try x=0.6:sin(0.6)≈0.5646, e^{-0.6}≈0.5488, so F(0.6)=0.5646 - 0.5488≈0.0158>0Close to zero. F(0.6)=~0.0158.F(0.59):sin(0.59)≈sin(0.59)≈0.5547, e^{-0.59}≈0.5535, so F(0.59)=0.5547 - 0.5535≈0.0012>0Almost zero.F(0.58):sin(0.58)≈0.5464, e^{-0.58}≈0.5583, so F(0.58)=0.5464 - 0.5583≈-0.0119<0So, between 0.58 and 0.59.Using linear approximation:At x=0.58, F=-0.0119At x=0.59, F=0.0012The change in x is 0.01, change in F is 0.0131.We need to find delta_x such that F=0:delta_x = 0.01 * (0 - (-0.0119)) / (0.0012 - (-0.0119)) ≈ 0.01 * (0.0119 / 0.0131) ≈0.01 *0.908≈0.00908So, approximate root at x≈0.58 + 0.00908≈0.5891Check F(0.5891):sin(0.5891)≈sin(0.5891)≈0.5525e^{-0.5891}≈e^{-0.5891}≈0.5543So, F≈0.5525 - 0.5543≈-0.0018Hmm, still slightly negative. Maybe need another iteration.Alternatively, let's use Newton-Raphson.Let me pick x0=0.59F(x)=sin(x) - e^{-x}F'(x)=cos(x) + e^{-x}Compute F(0.59)=0.5547 - 0.5535≈0.0012F'(0.59)=cos(0.59) + e^{-0.59}≈0.8312 + 0.5535≈1.3847Next approximation: x1 = x0 - F(x0)/F'(x0) ≈0.59 - (0.0012)/1.3847≈0.59 - 0.000866≈0.5891So, same as before. Then, F(0.5891)=sin(0.5891) - e^{-0.5891}≈0.5525 - 0.5543≈-0.0018So, F(x1)= -0.0018F'(x1)=cos(0.5891) + e^{-0.5891}≈0.8312 + 0.5543≈1.3855Next iteration: x2 = x1 - F(x1)/F'(x1)≈0.5891 - (-0.0018)/1.3855≈0.5891 + 0.001299≈0.5904Compute F(0.5904)=sin(0.5904) - e^{-0.5904}≈sin(0.5904)≈0.5535, e^{-0.5904}≈0.5535So, F≈0.5535 - 0.5535≈0. So, x≈0.5904 is a good approximation for c1.So, c1≈0.5904Similarly, let's find c2 in (π/2, π). Let's compute F(π/2)=1 - e^{-π/2}≈1 - 0.2079≈0.7921>0F(π)=0 - e^{-π}≈-0.0432<0So, F(x) goes from ~0.7921 to ~-0.0432, so there's a root between π/2 and π.Let's try x=2:F(2)=sin(2) - e^{-2}≈0.9093 - 0.1353≈0.774>0F(3)=sin(3) - e^{-3}≈0.1411 - 0.0498≈0.0913>0F(3.14)=sin(3.14)≈0.0016, e^{-3.14}≈0.0432, so F≈0.0016 - 0.0432≈-0.0416<0So, the root is between 3 and 3.14.Wait, but π is approximately 3.1416, so let's see:F(3)=0.1411 - 0.0498≈0.0913>0F(3.1)=sin(3.1)≈0.0416, e^{-3.1}≈0.0450, so F≈0.0416 - 0.0450≈-0.0034<0So, the root is between 3 and 3.1.Let's try x=3.05:sin(3.05)≈sin(3.05). Let's compute 3.05 radians is about 174.7 degrees.sin(3.05)≈sin(π - 0.0916)=sin(0.0916)≈0.0915e^{-3.05}≈e^{-3} * e^{-0.05}≈0.0498 * 0.9512≈0.0473So, F(3.05)=0.0915 - 0.0473≈0.0442>0F(3.05)=0.0442>0F(3.075):sin(3.075). 3.075 radians is about 176.3 degrees.sin(3.075)=sin(π - 0.0666)=sin(0.0666)≈0.0665e^{-3.075}=e^{-3} * e^{-0.075}≈0.0498 * 0.9281≈0.0462So, F(3.075)=0.0665 - 0.0462≈0.0203>0F(3.09):sin(3.09)=sin(π - 0.0516)=sin(0.0516)≈0.0515e^{-3.09}=e^{-3} * e^{-0.09}≈0.0498 * 0.9139≈0.0456F(3.09)=0.0515 - 0.0456≈0.0059>0F(3.095):sin(3.095)=sin(π - 0.0466)=sin(0.0466)≈0.0465e^{-3.095}=e^{-3} * e^{-0.095}≈0.0498 * 0.9089≈0.0453F(3.095)=0.0465 - 0.0453≈0.0012>0F(3.0975):sin(3.0975)=sin(π - 0.0441)=sin(0.0441)≈0.0440e^{-3.0975}=e^{-3} * e^{-0.0975}≈0.0498 * 0.9058≈0.0451F(3.0975)=0.0440 - 0.0451≈-0.0011<0So, the root is between 3.095 and 3.0975.Using linear approximation:At x=3.095, F=0.0012At x=3.0975, F=-0.0011Change in x=0.0025, change in F=-0.0023We need delta_x such that F=0:delta_x=0.0025*(0 - 0.0012)/(-0.0023 - 0.0012)=0.0025*(-0.0012)/(-0.0035)=0.0025*(0.0012/0.0035)=0.0025*0.3429≈0.000857So, approximate root at x≈3.095 + 0.000857≈3.0959Check F(3.0959):sin(3.0959)=sin(π - 0.0457)=sin(0.0457)≈0.0456e^{-3.0959}=e^{-3} * e^{-0.0959}≈0.0498 * 0.9073≈0.0452So, F≈0.0456 - 0.0452≈0.0004>0Almost there. Let's do another iteration.Using Newton-Raphson:x0=3.0959F(x0)=0.0004F'(x)=cos(x) + e^{-x}At x0=3.0959:cos(3.0959)=cos(π - 0.0457)= -cos(0.0457)≈-0.9989e^{-3.0959}≈0.0452So, F'(x0)= -0.9989 + 0.0452≈-0.9537Next approximation: x1 = x0 - F(x0)/F'(x0)=3.0959 - (0.0004)/(-0.9537)=3.0959 + 0.000419≈3.0963Compute F(3.0963):sin(3.0963)=sin(π - 0.0453)=sin(0.0453)≈0.0452e^{-3.0963}=e^{-3} * e^{-0.0963}≈0.0498 * 0.9069≈0.0452So, F≈0.0452 - 0.0452≈0. So, x≈3.0963 is a good approximation for c2.Therefore, c1≈0.5904 and c2≈3.0963.Now, with these points, we can set up the integrals.The area A is:A = ∫₀^{c1} [k(x) - h(x)] dx + ∫_{c1}^{c2} [h(x) - k(x)] dx + ∫_{c2}^{π} [k(x) - h(x)] dxBut actually, since from 0 to c1, k(x) > h(x); from c1 to c2, h(x) > k(x); and from c2 to π, k(x) > h(x). So, yes, that's correct.So, let's compute each integral separately.First integral: A1 = ∫₀^{c1} (e^{-x} - sin(x)) dxSecond integral: A2 = ∫_{c1}^{c2} (sin(x) - e^{-x}) dxThird integral: A3 = ∫_{c2}^{π} (e^{-x} - sin(x)) dxCompute each one:Compute A1:∫ (e^{-x} - sin(x)) dx = -e^{-x} + cos(x) + CEvaluate from 0 to c1:[-e^{-c1} + cos(c1)] - [-e^{0} + cos(0)] = (-e^{-c1} + cos(c1)) - (-1 + 1) = (-e^{-c1} + cos(c1)) - 0 = -e^{-c1} + cos(c1)Similarly, A2:∫ (sin(x) - e^{-x}) dx = -cos(x) - e^{-x} + CEvaluate from c1 to c2:[-cos(c2) - e^{-c2}] - [-cos(c1) - e^{-c1}] = (-cos(c2) - e^{-c2}) - (-cos(c1) - e^{-c1}) = (-cos(c2) - e^{-c2}) + cos(c1) + e^{-c1}A3:∫ (e^{-x} - sin(x)) dx = -e^{-x} + cos(x) + CEvaluate from c2 to π:[-e^{-π} + cos(π)] - [-e^{-c2} + cos(c2)] = (-e^{-π} -1) - (-e^{-c2} + cos(c2)) = (-e^{-π} -1) + e^{-c2} - cos(c2)Now, let's compute each part numerically.First, compute A1:A1 = -e^{-c1} + cos(c1)c1≈0.5904Compute e^{-c1}=e^{-0.5904}≈0.5543cos(c1)=cos(0.5904)≈0.8312So, A1≈-0.5543 + 0.8312≈0.2769Next, compute A2:A2 = (-cos(c2) - e^{-c2}) + cos(c1) + e^{-c1}c2≈3.0963Compute cos(c2)=cos(3.0963). Since 3.0963 is in radians, which is π - 0.0453≈3.0963.cos(3.0963)=cos(π - 0.0453)= -cos(0.0453)≈-0.9989e^{-c2}=e^{-3.0963}≈0.0452cos(c1)=0.8312e^{-c1}=0.5543So, A2≈(-(-0.9989) - 0.0452) + 0.8312 + 0.5543≈(0.9989 - 0.0452) + 0.8312 + 0.5543≈0.9537 + 0.8312 + 0.5543≈2.3392Wait, let me double-check:A2 = (-cos(c2) - e^{-c2}) + cos(c1) + e^{-c1}= (-(-0.9989) - 0.0452) + 0.8312 + 0.5543= (0.9989 - 0.0452) + 0.8312 + 0.5543= 0.9537 + 0.8312 + 0.5543= 0.9537 + 1.3855≈2.3392Yes, that's correct.Now, compute A3:A3 = (-e^{-π} -1) + e^{-c2} - cos(c2)Compute each term:e^{-π}≈0.0432So, -e^{-π} -1≈-0.0432 -1≈-1.0432e^{-c2}=0.0452cos(c2)= -0.9989So, A3≈(-1.0432) + 0.0452 - (-0.9989)= -1.0432 + 0.0452 + 0.9989≈(-1.0432 + 1.0441)≈0.0009Wait, that seems very small. Let me check:A3 = (-e^{-π} + cos(π)) - (-e^{-c2} + cos(c2))Wait, no, wait:Wait, the integral from c2 to π is:[-e^{-π} + cos(π)] - [-e^{-c2} + cos(c2)]= (-e^{-π} -1) - (-e^{-c2} + cos(c2))= -e^{-π} -1 + e^{-c2} - cos(c2)So, plugging in:-e^{-π}= -0.0432-1+ e^{-c2}= +0.0452- cos(c2)= -(-0.9989)= +0.9989So, total:-0.0432 -1 +0.0452 +0.9989≈(-0.0432 +0.0452) + (-1 +0.9989)≈0.002 + (-0.0011)≈0.0009Yes, so A3≈0.0009Therefore, the total area A≈A1 + A2 + A3≈0.2769 + 2.3392 + 0.0009≈2.617Wait, that seems a bit high. Let me check my calculations again.Wait, A1≈0.2769A2≈2.3392A3≈0.0009Total≈0.2769 + 2.3392 + 0.0009≈2.617But let me think about the functions. From 0 to c1≈0.59, e^{-x} is above sin(x). The area there is about 0.2769.From c1 to c2≈3.0963, sin(x) is above e^{-x}, and the area there is about 2.3392.From c2 to π≈3.1416, e^{-x} is slightly above sin(x), but the area is negligible, about 0.0009.So, total area≈2.617.But let me cross-verify by computing the integral numerically using another method.Alternatively, perhaps I can compute the integral ∫₀^π |sin(x) - e^{-x}| dx numerically.But since we've already split it into three parts and computed each, let's see.Alternatively, let's compute the integrals without splitting and see if we get the same result.Wait, but actually, the total area should be the sum of the absolute differences, which is what we've done.But let me compute A1 + A2 + A3≈0.2769 + 2.3392 + 0.0009≈2.617But let me compute the integral numerically using another approach.Alternatively, perhaps I can use numerical integration for the entire interval, considering the absolute value.But since we've already split it into regions where the functions cross, and computed each integral, I think 2.617 is a reasonable approximation.But let me check my calculations again for A2.A2 = ∫_{c1}^{c2} (sin(x) - e^{-x}) dxWhich is [-cos(x) - e^{-x}] from c1 to c2= (-cos(c2) - e^{-c2}) - (-cos(c1) - e^{-c1})= (-cos(c2) - e^{-c2}) + cos(c1) + e^{-c1}Plugging in:cos(c2)=cos(3.0963)=cos(π - 0.0453)= -cos(0.0453)≈-0.9989e^{-c2}=e^{-3.0963}≈0.0452cos(c1)=cos(0.5904)≈0.8312e^{-c1}=e^{-0.5904}≈0.5543So,(-(-0.9989) - 0.0452) + 0.8312 + 0.5543= (0.9989 - 0.0452) + 0.8312 + 0.5543= 0.9537 + 0.8312 + 0.5543≈2.3392Yes, that's correct.Similarly, A1≈0.2769 and A3≈0.0009.So, total area≈2.617.But let me compute the integrals using numerical methods to verify.Alternatively, perhaps I can use Simpson's rule or another method, but that might be time-consuming.Alternatively, let's compute the integrals symbolically and then plug in the approximate values.Wait, but we already did that.Alternatively, perhaps I can compute the exact expressions and then evaluate them numerically.Wait, let's see:A1 = -e^{-c1} + cos(c1)A2 = (-cos(c2) - e^{-c2}) + cos(c1) + e^{-c1}A3 = (-e^{-π} -1) + e^{-c2} - cos(c2)So, let's compute each term:A1:-e^{-c1} + cos(c1)≈-0.5543 + 0.8312≈0.2769A2:(-cos(c2) - e^{-c2}) + cos(c1) + e^{-c1}≈(0.9989 - 0.0452) + 0.8312 + 0.5543≈0.9537 + 0.8312 + 0.5543≈2.3392A3:(-e^{-π} -1) + e^{-c2} - cos(c2)≈(-0.0432 -1) + 0.0452 - (-0.9989)≈-1.0432 + 0.0452 + 0.9989≈-1.0432 + 1.0441≈0.0009So, same as before.Therefore, the total area is approximately 0.2769 + 2.3392 + 0.0009≈2.617But let me check if this makes sense.From 0 to c1≈0.59, the area is about 0.2769.From c1 to c2≈3.0963, the area is about 2.3392.From c2 to π≈3.1416, the area is negligible.So, total area≈2.617.But let me compute the integral numerically using another method to verify.Alternatively, perhaps I can use the fact that the area between two curves is the integral of the absolute difference.But since we've already split it into regions where one function is above the other, and computed each integral, I think 2.617 is correct.But let me compute the integral numerically using a calculator or software.Wait, since I don't have access to that right now, I'll proceed with the approximation.Therefore, the area bounded by h(x)=sin(x) and k(x)=e^{-x} over [0, π] is approximately 2.617.But let me check the units and the functions again.Wait, sin(x) and e^{-x} are both dimensionless functions, so the area is in square units, but since x is in radians, it's just a number.But let me see if 2.617 makes sense.From 0 to π, the integral of sin(x) is 2, and the integral of e^{-x} is (1 - e^{-π})≈1 - 0.0432≈0.9568.So, the area between them would be the integral of |sin(x) - e^{-x}|, which is more than the difference between these two integrals.Wait, the integral of sin(x) from 0 to π is 2.The integral of e^{-x} from 0 to π is (1 - e^{-π})≈0.9568.So, the difference between these integrals is 2 - 0.9568≈1.0432.But our computed area is 2.617, which is larger than this difference.Wait, that seems contradictory.Wait, no, because the area between the curves is not just the difference of the integrals, but the integral of the absolute difference.So, if the two functions cross, the area is the sum of the areas where one is above the other.So, it's possible that the total area is larger than the difference of the integrals.Wait, let me compute the integral of sin(x) from 0 to π: ∫₀^π sin(x) dx = 2Integral of e^{-x} from 0 to π: ∫₀^π e^{-x} dx = 1 - e^{-π}≈0.9568So, the integral of sin(x) - e^{-x} from 0 to π is 2 - 0.9568≈1.0432But the area between the curves is the integral of |sin(x) - e^{-x}|, which is larger because it accounts for the regions where sin(x) is below e^{-x} as positive areas.So, our computed area of≈2.617 is reasonable because it's larger than 1.0432.Therefore, I think 2.617 is a good approximation.But let me check if I can compute it more accurately.Alternatively, perhaps I can use the exact expressions and evaluate them more precisely.But since we've already used approximate values for c1 and c2, it's difficult to get a more precise result without more accurate values for c1 and c2.Alternatively, perhaps I can use more precise approximations for c1 and c2.But for the purposes of this problem, I think 2.617 is a reasonable approximation.Therefore, the area bounded by h(x)=sin(x) and k(x)=e^{-x} over [0, π] is approximately 2.617 square units.But let me write it as 2.617, but perhaps rounding to three decimal places, it's 2.617.Alternatively, if I use more precise values for c1 and c2, the area might be slightly different, but for now, I'll stick with 2.617.So, summarizing:1. The points of intersection are at x = [7 ± sqrt(33)] / 2, with corresponding y-values.2. The area bounded by sin(x) and e^{-x} over [0, π] is approximately 2.617.But wait, let me check if I can express the area in terms of exact expressions.Wait, for the first problem, we have exact points of intersection, but for the second problem, since the functions are transcendental, we can't express the area in terms of elementary functions, so we have to approximate it numerically.Therefore, the final answers are:1. The points of intersection are ([7 + sqrt(33)] / 2, (25 + 3sqrt(33))/2) and ([7 - sqrt(33)] / 2, (25 - 3sqrt(33))/2).2. The area is approximately 2.617.But let me check if I can write the area more precisely.Alternatively, perhaps I can use more decimal places for c1 and c2 to get a better approximation.But for the sake of time, I'll proceed with the approximation of 2.617.Therefore, the final answers are:1. The points of intersection are at x = [7 ± sqrt(33)] / 2, with corresponding y-values as calculated.2. The area is approximately 2.617.</think>

🧠:<think>Okay, so I have this problem where a social scientist is evaluating a development project that includes a literacy program and an economic empowerment initiative. The goal is to improve literacy rates and economic outcomes in a rural community. They've collected data over five years, measuring literacy rates (L) and average household income (I) before and after the project.The problem is divided into two parts. The first part is about determining the stability of the equilibrium points of a system of nonlinear differential equations. The second part is about maximizing a combined metric M, which is a weighted sum of literacy and income, subject to some constraints.Starting with part 1: The system of differential equations is given as:[frac{dL}{dt} = k_1Lleft(1 - frac{L}{L_{text{max}}}right) - aI][frac{dI}{dt} = k_2Ileft(1 - frac{I}{I_{text{max}}}right) + bL]Here, (k_1, k_2, a, b) are positive constants, and (L_{text{max}}, I_{text{max}}) are the maximum possible literacy rate and maximum sustainable income level, respectively.I need to find the equilibrium points of this system and determine their stability.First, equilibrium points occur where both (frac{dL}{dt} = 0) and (frac{dI}{dt} = 0). So I need to solve the system:1. (k_1Lleft(1 - frac{L}{L_{text{max}}}right) - aI = 0)2. (k_2Ileft(1 - frac{I}{I_{text{max}}}right) + bL = 0)Let me write these equations more clearly:Equation (1): (k_1L - frac{k_1}{L_{text{max}}} L^2 - aI = 0)Equation (2): (k_2I - frac{k_2}{I_{text{max}}} I^2 + bL = 0)So, I have two equations:1. (- frac{k_1}{L_{text{max}}} L^2 + k_1L - aI = 0)2. (- frac{k_2}{I_{text{max}}} I^2 + k_2I + bL = 0)This is a system of nonlinear equations. Let me try to solve for L and I.Let me denote:Equation (1): (A L^2 + B L + C I = 0), where (A = - frac{k_1}{L_{text{max}}}), (B = k_1), (C = -a)Equation (2): (D I^2 + E I + F L = 0), where (D = - frac{k_2}{I_{text{max}}}), (E = k_2), (F = b)So, I can write:From equation (1):(A L^2 + B L + C I = 0) => (I = frac{ - A L^2 - B L }{C })Similarly, from equation (2):(D I^2 + E I + F L = 0)So, substitute I from equation (1) into equation (2):(D left( frac{ - A L^2 - B L }{C } right)^2 + E left( frac{ - A L^2 - B L }{C } right) + F L = 0)This will result in a quartic equation in terms of L, which can be quite complicated. Maybe there's a smarter way.Alternatively, perhaps I can find the equilibrium points by considering possible cases.First, let's consider the trivial equilibrium points.Case 1: L = 0, I = 0Plugging into equation (1): 0 - 0 - a*0 = 0, which holds.Equation (2): 0 - 0 + b*0 = 0, which holds.So, (0,0) is an equilibrium point.Case 2: L = 0, I ≠ 0From equation (1): 0 - 0 - aI = 0 => I = 0. So, only (0,0) is possible here.Case 3: I = 0, L ≠ 0From equation (2): 0 - 0 + bL = 0 => L = 0. So, again only (0,0) is possible.Case 4: Both L and I are non-zero.So, we need to solve the system:1. (k_1Lleft(1 - frac{L}{L_{text{max}}}right) = aI)2. (k_2Ileft(1 - frac{I}{I_{text{max}}}right) = -bL)Wait, equation (2) is (k_2I(1 - I/I_{text{max}}) + bL = 0), so (k_2I(1 - I/I_{text{max}}) = -bL). Since (k_2, I, L) are positive, the right-hand side is negative, so (k_2I(1 - I/I_{text{max}})) must be negative.Therefore, (1 - I/I_{text{max}} < 0), which implies that (I > I_{text{max}}). But in the context, I is the average household income, which is constrained by (I leq I_{text{max}}). So, this suggests that in the feasible region, (I leq I_{text{max}}), so (1 - I/I_{text{max}} geq 0). Therefore, (k_2I(1 - I/I_{text{max}})) is non-negative, since all terms are positive.But from equation (2), (k_2I(1 - I/I_{text{max}}) = -bL). The left-hand side is non-negative, and the right-hand side is negative because (b, L) are positive. So, this can only hold if both sides are zero.Therefore, (k_2I(1 - I/I_{text{max}}) = 0) and (-bL = 0). So, either (I = 0) or (I = I_{text{max}}), and (L = 0).So, if (I = I_{text{max}}), then from equation (1):(k_1L(1 - L/L_{text{max}}) = aI_{text{max}})But (I = I_{text{max}}), so:(k_1L(1 - L/L_{text{max}}) = aI_{text{max}})This is a quadratic equation in L:(- frac{k_1}{L_{text{max}}} L^2 + k_1 L - a I_{text{max}} = 0)Multiply both sides by (-L_{text{max}} / k_1) to simplify:(L^2 - L_{text{max}} L + frac{a I_{text{max}} L_{text{max}}}{k_1} = 0)The discriminant is:(D = L_{text{max}}^2 - 4 cdot 1 cdot frac{a I_{text{max}} L_{text{max}}}{k_1})For real solutions, D must be non-negative:(L_{text{max}}^2 - frac{4 a I_{text{max}} L_{text{max}}}{k_1} geq 0)Divide both sides by (L_{text{max}}) (positive):(L_{text{max}} - frac{4 a I_{text{max}}}{k_1} geq 0)So, (L_{text{max}} geq frac{4 a I_{text{max}}}{k_1})If this holds, then we have two solutions:(L = frac{ L_{text{max}} pm sqrt{ L_{text{max}}^2 - frac{4 a I_{text{max}} L_{text{max}}}{k_1} } }{2})Simplify:(L = frac{ L_{text{max}} }{2 } pm frac{ sqrt{ L_{text{max}}^2 - frac{4 a I_{text{max}} L_{text{max}}}{k_1} } }{2})But since (L) must be non-negative, both solutions could be valid depending on the discriminant.However, if (L_{text{max}} < frac{4 a I_{text{max}}}{k_1}), then there are no real solutions, meaning the only equilibrium points are (0,0) and (L, I) where I = I_{text{max}} and L is as above.Wait, but earlier, we saw that equation (2) can only be satisfied if I = 0 or I = I_{text{max}}, but if I = I_{text{max}}, then equation (1) gives us L as above.But wait, if I = I_{text{max}}, then equation (2) gives:(k_2 I_{text{max}} (1 - 1) + b L = 0) => 0 + b L = 0 => L = 0.But from equation (1), if I = I_{text{max}}, then (k_1 L (1 - L / L_{text{max}}) = a I_{text{max}}). If L = 0, then the left-hand side is 0, which would require (a I_{text{max}} = 0). But a and I_{text{max}} are positive constants, so this is impossible.Therefore, there is a contradiction here. So, if I = I_{text{max}}, equation (2) requires L = 0, but equation (1) cannot be satisfied because it would require 0 = a I_{text{max}}, which is not possible. Therefore, the only equilibrium points are (0,0) and potentially other points where both L and I are non-zero but not at their maximums.Wait, but earlier, we saw that equation (2) requires that if I is non-zero, then (k_2 I (1 - I / I_{text{max}}) = -b L). Since the left side is non-negative (because I <= I_{text{max}} implies 1 - I / I_{text{max}} >= 0, and k2, I are positive), and the right side is negative (because b and L are positive). So, the only way this can hold is if both sides are zero. Therefore, either I = 0 or I = I_{text{max}}, but as we saw, I = I_{text{max}} leads to a contradiction. Therefore, the only equilibrium point is (0,0).Wait, that can't be right. Because if both L and I are zero, that's an equilibrium, but perhaps there are other equilibrium points where both L and I are positive.Wait, maybe I made a mistake in my earlier reasoning. Let me go back.From equation (2):(k_2 I (1 - I / I_{text{max}}) + b L = 0)Since (k_2, I, L) are positive, the term (k_2 I (1 - I / I_{text{max}})) is non-negative (because (I leq I_{text{max}})), and (b L) is positive. Therefore, the sum of two non-negative terms is zero only if both are zero.Therefore, (k_2 I (1 - I / I_{text{max}}) = 0) and (b L = 0). So, either I = 0 or I = I_{text{max}}, and L = 0.But if I = I_{text{max}}, then from equation (1):(k_1 L (1 - L / L_{text{max}}) = a I_{text{max}})But from equation (2), if I = I_{text{max}}, then L must be zero, which would require (a I_{text{max}} = 0), which is impossible. Therefore, the only possible equilibrium point is (0,0).Wait, that seems to suggest that the only equilibrium is (0,0), but that can't be right because the system is supposed to model growth in L and I. Maybe I'm missing something.Alternatively, perhaps I made a mistake in interpreting the equations. Let me check the original equations again.The equations are:[frac{dL}{dt} = k_1 L (1 - L / L_{text{max}}) - a I][frac{dI}{dt} = k_2 I (1 - I / I_{text{max}}) + b L]So, both L and I have logistic growth terms, but they are also coupled: L decreases due to I (term -aI) and I increases due to L (term +bL).So, perhaps there is another equilibrium point where both L and I are positive.Let me try to solve the system again.From equation (1):(k_1 L (1 - L / L_{text{max}}) = a I)From equation (2):(k_2 I (1 - I / I_{text{max}}) = -b L)But from equation (2), since the left side is non-negative (because I <= I_{text{max}}), and the right side is negative (because b and L are positive), the only way this can hold is if both sides are zero. Therefore, (k_2 I (1 - I / I_{text{max}}) = 0) and (-b L = 0). So, I must be 0 or I_{text{max}}, and L must be 0.But if I = I_{text{max}}, then from equation (1):(k_1 L (1 - L / L_{text{max}}) = a I_{text{max}})But from equation (2), if I = I_{text{max}}, then L must be 0, which would require (a I_{text{max}} = 0), which is impossible. Therefore, the only equilibrium point is (0,0).Wait, that can't be right because the system is supposed to model growth in L and I. Maybe I made a mistake in the sign.Wait, in equation (2), it's (k_2 I (1 - I / I_{text{max}}) + b L = 0). So, the term (k_2 I (1 - I / I_{text{max}})) is non-negative, and (b L) is positive, so their sum is positive, which cannot be zero unless both are zero. Therefore, the only equilibrium is (0,0).But that seems counterintuitive because the system is supposed to model growth. Maybe the model is set up such that without any external intervention, the system remains at zero, but with the project, it grows. But in the model, the project is represented by the terms -aI and +bL. Wait, no, the project is the entire system. So, perhaps the equilibrium points are only (0,0) and another point where both L and I are positive.Wait, perhaps I made a mistake in the earlier step. Let me try to solve the system again.From equation (1):(k_1 L (1 - L / L_{text{max}}) = a I) => (I = frac{k_1}{a} L (1 - L / L_{text{max}}))From equation (2):(k_2 I (1 - I / I_{text{max}}) = -b L)Substitute I from equation (1) into equation (2):(k_2 left( frac{k_1}{a} L (1 - L / L_{text{max}}) right) left(1 - frac{ frac{k_1}{a} L (1 - L / L_{text{max}}) }{ I_{text{max}} } right) = -b L)This is a complicated equation, but let's try to simplify it.Let me denote (I = frac{k_1}{a} L (1 - L / L_{text{max}})) as before.Then, equation (2) becomes:(k_2 I (1 - I / I_{text{max}}) = -b L)Substituting I:(k_2 cdot frac{k_1}{a} L (1 - L / L_{text{max}}) cdot left(1 - frac{ frac{k_1}{a} L (1 - L / L_{text{max}}) }{ I_{text{max}} } right) = -b L)Let me factor out L on both sides. Assuming L ≠ 0 (since we already considered L=0), we can divide both sides by L:(k_2 cdot frac{k_1}{a} (1 - L / L_{text{max}}) cdot left(1 - frac{ frac{k_1}{a} L (1 - L / L_{text{max}}) }{ I_{text{max}} } right) = -b)This is a nonlinear equation in L. Let me denote (x = L / L_{text{max}}), so x is between 0 and 1.Then, L = x L_{text{max}}, and 1 - L / L_{text{max}} = 1 - x.Substituting:(k_2 cdot frac{k_1}{a} (1 - x) cdot left(1 - frac{ frac{k_1}{a} x L_{text{max}} (1 - x) }{ I_{text{max}} } right) = -b)Let me simplify the term inside the second parenthesis:(1 - frac{ frac{k_1}{a} x L_{text{max}} (1 - x) }{ I_{text{max}} } = 1 - frac{k_1 x L_{text{max}} (1 - x)}{a I_{text{max}}})Let me denote (C = frac{k_1 L_{text{max}}}{a I_{text{max}}}), so the term becomes:(1 - C x (1 - x))Therefore, the equation becomes:(k_2 cdot frac{k_1}{a} (1 - x) (1 - C x (1 - x)) = -b)Let me compute the left-hand side:(k_2 cdot frac{k_1}{a} (1 - x) (1 - C x + C x^2) = -b)Expanding:(k_2 cdot frac{k_1}{a} [ (1)(1 - C x + C x^2) - x(1 - C x + C x^2) ] = -b)Simplify inside the brackets:= ( (1 - C x + C x^2) - x + C x^2 - C x^3 )= (1 - C x + C x^2 - x + C x^2 - C x^3)Combine like terms:= (1 - (C + 1) x + 2 C x^2 - C x^3)Therefore, the equation is:(k_2 cdot frac{k_1}{a} [1 - (C + 1) x + 2 C x^2 - C x^3] = -b)Bring all terms to one side:(k_2 cdot frac{k_1}{a} [1 - (C + 1) x + 2 C x^2 - C x^3] + b = 0)This is a cubic equation in x:(- C k_2 frac{k_1}{a} x^3 + 2 C k_2 frac{k_1}{a} x^2 - (C + 1) k_2 frac{k_1}{a} x + k_2 frac{k_1}{a} + b = 0)Let me denote the coefficients as follows:A = - C k_2 frac{k_1}{a}B = 2 C k_2 frac{k_1}{a}C' = - (C + 1) k_2 frac{k_1}{a}D = k_2 frac{k_1}{a} + bSo, the equation is:A x^3 + B x^2 + C' x + D = 0This is a cubic equation, which can have up to three real roots. Each real root corresponds to a possible equilibrium point (L, I) where L = x L_{text{max}}, and I is given by equation (1).However, solving this cubic equation analytically is quite involved, and the roots may not be expressible in a simple form. Therefore, perhaps it's better to analyze the stability of the equilibrium points without explicitly solving for them.But wait, earlier, we saw that the only possible equilibrium points are (0,0) and potentially other points where both L and I are positive, but due to the constraints from equation (2), it's unclear.Alternatively, perhaps the system has only the trivial equilibrium (0,0) and another equilibrium where both L and I are positive. Let me assume that there is another equilibrium point (L*, I*) where both are positive.To determine the stability, I need to linearize the system around the equilibrium points and analyze the eigenvalues of the Jacobian matrix.The Jacobian matrix J is given by:[J = begin{bmatrix}frac{partial}{partial L} frac{dL}{dt} & frac{partial}{partial I} frac{dL}{dt} frac{partial}{partial L} frac{dI}{dt} & frac{partial}{partial I} frac{dI}{dt}end{bmatrix}]Compute each partial derivative:1. (frac{partial}{partial L} frac{dL}{dt} = k_1 (1 - frac{L}{L_{text{max}}}) - k_1 frac{L}{L_{text{max}}}) = (k_1 (1 - frac{2L}{L_{text{max}}}))2. (frac{partial}{partial I} frac{dL}{dt} = -a)3. (frac{partial}{partial L} frac{dI}{dt} = b)4. (frac{partial}{partial I} frac{dI}{dt} = k_2 (1 - frac{I}{I_{text{max}}}) - k_2 frac{I}{I_{text{max}}}) = (k_2 (1 - frac{2I}{I_{text{max}}}))So, the Jacobian matrix is:[J = begin{bmatrix}k_1 (1 - frac{2L}{L_{text{max}}}) & -a b & k_2 (1 - frac{2I}{I_{text{max}}})end{bmatrix}]Now, evaluate J at each equilibrium point.First, at (0,0):J(0,0) = [begin{bmatrix}k_1 (1 - 0) & -a b & k_2 (1 - 0)end{bmatrix}= begin{bmatrix}k_1 & -a b & k_2end{bmatrix}]The eigenvalues of this matrix determine the stability. The trace is (k_1 + k_2), which is positive, and the determinant is (k_1 k_2 - (-a)(b) = k_1 k_2 + a b), which is positive. Since both trace and determinant are positive, the eigenvalues have positive real parts, meaning the equilibrium (0,0) is an unstable node.Next, consider the other equilibrium point (L*, I*), assuming it exists. At this point, both L* and I* are positive.The Jacobian at (L*, I*) is:[J(L*, I*) = begin{bmatrix}k_1 (1 - frac{2L*}{L_{text{max}}}) & -a b & k_2 (1 - frac{2I*}{I_{text{max}}})end{bmatrix}]To determine stability, we need to find the eigenvalues of this matrix. The eigenvalues λ satisfy:[lambda^2 - text{tr}(J) lambda + det(J) = 0]Where:- tr(J) = (k_1 (1 - frac{2L*}{L_{text{max}}}) + k_2 (1 - frac{2I*}{I_{text{max}}}))- det(J) = [k_1 (1 - frac{2L*}{L_{text{max}}})][k_2 (1 - frac{2I*}{I_{text{max}}})] + a bThe stability depends on the signs of tr(J) and det(J). If det(J) > 0 and tr(J) < 0, the equilibrium is a stable node. If det(J) > 0 and tr(J) > 0, it's an unstable node. If det(J) < 0, it's a saddle point.But without knowing the exact values of L* and I*, it's hard to determine the signs. However, we can reason about the possible signs.At the equilibrium (L*, I*), from equation (1):(k_1 L* (1 - L* / L_{text{max}}) = a I*)From equation (2):(k_2 I* (1 - I* / I_{text{max}}) = -b L*)But earlier, we saw that equation (2) implies that (k_2 I* (1 - I* / I_{text{max}})) must be negative, which is only possible if (I* > I_{text{max}}). But since I* is constrained by (I leq I_{text{max}}), this suggests that the only possible equilibrium is (0,0). Therefore, perhaps there are no other equilibrium points, and (0,0) is the only equilibrium, which is unstable.But this contradicts the intuition that the system should have another equilibrium where both L and I are positive. Maybe the model is set up such that the coupling terms cause the system to have another equilibrium.Alternatively, perhaps the system can have multiple equilibrium points depending on the parameter values.Given the complexity, perhaps it's better to conclude that the only equilibrium is (0,0), which is unstable, and the system may approach another equilibrium depending on the initial conditions.But wait, let's consider the possibility that there is another equilibrium where both L and I are positive. Let's assume that such a point exists.At (L*, I*), the trace of J is:tr(J) = (k_1 (1 - 2L*/L_{text{max}}) + k_2 (1 - 2I*/I_{text{max}}))The determinant is:det(J) = [k_1 (1 - 2L*/L_{text{max}})][k_2 (1 - 2I*/I_{text{max}})] + a bIf tr(J) < 0 and det(J) > 0, then the equilibrium is stable.Given that k1, k2, a, b are positive, and L*, I* are positive, it's possible that tr(J) is negative if the terms (1 - 2L*/L_{text{max}}) and (1 - 2I*/I_{text{max}}) are negative, i.e., if L* > L_{text{max}}/2 and I* > I_{text{max}}/2.But from equation (2), we have:(k_2 I* (1 - I*/I_{text{max}}) = -b L*)Since the left side is non-positive (because I* <= I_{text{max}} implies 1 - I*/I_{text{max}} >= 0, but the right side is negative), this implies that I* must be greater than I_{text{max}} to make the left side negative. But I* is constrained by I <= I_{text{max}}, so this is impossible. Therefore, the only equilibrium is (0,0), which is unstable.Therefore, the system has only one equilibrium point at (0,0), which is unstable. This suggests that the system does not settle into a stable equilibrium but rather grows without bound, which may not be realistic. Therefore, perhaps the model is missing some terms or the parameters are such that another equilibrium exists.Alternatively, perhaps the system has a limit cycle or other behavior, but without further analysis, it's hard to say.Given the time constraints, I think the answer is that the only equilibrium is (0,0), which is unstable.Now, moving to part 2: The project aims to maximize the combined metric M = α L + β I, subject to L <= L_{text{max}} and I <= I_{text{max}}.This is a constrained optimization problem. We need to maximize M = α L + β I, with L <= L_{text{max}}, I <= I_{text{max}}.Assuming that L and I can be controlled independently, the maximum of M would occur at the maximum possible L and I, i.e., L = L_{text{max}}, I = I_{text{max}}, since α and β are positive weights.But perhaps the problem is more involved, considering the dynamics of the system. However, the question says "given the constraints L <= L_{text{max}} and I <= I_{text{max}}", so it's a simple linear optimization problem.In linear optimization, the maximum of a linear function over a convex polygon occurs at a vertex. The feasible region is defined by L >= 0, I >= 0, L <= L_{text{max}}, I <= I_{text{max}}.The vertices are (0,0), (L_{text{max}}, 0), (0, I_{text{max}}), and (L_{text{max}}, I_{text{max}}).Evaluating M at each vertex:- At (0,0): M = 0- At (L_{text{max}}, 0): M = α L_{text{max}}- At (0, I_{text{max}}): M = β I_{text{max}}- At (L_{text{max}}, I_{text{max}}): M = α L_{text{max}} + β I_{text{max}}Therefore, the maximum M occurs at (L_{text{max}}, I_{text{max}}), so the optimal values are L = L_{text{max}}, I = I_{text{max}}.But wait, in the context of the problem, the project is aimed at improving L and I, so the maximum possible values would be the optimal. However, in the first part, we saw that the system may not reach these points due to instability. But in the optimization part, we're just maximizing M given the constraints, regardless of the system dynamics.Therefore, the optimal values are L = L_{text{max}}, I = I_{text{max}}.But perhaps the problem expects a more nuanced answer, considering the trade-offs between L and I. For example, if increasing L requires decreasing I or vice versa, but in this case, the constraints are independent, so the maximum is achieved at the upper bounds.Alternatively, if the problem is to maximize M subject to the system's dynamics, then it's a different story, but the question doesn't specify that. It just says "given the constraints L <= L_{text{max}} and I <= I_{text{max}}", so I think the answer is L = L_{text{max}}, I = I_{text{max}}.</think>

🧠:<think>Alright, so I've got this problem about a film director and their movie's structure represented by a directed graph. It's split into two parts. Let me try to tackle them one by one.Starting with part 1: The director wants exactly one unique path from the opening scene (node 1) to the final scene (node n). The structure is represented by an adjacency matrix A. I need to derive the conditions on A that guarantee this unique path.Hmm, okay. So, in graph theory terms, we're dealing with a directed graph where each node is a scene, and edges represent possible transitions. The adjacency matrix A has 1s where there's a direct transition and 0s otherwise.To have exactly one unique path from node 1 to node n, the graph must be structured in a way that there's no ambiguity in the transitions. That is, from each node, there should be only one outgoing edge that leads towards the final node n. Otherwise, if there are multiple edges from a node, that could create multiple paths.Wait, but it's not just about outgoing edges from each node. It's about the entire graph's structure ensuring that any path starting at 1 must follow a single route to n without any branches. So, the graph should be a directed acyclic graph (DAG) with a single topological order from 1 to n.But more specifically, each node along the path from 1 to n should have exactly one outgoing edge that leads towards n, and all other edges from these nodes should either lead to nodes that are not on the path or create cycles that don't interfere with the unique path.Wait, no, actually, if there are cycles, that could complicate things. Because cycles could create multiple paths if you can loop around and come back. So, perhaps the graph needs to be a DAG without any cycles, and the unique path is the only path from 1 to n.But even in a DAG, there can be multiple paths. So, to have exactly one unique path, the graph must be such that for every node on the path from 1 to n, there's only one outgoing edge that is part of the unique path. Additionally, any other edges from these nodes must not lead to nodes that could create alternative paths to n.Alternatively, perhaps the graph must be a tree with node 1 as the root and node n as the only leaf, but that might be too restrictive because in a tree, each node except the root has exactly one parent, but here we're dealing with directed edges.Wait, maybe it's better to think in terms of adjacency matrix properties. For there to be exactly one path from 1 to n, the number of paths from 1 to n in the graph must be exactly one. In terms of linear algebra, the number of paths can be found by raising the adjacency matrix to powers. Specifically, (A^k)_{1n} gives the number of paths of length k from 1 to n. So, if we sum over all k, we get the total number of paths.But we want this sum to be exactly 1. So, the adjacency matrix must be such that when we compute the total number of paths from 1 to n, it equals 1.But how can we ensure that? It must be that the graph is such that there's only one simple path from 1 to n, and no cycles, because cycles could create infinitely many paths if they're reachable from 1 and can reach n, but since we're dealing with finite matrices, it's more about the structure.Wait, but in a finite graph, if there's a cycle on the path from 1 to n, you could loop around the cycle multiple times, creating infinitely many paths, but in our case, the adjacency matrix is finite, so perhaps we need to ensure that there are no cycles that are on any path from 1 to n. So, the subgraph reachable from 1 and that can reach n must be a DAG.Moreover, within this DAG, the number of paths from 1 to n must be exactly one. So, the DAG must have a unique topological order from 1 to n.Alternatively, the graph must be such that it's a directed tree with node 1 as the root and node n as the only node with no outgoing edges, but that might not necessarily be the case.Wait, perhaps a better way is to consider that the graph must be a directed acyclic graph where each node on the unique path from 1 to n has exactly one outgoing edge that is part of the path, and any other outgoing edges must lead to nodes that are not on any path to n.So, for each node i on the unique path from 1 to n, A_{i,j} = 1 only for the next node in the path, and all other A_{i,k} = 0 for k not on the path. But that might be too restrictive because nodes not on the path can have outgoing edges, but those edges shouldn't create alternative paths to n.Wait, no. If a node not on the path has an outgoing edge to a node on the path, that could create an alternative path. So, actually, nodes not on the path should not have any edges that can reach n. That is, the subgraph induced by nodes not on the path should be disconnected from n.Alternatively, the graph should be such that the only path from 1 to n is the unique path, and all other edges either go from nodes on the path to nodes not on the path or from nodes not on the path to other nodes not on the path, but not creating any alternative routes to n.So, in terms of the adjacency matrix, for the unique path, say 1 → a → b → ... → n, each node in this path must have exactly one outgoing edge that is part of the path. Additionally, any other outgoing edges from these nodes must go to nodes that are not on any path to n. That is, for each node i on the path, if A_{i,j} = 1 for some j not on the path, then j must not be able to reach n.Moreover, nodes not on the path should not have any edges that can reach n. So, for any node k not on the path, there should be no directed path from k to n.Therefore, the adjacency matrix must satisfy:1. There exists a unique path from 1 to n.2. For each node i on this path, all outgoing edges except the one leading to the next node in the path must go to nodes that cannot reach n.3. All nodes not on the path must not have any outgoing edges that can reach n.Alternatively, in terms of the adjacency matrix, if we denote the unique path as 1 = v1, v2, ..., vk = n, then for each vi (i < k), A_{vi, vi+1} = 1, and for all other j ≠ vi+1, A_{vi,j} = 0 or if A_{vi,j} = 1, then j cannot reach n.But how do we express this in terms of the adjacency matrix? Maybe by considering the reachability matrix. The reachability matrix can be computed as (I + A + A^2 + ... + A^{n-1}), where I is the identity matrix. If we denote R as the reachability matrix, then R_{i,j} = 1 if there's a path from i to j.So, for nodes not on the unique path, their reachability to n must be 0. That is, for any node k not on the path, R_{k,n} = 0.Additionally, for nodes on the path, their outgoing edges must only go to the next node in the path or to nodes that cannot reach n.So, in terms of the adjacency matrix, for each node vi on the path (except the last one), A_{vi, vi+1} = 1, and for all other j, if A_{vi,j} = 1, then R_{j,n} = 0.Moreover, the number of paths from 1 to n must be exactly 1, which can be checked by computing the (1,n) entry of the matrix (I - A)^{-1}, which gives the total number of paths. So, (I - A)^{-1}_{1,n} = 1.But inverting (I - A) might not be straightforward, especially if the graph has cycles. However, since we need exactly one path, the graph must be a DAG, so (I - A) is invertible, and the (1,n) entry must be 1.Alternatively, since it's a DAG, we can perform a topological sort and ensure that the path from 1 to n is unique.Wait, maybe a better approach is to consider that the adjacency matrix must be such that the graph is a directed tree with node 1 as the root and node n as the only node with no children, but that might not be the case because nodes can have multiple children as long as they don't create alternative paths to n.Wait, no. If a node on the path has multiple children, but only one of them is on the path to n, then that's okay as long as the other children don't have paths to n. So, the adjacency matrix must have exactly one child on the path for each node on the path, and all other children must not lead to n.So, in summary, the conditions on A are:1. The graph must be a directed acyclic graph (DAG).2. There must be exactly one path from node 1 to node n.3. For each node on the unique path from 1 to n, all outgoing edges except the one leading to the next node in the path must go to nodes that cannot reach n.4. All nodes not on the unique path must not have any outgoing edges that can reach n.Alternatively, in terms of the adjacency matrix:- For each node i on the unique path, A_{i,j} = 1 only for j = next node in the path, or for j such that R_{j,n} = 0.- For all nodes not on the path, A_{i,j} = 0 for all j that can reach n.But perhaps a more precise condition is:- The adjacency matrix must be such that the graph is a DAG with a unique topological order from 1 to n, and all other edges (if any) must go from nodes in the path to nodes that are not in the path and cannot reach n.So, to formalize this, let's denote the unique path as P = [v1, v2, ..., vk], where v1 = 1 and vk = n.Then, for each vi in P (i < k), A_{vi, vi+1} = 1, and for all other j, A_{vi,j} = 0 or if A_{vi,j} = 1, then j is not in P and j cannot reach n.Additionally, for all nodes not in P, A_{j,l} = 0 for all l that can reach n.So, in terms of the adjacency matrix, these conditions must be satisfied.Now, moving on to part 2: The director wants a hidden subplot represented by a Hamiltonian path from scene k to scene m, where 1 < k < m < n. Given the adjacency matrix A, we need to determine if such a Hamiltonian path exists and describe how to find it.A Hamiltonian path is a path that visits every node exactly once. So, in this case, the subgraph must include all nodes from k to m, but wait, no. Wait, the subgraph is a Hamiltonian path starting at k and ending at m, which means it must visit every node exactly once in a sequence starting at k and ending at m.But wait, the problem says it's a subgraph of the original graph, so the Hamiltonian path must use edges that exist in A. So, we need to check if there's a directed path from k to m that visits every node exactly once.Wait, but Hamiltonian path usually refers to a path that visits every node in the graph exactly once. But in this case, the subgraph is a Hamiltonian path, which would imply that it's a path that includes all nodes. But the original graph has n nodes, so the Hamiltonian path would have to include all n nodes? Or is it a Hamiltonian path in the subgraph induced by some subset?Wait, the problem says "a Hamiltonian path that starts from scene k and ends in scene m". So, it's a path that starts at k, ends at m, and visits every node exactly once. So, it's a Hamiltonian path in the entire graph, but starting at k and ending at m.But wait, the original graph has n nodes, so a Hamiltonian path would have n-1 edges and visit all n nodes. So, the path would be k → ... → m, covering all nodes in between.But the problem says "another directed graph, which is a subgraph of the original graph". So, the hidden subplot is a subgraph that is a Hamiltonian path from k to m.So, the steps to determine if such a path exists would involve checking if there's a directed Hamiltonian path from k to m in the original graph.To find such a path, one approach is to perform a depth-first search (DFS) or backtracking algorithm, trying all possible paths from k to m and checking if any of them visit all nodes exactly once.However, since the problem is about formulating a method, not necessarily an efficient one, we can describe it as follows:1. Check if the graph is strongly connected, but since it's a directed graph, even if it's not strongly connected, there might still be a Hamiltonian path from k to m.2. Perform a search (like DFS) starting from k, keeping track of visited nodes, and trying to reach m while visiting all nodes exactly once.3. If such a path is found, return it; otherwise, conclude that no such Hamiltonian path exists.Alternatively, we can use dynamic programming or memoization to keep track of visited nodes and current position, but the basic idea is similar.So, in terms of the adjacency matrix, for each node, we can look at its outgoing edges and recursively explore each possibility, ensuring that each node is visited exactly once.But since the problem is about formulating the method, not implementing it, the steps are:- Start at node k.- At each step, choose an outgoing edge to a node that hasn't been visited yet.- Continue until either m is reached with all nodes visited, or all possibilities are exhausted.If during the search, we reach m having visited all nodes, then the Hamiltonian path exists.So, summarizing:To determine if a Hamiltonian path from k to m exists in the directed graph represented by A:1. Perform a depth-first search starting at node k.2. At each step, only traverse edges that exist in A (i.e., where A_{i,j} = 1).3. Keep track of visited nodes to ensure each node is visited exactly once.4. If during the search, node m is reached and all nodes have been visited, then such a Hamiltonian path exists.5. If all possible paths are exhausted without finding such a path, then it does not exist.This is a brute-force approach, but it's a valid method to determine the existence of the Hamiltonian path.So, putting it all together, the conditions for part 1 are that the graph is a DAG with a unique path from 1 to n, and all other edges from nodes on this path do not lead to nodes that can reach n, and nodes not on the path cannot reach n. For part 2, we can use a DFS-based approach to search for a Hamiltonian path from k to m.</think>