🧠:<think>Okay, so I've got these two problems here about Dr. Patel's study on an immigrant community. Let me try to work through them step by step. I'm a bit nervous because I haven't done much with logistic growth models or Gaussian distributions before, but I'll give it a shot.Starting with the first problem: calculating the population after 10 years using the logistic growth model. The formula given is:[ P(t) = frac{K}{1 + frac{K - P_0}{P_0} e^{-rt}} ]Alright, so I need to plug in the values they've given. Let me list them out:- Initial population, ( P_0 = 5000 )- Carrying capacity, ( K = 50000 )- Growth rate, ( r = 0.05 ) per year- Time, ( t = 10 ) yearsSo, substituting these into the formula:First, let me compute the numerator, which is just ( K ). That's straightforward: 50,000.Now, the denominator is a bit more involved. It's ( 1 + frac{K - P_0}{P_0} e^{-rt} ). Let me break this down.Compute ( K - P_0 ): 50,000 - 5,000 = 45,000.Then, divide that by ( P_0 ): 45,000 / 5,000 = 9.So now, the denominator becomes ( 1 + 9 e^{-rt} ).Next, I need to compute ( e^{-rt} ). Let's calculate the exponent first: ( -r t = -0.05 * 10 = -0.5 ).So, ( e^{-0.5} ). Hmm, I remember that ( e^{-0.5} ) is approximately 0.6065. Let me double-check that. Yeah, since ( e^{-1} ) is about 0.3679, so ( e^{-0.5} ) should be around 0.6065. I think that's correct.So, multiplying 9 by 0.6065: 9 * 0.6065. Let me do that. 9 * 0.6 is 5.4, and 9 * 0.0065 is approximately 0.0585. So, adding those together: 5.4 + 0.0585 = 5.4585.Therefore, the denominator is 1 + 5.4585 = 6.4585.Now, the entire population ( P(10) ) is ( 50,000 / 6.4585 ). Let me compute that.Dividing 50,000 by 6.4585. Hmm, let's see. 6.4585 goes into 50,000 how many times?First, approximate 6.4585 * 7,750 = 50,000? Wait, maybe I should do this division step by step.Alternatively, I can use a calculator approach. 50,000 divided by 6.4585.Let me compute 50,000 / 6.4585.Well, 6.4585 * 7,750 = 50,000? Let me check:6.4585 * 7,000 = 45,209.56.4585 * 750 = 4,843.875Adding those together: 45,209.5 + 4,843.875 = 50,053.375Oh, that's a bit over 50,000. So, maybe 7,750 is a bit high.Alternatively, let's see: 6.4585 * 7,740.Compute 6.4585 * 7,740.Wait, maybe it's easier to just do the division.50,000 divided by 6.4585.Let me write it as 50,000 / 6.4585 ≈ ?Well, 6.4585 is approximately 6.4585.So, 50,000 divided by 6.4585 is approximately equal to:Let me compute 50,000 / 6.4585.First, 6.4585 * 7,750 ≈ 50,053 as above, which is a bit over.So, 7,750 gives us about 50,053, which is 53 over 50,000.So, to get 50,000, we need to subtract a little bit from 7,750.How much is 53 / 6.4585 ≈ 8.2.So, 7,750 - 8.2 ≈ 7,741.8.So, approximately 7,741.8.But let me check that.6.4585 * 7,741.8 ≈ ?Compute 6.4585 * 7,741.8.Well, 6.4585 * 7,740 = ?Let me compute 6.4585 * 7,000 = 45,209.56.4585 * 740 = ?Compute 6.4585 * 700 = 4,520.956.4585 * 40 = 258.34So, 4,520.95 + 258.34 = 4,779.29So, total 45,209.5 + 4,779.29 = 49,988.79Then, 6.4585 * 1.8 = approximately 11.6253So, total is 49,988.79 + 11.6253 ≈ 49,999.4153Wow, that's really close to 50,000. So, 7,741.8 gives us approximately 49,999.4153, which is just about 50,000.So, ( P(10) ) is approximately 7,741.8.But wait, that seems low. Wait, the initial population is 5,000, and the carrying capacity is 50,000, so after 10 years, it's only 7,741? That doesn't seem right because the growth rate is 0.05, which is moderate.Wait, maybe I made a mistake in my calculations.Wait, let me double-check the denominator.Denominator is 1 + 9 * e^{-0.5}.e^{-0.5} is approximately 0.6065, so 9 * 0.6065 is approximately 5.4585.So, 1 + 5.4585 is 6.4585.So, 50,000 / 6.4585 ≈ 7,741.8.Hmm, that seems correct mathematically, but intuitively, starting from 5,000, with a growth rate of 0.05, after 10 years, is it reasonable to be around 7,700?Wait, let's think about the logistic growth model. It starts off exponentially, but as it approaches the carrying capacity, the growth slows down.Given that the carrying capacity is 50,000, which is 10 times the initial population, and the growth rate is 0.05, which isn't extremely high.Wait, let me compute the population at t=10.Alternatively, maybe I can compute it step by step.Alternatively, perhaps I can use another approach or formula.Wait, another way to write the logistic growth model is:[ P(t) = frac{K P_0 e^{rt}}{K + P_0 (e^{rt} - 1)} ]Is that equivalent? Let me check.Yes, because starting from the original formula:[ P(t) = frac{K}{1 + frac{K - P_0}{P_0} e^{-rt}} ]Multiply numerator and denominator by ( e^{rt} ):[ P(t) = frac{K e^{rt}}{e^{rt} + frac{K - P_0}{P_0}} ]Which can be rewritten as:[ P(t) = frac{K P_0 e^{rt}}{K + P_0 (e^{rt} - 1)} ]Yes, that's correct.So, let's try plugging in the numbers this way.Compute ( e^{rt} = e^{0.05 * 10} = e^{0.5} approx 1.64872 ).So, numerator: ( K P_0 e^{rt} = 50,000 * 5,000 * 1.64872 ).Wait, that seems huge. Wait, no, hold on.Wait, no, the formula is:[ P(t) = frac{K P_0 e^{rt}}{K + P_0 (e^{rt} - 1)} ]So, numerator is ( K P_0 e^{rt} ), which is 50,000 * 5,000 * 1.64872.Wait, that's 250,000,000 * 1.64872, which is 412,180,000.Denominator is ( K + P_0 (e^{rt} - 1) ).Compute ( e^{rt} - 1 = 1.64872 - 1 = 0.64872 ).So, ( P_0 (e^{rt} - 1) = 5,000 * 0.64872 = 3,243.6 ).So, denominator is ( 50,000 + 3,243.6 = 53,243.6 ).Therefore, ( P(t) = 412,180,000 / 53,243.6 ).Compute that division: 412,180,000 / 53,243.6.Let me approximate.53,243.6 * 7,740 ≈ 53,243.6 * 7,000 = 372,705,20053,243.6 * 740 ≈ 53,243.6 * 700 = 37,270,52053,243.6 * 40 = 2,129,744So, total is 372,705,200 + 37,270,520 = 409,975,720 + 2,129,744 ≈ 412,105,464So, 53,243.6 * 7,740 ≈ 412,105,464Which is very close to the numerator, 412,180,000.So, the difference is 412,180,000 - 412,105,464 ≈ 74,536.So, how much more do we need? 74,536 / 53,243.6 ≈ 1.399.So, approximately 7,740 + 1.399 ≈ 7,741.399.So, approximately 7,741.4.So, that's consistent with the previous result.So, even though it seems low, mathematically, it's correct.Wait, but let me think about the time it takes to reach carrying capacity. The logistic model has an inflection point at t = (ln((K - P0)/P0))/r.Wait, let me compute that.t_inflection = (ln((K - P0)/P0)) / rCompute (K - P0)/P0 = (50,000 - 5,000)/5,000 = 45,000 / 5,000 = 9.ln(9) ≈ 2.1972.So, t_inflection ≈ 2.1972 / 0.05 ≈ 43.94 years.So, the inflection point is around 44 years. That means that at t=10, the population is still in the early growth phase, but not yet halfway to the carrying capacity.Wait, but 7,741 is only about 15% of the carrying capacity. That seems a bit low, but considering the inflection point is at 44 years, maybe it's correct.Alternatively, let's compute the population at t=10 using another method.Wait, perhaps I can compute the population incrementally, using the logistic equation.But that might be time-consuming.Alternatively, perhaps I can use the formula:[ P(t) = frac{K}{1 + left( frac{K - P_0}{P_0} right) e^{-rt}} ]Which is the original formula.So, plugging in the numbers:K = 50,000(K - P0)/P0 = 9e^{-rt} = e^{-0.5} ≈ 0.6065So, denominator is 1 + 9 * 0.6065 ≈ 1 + 5.4585 ≈ 6.4585So, P(t) = 50,000 / 6.4585 ≈ 7,741.8So, that's consistent.Therefore, despite my initial hesitation, the calculation seems correct.So, the population after 10 years is approximately 7,741.8, which we can round to 7,742 individuals.Wait, but let me check if the formula is correctly applied.Yes, the formula is correct. So, unless I made a mistake in the arithmetic, which I don't think I did, this should be the answer.So, moving on to the second problem.The average income follows a Gaussian distribution with mean μ = 50,000 and standard deviation σ = 10,000. We need to find the probability that a randomly selected individual has an income between 40,000 and 60,000.Alright, so in a Gaussian distribution, the probability between two points can be found using the Z-scores and the standard normal distribution table.First, let's compute the Z-scores for 40,000 and 60,000.Z-score formula is:[ Z = frac{X - mu}{sigma} ]So, for X = 40,000:Z1 = (40,000 - 50,000) / 10,000 = (-10,000)/10,000 = -1For X = 60,000:Z2 = (60,000 - 50,000)/10,000 = 10,000/10,000 = 1So, we need the probability that Z is between -1 and 1.In the standard normal distribution, the probability between Z = -1 and Z = 1 is approximately 0.6827, or 68.27%.I remember that about 68% of the data lies within one standard deviation of the mean in a normal distribution.So, that should be the probability.But let me verify that.Using the Z-table, the area from Z = -1 to Z = 1 is the area from -1 to 0 plus the area from 0 to 1.Looking up Z = 1, the cumulative probability is 0.8413.Looking up Z = -1, the cumulative probability is 0.1587.So, the area between -1 and 1 is 0.8413 - 0.1587 = 0.6826, which is approximately 0.6827.So, that's correct.Therefore, the probability is approximately 68.27%.So, summarizing:1. The population after 10 years is approximately 7,742 individuals.2. The probability of an income between 40,000 and 60,000 is approximately 68.27%.Wait, just to make sure I didn't make any calculation errors.For the first problem, let me recompute the denominator:1 + (K - P0)/P0 * e^{-rt} = 1 + 9 * e^{-0.5} ≈ 1 + 9 * 0.6065 ≈ 1 + 5.4585 ≈ 6.4585So, 50,000 / 6.4585 ≈ 7,741.8, which is approximately 7,742.Yes, that seems correct.For the second problem, the Z-scores are -1 and 1, which correspond to 15.87% and 84.13% cumulative probabilities, so the difference is 68.26%, which is about 68.27%.So, I think both answers are correct.Final Answer1. The population after 10 years is boxed{7742} individuals.2. The probability of an income between 40,000 and 60,000 is boxed{0.6827}.</think>

🧠:<think>Alright, so I have this problem where a politician wants to maximize the impact of their policies across different districts. The city is divided into n districts, each with its own population p_i. They've got this effectiveness matrix E, where E_ij tells us how effective a policy in district i is on district j. The goal is to allocate a fixed budget B to these districts in a way that maximizes the total influence, which is calculated as the sum over all i and j of b_i * E_ij * p_j. And we have to do this subject to the constraint that the total budget allocated is exactly B.Hmm, okay. So, let me break this down. The total influence is a double sum: for each district i, we're allocating some budget b_i, and then for each district j, we're considering how effective that allocation is on district j, multiplied by the population of district j. So, it's like the influence is spread across all districts, both where the budget is allocated and where the effects are felt.I think this is an optimization problem. We need to maximize a function subject to a constraint. The function to maximize is the total influence, and the constraint is the total budget. So, this sounds like a linear optimization problem, maybe using Lagrange multipliers or something like that.Let me write down the problem formally. We need to maximize:Total Influence = Σ_{i=1 to n} Σ_{j=1 to n} b_i * E_ij * p_jSubject to:Σ_{i=1 to n} b_i = BAnd b_i ≥ 0, I assume, since you can't allocate negative budget.So, this is a linear program because the objective function is linear in terms of b_i, and the constraints are linear as well.To solve this, I can use the method of Lagrange multipliers. Let me set up the Lagrangian:L = Σ_{i,j} b_i E_ij p_j - λ (Σ_i b_i - B)Where λ is the Lagrange multiplier for the budget constraint.To find the maximum, we take the partial derivatives of L with respect to each b_i and set them equal to zero.So, for each i:∂L/∂b_i = Σ_j E_ij p_j - λ = 0Which implies:Σ_j E_ij p_j = λSo, for each i, the sum of E_ij p_j across all j must equal λ.Wait, but λ is the same for all i. That suggests that all the sums Σ_j E_ij p_j must be equal for each i. But that can't be right unless all rows of E multiplied by p are equal.Hmm, maybe I made a mistake here. Let me double-check.Wait, no, actually, the partial derivative with respect to b_i is Σ_j E_ij p_j - λ = 0, so each of these must equal λ. Therefore, all the Σ_j E_ij p_j must be equal. So, unless the rows of E weighted by p are all equal, which is probably not the case, this suggests that the optimal solution is to set all b_i such that the marginal influence per unit budget is equal across all districts.But wait, in linear programming, when the objective function is linear, the optimal solution occurs at the vertices of the feasible region. So, perhaps we should allocate as much as possible to the district that gives the highest marginal return.Wait, let me think again. The objective function is linear, so the gradient is constant. Therefore, the optimal solution is to allocate all the budget to the district with the highest coefficient in the objective function.So, the coefficient for each b_i is Σ_j E_ij p_j. So, the total influence contributed by each district i is b_i multiplied by Σ_j E_ij p_j. Therefore, to maximize the total influence, we should allocate the entire budget B to the district i that has the maximum value of Σ_j E_ij p_j.Is that correct? Because in linear programming, when the objective is linear, the maximum occurs at an extreme point, which in this case would be putting all the budget into the district with the highest coefficient.So, let me test this with a simple example. Suppose we have two districts, n=2. Let's say p1 and p2 are the populations, and E is a 2x2 matrix.Suppose E11=1, E12=2, E21=3, E22=4. And p1=100, p2=200.Then, for district 1, the coefficient is E11*p1 + E12*p2 = 1*100 + 2*200 = 100 + 400 = 500.For district 2, it's E21*p1 + E22*p2 = 3*100 + 4*200 = 300 + 800 = 1100.So, district 2 has a higher coefficient, so we should allocate all B to district 2. That would give a total influence of 1100*B.If we split the budget, say allocate B1 to district 1 and B2 to district 2, with B1 + B2 = B, the total influence would be 500*B1 + 1100*B2. Since 1100 > 500, any allocation to district 1 would reduce the total influence compared to putting all into district 2.Therefore, yes, the optimal strategy is to allocate all the budget to the district with the highest Σ_j E_ij p_j.So, in general, the optimal allocation is to find the district i* that maximizes Σ_j E_ij p_j, and set b_i* = B, and all other b_i = 0.Wait, but what if multiple districts have the same maximum coefficient? Then, we can allocate to any of them, or split between them, but since the objective is linear, it doesn't matter; the total influence will be the same.So, in conclusion, for Sub-problem 1, the optimal allocation is to allocate the entire budget B to the district(s) with the highest value of Σ_j E_ij p_j.Now, moving on to Sub-problem 2. The effectiveness matrix E is evolving over time according to E(t) = E0 * e^{Rt}, where E0 is the initial effectiveness matrix and R is a matrix representing the rate of change. We need to determine the long-term behavior of the influence scores as t approaches infinity and identify conditions under which the influence stabilizes, and describe the impact on the budget allocation strategy.Hmm, so E(t) is given by matrix exponential. The behavior as t→infty depends on the eigenvalues of R. If the eigenvalues of R have negative real parts, then E(t) will decay to zero. If they have positive real parts, E(t) will grow without bound. If there are eigenvalues with zero real parts, it could lead to constant or oscillatory behavior.But in this context, since E(t) is an effectiveness matrix, it's probably a non-negative matrix, and R might be such that E(t) remains non-negative. But regardless, the long-term behavior depends on the dominant eigenvalues of R.Wait, but E(t) is E0 multiplied by e^{Rt}. So, the matrix exponential e^{Rt} can be expressed in terms of its eigenvalues and eigenvectors. If R has eigenvalues λ_k with corresponding eigenvectors v_k, then e^{Rt} = Σ_k e^{λ_k t} v_k v_k^T (assuming R is diagonalizable).Therefore, as t→infty, the term with the largest real part will dominate. So, if the dominant eigenvalue of R has a positive real part, E(t) will grow exponentially. If it's negative, E(t) will decay. If it's zero, it will remain constant or grow polynomially.But in terms of influence, which is Σ_i Σ_j b_i E_ij(t) p_j, if E(t) is growing without bound, then the influence will also grow without bound, unless the budget allocation is adjusted.But in our initial problem, the budget allocation was fixed based on E0. So, if E(t) is changing, the optimal allocation might change over time.Wait, but in Sub-problem 2, after solving the allocation problem (which was based on E0), the matrix E starts evolving. So, the initial allocation was based on E0, but as time goes on, E(t) changes. We need to see what happens to the influence over time.But actually, the influence is calculated based on the current E(t). So, if E(t) is changing, the influence will change accordingly.But the question is about the long-term behavior as t→infty. So, we need to see whether the influence stabilizes or not.If E(t) stabilizes, meaning that e^{Rt} approaches a limit as t→infty, then the influence would stabilize. Otherwise, it might grow or decay indefinitely.So, when does e^{Rt} stabilize? That happens if all eigenvalues of R have negative real parts, leading e^{Rt} to decay to zero, or if R is nilpotent, but that's less likely. Alternatively, if R has eigenvalues with zero real parts, then e^{Rt} might approach a constant matrix or oscillate.But more formally, the limit of e^{Rt} as t→infty exists if and only if all eigenvalues of R have negative real parts or zero real parts with certain conditions. If any eigenvalue has a positive real part, e^{Rt} will grow without bound.So, for the influence to stabilize, we need that e^{Rt} approaches a finite limit as t→infty. That requires that all eigenvalues of R have non-positive real parts, and for those with zero real parts, they must be Jordan blocks of size 1 (i.e., R is diagonalizable for those eigenvalues).But in practice, for the influence to stabilize, we probably need that the dominant eigenvalue of R has a negative real part, so that e^{Rt} decays to zero, or that the dominant eigenvalue is zero, but with finite influence.Wait, but if R has eigenvalues with zero real parts, then e^{Rt} might not decay but could stay constant or oscillate. For example, if R is purely imaginary, e^{Rt} would be oscillatory.But in the context of effectiveness matrices, which are likely to be non-negative, having eigenvalues with positive real parts would mean that the influence could grow indefinitely, which might not be desirable.So, conditions for stabilization would be that all eigenvalues of R have non-positive real parts, and that any eigenvalues with zero real parts are semisimple (i.e., their geometric multiplicity equals their algebraic multiplicity), so that e^{Rt} doesn't blow up.In terms of the budget allocation strategy, if the influence is stabilizing, then the optimal allocation might also stabilize. However, if the influence is growing or decaying, the optimal allocation might need to be adjusted over time.But since in Sub-problem 1, the allocation was based on E0, and now E is changing, the initial allocation might not remain optimal as time goes on.So, if E(t) stabilizes, then after some time, the influence would reach a steady state, and the optimal allocation could be recalculated based on the stabilized E(t). However, if E(t) doesn't stabilize, the allocation might need to be adjusted dynamically.But the question is about the long-term behavior as t→infty. So, if E(t) stabilizes, then the influence would stabilize, and the optimal allocation would be based on the stabilized E(t). If E(t) doesn't stabilize, then the influence might not have a limit, and the budget allocation strategy would need to account for that.So, in summary, the influence will stabilize in the long term if all eigenvalues of R have non-positive real parts, and any eigenvalues with zero real parts are semisimple. Under these conditions, the effectiveness matrix E(t) will approach a limit, and the optimal budget allocation can be determined based on this limit. Otherwise, if R has eigenvalues with positive real parts, the influence will grow without bound, and the budget allocation strategy would need to adapt accordingly, possibly by continuously reallocating resources to districts whose influence is growing.Wait, but in the initial allocation, we allocated all the budget to the district with the highest Σ_j E_ij p_j. If E(t) is changing, this district might change over time. So, if E(t) stabilizes, then the optimal district would stabilize as well. If E(t) doesn't stabilize, the optimal district might change, requiring dynamic allocation.But since the problem is about the long-term behavior, we can say that if E(t) stabilizes, then the optimal allocation stabilizes, otherwise, it doesn't.So, putting it all together:For Sub-problem 1, the optimal allocation is to put all the budget into the district with the highest Σ_j E_ij p_j.For Sub-problem 2, the influence will stabilize as t→infty if all eigenvalues of R have non-positive real parts, and any eigenvalues with zero real parts are semisimple. In this case, the optimal budget allocation can be recalculated based on the stabilized E(t). If R has eigenvalues with positive real parts, the influence will grow indefinitely, and the allocation strategy would need to be adjusted dynamically.But wait, in the initial allocation, we used E0. If E(t) is changing, the optimal allocation might change. So, even if E(t) stabilizes, the initial allocation might not be optimal anymore. Therefore, the politician should monitor the effectiveness matrix over time and adjust the budget allocation accordingly.However, the question is about the impact on the budget allocation strategy. So, if E(t) stabilizes, the strategy can be recalibrated once it stabilizes. If it doesn't, the strategy needs to be more dynamic.So, in conclusion, the conditions for stabilization are related to the eigenvalues of R, and the budget allocation strategy must account for whether E(t) stabilizes or not.</think>

🧠:<think>Okay, so I have this problem about a mineral deposit shaped like an ellipsoid, and I need to find the total mass of two minerals, A and B, inside it. Let me try to break this down step by step.First, the ellipsoid has semi-principal axes of 4 meters, 3 meters, and 2 meters. That means its equation in standard form should be something like (frac{x^2}{4^2} + frac{y^2}{3^2} + frac{z^2}{2^2} leq 1). So, (frac{x^2}{16} + frac{y^2}{9} + frac{z^2}{4} leq 1). Got that.Now, for mineral A, the density is given by (rho_A(x, y, z) = k(x^2 + y^2)). Hmm, so the density depends on the x and y coordinates but not on z. Interesting. And for mineral B, the density is uniform at 5 kg/m³. So, I need to compute the total mass for each mineral.Starting with mineral A. The total mass should be the triple integral of the density function over the volume of the ellipsoid. So, mathematically, that's:[M_A = iiint_{text{ellipsoid}} rho_A(x, y, z) , dV = iiint_{text{ellipsoid}} k(x^2 + y^2) , dV]Right. So, I need to set up this integral. Since the ellipsoid is symmetric, maybe I can use a coordinate transformation to simplify the integral. Ellipsoids can be transformed into spheres using scaling, which might make the integration easier.Let me recall that if I have an ellipsoid defined by (frac{x^2}{a^2} + frac{y^2}{b^2} + frac{z^2}{c^2} leq 1), I can use the substitution (x = a u), (y = b v), (z = c w), which transforms the ellipsoid into a unit sphere (u^2 + v^2 + w^2 leq 1). The Jacobian determinant of this transformation is (abc), so the volume element (dV) becomes (abc , du , dv , dw).Applying this to our ellipsoid, (a = 4), (b = 3), (c = 2). So, substituting:(x = 4u), (y = 3v), (z = 2w).Then, (dV = 4 times 3 times 2 , du , dv , dw = 24 , du , dv , dw).Now, let's express (rho_A) in terms of (u, v, w):[rho_A = k(x^2 + y^2) = k((4u)^2 + (3v)^2) = k(16u^2 + 9v^2)]So, the integral becomes:[M_A = iiint_{u^2 + v^2 + w^2 leq 1} k(16u^2 + 9v^2) times 24 , du , dv , dw]Simplify that:[M_A = 24k iiint_{u^2 + v^2 + w^2 leq 1} (16u^2 + 9v^2) , du , dv , dw]Now, since the integrand is a function of u and v only, and the region is symmetric in all variables, we can separate the integrals. Let me write this as:[M_A = 24k left[ 16 iiint u^2 , du , dv , dw + 9 iiint v^2 , du , dv , dw right]]But wait, actually, since the integrals over the sphere can be separated into radial and angular parts, or perhaps even better, using spherical coordinates. But maybe it's easier to note that the integrals of (u^2) and (v^2) over the sphere are equal due to symmetry.In a unit sphere, the integral of (u^2) over the entire sphere is the same as the integral of (v^2) or (w^2). Let me recall that the integral of (x^2) over the unit ball in 3D is (frac{4pi}{15}). Wait, let me verify that.The integral of (x^2) over the unit sphere can be computed using spherical coordinates. Let me recall that in spherical coordinates, (x = r sintheta cosphi), (y = r sintheta sinphi), (z = r costheta), and the volume element is (r^2 sintheta , dr , dtheta , dphi).So, the integral of (x^2) over the unit sphere is:[int_0^{2pi} int_0^pi int_0^1 (r^2 sin^2theta cos^2phi) r^2 sintheta , dr , dtheta , dphi]Wait, no. Wait, (x^2 = r^2 sin^2theta cos^2phi), so the integral becomes:[int_0^{2pi} cos^2phi , dphi int_0^pi sin^3theta , dtheta int_0^1 r^4 , dr]Compute each integral separately.First, (int_0^{2pi} cos^2phi , dphi). Using the identity (cos^2phi = frac{1 + cos 2phi}{2}), so:[int_0^{2pi} frac{1 + cos 2phi}{2} , dphi = frac{1}{2} left[ int_0^{2pi} 1 , dphi + int_0^{2pi} cos 2phi , dphi right] = frac{1}{2} [2pi + 0] = pi]Second, (int_0^pi sin^3theta , dtheta). Let me use substitution. Let (u = costheta), then (du = -sintheta dtheta). So,[int sin^3theta dtheta = int (1 - cos^2theta) sintheta dtheta = -int (1 - u^2) du = -left( u - frac{u^3}{3} right) + C = -costheta + frac{cos^3theta}{3} + C]Evaluating from 0 to (pi):At (pi): (-cospi + frac{cos^3pi}{3} = -(-1) + frac{(-1)^3}{3} = 1 - frac{1}{3} = frac{2}{3})At 0: (-cos0 + frac{cos^3 0}{3} = -1 + frac{1}{3} = -frac{2}{3})So, the integral is (frac{2}{3} - (-frac{2}{3}) = frac{4}{3})Third, (int_0^1 r^4 dr = frac{1}{5})So, putting it all together:[int x^2 dV = pi times frac{4}{3} times frac{1}{5} = frac{4pi}{15}]Similarly, the integral of (y^2) and (z^2) over the unit sphere are also (frac{4pi}{15}). So, in our case, the integral of (u^2) over the unit sphere is (frac{4pi}{15}), same for (v^2).Therefore, going back to (M_A):[M_A = 24k [16 times frac{4pi}{15} + 9 times frac{4pi}{15}] = 24k left( frac{64pi}{15} + frac{36pi}{15} right ) = 24k times frac{100pi}{15}]Simplify:[24k times frac{100pi}{15} = 24k times frac{20pi}{3} = 8k times 20pi = 160kpi]Wait, hold on. Let me check that calculation again.Wait, 24 multiplied by (100π/15). 24 divided by 15 is 8/5, so 24*(100π/15) = (24/15)*100π = (8/5)*100π = 160π. So, yes, 160kπ.Wait, but hold on, 24*(100π/15) = (24/15)*100π = (8/5)*100π = 160π. So, 160kπ.Therefore, the total mass of mineral A is (160kpi) kg.Wait, but let me think again. The integral was over the unit sphere, but the original ellipsoid was scaled by 4, 3, 2. So, the substitution was x=4u, y=3v, z=2w, and dV=24 du dv dw. So, the integral in terms of u, v, w is over the unit sphere, which is correct.So, the integral of u² over the unit sphere is 4π/15, so 16*(4π/15) is 64π/15, and 9*(4π/15) is 36π/15. Adding those gives 100π/15, which is 20π/3. Then, 24k*(20π/3) is 24*(20π/3)*k = 8*20π*k = 160πk. So, yes, that seems correct.So, part 1 answer is 160πk kg.Now, moving on to part 2: total mass of mineral B.Since mineral B has a uniform density of 5 kg/m³, the total mass is just the density multiplied by the volume of the ellipsoid.So, first, let me compute the volume of the ellipsoid. The formula for the volume of an ellipsoid is (frac{4}{3}pi abc), where a, b, c are the semi-principal axes.Here, a=4, b=3, c=2. So,[V = frac{4}{3}pi times 4 times 3 times 2 = frac{4}{3}pi times 24 = 32pi , text{m}^3]So, the volume is 32π cubic meters.Therefore, the total mass of mineral B is:[M_B = rho_B times V = 5 times 32pi = 160pi , text{kg}]Wait, that's interesting. Both masses are 160π times something. For A, it's 160πk, and for B, it's 160π. So, unless k=1, they are different.But in the problem statement, it just says "the density of mineral A varies within the ellipsoid and is given by the function ρ_A(x, y, z) = k(x² + y²) where k is a constant." So, k is just a constant, which is given, but we don't know its value. So, in the answer for part 1, we have to leave it in terms of k.So, to recap:1. Total mass of mineral A is (160kpi) kg.2. Total mass of mineral B is (160pi) kg.Wait, that seems too similar. Let me double-check my calculations.For mineral A:- The integral was transformed into the unit sphere, with substitution scaling x, y, z.- The density became k(16u² + 9v²).- The integral over the unit sphere of 16u² + 9v² is 16*(4π/15) + 9*(4π/15) = (64 + 36)*(4π/15) = 100*(4π/15) = 400π/15 = 80π/3.Wait, hold on, earlier I thought it was 100π/15, but 16*(4π/15) is 64π/15, 9*(4π/15) is 36π/15, so total is 100π/15, which simplifies to 20π/3. Then, 24k*(20π/3) = 24k*(20π)/3 = 8k*20π = 160kπ. So, that's correct.For mineral B:- Volume of ellipsoid is (4/3)πabc = (4/3)π*4*3*2 = (4/3)*24π = 32π.- Multiply by density 5 kg/m³: 5*32π = 160π kg.So, both masses are 160π times something, but for A, it's multiplied by k, and for B, it's just 160π.So, unless k is given, we can't compute a numerical value for A, but the problem doesn't specify k, so we have to leave it in terms of k.Therefore, the answers are:1. (160kpi) kg2. (160pi) kgWait, but let me think again. Did I make a mistake in the integral for mineral A?Because the integral of u² over the unit sphere is 4π/15, so 16u² would be 16*(4π/15) = 64π/15, and 9v² would be 9*(4π/15) = 36π/15. Adding them gives 100π/15 = 20π/3. Then, multiplied by 24k gives 24k*(20π/3) = 160kπ. So, yes, that seems correct.Alternatively, maybe I can compute the integral without changing variables, just to verify.The original integral for M_A is:[M_A = iiint_{frac{x^2}{16} + frac{y^2}{9} + frac{z^2}{4} leq 1} k(x^2 + y^2) , dx , dy , dz]We can use a coordinate transformation to make this integral easier. Let me set u = x/4, v = y/3, w = z/2. Then, the ellipsoid becomes the unit sphere u² + v² + w² ≤ 1.The Jacobian determinant is 4*3*2 = 24, so dV = 24 du dv dw.Expressing x = 4u, y = 3v, so x² + y² = 16u² + 9v².Therefore, the integral becomes:[M_A = iiint_{u^2 + v^2 + w^2 leq 1} k(16u² + 9v²) * 24 du dv dw]Which is the same as before. So, same result.Alternatively, maybe using spherical coordinates in the original ellipsoid.But that might complicate things because the ellipsoid isn't a sphere. So, scaling it to a sphere is the right approach.Therefore, I think my calculations are correct.So, summarizing:1. The total mass of mineral A is (160kpi) kg.2. The total mass of mineral B is (160pi) kg.I think that's it.Final Answer1. The total mass of mineral A is boxed{160pi k} kilograms.2. The total mass of mineral B is boxed{160pi} kilograms.</think>

🧠:<think>Alright, so I've got these two math problems to solve, and they're both pretty interesting. Let me start with the first one about complex numbers. The first message is encoded using the sequence ( z_n = (1 + i)^n + (1 - i)^n ). I need to find the general form of ( z_n ) and then compute ( z_{10} ). Hmm, okay. I remember that complex numbers can be expressed in polar form, which might make exponentiation easier. Let me try that.First, let's convert ( 1 + i ) and ( 1 - i ) into polar form. The modulus of ( 1 + i ) is ( sqrt{1^2 + 1^2} = sqrt{2} ). The argument is ( pi/4 ) because it's in the first quadrant and both real and imaginary parts are equal. Similarly, ( 1 - i ) has the same modulus ( sqrt{2} ) but its argument is ( -pi/4 ) since it's in the fourth quadrant.So, ( 1 + i = sqrt{2} left( cos frac{pi}{4} + i sin frac{pi}{4} right) ) and ( 1 - i = sqrt{2} left( cos left( -frac{pi}{4} right) + i sin left( -frac{pi}{4} right) right) ).Now, raising these to the power of ( n ) using De Moivre's theorem:( (1 + i)^n = (sqrt{2})^n left( cos frac{npi}{4} + i sin frac{npi}{4} right) )( (1 - i)^n = (sqrt{2})^n left( cos left( -frac{npi}{4} right) + i sin left( -frac{npi}{4} right) right) )Since cosine is even and sine is odd, this simplifies to:( (1 - i)^n = (sqrt{2})^n left( cos frac{npi}{4} - i sin frac{npi}{4} right) )Adding these two together:( z_n = (1 + i)^n + (1 - i)^n = (sqrt{2})^n left[ left( cos frac{npi}{4} + i sin frac{npi}{4} right) + left( cos frac{npi}{4} - i sin frac{npi}{4} right) right] )The imaginary parts cancel out, leaving:( z_n = 2 (sqrt{2})^n cos frac{npi}{4} )So, that's the general form. Now, let's compute ( z_{10} ).First, ( (sqrt{2})^{10} = (2^{1/2})^{10} = 2^{5} = 32 ).Next, ( cos frac{10pi}{4} = cos frac{5pi}{2} ). Wait, ( frac{5pi}{2} ) is equivalent to ( 2pi + frac{pi}{2} ), so cosine of that is 0. Because cosine has a period of ( 2pi ), so ( cos frac{5pi}{2} = cos frac{pi}{2} = 0 ).Therefore, ( z_{10} = 2 times 32 times 0 = 0 ).Hmm, that seems straightforward. Let me double-check. Maybe I made a mistake with the angle.Wait, ( frac{10pi}{4} = frac{5pi}{2} ). Yes, that's correct. And ( cos frac{5pi}{2} ) is indeed 0 because it's at the point (0,1) on the unit circle, where cosine is 0. So, yeah, ( z_{10} = 0 ).Okay, moving on to the second problem. It's about a matrix transformation. The matrix ( A ) is a 2x2 matrix with trace 6, and it satisfies the equation ( A^3 - 6A^2 + 11A - 6I = 0 ). I need to find the possible eigenvalues of ( A ).I remember that for a matrix equation like this, the eigenvalues must satisfy the same polynomial equation. So, if ( lambda ) is an eigenvalue of ( A ), then it must satisfy ( lambda^3 - 6lambda^2 + 11lambda - 6 = 0 ).Let me factor this polynomial. Let's try possible rational roots using the Rational Root Theorem. The possible roots are factors of 6 over factors of 1, so ±1, ±2, ±3, ±6.Testing ( lambda = 1 ):( 1 - 6 + 11 - 6 = 0 ). Yes, 0. So, ( lambda = 1 ) is a root.Now, perform polynomial division or factor it out. Let's divide ( lambda^3 - 6lambda^2 + 11lambda - 6 ) by ( lambda - 1 ).Using synthetic division:1 | 1 -6 11 -6Bring down 1.Multiply by 1: 1Add to next coefficient: -6 + 1 = -5Multiply by 1: -5Add to next coefficient: 11 + (-5) = 6Multiply by 1: 6Add to last coefficient: -6 + 6 = 0So, the polynomial factors as ( (lambda - 1)(lambda^2 - 5lambda + 6) ).Now, factor ( lambda^2 - 5lambda + 6 ):Looking for two numbers that multiply to 6 and add to -5. Those are -2 and -3.So, it factors further into ( (lambda - 1)(lambda - 2)(lambda - 3) ).Therefore, the eigenvalues must be 1, 2, or 3.But wait, ( A ) is a 2x2 matrix, so it can have at most two eigenvalues (counting multiplicities). Also, the trace of ( A ) is 6, which is the sum of its eigenvalues.So, the possible pairs of eigenvalues (considering they can be repeated) are:1. 1 and 5: But 5 isn't a root, so no.2. 2 and 4: 4 isn't a root, so no.3. 3 and 3: 3 is a root, and 3 + 3 = 6. That works.4. 1 and 5: Again, 5 isn't a root.Wait, hold on. The eigenvalues must be among 1, 2, 3. So, possible pairs:- 1 and 5: 5 is not a root.- 2 and 4: 4 is not a root.- 3 and 3: Both are roots.- 1 and 2: 1 + 2 = 3 ≠ 6.- 1 and 3: 1 + 3 = 4 ≠ 6.- 2 and 3: 2 + 3 = 5 ≠ 6.Wait, so the only possible pair is 3 and 3 because 3 + 3 = 6. So, the eigenvalues must both be 3.But hold on, is that the only possibility? Let me think again.The characteristic equation of the matrix ( A ) is ( lambda^2 - text{tr}(A)lambda + det(A) = 0 ). Since the trace is 6, it's ( lambda^2 - 6lambda + det(A) = 0 ).But the minimal polynomial of ( A ) divides the given polynomial ( (lambda - 1)(lambda - 2)(lambda - 3) ). So, the minimal polynomial could be any product of these factors.But since ( A ) is 2x2, its minimal polynomial can be at most degree 2. So, possible minimal polynomials are:- ( (lambda - 1)(lambda - 2) )- ( (lambda - 1)(lambda - 3) )- ( (lambda - 2)(lambda - 3) )- ( (lambda - 1)^2 )- ( (lambda - 2)^2 )- ( (lambda - 3)^2 )But the minimal polynomial must divide the given polynomial ( (lambda - 1)(lambda - 2)(lambda - 3) ). So, the minimal polynomial can be any product of distinct linear factors or a square of a single linear factor.But in our case, since the trace is 6, which is the sum of eigenvalues, and the determinant is the product.If the minimal polynomial is ( (lambda - 3)^2 ), then the matrix is a Jordan block with eigenvalue 3, repeated. So, the trace would be 6, and determinant would be 9.Alternatively, if the minimal polynomial is ( (lambda - 2)(lambda - 3) ), then the eigenvalues are 2 and 3, which sum to 5, not 6. So that's not possible.Similarly, if minimal polynomial is ( (lambda - 1)(lambda - 3) ), eigenvalues 1 and 3, sum to 4. Not 6.If minimal polynomial is ( (lambda - 1)(lambda - 2) ), sum is 3. Not 6.If minimal polynomial is ( (lambda - 1)^2 ), trace is 2, which is not 6.If minimal polynomial is ( (lambda - 2)^2 ), trace is 4, not 6.Therefore, the only possibility is that the minimal polynomial is ( (lambda - 3)^2 ), meaning both eigenvalues are 3.Hence, the possible eigenvalues are both 3.Wait, but hold on. Another thought: the given polynomial equation is ( A^3 - 6A^2 + 11A - 6I = 0 ). So, the minimal polynomial must divide this polynomial. So, if the minimal polynomial is ( (lambda - 3)^2 ), does it divide the given polynomial?Let me check. Let me perform polynomial division.Divide ( lambda^3 - 6lambda^2 + 11lambda - 6 ) by ( (lambda - 3)^2 = lambda^2 - 6lambda + 9 ).Using polynomial long division:Divide ( lambda^3 - 6lambda^2 + 11lambda - 6 ) by ( lambda^2 - 6lambda + 9 ).First term: ( lambda^3 / lambda^2 = lambda ).Multiply ( lambda ) by ( lambda^2 - 6lambda + 9 ): ( lambda^3 - 6lambda^2 + 9lambda ).Subtract from the original polynomial:( (lambda^3 - 6lambda^2 + 11lambda - 6) - (lambda^3 - 6lambda^2 + 9lambda) = 0 + 0 + 2lambda - 6 ).Now, the remainder is ( 2lambda - 6 ). Since the degree of the remainder (1) is less than the degree of the divisor (2), we stop here.So, the division gives:( lambda^3 - 6lambda^2 + 11lambda - 6 = (lambda)(lambda^2 - 6lambda + 9) + (2lambda - 6) ).But since ( A ) satisfies ( A^3 - 6A^2 + 11A - 6I = 0 ), substituting ( A ) into the equation, we get:( A^3 - 6A^2 + 11A - 6I = 0 ).If the minimal polynomial were ( (lambda - 3)^2 ), then ( (A - 3I)^2 = 0 ), which implies ( A^2 - 6A + 9I = 0 ). Then, ( A^2 = 6A - 9I ).Let me compute ( A^3 ):( A^3 = A cdot A^2 = A(6A - 9I) = 6A^2 - 9A = 6(6A - 9I) - 9A = 36A - 54I - 9A = 27A - 54I ).Now, substitute into the original equation:( A^3 - 6A^2 + 11A - 6I = (27A - 54I) - 6(6A - 9I) + 11A - 6I )Compute each term:- ( 27A - 54I )- ( -6(6A - 9I) = -36A + 54I )- ( 11A )- ( -6I )Combine all terms:( 27A - 54I - 36A + 54I + 11A - 6I )Combine like terms:- ( 27A - 36A + 11A = (27 - 36 + 11)A = 2A )- ( -54I + 54I - 6I = (-54 + 54 - 6)I = -6I )So, overall: ( 2A - 6I ). But the original equation is supposed to be 0. So, ( 2A - 6I = 0 ) implies ( A = 3I ).Wait, so if ( A ) satisfies ( (A - 3I)^2 = 0 ), then ( A ) must be ( 3I ). Because if ( (A - 3I)^2 = 0 ), then ( A ) is a Jordan block, but if ( A ) is diagonalizable, then ( A = 3I ). However, if ( A ) is not diagonalizable, it could be a Jordan block. But in our case, since ( A ) satisfies ( 2A - 6I = 0 ), which implies ( A = 3I ). So, actually, ( A ) must be the scalar matrix ( 3I ).Therefore, the eigenvalues are both 3.But wait, hold on. If ( A = 3I ), then the minimal polynomial is ( lambda - 3 ), not ( (lambda - 3)^2 ). So, perhaps my earlier assumption was wrong.Wait, let's re-examine. If ( A ) satisfies ( (A - 3I)^2 = 0 ), but when we substitute back into the original equation, we get ( 2A - 6I = 0 ), which implies ( A = 3I ). So, actually, ( A ) must be diagonalizable, and hence, the minimal polynomial is ( lambda - 3 ), not ( (lambda - 3)^2 ).Therefore, the only eigenvalue is 3, with multiplicity 2.So, regardless, the eigenvalues are both 3.Alternatively, another approach: since the trace is 6, and the eigenvalues must satisfy the equation ( lambda^3 - 6lambda^2 + 11lambda - 6 = 0 ). So, possible eigenvalues are 1, 2, 3.But since the trace is 6, the sum of eigenvalues is 6. So, possible combinations:- 3 and 3: 3 + 3 = 6.- 2 and 4: 4 is not a root.- 1 and 5: 5 is not a root.So, only 3 and 3.Hence, the eigenvalues must both be 3.Therefore, the possible eigenvalues are 3 and 3.I think that's solid. So, the eigenvalues are both 3.Final Answer1. The general form of ( z_n ) is ( 2 (sqrt{2})^n cos frac{npi}{4} ) and ( z_{10} = boxed{0} ).2. The possible eigenvalues of matrix ( A ) are both ( boxed{3} ).</think>