🧠:<think>Okay, so I have this problem where a writer is analyzing archaeological sites. They've visited 15 sites over 10 years and recorded their coordinates. The first task is to find the centroid of these points. Hmm, I remember that the centroid is like the average position of all the points. So, for a set of points on a Cartesian plane, the centroid's x-coordinate is the average of all the x-coordinates, and similarly, the y-coordinate is the average of all the y-coordinates. Let me write that down. If I have points (x₁, y₁), (x₂, y₂), ..., (x₁₅, y₁₅), then the centroid (C_x, C_y) would be:C_x = (x₁ + x₂ + ... + x₁₅) / 15C_y = (y₁ + y₂ + ... + y₁₅) / 15So, I just need to sum up all the x-values and divide by 15, and do the same for the y-values. That should give me the centroid.Now, moving on to the second part. The writer thinks the sites aren't random and might follow a specific pattern. They want to fit a two-dimensional Gaussian distribution to the data. I need to write the probability density function (pdf) for a bivariate Gaussian distribution and explain how it can be used to assess the pattern.I recall that the bivariate Gaussian pdf is given by:f(x, y) = (1 / (2π√|Σ|)) * exp(-0.5 * (x - μ)^T Σ⁻¹ (x - μ))Where Σ is the covariance matrix, μ is the mean vector, and |Σ| is the determinant of Σ. Let me make sure I remember that correctly. The exponent part is the Mahalanobis distance squared, which measures how far a point is from the mean in terms of the covariance structure.So, to explain how the writer can use this pdf: once they fit the Gaussian distribution to their data, they can calculate the likelihood of each data point under this distribution. If the sites follow the hypothesized pattern, the points should have higher probabilities (or likelihoods) under this pdf. Conversely, if the points are randomly distributed, the likelihoods might be lower or more spread out.They could also visualize the data by plotting the contours of the Gaussian distribution and see if the points cluster around the mean in a way that aligns with the expected pattern. Additionally, statistical tests like the Kolmogorov-Smirnov test or the Anderson-Darling test could be used to formally test whether the data fits the Gaussian distribution, which would support the hypothesis of a specific pattern.Wait, but the problem doesn't provide specific coordinates or the covariance matrix and mean vector. So, for the first part, I can't compute the exact centroid without the data. Maybe the question is more about the method rather than the actual calculation. Similarly, for the pdf, it's about writing the formula and explaining its use rather than computing it.So, summarizing my thoughts:1. The centroid is calculated by averaging all x and y coordinates.2. The bivariate Gaussian pdf is as I wrote above, and it can be used to assess how well the data fits a specific distribution, thereby testing the hypothesis of a pattern.I think that's the gist of it. I should make sure I didn't miss any steps or formulas. Let me double-check the centroid formula. Yes, it's the mean of the coordinates. For the Gaussian, the formula includes the normalization factor, the exponential of the negative Mahalanobis distance, and the determinant of the covariance matrix in the denominator. That seems right.I might also consider that the covariance matrix captures the spread and correlation between x and y. So, if the sites are clustered in a certain way, the covariance matrix will reflect that, and the pdf will show higher density in those areas.Alright, I think I'm ready to write up the answers formally.</think>

🧠:<think>Okay, so I have this problem about analyzing the time complexity of a recursive function that processes a binary tree. The function processes each node, and each node has exactly two children except the leaves. The time complexity for processing each node is O(log n), where n is the number of nodes in the subtree rooted at that node. I need to derive an expression for the overall time complexity in terms of the depth d.Hmm, let's break this down. First, I should probably figure out how many nodes there are at each level of the tree. Since it's a binary tree with depth d, each level has 2^i nodes, where i is the level number starting from 0 at the root. So, the root is level 0 with 1 node, level 1 has 2 nodes, level 2 has 4 nodes, and so on up to level d, which has 2^d nodes. But wait, actually, if the tree has depth d, that means the height is d, so the number of levels is d+1. So, the total number of nodes in the tree is 2^(d+1) - 1. But I don't know if I need the total number of nodes right away.Each node has a processing time of O(log n), where n is the number of nodes in its subtree. So, for each node, I need to figure out the size of its subtree and then take the logarithm of that.Starting from the root, the subtree size is the entire tree, which is 2^(d+1) - 1. So, the processing time for the root is O(log(2^(d+1) - 1)). That simplifies to O(d+1), since log(2^x) is x. So, O(d) time for the root.Then, moving to the next level, each of the two children of the root has a subtree size of 2^d - 1. Wait, is that right? Let me think. If the root is level 0, then each child is the root of a subtree of depth d-1. So, the number of nodes in each subtree would be 2^(d) - 1. So, the processing time for each of these two nodes is O(log(2^d - 1)) which is O(d). So, each of these two nodes contributes O(d) time.Moving on to level 2, each node there has a subtree of depth d-2, so the number of nodes is 2^(d-1) - 1. The log of that is O(d-1). So, each node at level 2 contributes O(d-1) time.Wait, so in general, for a node at level i, its subtree has depth d - i, so the number of nodes is 2^(d - i + 1) - 1. Therefore, the log of that is O(d - i + 1). So, each node at level i contributes O(d - i + 1) time.But how many nodes are at each level i? It's 2^i nodes. So, the total time contributed by level i is 2^i * O(d - i + 1).Therefore, the total time complexity T(d) is the sum from i=0 to d of 2^i * O(d - i + 1). So, T(d) = sum_{i=0}^d 2^i * (d - i + 1).Hmm, that seems a bit complicated. Maybe I can find a closed-form expression for this sum.Let me write it out:T(d) = sum_{i=0}^d 2^i * (d - i + 1)Let me make a substitution: let j = d - i. Then when i=0, j=d; when i=d, j=0. So, the sum becomes:sum_{j=0}^d 2^{d - j} * (j + 1)Which is the same as:2^d * sum_{j=0}^d (j + 1)/2^jHmm, okay, so I need to compute sum_{j=0}^d (j + 1)/2^j.I remember that sum_{j=0}^infty (j + 1)/2^j is a known series. Let me recall. The sum of (j + 1) x^j from j=0 to infinity is 1/(1 - x)^2, for |x| < 1. So, substituting x = 1/2, we get sum_{j=0}^infty (j + 1)/2^j = 1/(1 - 1/2)^2 = 1/(1/2)^2 = 4.But in our case, the sum is up to j=d, not infinity. So, the finite sum will be less than 4. But for large d, it approaches 4. So, maybe for the purposes of asymptotic analysis, we can approximate the sum as 4, but I need to be careful.Wait, but actually, for the purposes of big O notation, constants don't matter, so if the sum is bounded by a constant, then the entire expression would be O(2^d). But let me check.Wait, no, because 2^d multiplied by a constant is still O(2^d). But let me see:If sum_{j=0}^d (j + 1)/2^j <= 4, then T(d) <= 2^d * 4 = 4*2^d = O(2^d). But is that tight?Wait, but let's compute the exact sum for finite d. Maybe I can find a closed-form expression.Let me denote S = sum_{j=0}^d (j + 1)/2^j.Let me consider S = sum_{j=0}^d (j + 1)/2^j.Let me write S as sum_{j=0}^d j/2^j + sum_{j=0}^d 1/2^j.I know that sum_{j=0}^infty 1/2^j = 2, and sum_{j=0}^infty j/2^j = 2.But again, for finite d, it's a bit different.Wait, let's compute sum_{j=0}^d 1/2^j. That's a finite geometric series. It equals 2 - 1/2^d.Similarly, sum_{j=0}^d j/2^j. There's a formula for this. Let me recall.Let me denote G = sum_{j=0}^d j x^j.I know that G = x(1 - (d + 1)x^d + d x^{d + 1}) / (1 - x)^2.So, substituting x = 1/2, we get:G = (1/2)(1 - (d + 1)(1/2)^d + d (1/2)^{d + 1}) / (1 - 1/2)^2Simplify denominator: (1/2)^2 = 1/4, so 1/(1/4) = 4.So, G = (1/2)(1 - (d + 1)/2^d + d / 2^{d + 1}) * 4Simplify:G = 2 [1 - (d + 1)/2^d + d / 2^{d + 1}]Simplify the terms inside:= 2 [1 - (d + 1)/2^d + d/(2*2^d)]= 2 [1 - (d + 1)/2^d + d/(2^{d + 1})]= 2 [1 - (2(d + 1) - d)/2^{d + 1}]Wait, maybe it's easier to compute each term:= 2 [1 - (d + 1)/2^d + d / 2^{d + 1}]= 2*1 - 2*(d + 1)/2^d + 2*d / 2^{d + 1}= 2 - (d + 1)/2^{d - 1} + d / 2^dHmm, not sure if that's helpful.Alternatively, maybe I can just compute S = sum_{j=0}^d (j + 1)/2^j = sum_{j=0}^d j/2^j + sum_{j=0}^d 1/2^j = G + (2 - 1/2^d).So, G is approximately 2 for large d, but for finite d, it's slightly less.But perhaps for the purposes of asymptotic analysis, the leading term is 2^d multiplied by a constant, so T(d) is O(2^d).But wait, let me think again. The sum S is bounded by 4, so T(d) = 2^d * S <= 4*2^d, so T(d) is O(2^d). But is it tight? Because each term in the sum is O(1), multiplied by 2^i, which is increasing exponentially.Wait, maybe I should approach this differently. Instead of trying to compute the exact sum, perhaps I can bound it.Note that for each level i, the processing time per node is O(d - i + 1), and there are 2^i nodes. So, the total time is sum_{i=0}^d 2^i * O(d - i + 1).Let me consider the maximum term in the sum. The term when i is small, say i=0, is 2^0 * O(d) = O(d). When i is around d/2, the term is 2^{d/2} * O(d/2). When i is near d, the term is 2^d * O(1).So, the largest term is when i is near d, which is 2^d * O(1). So, the sum is dominated by the last few terms, each contributing O(2^d). Therefore, the total time complexity is O(2^d).Wait, but let me check for small d. For example, if d=1, the tree has root and two leaves. The root has O(log 3) time, which is O(1). Each leaf has O(log 1) = O(0) time. So, total time is O(1). But 2^1 = 2, so O(2^1) would be O(2), which is consistent.If d=2, the tree has root, two children, and four leaves. Root: O(log 7) = O(3). Each child: O(log 3) = O(2). Each leaf: O(log 1) = 0. So, total time is O(3 + 2*2) = O(7). 2^2=4, so O(4) is less than 7, but 7 is O(4) since constants don't matter. Wait, no, 7 is O(4) because 7 <= c*4 for some c>0.Wait, actually, 7 is O(4), but 4 is O(7) as well. So, in big O terms, they are equivalent because big O is about asymptotic upper bounds, not tight bounds. So, O(7) is the same as O(4). So, for d=2, the total time is O(4), which is O(2^2).Similarly, for d=3, the total time would be O(8), which is O(2^3). So, it seems that for each d, the total time is O(2^d). Therefore, the overall time complexity is O(2^d).But wait, let me think again. The sum T(d) = sum_{i=0}^d 2^i * (d - i + 1). Let's see for d=3:i=0: 1*4=4i=1: 2*3=6i=2: 4*2=8i=3: 8*1=8Total: 4+6+8+8=26. 2^3=8, 26 is O(8). Wait, 26 is 3.25*8, so it's O(8). Similarly, for d=4:i=0:1*5=5i=1:2*4=8i=2:4*3=12i=3:8*2=16i=4:16*1=16Total:5+8+12+16+16=57. 2^4=16, 57 is O(16). 57 is about 3.56*16.So, it seems that T(d) is roughly around 3*2^{d+1} or something. Wait, 26 is about 3*8 + 2, 57 is about 3*16 + 9. Hmm, maybe it's approaching 4*2^d as d increases.Wait, because as d increases, the sum S approaches 4, so T(d) approaches 4*2^d. So, for large d, T(d) is approximately 4*2^d, which is O(2^d).Therefore, the overall time complexity is O(2^d).Wait, but let me confirm with another approach. Maybe using recurrence relations.Let me denote T(d) as the total time for a tree of depth d. The root contributes O(log n), where n is the total number of nodes, which is 2^{d+1} -1, so O(d). Then, the root has two children, each of which is the root of a subtree of depth d-1. So, the total time is O(d) + 2*T(d-1).So, the recurrence is T(d) = O(d) + 2*T(d-1), with T(0) = O(1).This is a linear recurrence. Let's solve it.First, write it as T(d) = 2*T(d-1) + c*d, where c is a constant.This is a nonhomogeneous linear recurrence. The homogeneous solution is T_h(d) = A*2^d.For the particular solution, since the nonhomogeneous term is c*d, we can try a particular solution of the form T_p(d) = a*d + b.Substitute into the recurrence:a*d + b = 2*(a*(d-1) + b) + c*dExpand RHS: 2a*d - 2a + 2b + c*dSo, LHS: a*d + bEquate coefficients:a*d + b = (2a + c)*d + (-2a + 2b)So,a = 2a + c => -a = c => a = -cAnd,b = -2a + 2b => b = -2a + 2b => 0 = -2a + b => b = 2aSince a = -c, then b = 2*(-c) = -2c.Therefore, the particular solution is T_p(d) = -c*d - 2c.So, the general solution is T(d) = T_h(d) + T_p(d) = A*2^d - c*d - 2c.Now, apply the initial condition T(0) = O(1). Let's say T(0) = k, where k is a constant.So, when d=0:k = A*2^0 - c*0 - 2c => k = A - 2c => A = k + 2c.Therefore, the solution is:T(d) = (k + 2c)*2^d - c*d - 2c.Simplify:T(d) = k*2^d + 2c*2^d - c*d - 2c= k*2^d + c*(2^{d+1} - d - 2)But since k and c are constants, we can write T(d) = O(2^d) + O(2^{d+1}) + O(d) + O(1). The dominant term is O(2^{d+1}), which is O(2^d). So, T(d) is O(2^d).Therefore, both approaches confirm that the overall time complexity is O(2^d).Now, moving on to part 2. The tree is now a k-ary tree, where each non-leaf node has k children, and k > 2. I need to reformulate the time complexity and analyze how increasing k affects it compared to the binary tree.So, similar to part 1, but now each node has k children instead of 2. Let's follow the same approach.First, the number of nodes at each level i is k^i. The total number of nodes in the tree is (k^{d+1} - 1)/(k - 1). But again, for each node, the processing time is O(log n), where n is the number of nodes in its subtree.For a node at level i, its subtree has depth d - i, so the number of nodes in its subtree is (k^{d - i + 1} - 1)/(k - 1). Therefore, the log of that is O(log(k^{d - i + 1})) = O((d - i + 1) log k). But since log k is a constant, it's O(d - i + 1).So, each node at level i contributes O(d - i + 1) time. There are k^i nodes at level i. Therefore, the total time contributed by level i is k^i * O(d - i + 1).Thus, the total time complexity T_k(d) is sum_{i=0}^d k^i * (d - i + 1).Again, let's try to find a closed-form expression or bound this sum.Let me make a substitution: let j = d - i. Then, when i=0, j=d; when i=d, j=0. So, the sum becomes:sum_{j=0}^d k^{d - j} * (j + 1)= k^d * sum_{j=0}^d (j + 1)/k^jSo, T_k(d) = k^d * sum_{j=0}^d (j + 1)/k^jAgain, for large d, the sum approaches sum_{j=0}^infty (j + 1)/k^j. Let's compute that.We know that sum_{j=0}^infty (j + 1)x^j = 1/(1 - x)^2 for |x| < 1. So, substituting x = 1/k, we get sum_{j=0}^infty (j + 1)/k^j = 1/(1 - 1/k)^2 = k^2/(k - 1)^2.Therefore, for large d, T_k(d) ≈ k^d * k^2/(k - 1)^2 = k^{d + 2}/(k - 1)^2.But for finite d, the sum is less than that. However, for asymptotic analysis, the dominant term is k^d multiplied by a constant, so T_k(d) is O(k^d).But let's check with the recurrence approach.Let T_k(d) be the total time for a k-ary tree of depth d. The root contributes O(log n), where n is the total number of nodes, which is (k^{d+1} - 1)/(k - 1). So, log n is O(d log k), which is O(d) since log k is a constant.Then, the root has k children, each of which is the root of a subtree of depth d-1. So, the recurrence is T_k(d) = O(d) + k*T_k(d-1), with T_k(0) = O(1).This is similar to the binary case but with k instead of 2.Solving this recurrence:T_k(d) = k*T_k(d-1) + c*d, with T_k(0) = k.The homogeneous solution is T_h(d) = A*k^d.For the particular solution, assume T_p(d) = a*d + b.Substitute:a*d + b = k*(a*(d - 1) + b) + c*dExpand RHS: k*a*d - k*a + k*b + c*dEquate coefficients:a*d + b = (k*a + c)*d + (-k*a + k*b)So,a = k*a + c => a(1 - k) = c => a = c/(1 - k)And,b = -k*a + k*b => b = -k*a + k*b => 0 = -k*a + (k - 1)*b => b = (k/(k - 1))*aSubstitute a = c/(1 - k):b = (k/(k - 1))*(c/(1 - k)) = -k c/(k - 1)^2Therefore, the particular solution is T_p(d) = (c/(1 - k))*d - (k c)/(k - 1)^2.So, the general solution is T_k(d) = A*k^d + (c/(1 - k))*d - (k c)/(k - 1)^2.Apply initial condition T_k(0) = k:k = A*k^0 + (c/(1 - k))*0 - (k c)/(k - 1)^2=> k = A - (k c)/(k - 1)^2=> A = k + (k c)/(k - 1)^2Therefore, T_k(d) = [k + (k c)/(k - 1)^2]*k^d + (c/(1 - k))*d - (k c)/(k - 1)^2Simplify:= k^{d + 1} + (k c k^d)/(k - 1)^2 + (c/(1 - k))*d - (k c)/(k - 1)^2But since c is a constant, the dominant term is k^{d + 1}, which is O(k^d). So, T_k(d) is O(k^d).Comparing this to the binary tree case, where T(d) was O(2^d), increasing k increases the base of the exponent, making the time complexity grow much faster. For example, for k=3, it's O(3^d), which is significantly larger than O(2^d) for large d.Therefore, increasing k from 2 to a larger value results in a higher time complexity, as the overall complexity becomes O(k^d), which grows exponentially with k.</think>

🧠:<think>Alright, so I have this problem about Dr. Elena Martinez assessing the impact of a new airport on local ecosystems. There are two parts to this problem. Let me tackle them one by one.Starting with the first part: I need to find the biodiversity index ( B(t) ) given the differential equation ( frac{dB}{dt} = -0.02B(t) + 0.5 sin(t) ) with the initial condition ( B(0) = 100 ). Hmm, okay. This looks like a linear first-order differential equation. I remember that for such equations, we can use an integrating factor to solve them.The general form of a linear differential equation is ( frac{dy}{dt} + P(t)y = Q(t) ). Comparing this with the given equation, let me rewrite it:( frac{dB}{dt} + 0.02B(t) = 0.5 sin(t) ).So here, ( P(t) = 0.02 ) and ( Q(t) = 0.5 sin(t) ).The integrating factor ( mu(t) ) is given by ( e^{int P(t) dt} ). So in this case, ( mu(t) = e^{int 0.02 dt} = e^{0.02t} ).Multiplying both sides of the differential equation by the integrating factor:( e^{0.02t} frac{dB}{dt} + 0.02 e^{0.02t} B(t) = 0.5 e^{0.02t} sin(t) ).The left side of this equation is the derivative of ( B(t) e^{0.02t} ) with respect to t. So, we can write:( frac{d}{dt} [B(t) e^{0.02t}] = 0.5 e^{0.02t} sin(t) ).Now, to find ( B(t) ), I need to integrate both sides with respect to t:( B(t) e^{0.02t} = int 0.5 e^{0.02t} sin(t) dt + C ).Okay, so I need to compute this integral. Hmm, integrating ( e^{at} sin(bt) ) dt is a standard integral, right? The formula is:( int e^{at} sin(bt) dt = frac{e^{at}}{a^2 + b^2} (a sin(bt) - b cos(bt)) + C ).Let me verify that. Yes, I think that's correct. So in this case, ( a = 0.02 ) and ( b = 1 ). So plugging these into the formula:( int e^{0.02t} sin(t) dt = frac{e^{0.02t}}{(0.02)^2 + 1^2} (0.02 sin(t) - 1 cos(t)) + C ).Simplify the denominator:( (0.02)^2 = 0.0004 ), so ( 0.0004 + 1 = 1.0004 ).So, the integral becomes:( frac{e^{0.02t}}{1.0004} (0.02 sin(t) - cos(t)) + C ).But remember, the integral we have is multiplied by 0.5:( int 0.5 e^{0.02t} sin(t) dt = 0.5 times frac{e^{0.02t}}{1.0004} (0.02 sin(t) - cos(t)) + C ).So, putting it back into our equation:( B(t) e^{0.02t} = 0.5 times frac{e^{0.02t}}{1.0004} (0.02 sin(t) - cos(t)) + C ).Now, let's simplify this expression. First, multiply through:( B(t) e^{0.02t} = frac{0.5 e^{0.02t}}{1.0004} (0.02 sin(t) - cos(t)) + C ).We can factor out ( e^{0.02t} ):( B(t) e^{0.02t} = frac{0.5 e^{0.02t}}{1.0004} (0.02 sin(t) - cos(t)) + C ).To solve for ( B(t) ), divide both sides by ( e^{0.02t} ):( B(t) = frac{0.5}{1.0004} (0.02 sin(t) - cos(t)) + C e^{-0.02t} ).Simplify the constants:( frac{0.5}{1.0004} ) is approximately ( 0.4998 ). Let me calculate that more precisely:( 1.0004 ) is approximately ( 1 + 0.0004 ), so ( frac{1}{1.0004} approx 1 - 0.0004 ) using the approximation ( frac{1}{1+x} approx 1 - x ) for small x.Thus, ( frac{0.5}{1.0004} approx 0.5 times (1 - 0.0004) = 0.5 - 0.0002 = 0.4998 ).So, ( B(t) approx 0.4998 (0.02 sin(t) - cos(t)) + C e^{-0.02t} ).But maybe it's better to keep it as ( frac{0.5}{1.0004} ) for exactness. Alternatively, since 1.0004 is very close to 1, we can approximate it as 1 for simplicity, but let's see.Wait, actually, 0.5 / 1.0004 is approximately 0.4998, which is very close to 0.5. So, perhaps for simplicity, we can write it as approximately 0.5, but I think it's better to keep it exact.So, let's write it as:( B(t) = frac{0.5}{1.0004} (0.02 sin(t) - cos(t)) + C e^{-0.02t} ).Now, apply the initial condition ( B(0) = 100 ). Let's plug t = 0 into the equation:( B(0) = frac{0.5}{1.0004} (0.02 sin(0) - cos(0)) + C e^{0} ).Simplify:( 100 = frac{0.5}{1.0004} (0 - 1) + C ).So,( 100 = -frac{0.5}{1.0004} + C ).Therefore,( C = 100 + frac{0.5}{1.0004} ).Calculate ( frac{0.5}{1.0004} ):Since 1.0004 is approximately 1.0004, so 0.5 / 1.0004 ≈ 0.4998.Thus,( C ≈ 100 + 0.4998 = 100.4998 ).So, approximately, C is 100.5.Therefore, the solution is:( B(t) ≈ frac{0.5}{1.0004} (0.02 sin(t) - cos(t)) + 100.5 e^{-0.02t} ).But let me express this more neatly. Let's compute ( frac{0.5}{1.0004} times 0.02 ) and ( frac{0.5}{1.0004} times (-1) ).First term: ( 0.5 / 1.0004 ≈ 0.4998 ). So,( 0.4998 times 0.02 ≈ 0.009996 ).Second term: ( 0.4998 times (-1) ≈ -0.4998 ).So, approximately,( B(t) ≈ 0.009996 sin(t) - 0.4998 cos(t) + 100.5 e^{-0.02t} ).But since 0.009996 is approximately 0.01, and 0.4998 is approximately 0.5, we can write:( B(t) ≈ 0.01 sin(t) - 0.5 cos(t) + 100.5 e^{-0.02t} ).But to be precise, let's carry out the exact calculations without approximating.So, ( frac{0.5}{1.0004} = frac{0.5}{1 + 0.0004} ). Using the exact division:( 0.5 ÷ 1.0004 ).Let me compute this:1.0004 × 0.4998 = 0.4998 + 0.4998×0.0004 ≈ 0.4998 + 0.00019992 ≈ 0.5.So, 0.4998 × 1.0004 ≈ 0.5, so 0.5 / 1.0004 ≈ 0.4998.Therefore, we can write:( B(t) = 0.4998 times (0.02 sin t - cos t) + 100.4998 e^{-0.02t} ).Simplify:( B(t) = (0.4998 × 0.02) sin t - 0.4998 cos t + 100.4998 e^{-0.02t} ).Calculating 0.4998 × 0.02:0.4998 × 0.02 = 0.009996 ≈ 0.01.So,( B(t) ≈ 0.01 sin t - 0.5 cos t + 100.5 e^{-0.02t} ).But let's check the exact value:0.4998 × 0.02 = 0.009996, which is 0.01 approximately.Similarly, 0.4998 is approximately 0.5.So, the solution is approximately:( B(t) ≈ 0.01 sin t - 0.5 cos t + 100.5 e^{-0.02t} ).But to be precise, let's carry the exact constants:( B(t) = frac{0.5}{1.0004} (0.02 sin t - cos t) + left(100 + frac{0.5}{1.0004}right) e^{-0.02t} ).Alternatively, we can factor out 0.5 / 1.0004:( B(t) = frac{0.5}{1.0004} (0.02 sin t - cos t) + left(100 + frac{0.5}{1.0004}right) e^{-0.02t} ).But perhaps it's better to leave it in terms of exact fractions. Alternatively, since 1.0004 is very close to 1, we can approximate the solution as:( B(t) ≈ 0.5 (0.02 sin t - cos t) + 100.5 e^{-0.02t} ).Which simplifies to:( B(t) ≈ 0.01 sin t - 0.5 cos t + 100.5 e^{-0.02t} ).But let me check if this satisfies the initial condition.At t=0:( B(0) ≈ 0.01 × 0 - 0.5 × 1 + 100.5 × 1 = -0.5 + 100.5 = 100 ).Yes, that works. So, the approximation is good.Therefore, the biodiversity index ( B(t) ) is approximately:( B(t) = 0.01 sin t - 0.5 cos t + 100.5 e^{-0.02t} ).But let me write it with more precise coefficients:( B(t) = frac{0.5 times 0.02}{1.0004} sin t - frac{0.5}{1.0004} cos t + left(100 + frac{0.5}{1.0004}right) e^{-0.02t} ).Calculating each term:1. ( frac{0.5 × 0.02}{1.0004} = frac{0.01}{1.0004} ≈ 0.009996 ).2. ( frac{0.5}{1.0004} ≈ 0.4998 ).3. ( 100 + frac{0.5}{1.0004} ≈ 100.4998 ).So, writing it out:( B(t) ≈ 0.009996 sin t - 0.4998 cos t + 100.4998 e^{-0.02t} ).Rounding to four decimal places:( B(t) ≈ 0.0100 sin t - 0.5000 cos t + 100.5000 e^{-0.02t} ).So, that's the solution for part 1.Moving on to part 2: Calculate the total increase in the carbon footprint due to deforestation over 3 years. The forest sequesters carbon at 10 metric tons per hectare per year. The entire 50 hectares is deforested immediately at the start of the project.So, deforestation leads to a loss of carbon sequestration. The rate is 10 metric tons per hectare per year. So, for 50 hectares, the annual sequestration would be 50 × 10 = 500 metric tons per year.But since the forest is deforested, this sequestration stops. So, the increase in carbon footprint is the amount of carbon that would have been sequestered if the forest remained.Therefore, over 3 years, the total increase in carbon footprint is 500 metric tons/year × 3 years = 1500 metric tons.Wait, but hold on. Is it that straightforward? Let me think.Carbon footprint is usually the amount of carbon emitted. Deforestation leads to loss of carbon sequestration, which effectively increases the net carbon in the atmosphere. So, yes, the increase in carbon footprint would be the amount that would have been sequestered.But sometimes, deforestation also releases stored carbon. So, depending on the context, the total carbon impact might include both the loss of future sequestration and the release of stored carbon. However, the problem states that the forest sequesters carbon at 10 metric tons per hectare per year, and asks for the increase in carbon footprint due to deforestation over the 3-year period, assuming the rate remains constant.So, I think it's just the loss of sequestration, not the release of stored carbon. Because the problem doesn't mention the stored carbon, only the sequestration rate.Therefore, the total increase in carbon footprint is 50 hectares × 10 metric tons/ha/year × 3 years = 1500 metric tons.So, the answer is 1500 metric tons.But let me double-check. The construction lasts 3 years, and the deforestation happens immediately. So, for each year, the forest would have sequestered 500 metric tons, but since it's deforested, that doesn't happen. So, over 3 years, 3 × 500 = 1500 metric tons increase in carbon footprint.Yes, that seems correct.So, summarizing:1. The biodiversity index ( B(t) ) is approximately ( 0.01 sin t - 0.5 cos t + 100.5 e^{-0.02t} ).2. The total increase in carbon footprint is 1500 metric tons.Final Answer1. The biodiversity index is ( boxed{B(t) = 0.01 sin(t) - 0.5 cos(t) + 100.5 e^{-0.02t}} ).2. The total increase in the carbon footprint is ( boxed{1500} ) metric tons.</think>

🧠:<think>Alright, so I've got this problem about a perpetual note with a varying interest rate. The judge needs to figure out the average annual interest rate over the first 8 years and then estimate the total interest accrued using continuous compounding. Hmm, okay, let's break this down step by step.First, let's understand the interest rate function. It's given by:[ r(t) = frac{1}{2} sinleft(frac{pi t}{2}right) + frac{1}{4} cosleft(frac{pi t}{4}right) + 0.03 ]So, this is a function that varies with time t in years. The principal is 1,000,000, but for the first part, we just need the average rate, so maybe the principal won't matter yet.Problem 1: Calculate the average annual interest rate over the first 8 years.Alright, average value of a function over an interval [a, b] is given by:[ text{Average} = frac{1}{b - a} int_{a}^{b} r(t) dt ]In this case, a is 0 and b is 8. So we need to compute:[ text{Average rate} = frac{1}{8} int_{0}^{8} left[ frac{1}{2} sinleft(frac{pi t}{2}right) + frac{1}{4} cosleft(frac{pi t}{4}right) + 0.03 right] dt ]Okay, so let's split this integral into three separate integrals for each term:1. Integral of (1/2) sin(πt/2) dt from 0 to 82. Integral of (1/4) cos(πt/4) dt from 0 to 83. Integral of 0.03 dt from 0 to 8Let me compute each of these one by one.First integral:[ int_{0}^{8} frac{1}{2} sinleft(frac{pi t}{2}right) dt ]Let me make a substitution. Let u = πt/2, so du = π/2 dt, which means dt = (2/π) du.Changing the limits, when t=0, u=0; when t=8, u= (π*8)/2 = 4π.So the integral becomes:[ frac{1}{2} times frac{2}{pi} int_{0}^{4pi} sin(u) du ]Simplify the constants:1/2 * 2/π = 1/πSo:[ frac{1}{pi} int_{0}^{4pi} sin(u) du ]The integral of sin(u) is -cos(u), so:[ frac{1}{pi} [ -cos(u) ]_{0}^{4pi} = frac{1}{pi} [ -cos(4π) + cos(0) ] ]We know that cos(4π) is 1, and cos(0) is also 1. So:[ frac{1}{pi} [ -1 + 1 ] = frac{1}{pi} (0) = 0 ]Interesting, the first integral is zero. That makes sense because sine is a periodic function with period 2π, and over an integer multiple of periods, the integral cancels out. Since 4π is two full periods, the integral is zero.Okay, moving on to the second integral:[ int_{0}^{8} frac{1}{4} cosleft(frac{pi t}{4}right) dt ]Again, substitution. Let u = πt/4, so du = π/4 dt, which means dt = (4/π) du.Limits: t=0, u=0; t=8, u= (π*8)/4 = 2π.So the integral becomes:[ frac{1}{4} times frac{4}{pi} int_{0}^{2pi} cos(u) du ]Simplify constants:1/4 * 4/π = 1/πSo:[ frac{1}{pi} int_{0}^{2pi} cos(u) du ]The integral of cos(u) is sin(u), so:[ frac{1}{pi} [ sin(u) ]_{0}^{2pi} = frac{1}{pi} [ sin(2π) - sin(0) ] ]Both sin(2π) and sin(0) are zero, so this integral is also zero.Hmm, so both oscillatory terms integrate to zero over their periods. That leaves us with the third integral:[ int_{0}^{8} 0.03 dt ]That's straightforward. The integral of a constant is the constant times the interval length.So:0.03 * (8 - 0) = 0.03 * 8 = 0.24Therefore, putting it all together, the average rate is:(1/8) * (0 + 0 + 0.24) = (1/8) * 0.24Compute that:0.24 divided by 8 is 0.03.So the average annual interest rate over the first 8 years is 0.03, or 3%.Wait, that seems surprisingly simple. Both the sine and cosine terms averaged out to zero over their periods, leaving only the constant term. So regardless of the oscillations, the average rate is just the constant part, 0.03. That makes sense because the sine and cosine functions have average values of zero over their periods.So, problem 1 is done. The average rate is 3%.Problem 2: Using the average interest rate found, estimate the total amount of interest accrued over 8 years, assuming continuous compounding.Alright, continuous compounding formula is:[ A = P e^{rt} ]Where:- A is the amount of money accumulated after t years, including interest.- P is the principal amount (1,000,000)- r is the annual interest rate (decimal)- t is the time the money is invested for (8 years)But wait, the question asks for the total amount of interest accrued, not the total amount. So we need to subtract the principal.So, total interest I is:[ I = A - P = P (e^{rt} - 1) ]Given that the average rate r is 0.03, and t is 8.So plug in the numbers:I = 1,000,000 * (e^{0.03*8} - 1)Compute 0.03*8 first:0.03 * 8 = 0.24So:I = 1,000,000 * (e^{0.24} - 1)Now, compute e^{0.24}. Let me recall that e^0.24 is approximately... Hmm, e^0.2 is about 1.2214, e^0.24 is a bit higher.Alternatively, use Taylor series or calculator approximation. Since I don't have a calculator here, maybe I can compute it step by step.Alternatively, remember that ln(2) ≈ 0.6931, so e^0.6931 ≈ 2. But 0.24 is less than that.Alternatively, use the approximation:e^x ≈ 1 + x + x^2/2 + x^3/6 + x^4/24For x = 0.24:e^{0.24} ≈ 1 + 0.24 + (0.24)^2 / 2 + (0.24)^3 / 6 + (0.24)^4 / 24Compute each term:1) 12) 0.243) (0.0576)/2 = 0.02884) (0.013824)/6 ≈ 0.0023045) (0.00331776)/24 ≈ 0.00013824Adding them up:1 + 0.24 = 1.241.24 + 0.0288 = 1.26881.2688 + 0.002304 ≈ 1.2711041.271104 + 0.00013824 ≈ 1.27124224So, e^{0.24} ≈ 1.27124224Therefore, I ≈ 1,000,000 * (1.27124224 - 1) = 1,000,000 * 0.27124224 ≈ 271,242.24So approximately 271,242.24 in interest.But wait, let me check if my approximation is accurate enough. Maybe I should use a better approximation for e^{0.24}.Alternatively, remember that e^{0.24} can be calculated as e^{0.2} * e^{0.04}.We know e^{0.2} ≈ 1.221402758e^{0.04} ≈ 1.040810774Multiplying these together:1.221402758 * 1.040810774 ≈ Let's compute this.First, 1.2214 * 1.0408:Compute 1 * 1.0408 = 1.04080.2214 * 1.0408 ≈ 0.2214*1 + 0.2214*0.0408 ≈ 0.2214 + 0.00903 ≈ 0.23043So total ≈ 1.0408 + 0.23043 ≈ 1.27123Which matches the previous approximation. So e^{0.24} ≈ 1.27123Therefore, I ≈ 1,000,000 * 0.27123 ≈ 271,230So approximately 271,230 in interest.But to be precise, let me use a calculator for e^{0.24}.Wait, since I don't have a calculator, but I can recall that e^{0.24} is approximately 1.271254.So, 1.271254 - 1 = 0.271254Therefore, 1,000,000 * 0.271254 = 271,254So, approximately 271,254.But depending on the precision, maybe we can write it as 271,254.40 or something, but since the question doesn't specify, maybe we can just round it to the nearest dollar or keep it as is.Alternatively, let's compute it more accurately.Wait, if I use the value of e^{0.24} as approximately 1.271254, then:I = 1,000,000 * (1.271254 - 1) = 1,000,000 * 0.271254 = 271,254So, approximately 271,254.Therefore, the total interest accrued over 8 years with continuous compounding is about 271,254.Wait, but let me double-check the continuous compounding formula. It is indeed A = P e^{rt}, so the interest is A - P, which is P(e^{rt} - 1). So that's correct.Alternatively, if the average rate was 3%, then over 8 years, the amount would be 1,000,000 * e^{0.03*8} = 1,000,000 * e^{0.24} ≈ 1,000,000 * 1.271254 ≈ 1,271,254. So the interest is 271,254.Yes, that's correct.So, summarizing:1. The average annual interest rate over the first 8 years is 3%.2. The total interest accrued over 8 years with continuous compounding is approximately 271,254.I think that's it. The key was recognizing that the oscillatory parts of the interest rate function average out to zero over their periods, leaving only the constant term. Then, applying the continuous compounding formula was straightforward.Final Answer1. The average annual interest rate is boxed{3%}.2. The total amount of interest accrued over 8 years is boxed{271254} dollars.</think>