🧠:<think>Okay, so I have this differential equation to solve: dL/dt = -0.1L(t) + 0.5cos(t). Hmm, it's a first-order linear non-homogeneous differential equation. I remember that to solve such equations, I need to find the integrating factor. Let me recall the standard form of a linear differential equation: dy/dt + P(t)y = Q(t). So, I should rewrite the given equation in that form.Starting with dL/dt = -0.1L + 0.5cos(t). If I move the -0.1L term to the left side, it becomes dL/dt + 0.1L = 0.5cos(t). Yeah, that looks right. So here, P(t) is 0.1 and Q(t) is 0.5cos(t). Since P(t) is a constant, the integrating factor will be e^(∫P(t)dt) which is e^(0.1t). Alright, so the integrating factor μ(t) is e^(0.1t). I need to multiply both sides of the differential equation by this factor. Let's do that:e^(0.1t) * dL/dt + 0.1e^(0.1t) * L = 0.5e^(0.1t)cos(t).The left side of this equation should now be the derivative of (μ(t) * L(t)). Let me check that:d/dt [e^(0.1t) * L(t)] = e^(0.1t) * dL/dt + 0.1e^(0.1t) * L(t). Yep, that's exactly what we have on the left side. So, we can rewrite the equation as:d/dt [e^(0.1t) * L(t)] = 0.5e^(0.1t)cos(t).Now, to find L(t), I need to integrate both sides with respect to t. Let's integrate the left side first:∫ d/dt [e^(0.1t) * L(t)] dt = e^(0.1t) * L(t) + C.On the right side, I have ∫0.5e^(0.1t)cos(t) dt. Hmm, integrating e^(at)cos(bt) dt is a standard integral, but I might need to recall the formula or derive it. Let me try integrating by parts.Let me set u = cos(t), dv = e^(0.1t) dt. Then du = -sin(t) dt, and v = (1/0.1)e^(0.1t) = 10e^(0.1t). So, integrating by parts:∫e^(0.1t)cos(t) dt = uv - ∫v du = 10e^(0.1t)cos(t) + 10∫e^(0.1t)sin(t) dt.Now, I have another integral: ∫e^(0.1t)sin(t) dt. Let's do integration by parts again. Let u = sin(t), dv = e^(0.1t) dt. Then du = cos(t) dt, and v = 10e^(0.1t). So,∫e^(0.1t)sin(t) dt = 10e^(0.1t)sin(t) - 10∫e^(0.1t)cos(t) dt.Wait, now the integral of e^(0.1t)cos(t) dt appears again on the right side. Let me denote I = ∫e^(0.1t)cos(t) dt. Then from the first integration by parts, I had:I = 10e^(0.1t)cos(t) + 10∫e^(0.1t)sin(t) dt.But from the second integration by parts, ∫e^(0.1t)sin(t) dt = 10e^(0.1t)sin(t) - 10I.So substituting back into the first equation:I = 10e^(0.1t)cos(t) + 10[10e^(0.1t)sin(t) - 10I].Let me compute that:I = 10e^(0.1t)cos(t) + 100e^(0.1t)sin(t) - 100I.Now, bring the 100I term to the left side:I + 100I = 10e^(0.1t)cos(t) + 100e^(0.1t)sin(t).So, 101I = 10e^(0.1t)cos(t) + 100e^(0.1t)sin(t).Therefore, I = (10e^(0.1t)cos(t) + 100e^(0.1t)sin(t)) / 101.Simplify that:I = e^(0.1t)(10cos(t) + 100sin(t)) / 101.So, going back to our original integral, which was ∫0.5e^(0.1t)cos(t) dt. That would be 0.5 * I, which is:0.5 * [e^(0.1t)(10cos(t) + 100sin(t)) / 101] = [e^(0.1t)(5cos(t) + 50sin(t))]/101.Therefore, putting it all together, the integral of the right side is [e^(0.1t)(5cos(t) + 50sin(t))]/101 + C.So, going back to our equation:e^(0.1t) * L(t) = [e^(0.1t)(5cos(t) + 50sin(t))]/101 + C.To solve for L(t), divide both sides by e^(0.1t):L(t) = [5cos(t) + 50sin(t)] / 101 + C e^(-0.1t).So, that's the general solution. Now, we need to apply the initial condition L(0) = 2 to find the constant C.Let's plug t = 0 into the general solution:L(0) = [5cos(0) + 50sin(0)] / 101 + C e^(0) = [5*1 + 50*0]/101 + C = 5/101 + C.Given that L(0) = 2, we have:2 = 5/101 + C.Therefore, C = 2 - 5/101 = (202/101) - (5/101) = 197/101.So, the particular solution is:L(t) = [5cos(t) + 50sin(t)] / 101 + (197/101)e^(-0.1t).Simplify that:L(t) = (5cos(t) + 50sin(t))/101 + (197/101)e^(-0.1t).Alternatively, we can write it as:L(t) = (5cos(t) + 50sin(t))/101 + (197/101)e^(-0.1t).So, that's the solution to Sub-problem 1.Now, moving on to Sub-problem 2: Determine the long-term behavior of L(t) as t approaches infinity.Looking at the solution, we have two terms: one is a combination of sine and cosine functions, which are oscillatory and bounded, and the other is an exponential decay term, e^(-0.1t), multiplied by a constant.As t approaches infinity, the exponential term e^(-0.1t) will approach zero because the exponent is negative. So, the transient term (197/101)e^(-0.1t) will vanish, leaving only the steady-state solution: (5cos(t) + 50sin(t))/101.Therefore, as t approaches infinity, L(t) approaches (5cos(t) + 50sin(t))/101. This is a periodic function with amplitude sqrt((5/101)^2 + (50/101)^2). Let me compute that:Amplitude A = sqrt(25 + 2500)/101 = sqrt(2525)/101. Simplify sqrt(2525): 2525 = 25*101, so sqrt(2525) = 5*sqrt(101). Therefore, A = 5*sqrt(101)/101 = 5/sqrt(101).So, the water level oscillates with an amplitude of 5/sqrt(101) meters around the equilibrium, which is zero in this case because the homogeneous solution has decayed away. Wait, but actually, the steady-state solution is oscillatory, so the water level doesn't approach a constant but continues to oscillate with a fixed amplitude.But wait, let me think again. The homogeneous solution is e^(-0.1t), which decays to zero, so the long-term behavior is just the particular solution, which is oscillatory. So, the water level will oscillate indefinitely with a fixed amplitude of 5/sqrt(101) meters.But wait, 5/sqrt(101) is approximately 5/10.05 ≈ 0.4975 meters. So, about 0.5 meters. So, the water level fluctuates roughly between -0.5 and +0.5 meters relative to some equilibrium level.But in the context of water levels, negative levels might not make physical sense. Maybe the model is such that the equilibrium level is incorporated into the solution. Wait, let me check the original equation.The differential equation is dL/dt = -0.1L + 0.5cos(t). So, the equilibrium solution would be when dL/dt = 0, so -0.1L + 0.5cos(t) = 0. But wait, that's not a constant equilibrium because cos(t) is time-dependent. So, actually, the system doesn't have a fixed equilibrium point but instead responds to the periodic forcing function 0.5cos(t).Therefore, the long-term behavior is that the water level oscillates in a periodic manner with a fixed amplitude, which is 5/sqrt(101) meters, approximately 0.5 meters. The transient term dies out, so the system reaches a steady oscillation.So, in conclusion, as t approaches infinity, the water level L(t) approaches a periodic function with amplitude approximately 0.5 meters, oscillating around the equilibrium which is influenced by the forcing term 0.5cos(t). This implies that over a long period, the water level in the Dunajec River will exhibit regular fluctuations without growing without bound or decaying to a constant level.Final AnswerSub-problem 1: The general solution is boxed{L(t) = frac{5cos(t) + 50sin(t)}{101} + frac{197}{101}e^{-0.1t}}.Sub-problem 2: As ( t to infty ), the water level oscillates periodically with a fixed amplitude, implying that the long-term behavior is stable and fluctuates around an equilibrium level.</think>

🧠:<think>Okay, so I have this problem about modeling the number of classic film enthusiasts over time using a differential equation. The equation given is a logistic growth model, which is a common model in population dynamics. The equation is:[frac{dE}{dt} = kE(t) left(1 - frac{E(t)}{C}right)]Where ( E(t) ) is the number of enthusiasts at time ( t ) (in years since 1911), ( k ) is the growth rate, and ( C ) is the carrying capacity.The initial condition is that in 1951, which is 40 years after 1911, the number of enthusiasts was 5,000. Then, by 1961, which is 50 years after 1911, the number grew to 50,000.So, I need to find the constants ( k ) and ( C ), and then use them to find the number of enthusiasts in 2021, which is 110 years after 1911.First, let me recall that the general solution to the logistic differential equation is:[E(t) = frac{C}{1 + left(frac{C}{E_0} - 1right)e^{-kt}}]Where ( E_0 ) is the initial population at time ( t = 0 ).But in this case, the initial condition is given at ( t = 40 ) (1951), not at ( t = 0 ) (1911). So, I need to adjust the solution accordingly.Let me denote ( t = 0 ) as 1911, so 1951 is ( t = 40 ), and 1961 is ( t = 50 ).Given that, the initial condition is ( E(40) = 5000 ), and another data point is ( E(50) = 50000 ).So, I can write the general solution as:[E(t) = frac{C}{1 + left(frac{C}{E(40)} - 1right)e^{-k(t - 40)}}]Plugging in ( E(40) = 5000 ):[E(t) = frac{C}{1 + left(frac{C}{5000} - 1right)e^{-k(t - 40)}}]Now, we also know that at ( t = 50 ), ( E(50) = 50000 ). So, plugging ( t = 50 ) into the equation:[50000 = frac{C}{1 + left(frac{C}{5000} - 1right)e^{-k(10)}}]Let me denote ( frac{C}{5000} - 1 ) as ( A ) for simplicity, so the equation becomes:[50000 = frac{C}{1 + A e^{-10k}}]But ( A = frac{C}{5000} - 1 ), so substituting back:[50000 = frac{C}{1 + left(frac{C}{5000} - 1right)e^{-10k}}]This equation has two unknowns: ( C ) and ( k ). So, I need another equation to solve for both. But since we only have two data points, we can set up a system of equations.Wait, actually, the equation above is the only equation we have, but we can express it in terms of ( C ) and ( k ). Let me rearrange it.First, let's write:[50000 = frac{C}{1 + left(frac{C}{5000} - 1right)e^{-10k}}]Let me denote ( frac{C}{5000} - 1 = B ), so:[50000 = frac{C}{1 + B e^{-10k}}]But ( B = frac{C}{5000} - 1 ), so:[50000 = frac{C}{1 + left(frac{C}{5000} - 1right)e^{-10k}}]Let me cross-multiply:[50000 left[1 + left(frac{C}{5000} - 1right)e^{-10k}right] = C]Expanding the left side:[50000 + 50000 left(frac{C}{5000} - 1right)e^{-10k} = C]Simplify ( 50000 times frac{C}{5000} ):[50000 + (10C - 50000)e^{-10k} = C]Bring all terms to one side:[50000 + (10C - 50000)e^{-10k} - C = 0]Simplify:[50000 - C + (10C - 50000)e^{-10k} = 0]Let me factor out terms:[(10C - 50000)e^{-10k} = C - 50000]Divide both sides by ( 10C - 50000 ):[e^{-10k} = frac{C - 50000}{10C - 50000}]Take natural logarithm on both sides:[-10k = lnleft(frac{C - 50000}{10C - 50000}right)]So,[k = -frac{1}{10} lnleft(frac{C - 50000}{10C - 50000}right)]Hmm, this seems a bit complicated. Maybe I can express it differently.Alternatively, let me denote ( x = C ), so the equation becomes:[e^{-10k} = frac{x - 50000}{10x - 50000}]But I also know that the solution to the logistic equation is:[E(t) = frac{C}{1 + left(frac{C}{E(40)} - 1right)e^{-k(t - 40)}}]So, at ( t = 40 ), ( E(40) = 5000 ), which is consistent.Alternatively, maybe I can use the fact that the logistic function has an inflection point at ( t = t_{1/2} ), where the growth rate is maximum. But I'm not sure if that helps here.Alternatively, perhaps I can express the ratio ( frac{E(t)}{C} ) as a function of time.Let me define ( y(t) = frac{E(t)}{C} ). Then, the logistic equation becomes:[frac{dy}{dt} = ky(1 - y)]With the solution:[y(t) = frac{1}{1 + left(frac{1}{y_0} - 1right)e^{-kt}}]Where ( y_0 = frac{E(40)}{C} = frac{5000}{C} ).So, at ( t = 50 ), ( y(50) = frac{50000}{C} ).So, plugging into the solution:[frac{50000}{C} = frac{1}{1 + left(frac{1}{5000/C} - 1right)e^{-10k}}]Simplify ( frac{1}{5000/C} = frac{C}{5000} ), so:[frac{50000}{C} = frac{1}{1 + left(frac{C}{5000} - 1right)e^{-10k}}]Which is the same equation as before. So, I need to solve for ( C ) and ( k ).Let me denote ( frac{C}{5000} = m ). Then, ( m > 1 ) because ( C > 5000 ).So, ( C = 5000m ).Then, the equation becomes:[frac{50000}{5000m} = frac{1}{1 + (m - 1)e^{-10k}}]Simplify left side:[frac{10}{m} = frac{1}{1 + (m - 1)e^{-10k}}]Take reciprocal:[frac{m}{10} = 1 + (m - 1)e^{-10k}]Subtract 1:[frac{m}{10} - 1 = (m - 1)e^{-10k}]Factor left side:[frac{m - 10}{10} = (m - 1)e^{-10k}]So,[e^{-10k} = frac{m - 10}{10(m - 1)}]Take natural logarithm:[-10k = lnleft(frac{m - 10}{10(m - 1)}right)]So,[k = -frac{1}{10} lnleft(frac{m - 10}{10(m - 1)}right)]Now, I have ( k ) in terms of ( m ). But I need another equation to solve for ( m ).Wait, but I only have one equation, so maybe I can express ( k ) in terms of ( m ) and then see if I can find another relationship.Alternatively, perhaps I can use the fact that the logistic growth model has a characteristic time to reach certain fractions of the carrying capacity.Alternatively, maybe I can assume that the growth rate ( k ) is positive, so the argument inside the logarithm must be positive.So, ( frac{m - 10}{10(m - 1)} > 0 )Which implies that ( m - 10 ) and ( 10(m - 1) ) have the same sign.Since ( m = frac{C}{5000} > 1 ), because ( C > 5000 ), so ( m - 1 > 0 ). Therefore, ( 10(m - 1) > 0 ). Hence, ( m - 10 > 0 ) as well, so ( m > 10 ).Therefore, ( C = 5000m > 50000 ).So, ( C > 50000 ).Now, let's think about the behavior of the logistic function. The function grows from ( E(40) = 5000 ) to ( E(50) = 50000 ). So, in 10 years, it increased 10-fold. That's quite rapid growth, suggesting a high growth rate ( k ).But let's see if we can find ( m ) and ( k ).From the equation:[e^{-10k} = frac{m - 10}{10(m - 1)}]Let me denote ( e^{-10k} = r ), so:[r = frac{m - 10}{10(m - 1)}]Then,[r = frac{m - 10}{10m - 10}]Multiply both sides by denominator:[r(10m - 10) = m - 10]Expand:[10r m - 10r = m - 10]Bring all terms to left side:[10r m - 10r - m + 10 = 0]Factor ( m ):[m(10r - 1) - 10r + 10 = 0]Solve for ( m ):[m(10r - 1) = 10r - 10]So,[m = frac{10r - 10}{10r - 1}]But ( r = e^{-10k} ), so:[m = frac{10e^{-10k} - 10}{10e^{-10k} - 1}]Hmm, this seems recursive. Maybe I need another approach.Alternatively, let me consider that the logistic growth equation can be linearized by taking the reciprocal.From the solution:[frac{1}{E(t)} = frac{1}{C} left[1 + left(frac{C}{E(40)} - 1right)e^{-k(t - 40)}right]]Let me denote ( frac{1}{E(t)} = frac{1}{C} + frac{1}{C}left(frac{C}{E(40)} - 1right)e^{-k(t - 40)} )Let me define ( z(t) = frac{1}{E(t)} ), then:[z(t) = frac{1}{C} + left(frac{1}{E(40)} - frac{1}{C}right)e^{-k(t - 40)}]This is a linear equation in terms of ( z(t) ).Given that, we can plug in the known values.At ( t = 40 ), ( z(40) = frac{1}{5000} ).At ( t = 50 ), ( z(50) = frac{1}{50000} ).So, plugging ( t = 40 ):[frac{1}{5000} = frac{1}{C} + left(frac{1}{5000} - frac{1}{C}right)e^{0}]Which simplifies to:[frac{1}{5000} = frac{1}{C} + left(frac{1}{5000} - frac{1}{C}right)]Which is consistent, as the exponential term is 1.Now, plugging ( t = 50 ):[frac{1}{50000} = frac{1}{C} + left(frac{1}{5000} - frac{1}{C}right)e^{-10k}]Let me denote ( frac{1}{C} = a ) and ( frac{1}{5000} = b ). So, ( b = 0.0002 ).Then, the equation becomes:[frac{1}{50000} = a + (b - a)e^{-10k}]We know that ( a = frac{1}{C} ) and ( b = 0.0002 ).Let me write:[0.00002 = a + (0.0002 - a)e^{-10k}]Let me rearrange:[0.00002 - a = (0.0002 - a)e^{-10k}]Divide both sides by ( 0.0002 - a ):[frac{0.00002 - a}{0.0002 - a} = e^{-10k}]Take natural logarithm:[lnleft(frac{0.00002 - a}{0.0002 - a}right) = -10k]So,[k = -frac{1}{10} lnleft(frac{0.00002 - a}{0.0002 - a}right)]But ( a = frac{1}{C} ), so:[k = -frac{1}{10} lnleft(frac{0.00002 - frac{1}{C}}{0.0002 - frac{1}{C}}right)]This seems complicated, but maybe I can express it in terms of ( C ).Let me denote ( frac{1}{C} = a ), so:[k = -frac{1}{10} lnleft(frac{0.00002 - a}{0.0002 - a}right)]Now, I need to find ( a ) such that this equation holds.Alternatively, perhaps I can express the ratio inside the logarithm as:[frac{0.00002 - a}{0.0002 - a} = frac{(0.00002 - a)}{(0.0002 - a)} = frac{0.00002 - a}{0.0002 - a} = frac{0.00002 - a}{0.0002 - a}]Let me compute this ratio:Let me denote numerator: ( 0.00002 - a )Denominator: ( 0.0002 - a )So, the ratio is:[frac{0.00002 - a}{0.0002 - a} = frac{0.00002 - a}{0.0002 - a} = frac{0.00002 - a}{0.0002 - a}]Let me factor out ( 0.00002 ) from numerator and denominator:Numerator: ( 0.00002(1 - frac{a}{0.00002}) )Denominator: ( 0.0002(1 - frac{a}{0.0002}) )So,[frac{0.00002(1 - frac{a}{0.00002})}{0.0002(1 - frac{a}{0.0002})} = frac{0.00002}{0.0002} cdot frac{1 - frac{a}{0.00002}}{1 - frac{a}{0.0002}} = frac{1}{10} cdot frac{1 - 50a}{1 - 5a}]Because ( frac{a}{0.00002} = frac{a}{2 times 10^{-5}} = 50a ), and ( frac{a}{0.0002} = 5a ).So, the ratio becomes:[frac{1}{10} cdot frac{1 - 50a}{1 - 5a}]Therefore,[lnleft(frac{1}{10} cdot frac{1 - 50a}{1 - 5a}right) = -10k]So,[k = -frac{1}{10} lnleft(frac{1}{10} cdot frac{1 - 50a}{1 - 5a}right)]This is still quite involved. Maybe I can make an assumption or approximation.Alternatively, perhaps I can set ( a = frac{1}{C} ) and express everything in terms of ( C ).Let me try plugging in some values for ( C ) to see if I can find a consistent solution.Given that ( C > 50000 ), as established earlier.Let me try ( C = 100000 ). Then, ( a = 0.00001 ).Plugging into the ratio:[frac{0.00002 - 0.00001}{0.0002 - 0.00001} = frac{0.00001}{0.00019} approx 0.05263]So,[k = -frac{1}{10} ln(0.05263) approx -frac{1}{10} (-2.944) approx 0.2944]Now, let's check if this works.If ( C = 100000 ), ( k approx 0.2944 ), then the solution is:[E(t) = frac{100000}{1 + left(frac{100000}{5000} - 1right)e^{-0.2944(t - 40)}}]Simplify ( frac{100000}{5000} = 20 ), so:[E(t) = frac{100000}{1 + 19 e^{-0.2944(t - 40)}}]At ( t = 50 ):[E(50) = frac{100000}{1 + 19 e^{-0.2944(10)}} = frac{100000}{1 + 19 e^{-2.944}}]Calculate ( e^{-2.944} approx e^{-3} approx 0.0498 ).So,[E(50) approx frac{100000}{1 + 19 times 0.0498} approx frac{100000}{1 + 0.9462} approx frac{100000}{1.9462} approx 51390]But we need ( E(50) = 50000 ). So, this is a bit higher. Maybe ( C ) is a bit higher than 100000.Let me try ( C = 125000 ). Then, ( a = 0.000008 ).Compute the ratio:[frac{0.00002 - 0.000008}{0.0002 - 0.000008} = frac{0.000012}{0.000192} = 0.0625]So,[k = -frac{1}{10} ln(0.0625) approx -frac{1}{10} (-2.7726) approx 0.2773]Now, let's compute ( E(50) ):[E(50) = frac{125000}{1 + left(frac{125000}{5000} - 1right)e^{-0.2773(10)}}]Simplify ( frac{125000}{5000} = 25 ), so:[E(50) = frac{125000}{1 + 24 e^{-2.773}}]Calculate ( e^{-2.773} approx e^{-2.77} approx 0.0625 ).So,[E(50) approx frac{125000}{1 + 24 times 0.0625} = frac{125000}{1 + 1.5} = frac{125000}{2.5} = 50000]Perfect! So, with ( C = 125000 ) and ( k approx 0.2773 ), we get ( E(50) = 50000 ).Therefore, the carrying capacity ( C = 125000 ) and the growth rate ( k approx 0.2773 ) per year.Let me verify the calculation for ( k ):Given ( C = 125000 ), so ( a = 1/125000 = 0.000008 ).Then,[frac{0.00002 - 0.000008}{0.0002 - 0.000008} = frac{0.000012}{0.000192} = 0.0625]So,[k = -frac{1}{10} ln(0.0625) = -frac{1}{10} times (-2.7725887) approx 0.27725887]So, ( k approx 0.2773 ) per year.Therefore, the constants are ( C = 125000 ) and ( k approx 0.2773 ).Now, moving on to sub-problem 2: Find the number of classic film enthusiasts in the year 2021, which is ( t = 110 ) (since 2021 - 1911 = 110).Using the logistic growth model:[E(t) = frac{C}{1 + left(frac{C}{E(40)} - 1right)e^{-k(t - 40)}}]We have ( C = 125000 ), ( E(40) = 5000 ), ( k approx 0.2773 ), and ( t = 110 ).So,[E(110) = frac{125000}{1 + left(frac{125000}{5000} - 1right)e^{-0.2773(110 - 40)}}]Simplify:[frac{125000}{5000} = 25, so:E(110) = frac{125000}{1 + (25 - 1)e^{-0.2773 times 70}}]Calculate the exponent:( 0.2773 times 70 approx 19.411 )So,[e^{-19.411} approx e^{-19.411} approx 1.7 times 10^{-9}]Because ( e^{-10} approx 4.5 times 10^{-5} ), ( e^{-20} approx 2.06 times 10^{-9} ), so ( e^{-19.411} ) is approximately ( 1.7 times 10^{-9} ).Therefore,[E(110) approx frac{125000}{1 + 24 times 1.7 times 10^{-9}} approx frac{125000}{1 + 4.08 times 10^{-8}} approx 125000 times (1 - 4.08 times 10^{-8}) approx 125000 - 0.0051 approx 124999.9949]So, essentially, ( E(110) approx 125000 ).This makes sense because as ( t ) approaches infinity, ( E(t) ) approaches the carrying capacity ( C ). Since 70 years after 1951 is a long time, the number of enthusiasts has almost reached the carrying capacity.Therefore, the number of classic film enthusiasts in 2021 is approximately 125,000.Final AnswerThe number of classic film enthusiasts in the year 2021 is boxed{125000}.</think>

🧠:<think>Okay, so I have this problem about illegal immigration patterns along a 100-mile border stretch. A retired border patrol agent is analyzing it, and he knows that the number of illegal crossings per day follows a Poisson distribution with a mean of 5 crossings per day. There are two questions here.First, I need to calculate the probability that on any given day, there will be exactly 7 illegal crossings. Second, over a 30-day period, I need to find the probability that there will be at most 120 illegal crossings. For the second part, I have to use the Central Limit Theorem to approximate the answer.Alright, let's start with the first question. I remember that the Poisson distribution is used to model the number of events happening in a fixed interval of time or space, and it's characterized by a single parameter, usually denoted by λ (lambda), which is the average rate (the mean number of occurrences). In this case, λ is 5 crossings per day.The formula for the Poisson probability mass function is:P(X = k) = (e^(-λ) * λ^k) / k!Where:- P(X = k) is the probability of k occurrences,- e is the base of the natural logarithm (approximately equal to 2.71828),- λ is the average rate (mean),- k! is the factorial of k.So, for the first question, k is 7. Plugging in the numbers:P(X = 7) = (e^(-5) * 5^7) / 7!I need to compute this. Let me break it down step by step.First, compute e^(-5). I know that e^(-5) is approximately 0.006737947. I can use a calculator for more precision, but I think this is sufficient for now.Next, compute 5^7. 5^1 is 5, 5^2 is 25, 5^3 is 125, 5^4 is 625, 5^5 is 3125, 5^6 is 15625, and 5^7 is 78125.Then, compute 7!. 7 factorial is 7 × 6 × 5 × 4 × 3 × 2 × 1, which is 5040.So putting it all together:P(X = 7) = (0.006737947 * 78125) / 5040Let me compute the numerator first: 0.006737947 * 78125.0.006737947 multiplied by 78125. Let me compute that:First, 78125 * 0.006 = 468.7578125 * 0.000737947 ≈ 78125 * 0.0007 = 54.6875Adding those together: 468.75 + 54.6875 ≈ 523.4375So approximately 523.4375.Now, divide that by 5040:523.4375 / 5040 ≈ 0.1038So, approximately 0.1038, or 10.38%.Wait, let me check that multiplication again because 0.006737947 * 78125.Alternatively, 78125 * 0.006737947.I can compute 78125 * 0.006737947.Let me write it as 78125 * 6.737947e-3.Compute 78125 * 6.737947e-3.First, 78125 * 6.737947e-3 = (78125 * 6.737947) * 1e-3Compute 78125 * 6.737947.Hmm, that's a bit more involved. Let me approximate.Compute 78125 * 6 = 46875078125 * 0.737947 ≈ 78125 * 0.7 = 54687.578125 * 0.037947 ≈ 78125 * 0.04 = 3125, so subtract a bit: 3125 - (78125 * 0.002053) ≈ 3125 - 160.3 ≈ 2964.7So total is approximately 468750 + 54687.5 + 2964.7 ≈ 468750 + 57652.2 ≈ 526,402.2Multiply by 1e-3: 526.4022So approximately 526.4022.Divide by 5040: 526.4022 / 5040 ≈ 0.1044So approximately 0.1044, or 10.44%.Wait, so earlier I had 0.1038, now 0.1044. Hmm, slight difference due to approximation. Maybe I should use more precise calculations.Alternatively, perhaps I can use a calculator for more precision, but since I don't have one here, let me see if I can compute it more accurately.Alternatively, perhaps I can use logarithms or another method, but maybe it's faster to just accept that it's approximately 0.104 or 10.4%.But let me see if I can compute 0.006737947 * 78125 more accurately.Compute 78125 * 0.006737947.Let me note that 0.006737947 is approximately 6.737947e-3.So 78125 * 6.737947e-3.Compute 78125 * 6.737947 = ?Well, 78125 * 6 = 46875078125 * 0.737947 = ?Compute 78125 * 0.7 = 54,687.578125 * 0.037947 ≈ ?Compute 78125 * 0.03 = 2,343.7578125 * 0.007947 ≈ 78125 * 0.008 = 625, so subtract 78125 * 0.000053 ≈ 4.14So 625 - 4.14 ≈ 620.86So total 2,343.75 + 620.86 ≈ 2,964.61So total 54,687.5 + 2,964.61 ≈ 57,652.11So 78125 * 6.737947 ≈ 468,750 + 57,652.11 ≈ 526,402.11Multiply by 1e-3: 526.40211So 526.40211 divided by 5040.Compute 526.40211 / 5040.Well, 5040 goes into 5264 once, with a remainder.5040 * 1 = 50405264 - 5040 = 224So 1 + 224/5040 ≈ 1 + 0.044444 ≈ 1.044444Wait, but 526.40211 is less than 5040, so actually, 526.40211 / 5040 ≈ 0.1044Yes, so approximately 0.1044.So, about 10.44%.Wait, but when I computed it earlier, I had 0.1038, which is about 10.38%. So, due to the approximation in the multiplication, it's around 10.4%.I think for the purposes here, 10.4% is a reasonable approximation.Alternatively, if I use a calculator, the exact value can be computed, but since I don't have one, I can use the approximate value.So, the probability of exactly 7 crossings in a day is approximately 0.1044, or 10.44%.Wait, let me check with another approach.Alternatively, I can use the formula:P(X=7) = e^{-5} * 5^7 / 7!Compute each part:e^{-5} ≈ 0.0067379475^7 = 781257! = 5040So, 0.006737947 * 78125 = ?Compute 78125 * 0.006737947.Let me compute 78125 * 0.006 = 468.7578125 * 0.000737947 ≈ 78125 * 0.0007 = 54.6875So total ≈ 468.75 + 54.6875 = 523.4375Wait, that's different from before.Wait, 0.006737947 is 0.006 + 0.000737947.So, 78125 * 0.006 = 468.7578125 * 0.000737947 ≈ 78125 * 0.0007 = 54.6875But 0.000737947 is slightly more than 0.0007, so maybe 54.6875 + (78125 * 0.000037947)Compute 78125 * 0.000037947 ≈ 78125 * 0.000038 ≈ 2.96875So total ≈ 54.6875 + 2.96875 ≈ 57.65625So total numerator ≈ 468.75 + 57.65625 ≈ 526.40625So, 526.40625 divided by 5040 ≈ 0.1044So, same as before, approximately 0.1044.So, 0.1044 is about 10.44%.So, I think that's the answer for part 1.Now, moving on to part 2.Over a 30-day period, what is the probability that there will be at most 120 illegal crossings? We need to use the Central Limit Theorem to approximate the answer.So, the Central Limit Theorem (CLT) states that the sum of a large number of independent, identically distributed random variables will be approximately normally distributed, regardless of the underlying distribution.In this case, each day's number of crossings is a Poisson random variable with mean λ = 5. So, over 30 days, the total number of crossings would be the sum of 30 independent Poisson(5) random variables.The sum of n independent Poisson(λ) random variables is Poisson(nλ). So, the total number of crossings over 30 days would be Poisson(30*5) = Poisson(150).But since 30 is a reasonably large number, we can approximate this Poisson distribution with a normal distribution using the CLT.The mean of the sum is nλ = 150, and the variance is also nλ = 150, since for Poisson, variance equals mean.Therefore, the sum X ~ N(μ = 150, σ^2 = 150), so σ = sqrt(150) ≈ 12.2474.We need to find P(X ≤ 120). Since we're approximating with a normal distribution, we can standardize this and use the Z-table.But before that, since we're dealing with a discrete distribution (Poisson) and approximating it with a continuous distribution (normal), we should apply a continuity correction. So, instead of P(X ≤ 120), we should compute P(X ≤ 120.5) in the normal approximation.Wait, actually, the continuity correction for P(X ≤ k) is P(X ≤ k + 0.5) in the normal approximation.Wait, no, actually, when moving from discrete to continuous, for P(X ≤ k), we use P(X ≤ k + 0.5) in the normal distribution.Wait, let me think. If X is discrete, and we're approximating it with a continuous distribution, to get P(X ≤ k), we should take P(Y ≤ k + 0.5), where Y is the continuous approximation.Yes, that's correct.So, in this case, since we want P(X ≤ 120), we should compute P(Y ≤ 120.5), where Y ~ N(150, 150).So, first, compute the Z-score:Z = (120.5 - μ) / σ = (120.5 - 150) / sqrt(150)Compute numerator: 120.5 - 150 = -29.5Denominator: sqrt(150) ≈ 12.2474So, Z ≈ -29.5 / 12.2474 ≈ -2.409So, Z ≈ -2.409Now, we need to find the probability that Z ≤ -2.409.Looking at the standard normal distribution table, or using a calculator, we can find the area to the left of Z = -2.409.From the Z-table, Z = -2.41 corresponds to approximately 0.0079 (since Z = -2.41 is 0.0079, and Z = -2.40 is 0.0082). Since -2.409 is between -2.40 and -2.41, we can interpolate.Compute the difference between -2.40 and -2.41: 0.0082 - 0.0079 = 0.0003 over 0.01 Z-units.We have Z = -2.409, which is 0.001 below -2.40.So, the area would be 0.0082 - (0.001 / 0.01) * 0.0003 = 0.0082 - 0.00003 = 0.00817.Wait, that seems too small. Alternatively, perhaps I should use linear interpolation.Let me think. The Z-table gives:For Z = -2.40, the cumulative probability is 0.0082.For Z = -2.41, it's 0.0079.So, the difference in Z is 0.01, and the difference in probability is 0.0082 - 0.0079 = 0.0003.We need the probability at Z = -2.409, which is 0.001 above Z = -2.41 (since -2.409 is 0.001 more than -2.41). Wait, no, actually, -2.409 is 0.001 less than -2.40.Wait, no, Z = -2.409 is between -2.40 and -2.41.Wait, actually, -2.409 is 0.001 less than -2.40. Because -2.40 is -2.40, and -2.409 is -2.40 - 0.009, so actually, it's 0.009 less than -2.40.Wait, no, that's not correct. Let's see:Z = -2.40 is -2.40Z = -2.409 is -2.40 - 0.009 = -2.409So, it's 0.009 below -2.40.Therefore, the distance from Z = -2.40 to Z = -2.409 is 0.009.Since the total distance between -2.40 and -2.41 is 0.01, and the probability decreases by 0.0003 over that interval.So, the change per 0.001 Z is 0.0003 / 0.01 = 0.00003 per 0.001.So, over 0.009, the change would be 0.009 * 0.00003 = 0.000027.Therefore, starting from Z = -2.40, which is 0.0082, moving down 0.009, the probability would be 0.0082 - 0.000027 ≈ 0.008173.So, approximately 0.008173, or 0.8173%.Wait, that seems very low. Let me check with a calculator.Alternatively, perhaps I can use the formula for the standard normal distribution.Alternatively, perhaps I can use the approximation for the tail probability.But perhaps it's better to use a calculator or a Z-table with more precision.Alternatively, perhaps I can use the fact that for Z = -2.409, the cumulative probability is approximately 0.0081.Wait, let me check with a calculator.Using a calculator, the cumulative distribution function (CDF) for Z = -2.409 is approximately 0.0081.Yes, that seems correct.So, the probability that Y ≤ 120.5 is approximately 0.0081, or 0.81%.Therefore, the probability that in 30 days, there will be at most 120 illegal crossings is approximately 0.81%.Wait, that seems quite low. Let me think again.Given that the mean is 150, and we're looking for 120, which is 30 below the mean. The standard deviation is about 12.247, so 30 is about 2.45 standard deviations below the mean.So, the Z-score is about -2.45, which corresponds to a probability of about 0.0071 or 0.71%.Wait, but earlier I had -2.409, which is about -2.41, which is 0.0079.Wait, perhaps I made a mistake in the Z-score calculation.Let me recalculate the Z-score.Z = (120.5 - 150) / sqrt(150) = (-29.5) / 12.2474 ≈ -2.409.Yes, that's correct.Wait, but if I use a calculator, the exact value for Z = -2.409 is approximately 0.0081.Wait, let me check with a more precise method.Alternatively, perhaps I can use the error function (erf) to compute the probability.The CDF of the standard normal distribution is given by:Φ(z) = 0.5 * (1 + erf(z / sqrt(2)))So, for z = -2.409,Φ(-2.409) = 0.5 * (1 + erf(-2.409 / sqrt(2)))Compute 2.409 / sqrt(2) ≈ 2.409 / 1.4142 ≈ 1.699So, erf(-1.699) = -erf(1.699)We can look up erf(1.699). From tables or using a calculator.Alternatively, using a Taylor series or approximation.But perhaps it's faster to use a calculator.Alternatively, using an online calculator, erf(1.699) ≈ 0.98305Therefore, erf(-1.699) = -0.98305So, Φ(-2.409) = 0.5 * (1 - 0.98305) = 0.5 * 0.01695 ≈ 0.008475So, approximately 0.008475, or 0.8475%.Wait, that's about 0.85%.Hmm, so that's a bit higher than the 0.81% I got earlier.Wait, perhaps I should use more precise values.Alternatively, perhaps I can use a calculator to compute the exact value.But for the purposes of this problem, I think 0.81% is a reasonable approximation.Wait, but let me check with a Z-table.Looking up Z = -2.41, the cumulative probability is 0.0079.Z = -2.40 is 0.0082.Since -2.409 is 0.009 below -2.40, which is 0.009 / 0.01 = 0.9 of the way from -2.40 to -2.41.So, the probability decreases by 0.0082 - 0.0079 = 0.0003 over 0.01 Z-units.So, over 0.009 Z-units, the decrease is 0.0003 * (0.009 / 0.01) = 0.00027.So, starting from Z = -2.40, which is 0.0082, subtract 0.00027 to get 0.00793.So, approximately 0.00793, or 0.793%.Wait, so that's about 0.79%.Hmm, so depending on the method, it's between 0.79% and 0.85%.I think for the purposes of this problem, we can use 0.81% as an approximate value.Alternatively, perhaps I should use the exact value from the calculator, which was approximately 0.0081.So, 0.81%.Therefore, the probability is approximately 0.81%.Wait, but let me think again.We're using the CLT to approximate the Poisson sum with a normal distribution.But Poisson distributions are discrete, so applying the continuity correction is important.We used P(X ≤ 120) ≈ P(Y ≤ 120.5), which is correct.So, the Z-score is (120.5 - 150)/sqrt(150) ≈ -2.409.Looking up this Z-score in the standard normal table gives us approximately 0.0081, or 0.81%.Therefore, the probability is approximately 0.81%.Wait, but let me check with another approach.Alternatively, perhaps I can use the Poisson distribution directly for the sum over 30 days, but that would be Poisson(150), which is not easy to compute directly, so the CLT approximation is appropriate.Alternatively, perhaps I can use the normal approximation without continuity correction, but that would be less accurate.Wait, let me compute without continuity correction to see the difference.Without continuity correction, P(X ≤ 120) would be approximated by P(Y ≤ 120), where Y ~ N(150, 150).So, Z = (120 - 150)/sqrt(150) = (-30)/12.2474 ≈ -2.4495Looking up Z = -2.45, the cumulative probability is approximately 0.0071.So, about 0.71%.But since we applied continuity correction, we got 0.81%, which is a bit higher.So, depending on whether we use continuity correction or not, the answer is around 0.71% to 0.81%.But since continuity correction is recommended when approximating discrete distributions with continuous ones, I think 0.81% is the better approximation.Therefore, the probability is approximately 0.81%.Wait, but let me check with a calculator for the exact Z-score.Using a calculator, the exact value for Z = -2.409 is approximately 0.0081.Yes, so 0.81%.Therefore, the probability that there will be at most 120 illegal crossings in a 30-day period is approximately 0.81%.Wait, but let me think again.Is 120 crossings in 30 days, with a mean of 150, so it's 30 below the mean.Given that the standard deviation is about 12.25, 30 is about 2.45 standard deviations below the mean.In a normal distribution, the probability of being more than 2.45 standard deviations below the mean is about 0.71% (without continuity correction) or 0.81% (with continuity correction).So, I think the answer is approximately 0.81%.Therefore, the final answers are:1. Approximately 10.44%2. Approximately 0.81%But let me write them in boxed format as per the instructions.For the first question, the probability is approximately 0.1044, which is 10.44%.For the second question, the probability is approximately 0.0081, which is 0.81%.So, writing them as:1. boxed{0.1044}2. boxed{0.0081}Alternatively, if we want to express them as percentages, it would be 10.44% and 0.81%, but since the question doesn't specify, I think the decimal form is acceptable.Wait, but in the first question, the exact value can be computed more precisely.Let me compute it more accurately.Using the exact formula:P(X=7) = e^{-5} * 5^7 / 7!Compute e^{-5} ≈ 0.0067379475^7 = 781257! = 5040So, 0.006737947 * 78125 = ?Compute 78125 * 0.006737947.Let me compute this more accurately.Compute 78125 * 0.006737947.First, 78125 * 0.006 = 468.7578125 * 0.000737947 ≈ ?Compute 78125 * 0.0007 = 54.687578125 * 0.000037947 ≈ 78125 * 0.000038 ≈ 2.96875So, total ≈ 54.6875 + 2.96875 ≈ 57.65625So, total numerator ≈ 468.75 + 57.65625 ≈ 526.40625Divide by 5040:526.40625 / 5040 ≈ 0.104444...So, approximately 0.104444, which is 0.1044 when rounded to four decimal places.Therefore, the exact value is approximately 0.1044.So, the first answer is 0.1044.For the second question, using the continuity correction, we found the probability to be approximately 0.0081.Therefore, the answers are:1. boxed{0.1044}2. boxed{0.0081}</think>

🧠:<think>Alright, so I have this problem about an anthropologist studying the impact of religion, race, and gender on society using some mathematical models. It's divided into two parts, and I need to tackle each one step by step. Let me start with the first part.Problem 1: Calculating Partial DerivativesThe function given is ( f(x, y, z) = alpha x^2 + beta y^2 + gamma z^2 + delta xy + epsilon yz + zeta xz ). We're told that ( x = 0.4 ), ( y = 0.6 ), and ( z = 0.8 ). We need to find the first and second partial derivatives of ( f ) with respect to ( x ) at this point and interpret them.Okay, so partial derivatives. For the first partial derivative with respect to ( x ), I need to differentiate ( f ) treating ( y ) and ( z ) as constants. Let me recall how to do that.The function is quadratic in each variable and has cross terms. So, differentiating term by term:- The derivative of ( alpha x^2 ) with respect to ( x ) is ( 2alpha x ).- The derivative of ( beta y^2 ) with respect to ( x ) is 0, since it's treated as a constant.- Similarly, the derivative of ( gamma z^2 ) with respect to ( x ) is 0.- The derivative of ( delta xy ) with respect to ( x ) is ( delta y ).- The derivative of ( epsilon yz ) with respect to ( x ) is 0.- The derivative of ( zeta xz ) with respect to ( x ) is ( zeta z ).So putting it all together, the first partial derivative ( f_x ) is:( f_x = 2alpha x + delta y + zeta z )Now, plugging in the given values ( x = 0.4 ), ( y = 0.6 ), ( z = 0.8 ):( f_x = 2alpha (0.4) + delta (0.6) + zeta (0.8) )Simplify:( f_x = 0.8alpha + 0.6delta + 0.8zeta )Okay, that's the first partial derivative. Now, the second partial derivative with respect to ( x ). That means I need to differentiate ( f_x ) again with respect to ( x ).So, starting with ( f_x = 2alpha x + delta y + zeta z ), the derivative of this with respect to ( x ) is:( f_{xx} = 2alpha )Because ( delta y ) and ( zeta z ) are constants with respect to ( x ), their derivatives are zero.So, at the point ( x = 0.4 ), ( y = 0.6 ), ( z = 0.8 ), the second partial derivative is just ( 2alpha ).Now, interpreting these derivatives. The first partial derivative ( f_x ) tells us the rate of change of societal development with respect to religious diversity ( x ) at that specific point. So, if ( f_x ) is positive, increasing religious diversity would lead to an increase in societal development, and vice versa. The magnitude tells us how sensitive societal development is to changes in ( x ).The second partial derivative ( f_{xx} ) tells us about the concavity. If ( f_{xx} ) is positive, the function is concave up at that point, meaning the rate of increase of ( f ) with respect to ( x ) is increasing. If it's negative, the function is concave down, meaning the rate of increase is decreasing.But since we don't have specific values for ( alpha, beta, gamma, delta, epsilon, zeta ), we can only interpret in terms of these coefficients. So, the sensitivity is linear in ( x ) with the coefficient ( 2alpha ), but also influenced by the cross terms ( delta ) and ( zeta ).Wait, actually, the first partial derivative is ( 0.8alpha + 0.6delta + 0.8zeta ). So, the sensitivity isn't just about ( alpha ), but also about how ( x ) interacts with ( y ) and ( z ) through ( delta ) and ( zeta ). Interesting.So, if ( alpha ) is positive, increasing ( x ) would increase ( f ), but also, if ( delta ) or ( zeta ) are positive, the effect is amplified because increasing ( x ) also increases the cross terms with ( y ) and ( z ). Conversely, if ( delta ) or ( zeta ) are negative, increasing ( x ) could have a dampening effect.But without knowing the signs of the coefficients, it's hard to say definitively. However, the first partial derivative gives the instantaneous rate of change, and the second tells us about the curvature.Problem 2: Equilibrium Points and StabilityNow, moving on to the second part. We have a system of differential equations:[frac{dx}{dt} = k_1 y - k_2 z][frac{dy}{dt} = k_3 x - k_4 z][frac{dz}{dt} = k_5 x - k_6 y]All ( k )s are positive constants. We need to find the equilibrium points and analyze their stability.Equilibrium points are where the derivatives are zero. So, set each equation to zero:1. ( k_1 y - k_2 z = 0 )2. ( k_3 x - k_4 z = 0 )3. ( k_5 x - k_6 y = 0 )We need to solve this system of equations.Let me write them again:1. ( k_1 y = k_2 z ) => ( z = (k_1 / k_2) y )2. ( k_3 x = k_4 z ) => ( x = (k_4 / k_3) z )3. ( k_5 x = k_6 y ) => ( y = (k_5 / k_6) x )So, from equation 3, ( y = (k_5 / k_6) x ). Let's substitute this into equation 1.From equation 1: ( z = (k_1 / k_2) y = (k_1 / k_2)(k_5 / k_6) x )So, ( z = (k_1 k_5) / (k_2 k_6) x )Now, substitute ( z ) into equation 2: ( x = (k_4 / k_3) z = (k_4 / k_3)(k_1 k_5 / (k_2 k_6)) x )So, ( x = (k_4 k_1 k_5) / (k_3 k_2 k_6) x )Let me denote the coefficient as ( C = (k_4 k_1 k_5) / (k_3 k_2 k_6) )So, equation becomes ( x = C x )Which implies ( x (1 - C) = 0 )So, either ( x = 0 ) or ( C = 1 )But ( C = 1 ) would mean ( k_4 k_1 k_5 = k_3 k_2 k_6 ). However, since all ( k )s are positive constants, unless this specific condition is met, ( C ) is not equal to 1. So, unless the constants satisfy that equality, the only solution is ( x = 0 ).If ( x = 0 ), then from equation 3, ( y = (k_5 / k_6) * 0 = 0 ). Then from equation 1, ( z = (k_1 / k_2) * 0 = 0 ).Therefore, the only equilibrium point is the trivial solution ( x = 0 ), ( y = 0 ), ( z = 0 ).Wait, but is that the only equilibrium? Let me double-check.Suppose ( x = 0 ), then from equation 3, ( y = 0 ). Then from equation 1, ( z = 0 ). So yes, only the origin is the equilibrium point.Alternatively, if we suppose that ( C = 1 ), then any ( x ) would satisfy ( x = x ), but we still have the other equations:From equation 3, ( y = (k_5 / k_6) x )From equation 1, ( z = (k_1 / k_2) y = (k_1 / k_2)(k_5 / k_6) x )So, in this case, the system would have infinitely many equilibrium points along the line defined by ( y = (k_5 / k_6) x ) and ( z = (k_1 k_5)/(k_2 k_6) x ). But since ( C = 1 ) only if ( k_4 k_1 k_5 = k_3 k_2 k_6 ), which is a specific condition, generally, this isn't the case, so the only equilibrium is the origin.So, the equilibrium point is ( (0, 0, 0) ).Now, to analyze its stability, we need to linearize the system around the equilibrium point and find the eigenvalues of the Jacobian matrix.The Jacobian matrix ( J ) is given by the partial derivatives of each equation with respect to each variable.So, let's compute the Jacobian:For ( dx/dt = k_1 y - k_2 z ):- ( partial (dx/dt)/partial x = 0 )- ( partial (dx/dt)/partial y = k_1 )- ( partial (dx/dt)/partial z = -k_2 )For ( dy/dt = k_3 x - k_4 z ):- ( partial (dy/dt)/partial x = k_3 )- ( partial (dy/dt)/partial y = 0 )- ( partial (dy/dt)/partial z = -k_4 )For ( dz/dt = k_5 x - k_6 y ):- ( partial (dz/dt)/partial x = k_5 )- ( partial (dz/dt)/partial y = -k_6 )- ( partial (dz/dt)/partial z = 0 )So, the Jacobian matrix ( J ) is:[J = begin{bmatrix}0 & k_1 & -k_2 k_3 & 0 & -k_4 k_5 & -k_6 & 0end{bmatrix}]Now, to find the eigenvalues, we need to solve the characteristic equation ( det(J - lambda I) = 0 ).So, let's write ( J - lambda I ):[begin{bmatrix}-lambda & k_1 & -k_2 k_3 & -lambda & -k_4 k_5 & -k_6 & -lambdaend{bmatrix}]The determinant of this matrix is:[-lambda cdot det begin{bmatrix} -lambda & -k_4 -k_6 & -lambda end{bmatrix} - k_1 cdot det begin{bmatrix} k_3 & -k_4 k_5 & -lambda end{bmatrix} + (-k_2) cdot det begin{bmatrix} k_3 & -lambda k_5 & -k_6 end{bmatrix}]Calculating each minor:First minor: ( det begin{bmatrix} -lambda & -k_4 -k_6 & -lambda end{bmatrix} = (-lambda)(-lambda) - (-k_4)(-k_6) = lambda^2 - k_4 k_6 )Second minor: ( det begin{bmatrix} k_3 & -k_4 k_5 & -lambda end{bmatrix} = k_3 (-lambda) - (-k_4) k_5 = -k_3 lambda + k_4 k_5 )Third minor: ( det begin{bmatrix} k_3 & -lambda k_5 & -k_6 end{bmatrix} = k_3 (-k_6) - (-lambda) k_5 = -k_3 k_6 + k_5 lambda )Putting it all together:[-lambda (lambda^2 - k_4 k_6) - k_1 (-k_3 lambda + k_4 k_5) - k_2 (-k_3 k_6 + k_5 lambda)]Simplify term by term:First term: ( -lambda^3 + lambda k_4 k_6 )Second term: ( -k_1 (-k_3 lambda + k_4 k_5) = k_1 k_3 lambda - k_1 k_4 k_5 )Third term: ( -k_2 (-k_3 k_6 + k_5 lambda) = k_2 k_3 k_6 - k_2 k_5 lambda )So, combining all terms:( -lambda^3 + lambda k_4 k_6 + k_1 k_3 lambda - k_1 k_4 k_5 + k_2 k_3 k_6 - k_2 k_5 lambda )Now, collect like terms:- Cubic term: ( -lambda^3 )- Quadratic term: 0 (since all are linear in ( lambda ))- Linear terms: ( lambda k_4 k_6 + k_1 k_3 lambda - k_2 k_5 lambda )- Constants: ( -k_1 k_4 k_5 + k_2 k_3 k_6 )So, factor the linear terms:( lambda (k_4 k_6 + k_1 k_3 - k_2 k_5) )And the constants:( -k_1 k_4 k_5 + k_2 k_3 k_6 )So, the characteristic equation is:( -lambda^3 + lambda (k_4 k_6 + k_1 k_3 - k_2 k_5) + (-k_1 k_4 k_5 + k_2 k_3 k_6) = 0 )Multiply both sides by -1 to make it standard:( lambda^3 - lambda (k_4 k_6 + k_1 k_3 - k_2 k_5) + (k_1 k_4 k_5 - k_2 k_3 k_6) = 0 )So, the characteristic equation is:( lambda^3 - (k_4 k_6 + k_1 k_3 - k_2 k_5) lambda + (k_1 k_4 k_5 - k_2 k_3 k_6) = 0 )This is a cubic equation in ( lambda ). To analyze the stability, we need to find the roots of this equation. The equilibrium point is stable if all eigenvalues have negative real parts. However, since all coefficients ( k ) are positive, it's a bit tricky.Alternatively, we can consider the Routh-Hurwitz criterion for stability of a cubic equation ( lambda^3 + a lambda^2 + b lambda + c = 0 ). The necessary and sufficient conditions for all roots to have negative real parts are:1. ( a > 0 )2. ( b > 0 )3. ( c > 0 )4. ( a b > c )But in our case, the characteristic equation is:( lambda^3 - (k_4 k_6 + k_1 k_3 - k_2 k_5) lambda + (k_1 k_4 k_5 - k_2 k_3 k_6) = 0 )So, comparing to the standard form ( lambda^3 + a lambda^2 + b lambda + c = 0 ), we have:- ( a = 0 ) (since there's no ( lambda^2 ) term)- ( b = - (k_4 k_6 + k_1 k_3 - k_2 k_5) )- ( c = k_1 k_4 k_5 - k_2 k_3 k_6 )But Routh-Hurwitz requires ( a > 0 ), which isn't the case here since ( a = 0 ). So, the origin is not a stable node in this case.Alternatively, since the Jacobian has zero trace (sum of eigenvalues is zero), it suggests that the system might have a center or a saddle point. But since all ( k ) are positive, the eigenvalues could be complex with zero real parts or have both positive and negative real parts.Alternatively, perhaps the system has a Hopf bifurcation, leading to oscillatory behavior.But without specific values, it's hard to determine the exact nature. However, given that the trace is zero, the system is conservative in some sense, but with all positive coefficients, it's more likely that the origin is an unstable equilibrium.Wait, actually, let's think about the eigenvalues. If we have a cubic equation with no ( lambda^2 ) term, the sum of the eigenvalues is zero. So, if all eigenvalues have negative real parts, their sum would be negative, which contradicts the trace being zero. Therefore, it's impossible for all eigenvalues to have negative real parts. Hence, the origin is not asymptotically stable.Moreover, if one eigenvalue is positive and the others are complex conjugates with negative real parts, the origin is a saddle point, which is unstable. Alternatively, if all eigenvalues are purely imaginary, the origin is a center, which is neutrally stable but not asymptotically stable.Given that all ( k ) are positive, it's possible that the system could have oscillatory solutions around the origin, but the origin itself is unstable or neutrally stable.Therefore, the equilibrium point at the origin is unstable. This suggests that small perturbations away from zero will not return to zero, implying that religious diversity, racial diversity, and gender equity might not settle at zero levels but could oscillate or grow, depending on the initial conditions.InterpretationFor the first part, the partial derivatives tell us how sensitive societal development is to changes in religious diversity at the given point. The first derivative gives the immediate rate of change, while the second derivative tells us about the curvature, indicating whether the effect is increasing or decreasing as ( x ) changes.For the second part, the system's equilibrium at zero is unstable, suggesting that the variables (religious diversity, racial diversity, gender equity) don't stabilize at zero. Instead, they might exhibit oscillatory behavior or diverge, indicating that these factors interact in a way that prevents them from settling at minimal levels, which could have implications for societal development over time.Final Answer1. The first partial derivative of ( f ) with respect to ( x ) at the point ( (0.4, 0.6, 0.8) ) is ( 0.8alpha + 0.6delta + 0.8zeta ), and the second partial derivative is ( 2alpha ). These indicate the sensitivity and curvature of societal development concerning religious diversity.2. The only equilibrium point is at the origin ( (0, 0, 0) ), which is unstable, suggesting that societal development factors do not stabilize at zero levels.The final answers are:1. First partial derivative: ( boxed{0.8alpha + 0.6delta + 0.8zeta} ), Second partial derivative: ( boxed{2alpha} ).2. The equilibrium point is ( boxed{(0, 0, 0)} ) and it is unstable.</think>