🧠:<think>Okay, so I have this differential equation modeling crop yields over time, and I need to solve it and analyze its stability. Let me try to break this down step by step.First, the equation is:[ frac{dY(t)}{dt} = alpha Y(t) - beta Y(t)^2 + gamma sin(omega t) ]This looks like a logistic growth model with an added sinusoidal term. The logistic part is because of the (alpha Y(t) - beta Y(t)^2) terms, which model growth with carrying capacity. The (gamma sin(omega t)) term probably represents periodic climatic effects, like seasonal changes or cyclical weather patterns affecting crop yields.Problem 1: Solving the Differential EquationAlright, so I need to solve this differential equation with the initial condition ( Y(0) = Y_0 ). Let me write it again:[ frac{dY}{dt} = alpha Y - beta Y^2 + gamma sin(omega t) ]This is a first-order nonlinear ordinary differential equation (ODE) because of the ( Y^2 ) term. Nonlinear ODEs can be tricky because they don't always have closed-form solutions. I remember that for logistic equations without the sinusoidal term, we can solve them analytically, but adding that sine term complicates things.Let me see if I can rewrite this equation in a more manageable form. Maybe it's a Bernoulli equation? Bernoulli equations have the form:[ frac{dy}{dt} + P(t)y = Q(t)y^n ]Comparing that to our equation:[ frac{dY}{dt} - alpha Y + beta Y^2 = gamma sin(omega t) ]Hmm, so if I rearrange terms:[ frac{dY}{dt} + (-alpha) Y = gamma sin(omega t) + beta Y^2 ]This is almost a Bernoulli equation with ( n = 2 ). The standard form for Bernoulli is:[ frac{dy}{dt} + P(t)y = Q(t)y^n ]So in our case, ( P(t) = -alpha ), ( Q(t) = gamma sin(omega t) ), and ( n = 2 ). To solve Bernoulli equations, we use the substitution ( v = y^{1 - n} ). For ( n = 2 ), this becomes ( v = y^{-1} ). Let me try that substitution.Let ( v = frac{1}{Y} ). Then, ( Y = frac{1}{v} ), and differentiating both sides with respect to t:[ frac{dY}{dt} = -frac{1}{v^2} frac{dv}{dt} ]Substituting into the original equation:[ -frac{1}{v^2} frac{dv}{dt} = alpha left( frac{1}{v} right) - beta left( frac{1}{v} right)^2 + gamma sin(omega t) ]Multiply both sides by ( -v^2 ):[ frac{dv}{dt} = -alpha v + beta + -gamma v^2 sin(omega t) ]Wait, let me double-check that:Left side: ( -frac{1}{v^2} frac{dv}{dt} times (-v^2) = frac{dv}{dt} )Right side: ( alpha frac{1}{v} times (-v^2) = -alpha v )( -beta frac{1}{v^2} times (-v^2) = +beta )( gamma sin(omega t) times (-v^2) = -gamma v^2 sin(omega t) )So putting it all together:[ frac{dv}{dt} = -alpha v + beta - gamma v^2 sin(omega t) ]Hmm, this doesn't seem to simplify things much. It's still a nonlinear equation because of the ( v^2 ) term. Maybe Bernoulli substitution isn't the way to go here.Alternatively, perhaps I can consider this as a Riccati equation. Riccati equations have the form:[ frac{dy}{dt} = q_0(t) + q_1(t) y + q_2(t) y^2 ]Comparing to our equation:[ frac{dY}{dt} = alpha Y - beta Y^2 + gamma sin(omega t) ]Yes, it is a Riccati equation with ( q_0(t) = gamma sin(omega t) ), ( q_1(t) = alpha ), and ( q_2(t) = -beta ).Riccati equations are also generally difficult to solve unless we can find a particular solution. If I can find one particular solution, then I can reduce the equation to a Bernoulli or linear equation.But finding a particular solution for this might be challenging because of the sine term. Maybe I can assume a particular solution of the form ( Y_p(t) = A sin(omega t) + B cos(omega t) ). Let me try that.Assume:[ Y_p(t) = A sin(omega t) + B cos(omega t) ]Compute its derivative:[ frac{dY_p}{dt} = A omega cos(omega t) - B omega sin(omega t) ]Plug ( Y_p ) and its derivative into the ODE:[ A omega cos(omega t) - B omega sin(omega t) = alpha (A sin(omega t) + B cos(omega t)) - beta (A sin(omega t) + B cos(omega t))^2 + gamma sin(omega t) ]Let me expand the right-hand side:First, compute ( alpha Y_p ):[ alpha A sin(omega t) + alpha B cos(omega t) ]Next, compute ( -beta Y_p^2 ):[ -beta (A^2 sin^2(omega t) + 2AB sin(omega t)cos(omega t) + B^2 cos^2(omega t)) ]And the last term is ( gamma sin(omega t) ).So putting it all together:Left side: ( A omega cos(omega t) - B omega sin(omega t) )Right side:[ alpha A sin(omega t) + alpha B cos(omega t) - beta (A^2 sin^2(omega t) + 2AB sin(omega t)cos(omega t) + B^2 cos^2(omega t)) + gamma sin(omega t) ]Now, let's collect like terms on the right side.First, the terms without sine or cosine squared:- The linear terms in sine and cosine: ( (alpha A + gamma) sin(omega t) + alpha B cos(omega t) )- The quadratic terms: ( -beta A^2 sin^2(omega t) - 2beta AB sin(omega t)cos(omega t) - beta B^2 cos^2(omega t) )So, the right side is:[ (alpha A + gamma) sin(omega t) + alpha B cos(omega t) - beta A^2 sin^2(omega t) - 2beta AB sin(omega t)cos(omega t) - beta B^2 cos^2(omega t) ]Now, let's write the equation:Left side: ( A omega cos(omega t) - B omega sin(omega t) )Right side: ( (alpha A + gamma) sin(omega t) + alpha B cos(omega t) - beta A^2 sin^2(omega t) - 2beta AB sin(omega t)cos(omega t) - beta B^2 cos^2(omega t) )For this equality to hold for all t, the coefficients of like terms must be equal on both sides.But on the left side, we have only terms with (sin(omega t)) and (cos(omega t)), while on the right side, we have quadratic terms as well. This suggests that unless the coefficients of the quadratic terms are zero, the equation can't hold. So, we need:1. Coefficient of (sin^2(omega t)): ( -beta A^2 = 0 ) => ( A = 0 )2. Coefficient of (cos^2(omega t)): ( -beta B^2 = 0 ) => ( B = 0 )3. Coefficient of (sin(omega t)cos(omega t)): ( -2beta AB = 0 ) which is automatically satisfied if A or B is zero.But if A and B are zero, then our particular solution is zero, which doesn't help because plugging Y_p = 0 into the equation gives:Left side: 0Right side: 0 + 0 + gamma sin(omega t) => gamma sin(omega t) ≠ 0So this approach doesn't work. Maybe assuming a particular solution with only sine and cosine isn't sufficient because the quadratic term complicates things. Perhaps I need a more complex form for Y_p, but that might get too complicated.Alternatively, maybe I can use perturbation methods if the parameters are small, but I don't know if that's applicable here.Wait, another thought: since the equation is nonlinear, maybe it's better to look for an integrating factor or see if it can be transformed into a linear equation. But given the ( Y^2 ) term, it's inherently nonlinear, so linear methods might not apply.Alternatively, perhaps I can use numerical methods to solve it, but the problem asks for an analytical solution, so I need to find another approach.Wait, let me think again about the substitution. Maybe instead of Bernoulli, I can use another substitution.Let me consider the substitution ( Z = Y + k ), where k is a constant to be determined. Maybe this can simplify the equation.But I'm not sure. Alternatively, perhaps I can write the equation as:[ frac{dY}{dt} = alpha Y - beta Y^2 + gamma sin(omega t) ]Let me rearrange terms:[ frac{dY}{dt} + beta Y^2 = alpha Y + gamma sin(omega t) ]This is a Riccati equation, as I thought earlier. Riccati equations are tough because they are nonlinear. However, if we can find a particular solution, we can reduce it to a Bernoulli equation.But as I tried earlier, assuming a particular solution of the form ( Y_p = A sin(omega t) + B cos(omega t) ) didn't work because it led to a contradiction unless A and B are zero, which doesn't help.Maybe I need to consider a different form for the particular solution. Perhaps including higher harmonics or something else.Alternatively, perhaps I can use the method of variation of parameters, but I think that applies to linear equations.Wait, another idea: maybe I can use the substitution ( Y = frac{u}{v} ), where u and v are functions to be determined. But I'm not sure.Alternatively, perhaps I can write the equation in terms of reciprocal, but that didn't help earlier.Wait, maybe I can write the equation as:[ frac{dY}{dt} = alpha Y - beta Y^2 + gamma sin(omega t) ]Let me divide both sides by ( Y^2 ):[ frac{1}{Y^2} frac{dY}{dt} = frac{alpha}{Y} - beta + frac{gamma}{Y^2} sin(omega t) ]Let me set ( v = frac{1}{Y} ), so ( Y = frac{1}{v} ), and ( frac{dY}{dt} = -frac{1}{v^2} frac{dv}{dt} ). Substituting:[ frac{1}{Y^2} frac{dY}{dt} = -frac{dv}{dt} ]So the equation becomes:[ -frac{dv}{dt} = alpha v - beta + gamma v^2 sin(omega t) ]Which rearranges to:[ frac{dv}{dt} = -alpha v + beta - gamma v^2 sin(omega t) ]Hmm, this is similar to what I had before. It's still a nonlinear equation because of the ( v^2 ) term. So, I'm back to square one.Maybe I can consider this as a perturbed logistic equation. If ( gamma ) is small, I could use perturbation methods, but since the problem doesn't specify that ( gamma ) is small, I can't assume that.Alternatively, perhaps I can look for an exact solution by finding an integrating factor, but again, the nonlinearity complicates things.Wait, another thought: maybe I can use the substitution ( w = Y e^{-alpha t} ). Let me try that.Let ( w = Y e^{-alpha t} ). Then, ( Y = w e^{alpha t} ), and:[ frac{dY}{dt} = frac{dw}{dt} e^{alpha t} + alpha w e^{alpha t} ]Substituting into the original equation:[ frac{dw}{dt} e^{alpha t} + alpha w e^{alpha t} = alpha (w e^{alpha t}) - beta (w e^{alpha t})^2 + gamma sin(omega t) ]Simplify:Left side: ( frac{dw}{dt} e^{alpha t} + alpha w e^{alpha t} )Right side: ( alpha w e^{alpha t} - beta w^2 e^{2alpha t} + gamma sin(omega t) )Subtract ( alpha w e^{alpha t} ) from both sides:[ frac{dw}{dt} e^{alpha t} = - beta w^2 e^{2alpha t} + gamma sin(omega t) ]Divide both sides by ( e^{alpha t} ):[ frac{dw}{dt} = - beta w^2 e^{alpha t} + gamma e^{-alpha t} sin(omega t) ]Hmm, this still looks complicated. The term ( e^{alpha t} ) is now in the equation, making it non-autonomous and still nonlinear.I don't think this substitution helps. Maybe another substitution?Alternatively, perhaps I can consider this equation as a forced logistic equation and look for solutions in terms of the forcing function. But I don't recall a standard form for that.Wait, maybe I can use the method of undetermined coefficients but for nonlinear equations. I don't think that's a standard method, though.Alternatively, perhaps I can use a Green's function approach, but I'm not sure how that would apply to a nonlinear equation.At this point, I'm stuck on finding an analytical solution. Maybe the problem expects a qualitative analysis rather than an explicit solution? But the question specifically says "solve the differential equation," so I must be missing something.Wait, perhaps I can linearize the equation around the equilibrium points? But that's more for stability analysis, which is part 2.Alternatively, maybe the equation can be transformed into a Bernoulli equation if I rearrange terms differently.Wait, let me go back to the original equation:[ frac{dY}{dt} = alpha Y - beta Y^2 + gamma sin(omega t) ]Let me write it as:[ frac{dY}{dt} + beta Y^2 = alpha Y + gamma sin(omega t) ]This is a Riccati equation, as I thought earlier. Riccati equations are tough, but sometimes they can be transformed into linear equations if we know a particular solution.But as I tried earlier, assuming a particular solution of the form ( Y_p = A sin(omega t) + B cos(omega t) ) didn't work because it led to a contradiction unless A and B are zero, which doesn't help.Wait, maybe I can consider a particular solution that includes exponential terms? Or perhaps a constant particular solution?Wait, if I set ( frac{dY}{dt} = 0 ), then the equilibrium points satisfy:[ 0 = alpha Y - beta Y^2 + gamma sin(omega t) ]But since ( sin(omega t) ) is time-dependent, the equilibrium points are also time-dependent, which complicates things.Alternatively, maybe I can consider the homogeneous equation:[ frac{dY}{dt} = alpha Y - beta Y^2 ]Which is the logistic equation, and its solution is:[ Y(t) = frac{alpha}{beta} cdot frac{1}{1 + C e^{-alpha t}} ]Where C is determined by the initial condition. But our equation has an additional forcing term ( gamma sin(omega t) ), so it's a nonhomogeneous logistic equation.I wonder if there's a way to find a particular solution using variation of parameters or something similar, but I don't recall such a method for Riccati equations.Alternatively, maybe I can use the method of integrating factors for Bernoulli equations, but since this is Riccati, I'm not sure.Wait, another idea: perhaps I can write the equation in terms of reciprocal, as I tried earlier, but then use an integrating factor.Let me recall that for Bernoulli equations, after substitution, we can make them linear. But in this case, after substitution, it's still nonlinear.Wait, perhaps I can use the substitution ( v = Y + k ), where k is chosen to eliminate the quadratic term? But I don't think that works because the quadratic term is essential to the logistic growth.Alternatively, perhaps I can use a substitution to make the equation separable. Let me try:[ frac{dY}{dt} = alpha Y - beta Y^2 + gamma sin(omega t) ]Let me rearrange:[ frac{dY}{alpha Y - beta Y^2} = dt + frac{gamma sin(omega t)}{alpha Y - beta Y^2} dt ]Hmm, that doesn't seem helpful because the right side still has Y in the denominator, making it difficult to integrate.Alternatively, perhaps I can write it as:[ frac{dY}{alpha Y - beta Y^2} - frac{gamma sin(omega t)}{alpha Y - beta Y^2} dY = dt ]No, that doesn't make sense.Wait, another approach: maybe use the substitution ( u = Y ), ( v = t ), and try to find an integrating factor. But I don't think that helps.Alternatively, perhaps I can use the method of characteristics, but that's more for PDEs.At this point, I'm stuck. Maybe the problem expects me to recognize that this is a Riccati equation and that without a particular solution, it's difficult to solve analytically, so perhaps the answer is that an explicit solution isn't feasible and we need to analyze it qualitatively or numerically.But the problem says "solve the differential equation," so maybe I'm missing a trick. Let me think again.Wait, perhaps I can write the equation as:[ frac{dY}{dt} = alpha Y - beta Y^2 + gamma sin(omega t) ]Let me factor out Y:[ frac{dY}{dt} = Y(alpha - beta Y) + gamma sin(omega t) ]This is a logistic growth term plus a sinusoidal forcing term. Maybe I can use the method of averaging or something similar for weakly nonlinear systems, but again, without knowing if the parameters are small, I can't assume that.Alternatively, perhaps I can use a Green's function approach for linear equations, but this is nonlinear.Wait, another thought: perhaps I can use the substitution ( Y = frac{alpha}{beta} - Z ), shifting the variable to eliminate the linear term. Let me try that.Let ( Y = frac{alpha}{beta} - Z ). Then:[ frac{dY}{dt} = -frac{dZ}{dt} ]Substitute into the equation:[ -frac{dZ}{dt} = alpha left( frac{alpha}{beta} - Z right) - beta left( frac{alpha}{beta} - Z right)^2 + gamma sin(omega t) ]Expand the terms:First, compute ( alpha Y ):[ alpha left( frac{alpha}{beta} - Z right) = frac{alpha^2}{beta} - alpha Z ]Next, compute ( -beta Y^2 ):[ -beta left( frac{alpha^2}{beta^2} - frac{2alpha}{beta} Z + Z^2 right) = -frac{alpha^2}{beta} + 2alpha Z - beta Z^2 ]So putting it all together:[ -frac{dZ}{dt} = left( frac{alpha^2}{beta} - alpha Z right) + left( -frac{alpha^2}{beta} + 2alpha Z - beta Z^2 right) + gamma sin(omega t) ]Simplify:The ( frac{alpha^2}{beta} ) and ( -frac{alpha^2}{beta} ) terms cancel.Combine the Z terms:( -alpha Z + 2alpha Z = alpha Z )So:[ -frac{dZ}{dt} = alpha Z - beta Z^2 + gamma sin(omega t) ]Multiply both sides by -1:[ frac{dZ}{dt} = -alpha Z + beta Z^2 - gamma sin(omega t) ]Hmm, this is similar to the original equation but with a sign change. It doesn't seem to have simplified the problem. Maybe this substitution isn't helpful.Wait, perhaps I can consider this as a perturbation around the equilibrium point. Let me find the equilibrium points first, which might help in part 2, but maybe also for solving the equation.Equilibrium points are found by setting ( frac{dY}{dt} = 0 ):[ 0 = alpha Y - beta Y^2 + gamma sin(omega t) ]But since ( sin(omega t) ) is time-dependent, the equilibrium points are also time-dependent, which complicates things. So, maybe instead, I can consider the system's behavior around the average equilibrium.Wait, if I average over time, the average of ( sin(omega t) ) is zero, so the average equilibrium would be where ( alpha Y - beta Y^2 = 0 ), which gives Y = 0 or Y = α/β. But this is just the equilibrium for the logistic equation without the forcing term.But with the forcing term, the system is periodically driven, so the solution might oscillate around these equilibria.But I'm not sure how to proceed analytically. Maybe I need to accept that an explicit solution isn't feasible and instead focus on the qualitative behavior, but the problem specifically asks to solve the differential equation.Wait, perhaps I can use the method of integrating factors for Bernoulli equations. Let me recall that for Bernoulli equations:[ frac{dy}{dt} + P(t) y = Q(t) y^n ]The substitution is ( v = y^{1 - n} ), which transforms it into a linear equation.In our case, n = 2, so ( v = y^{-1} ). Let me try that again.So, ( v = 1/Y ), then ( Y = 1/v ), and ( dY/dt = -v^{-2} dv/dt ).Substitute into the original equation:[ -v^{-2} frac{dv}{dt} = alpha (1/v) - beta (1/v)^2 + gamma sin(omega t) ]Multiply both sides by ( -v^2 ):[ frac{dv}{dt} = -alpha v + beta - gamma v^2 sin(omega t) ]This is the same equation I had earlier. So, it's a Riccati equation in terms of v.Riccati equations are difficult, but maybe I can find a particular solution for v.Assume a particular solution ( v_p(t) = C sin(omega t) + D cos(omega t) ).Compute its derivative:[ frac{dv_p}{dt} = C omega cos(omega t) - D omega sin(omega t) ]Substitute into the equation:[ C omega cos(omega t) - D omega sin(omega t) = -alpha (C sin(omega t) + D cos(omega t)) + beta - gamma (C sin(omega t) + D cos(omega t))^2 sin(omega t) ]This looks complicated, but let's try to expand and collect like terms.First, expand the right side:1. ( -alpha C sin(omega t) - alpha D cos(omega t) )2. ( + beta )3. ( -gamma (C^2 sin^2(omega t) + 2CD sin(omega t)cos(omega t) + D^2 cos^2(omega t)) sin(omega t) )So, expanding the last term:[ -gamma C^2 sin^3(omega t) - 2gamma CD sin^2(omega t)cos(omega t) - gamma D^2 sin(omega t)cos^2(omega t) ]Now, collect all terms:Left side:[ C omega cos(omega t) - D omega sin(omega t) ]Right side:1. ( -alpha C sin(omega t) - alpha D cos(omega t) )2. ( + beta )3. ( -gamma C^2 sin^3(omega t) - 2gamma CD sin^2(omega t)cos(omega t) - gamma D^2 sin(omega t)cos^2(omega t) )For the equation to hold for all t, the coefficients of like terms must be equal.But on the left side, we have terms up to first order in sine and cosine, while on the right side, we have up to third order. This suggests that the coefficients of the higher-order terms must be zero.So, set the coefficients of ( sin^3(omega t) ), ( sin^2(omega t)cos(omega t) ), and ( sin(omega t)cos^2(omega t) ) to zero:1. ( -gamma C^2 = 0 ) => ( C = 0 )2. ( -2gamma CD = 0 ) => automatically satisfied if C = 03. ( -gamma D^2 = 0 ) => ( D = 0 )But if C = 0 and D = 0, then our particular solution is zero, which doesn't help because plugging back into the equation gives:Left side: 0Right side: ( -alpha (0) + beta - 0 = beta )Which is not equal to zero unless β = 0, which isn't the case. So, this approach doesn't work.Maybe I need a different form for the particular solution, perhaps including higher harmonics or exponential terms, but that might get too complicated.Alternatively, perhaps I can use a Fourier series approach, assuming the solution can be expressed as a sum of sine and cosine terms with multiple frequencies. But that's a more advanced technique and might not be feasible here.Given that I'm stuck on finding an analytical solution, maybe the problem expects me to recognize that the equation is a Riccati equation and that without a particular solution, it's difficult to solve explicitly. Therefore, perhaps the answer is that an explicit solution isn't possible with elementary functions, and we need to analyze it numerically or qualitatively.But the problem specifically asks to solve the differential equation, so I must be missing something. Maybe I can use an integrating factor for the logistic part and then handle the forcing term separately.Wait, another idea: perhaps I can write the equation as:[ frac{dY}{dt} = alpha Y - beta Y^2 + gamma sin(omega t) ]Let me consider the homogeneous equation:[ frac{dY}{dt} = alpha Y - beta Y^2 ]Which has the solution:[ Y(t) = frac{alpha}{beta} cdot frac{1}{1 + C e^{-alpha t}} ]Now, perhaps I can use variation of parameters to find a particular solution. Let me recall that for linear equations, we can use variation of parameters, but this is a nonlinear equation, so I'm not sure.Wait, but if I consider the homogeneous solution as ( Y_h(t) = frac{alpha}{beta} cdot frac{1}{1 + C e^{-alpha t}} ), maybe I can assume a particular solution of the form ( Y_p(t) = Y_h(t) cdot u(t) ), where u(t) is a function to be determined.Let me try that substitution.Let ( Y_p = Y_h u ). Then:[ frac{dY_p}{dt} = frac{dY_h}{dt} u + Y_h frac{du}{dt} ]Substitute into the original equation:[ frac{dY_h}{dt} u + Y_h frac{du}{dt} = alpha Y_h u - beta (Y_h u)^2 + gamma sin(omega t) ]But ( frac{dY_h}{dt} = alpha Y_h - beta Y_h^2 ), so substitute that:[ (alpha Y_h - beta Y_h^2) u + Y_h frac{du}{dt} = alpha Y_h u - beta Y_h^2 u^2 + gamma sin(omega t) ]Simplify the left side:[ alpha Y_h u - beta Y_h^2 u + Y_h frac{du}{dt} ]Right side:[ alpha Y_h u - beta Y_h^2 u^2 + gamma sin(omega t) ]Subtract ( alpha Y_h u ) from both sides:[ - beta Y_h^2 u + Y_h frac{du}{dt} = - beta Y_h^2 u^2 + gamma sin(omega t) ]Rearrange:[ Y_h frac{du}{dt} = - beta Y_h^2 u^2 + beta Y_h^2 u + gamma sin(omega t) ]Factor out ( beta Y_h^2 ):[ Y_h frac{du}{dt} = beta Y_h^2 (u - u^2) + gamma sin(omega t) ]Divide both sides by ( Y_h ):[ frac{du}{dt} = beta Y_h (u - u^2) + frac{gamma}{Y_h} sin(omega t) ]This is still a nonlinear equation in u, so it doesn't help much. I was hoping to reduce it to a linear equation, but it's still nonlinear.At this point, I think I've exhausted my methods for finding an analytical solution. It seems that without a particular solution, it's difficult to solve this Riccati equation explicitly. Therefore, I might need to conclude that an explicit solution isn't feasible with elementary functions and that numerical methods or qualitative analysis would be more appropriate.However, since the problem specifically asks to solve the differential equation, perhaps I'm missing a trick or there's a substitution I haven't considered. Let me think one more time.Wait, perhaps I can use the substitution ( Y = frac{alpha}{beta} cdot frac{1}{1 + Z(t)} ), which is similar to the logistic solution. Let me try that.Let ( Y = frac{alpha}{beta} cdot frac{1}{1 + Z} ). Then:[ frac{dY}{dt} = -frac{alpha}{beta} cdot frac{Z'}{(1 + Z)^2} ]Substitute into the original equation:[ -frac{alpha}{beta} cdot frac{Z'}{(1 + Z)^2} = alpha cdot frac{alpha}{beta} cdot frac{1}{1 + Z} - beta cdot left( frac{alpha}{beta} cdot frac{1}{1 + Z} right)^2 + gamma sin(omega t) ]Simplify each term:1. Left side: ( -frac{alpha}{beta} cdot frac{Z'}{(1 + Z)^2} )2. First term on right: ( frac{alpha^2}{beta} cdot frac{1}{1 + Z} )3. Second term on right: ( -beta cdot frac{alpha^2}{beta^2} cdot frac{1}{(1 + Z)^2} = -frac{alpha^2}{beta} cdot frac{1}{(1 + Z)^2} )4. Third term: ( gamma sin(omega t) )So, putting it all together:[ -frac{alpha}{beta} cdot frac{Z'}{(1 + Z)^2} = frac{alpha^2}{beta} cdot frac{1}{1 + Z} - frac{alpha^2}{beta} cdot frac{1}{(1 + Z)^2} + gamma sin(omega t) ]Multiply both sides by ( -beta / alpha ):[ frac{Z'}{(1 + Z)^2} = -alpha cdot frac{1}{1 + Z} + alpha cdot frac{1}{(1 + Z)^2} - frac{beta gamma}{alpha} sin(omega t) ]Let me write this as:[ frac{Z'}{(1 + Z)^2} = alpha left( frac{1}{(1 + Z)^2} - frac{1}{1 + Z} right) - frac{beta gamma}{alpha} sin(omega t) ]Simplify the terms inside the parentheses:[ frac{1}{(1 + Z)^2} - frac{1}{1 + Z} = frac{1 - (1 + Z)}{(1 + Z)^2} = frac{-Z}{(1 + Z)^2} ]So, the equation becomes:[ frac{Z'}{(1 + Z)^2} = -alpha cdot frac{Z}{(1 + Z)^2} - frac{beta gamma}{alpha} sin(omega t) ]Multiply both sides by ( (1 + Z)^2 ):[ Z' = -alpha Z - frac{beta gamma}{alpha} (1 + Z)^2 sin(omega t) ]This is still a nonlinear equation because of the ( (1 + Z)^2 ) term. It doesn't seem to have simplified the problem.At this point, I think it's safe to conclude that an explicit analytical solution isn't feasible with elementary functions. Therefore, the answer to part 1 is that the differential equation doesn't have a closed-form solution in terms of elementary functions and must be solved numerically or analyzed qualitatively.Problem 2: Stability AnalysisNow, moving on to part 2, which asks to analyze the stability of the system by finding the equilibrium points and determining their nature for given parameter values: ( alpha = 0.1 ), ( beta = 0.02 ), ( gamma = 0.5 ), and ( omega = pi ).First, let's recall that equilibrium points are solutions where ( frac{dY}{dt} = 0 ). However, in this case, the equation is time-dependent due to the ( sin(omega t) ) term, so the equilibrium points are also time-dependent. This complicates the analysis because the system doesn't have fixed equilibrium points but rather oscillating ones.However, for the sake of stability analysis, perhaps we can consider the system's behavior around the average equilibrium. The average of ( sin(omega t) ) over a period is zero, so the average equilibrium would be where ( alpha Y - beta Y^2 = 0 ), which gives Y = 0 or Y = α/β.Given ( alpha = 0.1 ) and ( beta = 0.02 ), the equilibrium points are Y = 0 and Y = 0.1 / 0.02 = 5.So, the average equilibria are at Y = 0 and Y = 5. Now, we can analyze the stability of these points by linearizing the system around them.To linearize, we'll compute the Jacobian matrix (which, in this case, is just the derivative of the right-hand side with respect to Y) and evaluate it at the equilibrium points.The right-hand side of the differential equation is:[ f(Y, t) = alpha Y - beta Y^2 + gamma sin(omega t) ]The derivative with respect to Y is:[ f_Y = alpha - 2beta Y ]Now, evaluate this at the equilibrium points.1. At Y = 0:[ f_Y(0) = alpha - 0 = alpha = 0.1 ]Since this is positive, the equilibrium at Y = 0 is unstable.2. At Y = 5:[ f_Y(5) = 0.1 - 2 * 0.02 * 5 = 0.1 - 0.2 = -0.1 ]Since this is negative, the equilibrium at Y = 5 is stable.However, this analysis is under the assumption that the forcing term ( gamma sin(omega t) ) has an average of zero. In reality, the forcing term can cause the system to oscillate around these equilibria. The stability analysis here is linear and pertains to the average behavior, but the actual system may exhibit more complex dynamics due to the periodic forcing.But given the parameters, the equilibrium at Y = 5 is stable, and Y = 0 is unstable.Wait, but let me double-check the derivative calculation. The derivative of f with respect to Y is indeed ( alpha - 2beta Y ). At Y = 5, that's 0.1 - 2*0.02*5 = 0.1 - 0.2 = -0.1, which is correct. So, the equilibrium at Y = 5 is stable, and Y = 0 is unstable.Therefore, the system has two equilibrium points: an unstable one at Y = 0 and a stable one at Y = 5. However, due to the periodic forcing term, the actual solution may oscillate around Y = 5 rather than settling exactly at it.But for the purpose of this analysis, we can say that the equilibrium at Y = 5 is stable, and Y = 0 is unstable.Final Answer1. The differential equation does not have a closed-form solution in terms of elementary functions and must be solved numerically.2. The equilibrium points are ( Y = 0 ) (unstable) and ( Y = 5 ) (stable).So, the final answers are:1. boxed{text{No closed-form solution; numerical methods required}}2. The equilibrium points are ( boxed{0} ) (unstable) and ( boxed{5} ) (stable).</think>

🧠:<think>First, I'll calculate the total area of the farmland by multiplying its length and width.Next, I'll determine the combined area of the corn and cattle sections, which is 60% of the total farmland area.Since the corn section is twice the area of the cattle section, I'll set up an equation to find the individual areas of each section.Once I have the areas, I'll assume that both sections have the same dimensions as the farmland but with different widths. I'll set up another equation to solve for the widths of each section.After finding the widths, I'll calculate the total fencing required for the entire farmland and each section, then multiply by the cost per meter to find the total cost.</think>

🧠:<think>Alright, so I have this problem about vehicle stability in crosswind. It's a bit complex, but let me try to break it down step by step. First, the scenario: a vehicle is moving at a constant speed ( v ) on a flat road, and there's a crosswind blowing perpendicular to the direction of motion at speed ( w ). The vehicle has dimensions length ( L ), width ( W ), height ( H ). The center of gravity (CG) is at height ( h ) above the ground and distance ( d ) from the front axle. The aerodynamic drag coefficient ( C_d ) and lift coefficient ( C_l ) depend on the Reynolds number ( Re = frac{rho v L}{mu} ), where ( rho ) is air density and ( mu ) is dynamic viscosity.The first part asks to derive the total aerodynamic force due to crosswind, considering both drag and lift. The given formulas are ( F_d = frac{1}{2} rho v^2 A C_d(Re) ) and ( F_l = frac{1}{2} rho v^2 A C_l(Re) ), where ( A ) is the frontal area.Hmm, okay. So, drag force is usually in the direction opposite to the motion, and lift is perpendicular. But since the crosswind is perpendicular, does that mean the lift force is now in the direction of the crosswind? Or is it still relative to the vehicle's motion?Wait, the vehicle is moving forward, and the crosswind is blowing from the side. So, relative to the vehicle, the wind has components both in the direction of motion and perpendicular. But since the vehicle is moving at constant speed, maybe we can consider the relative wind as a combination of the vehicle's velocity and the crosswind.But the problem says the crosswind is blowing perpendicular, so maybe the relative wind is a vector sum of the vehicle's velocity and the crosswind. However, since the vehicle is moving forward, the crosswind is perpendicular, so the relative wind direction is at some angle. But for simplicity, maybe we can consider the crosswind as the only wind component affecting the vehicle's sideslip angle.Wait, perhaps I'm overcomplicating. The problem states that the crosswind is blowing perpendicular, so maybe the aerodynamic forces are calculated based on the crosswind speed ( w ) instead of the vehicle's speed ( v ). Or is it a combination?Wait, the drag and lift coefficients are functions of the Reynolds number based on vehicle speed ( v ). So, ( Re = frac{rho v L}{mu} ). So, the aerodynamic forces are dependent on the vehicle's speed, not the wind speed. Hmm, that seems a bit counterintuitive because crosswind would typically induce side forces.But maybe in this case, the crosswind causes a sideslip angle, which affects the lift and drag coefficients. However, the problem gives ( F_d ) and ( F_l ) as functions of ( v ) and ( A ). So, perhaps the crosswind is considered as a lateral wind, and the vehicle's motion is forward, so the relative wind is a combination of the two.Wait, perhaps the total aerodynamic force is a vector sum of the drag force in the direction of motion and the lift force perpendicular to it. But since the crosswind is blowing perpendicular, maybe the lift force is in the direction of the crosswind.Alternatively, maybe the crosswind induces a side force which is a lift force. So, perhaps the total aerodynamic force is a combination of drag and lift, but in different directions.Wait, the problem says "due to the crosswind," so maybe the crosswind induces a side force, which is a lift force, and also affects the drag. Hmm, but the given formulas for ( F_d ) and ( F_l ) are based on the vehicle's speed ( v ). So, perhaps the crosswind doesn't directly affect the drag and lift coefficients, but the direction of the force is altered.Wait, maybe the crosswind causes the vehicle to experience a sideslip angle, which would change the effective angle of attack, thereby affecting both lift and drag. But this might be more complicated than what's required here.Wait, the problem says "due to the crosswind," so perhaps the crosswind induces a lateral force, which is a lift force. So, maybe the total aerodynamic force is just the lift force in the direction of the crosswind, and the drag is in the direction of the vehicle's motion.But the question says "taking into account both the drag and lift forces." So, perhaps the total aerodynamic force is a vector sum of drag and lift. But drag is in the direction opposite to the vehicle's motion, and lift is perpendicular. Since the crosswind is perpendicular, maybe the lift is in the direction of the crosswind.Wait, maybe I need to model the relative wind. The vehicle is moving forward at speed ( v ), and there's a crosswind of speed ( w ). So, the relative wind speed is the vector sum of ( v ) and ( w ). The angle of this relative wind with respect to the vehicle's direction would be ( theta = arctan(w / v) ). Therefore, the effective angle of attack for the vehicle is ( theta ), which would cause both lift and drag forces. So, the total aerodynamic force would have components in the direction of the relative wind, which is at an angle ( theta ) from the vehicle's direction.But the problem gives ( F_d ) and ( F_l ) as functions of ( v ). So, perhaps we can consider that the crosswind causes a sideslip, and the aerodynamic forces are calculated based on the vehicle's speed, but the direction is altered.Alternatively, maybe the crosswind adds a lateral component to the aerodynamic force, so the total force is a combination of drag in the forward direction and lift in the lateral direction.Wait, perhaps the total aerodynamic force is the vector sum of the drag force ( F_d ) in the direction of vehicle motion and the lift force ( F_l ) perpendicular to it. So, the total force would have components ( F_d ) and ( F_l ), and the magnitude would be ( sqrt{F_d^2 + F_l^2} ). But the problem says "due to the crosswind," so maybe only the lift component is relevant?Wait, no, because the crosswind would cause both a lift force (perpendicular to the direction of motion) and a drag force (opposite to the direction of motion). But the vehicle is moving forward, so the drag is in the direction opposite to the vehicle's motion, and the lift is in the direction of the crosswind.But the problem says "due to the crosswind," so perhaps the crosswind induces a lateral force, which is the lift force, and the drag is due to the vehicle's motion. But both are aerodynamic forces, so both should be considered.Wait, maybe the crosswind causes a sideslip, so the vehicle experiences both a drag force in the direction opposite to the relative wind and a lift force perpendicular to it. But the relative wind is a combination of the vehicle's forward speed and the crosswind.This is getting a bit complicated. Maybe I should consider the relative wind speed as ( sqrt{v^2 + w^2} ), but the problem gives ( F_d ) and ( F_l ) as functions of ( v ), not the relative speed. So, perhaps the crosswind only affects the direction of the aerodynamic force, not its magnitude.Alternatively, maybe the crosswind causes a lateral force which is a lift force, and the drag is still in the forward direction. So, the total aerodynamic force would have two components: drag in the forward direction and lift in the lateral direction.But the problem says "due to the crosswind," so maybe only the lift force is considered? Or both?Wait, the problem says "taking into account both the drag and lift forces." So, both should be included. So, the total aerodynamic force is the vector sum of drag and lift.But drag is in the direction opposite to the vehicle's motion, and lift is perpendicular to it. So, the total force would have components in both the forward and lateral directions.But the crosswind is blowing perpendicular, so the lift force is in the direction of the crosswind, which is lateral. The drag is still in the forward direction.Wait, but the vehicle is moving forward, so the drag is opposite to the forward direction, and the lift is in the lateral direction. So, the total aerodynamic force is a combination of these two.But the problem says "due to the crosswind," so maybe the crosswind induces a lateral force, which is lift, and the drag is due to the vehicle's motion. So, the total force is the vector sum of these two.But I'm not sure. Maybe I should think about the reference frame. In the vehicle's frame, the wind is coming at an angle, so the relative wind has both forward and lateral components. Therefore, the aerodynamic forces would have both drag and lift components.But since the vehicle is moving at constant speed, the drag force is balanced by the engine's thrust, but the lift force is a new force due to the crosswind.Wait, but the problem is about the total aerodynamic force, so it's just the sum of drag and lift, regardless of what balances them. So, the total force would be ( F_{total} = sqrt{F_d^2 + F_l^2} ), but in vector form, it's ( F_d ) in the forward direction and ( F_l ) in the lateral direction.But the problem says "due to the crosswind," so maybe only the lateral component is considered? Or both?Wait, the crosswind causes a lateral force (lift) and also affects the drag because the vehicle is now effectively moving at an angle, so the drag might increase. But the problem gives ( F_d ) as a function of ( v ), so maybe the drag is still based on the vehicle's speed, not the relative wind speed.Alternatively, maybe the crosswind doesn't affect the drag coefficient, but only induces a lift force. So, the total aerodynamic force is just the lift force in the lateral direction.But the problem says "taking into account both the drag and lift forces," so I think both should be included. So, the total force is a vector with components ( F_d ) in the forward direction and ( F_l ) in the lateral direction.But wait, the crosswind is blowing perpendicular, so the lift force is in the direction of the crosswind, which is lateral. The drag force is opposite to the vehicle's motion, which is forward. So, the total aerodynamic force is a combination of these two.But the problem is asking for the total aerodynamic force due to the crosswind. So, maybe the crosswind only induces the lift force, and the drag is due to the vehicle's motion. So, the total force is just the lift force.Wait, no, because the crosswind would also cause a change in the drag force. For example, if the vehicle is moving forward and there's a crosswind, the effective angle of attack changes, which affects both lift and drag.But the problem gives ( F_d ) and ( F_l ) as functions of ( v ), so maybe the crosswind doesn't change the drag and lift coefficients, but only the direction of the force.Wait, this is confusing. Maybe I should look up how crosswind affects aerodynamic forces. From what I remember, crosswind induces a lateral force, which is lift, and also a drag force due to the sideslip angle.But in this problem, the drag and lift coefficients are given as functions of ( Re ), which is based on ( v ). So, maybe the crosswind doesn't affect the coefficients, but only the direction of the force.Alternatively, maybe the crosswind adds a lateral component to the aerodynamic force, so the total force is the vector sum of the drag in the forward direction and the lift in the lateral direction.But the problem says "due to the crosswind," so perhaps the crosswind is the cause of both the lift and an additional drag component. Hmm.Wait, maybe I'm overcomplicating. Let's think about it this way: the vehicle is moving forward, so it experiences drag in the opposite direction. The crosswind is blowing perpendicular, so it causes a lift force in the direction of the wind. So, the total aerodynamic force is the vector sum of these two forces.Therefore, the total force ( F_{total} ) would have components:- Drag: ( F_d = frac{1}{2} rho v^2 A C_d(Re) ) in the forward direction (opposite to motion)- Lift: ( F_l = frac{1}{2} rho v^2 A C_l(Re) ) in the lateral direction (direction of crosswind)So, the total force is the vector sum of these two, which can be represented as:( F_{total} = sqrt{F_d^2 + F_l^2} )But since the problem asks for the expression, not the magnitude, it's better to represent it as a vector with components ( F_d ) and ( F_l ).But the problem says "due to the crosswind," so maybe the crosswind is the cause of the lift force, and the drag is due to the vehicle's motion. So, perhaps the total aerodynamic force is just the lift force, but that doesn't make sense because drag is also an aerodynamic force.Wait, maybe the crosswind causes a sideslip, which results in both lift and drag. So, the total aerodynamic force is the vector sum of these two.But the problem gives ( F_d ) and ( F_l ) as functions of ( v ), so maybe the crosswind doesn't change the magnitude of the forces, just their direction.Alternatively, maybe the crosswind adds to the relative wind, so the effective wind speed is higher, but the problem gives ( F_d ) and ( F_l ) based on ( v ), so perhaps the crosswind only affects the direction.Wait, I'm getting stuck here. Maybe I should proceed with the assumption that the total aerodynamic force is the vector sum of drag and lift, with drag in the forward direction and lift in the lateral direction.So, the total force ( vec{F} ) would be:( vec{F} = -F_d hat{i} + F_l hat{j} )Where ( hat{i} ) is the forward direction and ( hat{j} ) is the lateral direction (direction of crosswind).But the problem says "due to the crosswind," so maybe only the lift component is considered? Or both?Wait, the crosswind is the cause of the lateral force, which is lift, and the drag is due to the vehicle's motion. So, the total aerodynamic force is the combination of both.Therefore, the expression for the total aerodynamic force is:( vec{F} = frac{1}{2} rho v^2 A left( -C_d(Re) hat{i} + C_l(Re) hat{j} right) )So, that's the vector form. If they want the magnitude, it would be ( sqrt{F_d^2 + F_l^2} ), but since they asked for the expression, probably the vector form is acceptable.Okay, that seems reasonable. So, part 1 is done.Now, part 2: Determine the conditions under which the vehicle will remain stable and not tip over due to the aerodynamic forces. Consider moments about the center of gravity and ground reaction forces at the tires. Assume tipping occurs when the sum of moments about one of the tires becomes zero.Hmm, stability in crosswind. So, when a vehicle is subjected to a crosswind, it can experience a lateral force which may cause it to tip over if the moment caused by this force is enough to lift one of the wheels off the ground.The vehicle's stability depends on the moments created by the aerodynamic forces about the center of gravity and the ground reaction forces.So, to find the condition for stability, we need to ensure that the moments caused by the aerodynamic forces do not cause the vehicle to tip over.First, let's model the vehicle. The vehicle has a center of gravity at height ( h ) above the ground and distance ( d ) from the front axle. The vehicle's length is ( L ), so the distance from the CG to the rear axle is ( L - d ).The aerodynamic force ( F_l ) acts at the center of pressure, which is typically around the center of gravity for many vehicles, but for simplicity, let's assume it acts at the CG. So, the lift force ( F_l ) acts at the CG, which is at height ( h ).Wait, but the lift force is a lateral force, so it acts horizontally. So, the moment caused by ( F_l ) about the ground will tend to rotate the vehicle around the point where the force is applied.Wait, no, the moment is about the center of gravity. Wait, the problem says "moments about the center of gravity." So, the moments caused by the aerodynamic forces and the ground reaction forces.Wait, the ground reaction forces are the normal forces at the tires. So, when the vehicle is subjected to a lateral force, the normal forces at the front and rear axles will shift, creating moments about the CG.Wait, perhaps it's better to consider the moments about the center of gravity. The aerodynamic force ( F_l ) acts at the CG, so it doesn't create a moment about the CG. Wait, no, because the force is applied at the CG, so the moment arm is zero. Hmm, that can't be right.Wait, maybe the aerodynamic force is applied at a different point, not the CG. Typically, the center of pressure is different from the CG. For a vehicle, the center of pressure is often around the midpoint of the vehicle, but it can vary.But the problem doesn't specify, so maybe we can assume that the aerodynamic force acts at the CG. If that's the case, then the moment about the CG due to ( F_l ) is zero. That doesn't make sense because then there would be no moment causing tipping.Alternatively, maybe the aerodynamic force acts at a different point, say, the midpoint of the vehicle's height or something. But without specific information, it's hard to say.Wait, perhaps the problem is considering the moment due to the ground reaction forces shifting. When a lateral force is applied, the normal forces at the front and rear axles change, creating a moment about the CG.So, let's think about it. The vehicle has two axles: front and rear. The distance between them is ( L ). The CG is at distance ( d ) from the front axle. The height of the CG is ( h ).When a lateral force ( F_l ) is applied, it causes a shift in the normal forces at the front and rear axles. The sum of the moments about the CG due to these normal forces and the aerodynamic force must be zero for equilibrium.Wait, but the problem says "the sum of the moments about one of the tires becomes zero." So, tipping occurs when the moment about one tire becomes zero, meaning that the normal force at that tire becomes zero.So, to find the condition for stability, we need to ensure that the moment about the front or rear tire does not cause the normal force at that tire to become zero.Let's model this.First, let's consider the forces acting on the vehicle:1. Aerodynamic lift force ( F_l ) acting laterally at the center of pressure. Let's assume it acts at the CG for simplicity, so at height ( h ) and distance ( d ) from the front axle.2. Normal forces at the front and rear axles: ( N_f ) and ( N_r ).3. The vehicle's weight ( W = mg ) acting downward at the CG.In the lateral direction, the sum of forces must be zero for equilibrium:( F_l = N_f cdot mu ) or something? Wait, no, in the lateral direction, the sum of forces is ( F_l = N_f cdot mu ) if considering friction, but since we're considering tipping, maybe friction isn't the issue here.Wait, no, in the lateral direction, the sum of forces is ( F_l = N_f cdot mu ) if the vehicle is on the verge of skidding, but tipping is a different condition.Wait, maybe I should consider the moments about one of the tires. Let's say we consider moments about the front tire.The moment due to the aerodynamic force ( F_l ) about the front tire is ( F_l cdot h ) (since the force is lateral and the height is ( h )).The moment due to the vehicle's weight ( W ) about the front tire is ( W cdot d ) (since the CG is at distance ( d ) from the front axle).The moment due to the normal force at the rear axle ( N_r ) about the front tire is ( N_r cdot L ) (since the rear axle is ( L ) distance from the front axle).For equilibrium, the sum of moments about the front tire must be zero:( F_l cdot h - W cdot d + N_r cdot L = 0 )But when the vehicle is about to tip, the normal force at the rear axle ( N_r ) becomes zero. So, setting ( N_r = 0 ):( F_l cdot h - W cdot d = 0 )Therefore, the condition for tipping about the front tire is:( F_l cdot h = W cdot d )Similarly, considering moments about the rear tire, the condition for tipping about the rear tire would be:( F_l cdot h = W cdot (L - d) )But which one is more critical? It depends on which condition is met first.So, the vehicle will tip over when either ( F_l cdot h = W cdot d ) or ( F_l cdot h = W cdot (L - d) ), whichever occurs first.Therefore, the condition for stability is that the aerodynamic lift force ( F_l ) must satisfy:( F_l < frac{W cdot min(d, L - d)}{h} )So, the vehicle remains stable if the lift force is less than ( frac{W cdot min(d, L - d)}{h} ).But let's write this more formally.The moment due to the aerodynamic force about the front tire is ( M_f = F_l cdot h ).The moment due to the vehicle's weight about the front tire is ( M_{gf} = W cdot d ).For the vehicle to not tip over the front tire, the moment due to ( F_l ) must not exceed the moment due to the weight:( F_l cdot h leq W cdot d )Similarly, for the rear tire:( F_l cdot h leq W cdot (L - d) )Therefore, the vehicle remains stable if:( F_l leq frac{W cdot d}{h} ) and ( F_l leq frac{W cdot (L - d)}{h} )Which can be written as:( F_l leq frac{W}{h} cdot min(d, L - d) )So, the condition is that the lift force must be less than or equal to ( frac{W}{h} cdot min(d, L - d) ).But let's express this in terms of the given variables.We know that ( F_l = frac{1}{2} rho v^2 A C_l(Re) ).So, substituting:( frac{1}{2} rho v^2 A C_l(Re) leq frac{W}{h} cdot min(d, L - d) )But ( W = mg ), so:( frac{1}{2} rho v^2 A C_l(Re) leq frac{mg}{h} cdot min(d, L - d) )Therefore, the condition for stability is:( frac{1}{2} rho v^2 A C_l(Re) leq frac{mg}{h} cdot min(d, L - d) )So, that's the condition under which the vehicle will remain stable.But let me double-check the moments. When considering moments about the front tire, the aerodynamic force ( F_l ) acts at the CG, which is at height ( h ) and distance ( d ) from the front axle. The moment due to ( F_l ) is ( F_l cdot h ) because the force is lateral and the moment arm is the height ( h ). The moment due to the weight ( W ) is ( W cdot d ) because the CG is ( d ) away from the front axle. The normal force at the rear axle ( N_r ) creates a moment ( N_r cdot L ) about the front axle.So, sum of moments about front axle:( F_l cdot h - W cdot d + N_r cdot L = 0 )When tipping occurs, ( N_r = 0 ), so:( F_l cdot h = W cdot d )Similarly, for the rear axle:Sum of moments about rear axle:( F_l cdot h - W cdot (L - d) + N_f cdot L = 0 )When tipping occurs, ( N_f = 0 ), so:( F_l cdot h = W cdot (L - d) )Therefore, the vehicle will tip over when either ( F_l cdot h = W cdot d ) or ( F_l cdot h = W cdot (L - d) ), whichever is smaller.So, the critical condition is when ( F_l ) reaches ( frac{W}{h} cdot min(d, L - d) ).Therefore, the condition for stability is:( F_l leq frac{W}{h} cdot min(d, L - d) )Substituting ( F_l ) and ( W ):( frac{1}{2} rho v^2 A C_l(Re) leq frac{mg}{h} cdot min(d, L - d) )So, that's the condition.But wait, in the problem statement, it's mentioned that the crosswind is blowing at speed ( w ). So, does the crosswind speed ( w ) affect the aerodynamic forces? Because in our derivation, we only considered ( F_l ) as a function of ( v ). But in reality, the crosswind speed ( w ) would affect the relative wind speed and hence the aerodynamic forces.Wait, this is a crucial point. The problem mentions a crosswind speed ( w ), but in our previous analysis, we only considered the vehicle's speed ( v ) for calculating ( F_l ). So, perhaps we need to consider the relative wind speed, which is a combination of ( v ) and ( w ).So, the relative wind speed ( v_{rel} ) is the vector sum of ( v ) and ( w ). Therefore, ( v_{rel} = sqrt{v^2 + w^2} ). But the problem gives ( F_d ) and ( F_l ) as functions of ( v ), not ( v_{rel} ). So, maybe the crosswind doesn't affect the Reynolds number, but only the direction of the force.Alternatively, perhaps the crosswind causes a sideslip angle, which affects the lift and drag coefficients. But the problem states that ( C_d ) and ( C_l ) are functions of ( Re ), which is based on ( v ). So, maybe the crosswind doesn't change the coefficients, but only the direction of the force.Wait, but the crosswind would change the effective angle of attack, which would change ( C_l ) and ( C_d ). However, the problem says ( C_d ) and ( C_l ) are functions of ( Re ), which is based on ( v ). So, perhaps the crosswind doesn't affect the coefficients, but only the direction of the force.This is a bit conflicting. If the crosswind affects the angle of attack, then ( C_l ) and ( C_d ) would change, but the problem says they are functions of ( Re ), which is based on ( v ). So, maybe we can assume that the crosswind doesn't change the coefficients, but only the direction of the force.Therefore, the lift force ( F_l ) is still calculated based on ( v ), but it's acting in the lateral direction due to the crosswind.But wait, in reality, the crosswind would cause a sideslip, which would increase the lift force. So, perhaps the lift force should be based on the relative wind speed ( v_{rel} ).But the problem gives ( F_l ) as a function of ( v ), so maybe we have to stick with that.Alternatively, maybe the crosswind adds to the lateral force, so the total lift force is ( F_l = frac{1}{2} rho w^2 A C_l(Re_w) ), where ( Re_w = frac{rho w L}{mu} ). But the problem doesn't specify this, so perhaps we can't assume that.This is a bit confusing. Maybe the problem assumes that the crosswind induces a lateral force ( F_l ) which is given by the same formula as the lift force due to the vehicle's motion, but in the lateral direction.Alternatively, perhaps the crosswind causes a lateral force ( F_l = frac{1}{2} rho w^2 A C_l(Re_w) ), but since the problem doesn't specify, maybe we have to assume that the lift force is due to the crosswind and is given by ( F_l = frac{1}{2} rho w^2 A C_l(Re_w) ).But the problem says "due to the crosswind," so maybe the lift force is due to the crosswind, and the drag is due to the vehicle's motion. So, the total aerodynamic force is the vector sum of both.But in our previous analysis, we considered ( F_l ) as a function of ( v ). So, perhaps the crosswind doesn't affect the lift coefficient, but only the direction.Wait, maybe the crosswind causes a lateral force which is a function of ( w ), so ( F_l = frac{1}{2} rho w^2 A C_l(Re_w) ), where ( Re_w = frac{rho w L}{mu} ).But the problem doesn't specify this, so maybe we have to stick with the given formulas, which are functions of ( v ).This is a bit of a dilemma. Since the problem states that ( C_d ) and ( C_l ) are functions of ( Re = frac{rho v L}{mu} ), which is based on ( v ), not ( w ), perhaps the crosswind doesn't affect the coefficients, but only the direction of the force.Therefore, the lift force ( F_l ) is still given by ( frac{1}{2} rho v^2 A C_l(Re) ), but it's acting in the lateral direction due to the crosswind.So, in that case, our previous analysis holds, and the condition for stability is:( frac{1}{2} rho v^2 A C_l(Re) leq frac{mg}{h} cdot min(d, L - d) )But wait, the crosswind speed ( w ) isn't appearing in this condition. That seems odd because a stronger crosswind should make the vehicle more likely to tip over.So, perhaps I made a mistake in assuming that ( F_l ) is based on ( v ). Maybe the crosswind causes a lateral force which is a function of ( w ), so ( F_l = frac{1}{2} rho w^2 A C_l(Re_w) ), where ( Re_w = frac{rho w L}{mu} ).But the problem doesn't specify this, so maybe we have to assume that the crosswind induces a lateral force which is a function of ( w ), but the problem gives ( F_l ) as a function of ( v ). So, perhaps the crosswind is considered as part of the relative wind, but the problem simplifies it by giving ( F_l ) as a function of ( v ).Alternatively, maybe the crosswind speed ( w ) is the cause of the lateral force, so ( F_l = frac{1}{2} rho w^2 A C_l(Re_w) ), but since the problem doesn't specify, maybe we have to leave it as ( F_l ) as given.Wait, the problem says "due to the crosswind," so perhaps the lift force is due to the crosswind, and the drag is due to the vehicle's motion. So, the total aerodynamic force is the vector sum of both.But in the stability condition, we only need to consider the lateral force, because the drag force is in the forward direction and doesn't contribute to lateral moments.Wait, no, the drag force acts opposite to the vehicle's motion, which is forward, so it doesn't create a lateral moment. The lift force, acting laterally, creates a moment about the CG.Wait, but earlier I thought that the lift force acts at the CG, so it doesn't create a moment about the CG. But if it acts elsewhere, it would.Wait, perhaps I made a mistake earlier. If the lift force acts at the center of pressure, which is not at the CG, then it would create a moment about the CG.But the problem doesn't specify where the aerodynamic forces act, so maybe we have to assume they act at the CG, making the moment zero.But that can't be right because then there would be no moment causing tipping.Alternatively, maybe the lift force acts at the midpoint of the vehicle's height, so at height ( H/2 ), creating a moment about the CG.Wait, but the problem doesn't specify, so maybe we have to make an assumption.Alternatively, perhaps the lift force acts at the center of pressure, which is typically around the midpoint of the vehicle's length, but height-wise, it's around the midpoint as well.But without specific information, it's hard to say. Maybe the problem assumes that the lift force acts at the CG, so no moment is created, but that doesn't make sense for tipping.Alternatively, maybe the lift force acts at a different point, say, the midpoint of the vehicle's height, so at height ( H/2 ), creating a moment about the CG.But the problem doesn't specify, so maybe we have to proceed with the assumption that the lift force acts at the CG, making the moment zero, which would mean that the only moments come from the shift in normal forces.Wait, that makes more sense. So, when a lateral force is applied, the normal forces at the front and rear axles shift, creating moments about the CG.So, let's model this.The vehicle has a weight ( W = mg ) acting downward at the CG. The aerodynamic lift force ( F_l ) acts laterally at the CG. The normal forces ( N_f ) and ( N_r ) act upward at the front and rear axles, respectively.In the vertical direction, the sum of forces is:( N_f + N_r = W )In the lateral direction, the sum of forces is:( F_l = N_f cdot mu ) (if considering friction), but since we're considering tipping, maybe friction isn't the issue here.Wait, no, tipping is about the moments, not the friction. So, the lateral force ( F_l ) causes a shift in the normal forces, creating moments about the CG.So, the moments about the CG due to the normal forces must balance the moment due to ( F_l ).Wait, but ( F_l ) acts at the CG, so it doesn't create a moment about the CG. Therefore, the only moments are due to the normal forces.Wait, that can't be right because then the moments would have to balance each other without any external moment, which is not the case.Wait, perhaps I'm misunderstanding. The lateral force ( F_l ) creates a moment about the ground, not about the CG. So, the moments about the ground (at the front and rear axles) must balance.Wait, let's think again.When a lateral force ( F_l ) is applied at the CG, it creates a moment about the ground. The moment about the front axle is ( F_l cdot h ), and the moment about the rear axle is ( F_l cdot h ) as well, but in the opposite direction.Wait, no, because the CG is at a distance ( d ) from the front axle and ( L - d ) from the rear axle. So, the moment about the front axle due to ( F_l ) is ( F_l cdot h ), and the moment about the rear axle is ( F_l cdot h ) as well, but the direction depends on the position.Wait, perhaps it's better to consider the moments about the front axle.The moment due to ( F_l ) about the front axle is ( F_l cdot h ) (since the force is lateral and the height is ( h )).The moment due to the normal force at the rear axle ( N_r ) about the front axle is ( N_r cdot L ).The moment due to the vehicle's weight ( W ) about the front axle is ( W cdot d ).For equilibrium, the sum of moments about the front axle must be zero:( F_l cdot h - W cdot d + N_r cdot L = 0 )Similarly, for the rear axle:( F_l cdot h - W cdot (L - d) + N_f cdot L = 0 )But when the vehicle is about to tip, one of the normal forces becomes zero. So, for tipping about the front axle, ( N_r = 0 ):( F_l cdot h = W cdot d )For tipping about the rear axle, ( N_f = 0 ):( F_l cdot h = W cdot (L - d) )Therefore, the vehicle will tip when either of these conditions is met. So, the critical condition is when ( F_l cdot h ) equals the lesser of ( W cdot d ) or ( W cdot (L - d) ).So, the condition for stability is:( F_l cdot h leq W cdot min(d, L - d) )Substituting ( F_l = frac{1}{2} rho v^2 A C_l(Re) ) and ( W = mg ):( frac{1}{2} rho v^2 A C_l(Re) cdot h leq mg cdot min(d, L - d) )Wait, no, earlier we had:( F_l cdot h = W cdot d ) or ( W cdot (L - d) )So, the condition is:( F_l leq frac{W}{h} cdot min(d, L - d) )Which is:( frac{1}{2} rho v^2 A C_l(Re) leq frac{mg}{h} cdot min(d, L - d) )So, that's the condition.But wait, in this analysis, the crosswind speed ( w ) doesn't appear. That seems odd because a stronger crosswind should increase the lateral force ( F_l ), making it more likely to tip.So, perhaps the crosswind speed ( w ) does affect ( F_l ). Maybe ( F_l ) is proportional to ( w^2 ) instead of ( v^2 ).But the problem gives ( F_l ) as a function of ( v ), so maybe we have to proceed with that.Alternatively, perhaps the crosswind causes a sideslip angle, which affects the lift coefficient ( C_l ). But the problem states that ( C_l ) is a function of ( Re ), which is based on ( v ), not ( w ).This is a bit conflicting. Maybe the problem assumes that the crosswind induces a lateral force which is given by ( F_l = frac{1}{2} rho w^2 A C_l(Re_w) ), but since the problem doesn't specify, maybe we have to stick with the given formula.Alternatively, perhaps the crosswind speed ( w ) is incorporated into the relative wind speed, so ( v_{rel} = sqrt{v^2 + w^2} ), and then ( Re = frac{rho v_{rel} L}{mu} ). But the problem gives ( Re ) as ( frac{rho v L}{mu} ), so maybe we have to assume that ( v ) is the relative wind speed.This is getting too complicated. Given the problem's constraints, I think the answer should be based on the given formulas, assuming that ( F_l ) is a function of ( v ), and the crosswind only affects the direction of the force.Therefore, the condition for stability is:( frac{1}{2} rho v^2 A C_l(Re) leq frac{mg}{h} cdot min(d, L - d) )So, that's the condition under which the vehicle remains stable.</think>

🧠:<think>Okay, so I have this problem where a social activist is looking at how inclusive hiring practices affect recidivism rates. She has data from five companies, each with their own recidivism rate and number of employees. I need to calculate two things: first, the weighted average recidivism rate across these five companies, and second, predict the overall recidivism rate if a new company is added. Let me start with the first part. The question asks for a weighted average, where the weight is the number of individuals employed at each company. Hmm, okay, so I remember that a weighted average is different from a simple average because each value is multiplied by its weight before adding them up. The formula for the weighted average should be the sum of each company's recidivism rate multiplied by the number of employees, divided by the total number of employees across all companies. Let me write that down to make sure I have it right. The formula should be:[text{Weighted Average} = frac{sum (R_i times E_i)}{sum E_i}]So, I need to compute each ( R_i times E_i ) for companies 1 through 5, add them all up, and then divide by the total number of employees.Looking at the table:Company 1: ( R_1 = 15% ), ( E_1 = 120 )Company 2: ( R_2 = 10% ), ( E_2 = 150 )Company 3: ( R_3 = 20% ), ( E_3 = 100 )Company 4: ( R_4 = 12% ), ( E_4 = 130 )Company 5: ( R_5 = 18% ), ( E_5 = 110 )First, I'll calculate each ( R_i times E_i ):1. Company 1: ( 15 times 120 ). Let me compute that. 15 times 100 is 1500, and 15 times 20 is 300, so total is 1800.2. Company 2: ( 10 times 150 ). That's straightforward, 10 times 150 is 1500.3. Company 3: ( 20 times 100 ). That's 2000.4. Company 4: ( 12 times 130 ). Hmm, 12 times 100 is 1200, and 12 times 30 is 360, so total is 1560.5. Company 5: ( 18 times 110 ). Let's see, 18 times 100 is 1800, and 18 times 10 is 180, so total is 1980.Now, adding up all these products:1800 (Company 1) + 1500 (Company 2) = 33003300 + 2000 (Company 3) = 53005300 + 1560 (Company 4) = 68606860 + 1980 (Company 5) = 8840So the total sum of ( R_i times E_i ) is 8840.Next, I need the total number of employees, which is the sum of all ( E_i ):120 + 150 = 270270 + 100 = 370370 + 130 = 500500 + 110 = 610So total employees ( sum E_i = 610 ).Therefore, the weighted average recidivism rate is:[frac{8840}{610}]Let me compute that. 610 goes into 8840 how many times? Let's see:610 x 14 = 8540 (since 610 x 10 = 6100, 610 x 4 = 2440; 6100 + 2440 = 8540)Subtracting that from 8840: 8840 - 8540 = 300So, 300 / 610 is approximately 0.4918.So total is 14 + 0.4918 ≈ 14.4918%.So, approximately 14.49%.Wait, let me double-check my calculations because 14.49 seems a bit low given the rates range from 10% to 20%. Let me verify each step.First, the products:1. 15 x 120 = 1800. Correct.2. 10 x 150 = 1500. Correct.3. 20 x 100 = 2000. Correct.4. 12 x 130 = 1560. Correct.5. 18 x 110 = 1980. Correct.Total products: 1800 + 1500 = 3300; 3300 + 2000 = 5300; 5300 + 1560 = 6860; 6860 + 1980 = 8840. That's correct.Total employees: 120 + 150 = 270; 270 + 100 = 370; 370 + 130 = 500; 500 + 110 = 610. Correct.So 8840 / 610. Let me compute this division more accurately.610 x 14 = 85408840 - 8540 = 300So, 300 / 610 = 0.4918...So, 14.4918%, which is approximately 14.49%.Wait, but let me check another way. Maybe I can compute each company's contribution as a percentage of total employees and then multiply by their recidivism rate.But that would be similar. Alternatively, maybe I can compute the weighted average step by step.Alternatively, perhaps I can compute the total recidivism as 8840, which is in percentage points times employees, but actually, wait, is that correct?Wait, hold on. Recidivism rate is a percentage, so when we multiply 15% by 120 employees, does that give us 1800%? That doesn't make sense. Wait, perhaps I need to convert the percentages to decimals first?Oh, wait, that's a good point. The recidivism rate is given in percentages, so 15% is 0.15 in decimal. So perhaps I should have converted them first before multiplying.Wait, hold on, maybe I made a mistake here. Because if I don't convert the percentages to decimals, then the multiplication would be incorrect.Let me think. If I have a recidivism rate of 15%, that means 15 per 100, so 0.15 in decimal. So when calculating the weighted average, I should use 0.15 instead of 15.Therefore, my previous calculation was wrong because I used 15 instead of 0.15.Oh no, that's a big mistake. So I need to correct that.So, let me redo the calculations with the correct decimal values.First, convert each ( R_i ) from percentage to decimal:Company 1: 15% = 0.15Company 2: 10% = 0.10Company 3: 20% = 0.20Company 4: 12% = 0.12Company 5: 18% = 0.18Now, compute each ( R_i times E_i ):1. Company 1: 0.15 x 120 = 182. Company 2: 0.10 x 150 = 153. Company 3: 0.20 x 100 = 204. Company 4: 0.12 x 130 = 15.65. Company 5: 0.18 x 110 = 19.8Now, sum these up:18 + 15 = 3333 + 20 = 5353 + 15.6 = 68.668.6 + 19.8 = 88.4So, the total is 88.4.Total employees is still 610.Therefore, the weighted average recidivism rate is:88.4 / 610Let me compute that.First, 610 goes into 88.4 how many times? Since 610 is larger than 88.4, it's less than 1. So, 88.4 / 610.Let me compute this as a decimal:88.4 ÷ 610Well, 610 x 0.14 = 85.4Subtract that from 88.4: 88.4 - 85.4 = 3.0So, 3.0 / 610 ≈ 0.004918So total is approximately 0.14 + 0.004918 ≈ 0.144918Converting back to percentage, that's 14.4918%.So, approximately 14.49%.Wait, that's the same result as before, but now I see that the intermediate steps are in decimal form. So, actually, my initial calculation was correct in terms of the numerical value, but I was confused because I didn't realize that 15% is 0.15, so when multiplied by 120, it's 18, not 1800. So, the total was 88.4, not 8840, but the division by 610 still gave me the same percentage because I was effectively scaling correctly.Wait, let me explain that again. If I had kept the percentages as whole numbers, 15, 10, etc., and multiplied by employees, I would have gotten 1800, 1500, etc., but then dividing by 610 would give me 14.49, which is the same as 14.49% because 8840 / 610 is 14.49, but since the original numbers were percentages, it's actually 14.49%.Wait, no, hold on. If I keep the percentages as whole numbers, 15, 10, etc., then 15 x 120 is 1800, which is 1800 percentage points. Then, dividing by 610, I get 14.49 percentage points, which is 14.49%. So, actually, whether I convert to decimal or not, the result is the same because I'm effectively scaling appropriately.So, in both cases, whether I convert to decimal or not, the result is 14.49%. So, my initial calculation was correct, but I just confused myself by thinking about the units.Therefore, the weighted average recidivism rate is approximately 14.49%.Now, moving on to the second part. The activist wants to predict the overall recidivism rate if a new company is added. The new company employs 140 individuals, and its recidivism rate is expected to be the arithmetic mean of the existing companies' recidivism rates.First, I need to compute the arithmetic mean of the existing companies' recidivism rates. The arithmetic mean is just the sum of all ( R_i ) divided by the number of companies.So, let's compute that.The recidivism rates are: 15, 10, 20, 12, 18.Sum: 15 + 10 = 25; 25 + 20 = 45; 45 + 12 = 57; 57 + 18 = 75.Total sum is 75.Number of companies is 5.So, arithmetic mean is 75 / 5 = 15%.Therefore, the new company's recidivism rate ( R_{text{new}} ) is 15%.Now, we need to calculate the new overall recidivism rate ( R_{text{overall}} ) after including this new company.Again, this is a weighted average, where the weights are the number of employees. So, we need to include the new company's recidivism rate multiplied by its number of employees, and then divide by the new total number of employees.So, let's compute the new total sum of ( R_i times E_i ) including the new company.Previously, the total was 88.4 (in decimal terms). Wait, no, hold on. Earlier, when I converted to decimal, the total was 88.4. But actually, in the first part, I had two different interpretations: one where I kept the percentages as whole numbers and got 8840, and another where I converted to decimal and got 88.4. But both gave the same percentage result because of scaling.Wait, perhaps I should stick to one method to avoid confusion. Let me clarify.If I keep everything in percentages as whole numbers, the total sum of ( R_i times E_i ) is 8840, as I initially calculated, and total employees is 610, so 8840 / 610 = 14.49%.Alternatively, converting to decimal, each ( R_i ) is divided by 100, so 0.15, 0.10, etc., and then multiplied by ( E_i ), giving 18, 15, etc., totaling 88.4. Then, 88.4 / 610 = 0.1449, which is 14.49%.So, both methods are consistent because 8840 is 88.4 x 100, and 610 is 6.1 x 100, so 8840 / 610 = (88.4 x 100) / (6.1 x 100) = 88.4 / 6.1 ≈ 14.49.Therefore, to avoid confusion, perhaps it's better to work in decimals.So, let's proceed with decimals.Previously, the total was 88.4 (sum of ( R_i times E_i ) in decimal form). Now, adding the new company:( R_{text{new}} = 15% = 0.15 )( E_{text{new}} = 140 )So, ( R_{text{new}} times E_{text{new}} = 0.15 times 140 = 21 )Therefore, the new total sum is 88.4 + 21 = 109.4The new total number of employees is 610 + 140 = 750Therefore, the new weighted average recidivism rate is:109.4 / 750Let me compute that.First, 750 goes into 109.4 how many times? Since 750 is larger than 109.4, it's less than 1. So, let's compute 109.4 / 750.Alternatively, we can write this as:109.4 ÷ 750 = ?Let me compute this step by step.First, 750 x 0.14 = 105Subtracting that from 109.4: 109.4 - 105 = 4.4So, 4.4 / 750 ≈ 0.00586666...Therefore, total is approximately 0.14 + 0.00586666 ≈ 0.14586666...Converting back to percentage, that's approximately 14.586666...%, which is roughly 14.59%.So, approximately 14.59%.Wait, let me verify that calculation again.750 x 0.14 = 105109.4 - 105 = 4.44.4 / 750 = 0.00586666...So, total is 0.14 + 0.00586666 ≈ 0.14586666...Multiply by 100 to get percentage: 14.586666...%, which is approximately 14.59%.Alternatively, let me compute 109.4 / 750 directly.109.4 ÷ 750Well, 750 goes into 1094 (moving decimal) 1.458 times because 750 x 1.458 ≈ 1093.5, which is close to 1094.So, approximately 1.458, but since we moved the decimal two places, it's 0.1458, which is 14.58%.So, approximately 14.58%.Wait, but earlier I had 14.59%. Hmm, slight difference due to rounding.But in any case, it's approximately 14.58% or 14.59%.To be precise, let's compute 109.4 divided by 750.109.4 ÷ 750:Convert 109.4 to 1094.0 by multiplying numerator and denominator by 10: 1094 / 7500.Now, 7500 goes into 10940 how many times?7500 x 1 = 7500Subtract: 10940 - 7500 = 3440Bring down a zero: 344007500 x 4 = 30000Subtract: 34400 - 30000 = 4400Bring down a zero: 440007500 x 5 = 37500Subtract: 44000 - 37500 = 6500Bring down a zero: 650007500 x 8 = 60000Subtract: 65000 - 60000 = 5000Bring down a zero: 500007500 x 6 = 45000Subtract: 50000 - 45000 = 5000So, we have 1.458666...So, 1.458666... is the decimal, which is 0.1458666... when considering the original division (since we multiplied numerator and denominator by 10 earlier). Wait, no, actually, we had 109.4 / 750 = 0.1458666...So, 0.1458666... is approximately 0.1458666, which is 14.58666...%.So, approximately 14.59%.Therefore, the overall recidivism rate after adding the new company is approximately 14.59%.Wait, but let me think again. The new company's recidivism rate is the arithmetic mean of the existing companies' rates, which was 15%. So, adding a company with a higher recidivism rate than the current weighted average (which was 14.49%) would slightly increase the overall rate.Indeed, from 14.49% to approximately 14.59%, which is a small increase, as expected.Alternatively, let me compute it using the original method without converting to decimals, just to see if I get the same result.Original total sum of ( R_i times E_i ) was 8840 (with R as percentages). Adding the new company: ( R_{text{new}} = 15% ), ( E_{text{new}} = 140 ). So, 15 x 140 = 2100.Therefore, new total sum is 8840 + 2100 = 10940.Total employees: 610 + 140 = 750.So, 10940 / 750.Compute that:750 x 14 = 10500Subtract: 10940 - 10500 = 440440 / 750 ≈ 0.586666...So, total is 14 + 0.586666 ≈ 14.586666...%, which is approximately 14.59%.Same result as before. So, that's consistent.Therefore, the overall recidivism rate after adding the new company is approximately 14.59%.So, summarizing:1. The weighted average recidivism rate across the five companies is approximately 14.49%.2. After adding the new company, the overall recidivism rate becomes approximately 14.59%.I think that's it. I've double-checked both methods, converting to decimals and keeping as percentages, and both give the same result. So, I feel confident about these answers.Final Answer1. The weighted average recidivism rate is boxed{14.49%}.2. The overall recidivism rate after adding the new company is boxed{14.59%}.</think>