🧠:<think>Alright, so I have this problem about an animal rights activist studying the correlation between youth participation in sports and their attitudes towards animals. The activist is looking into whether certain sports might promote aggression, which could negatively affect empathy towards animals. She collected data from 200 students, split into two groups: those who play team sports and those who don't. They used a standardized empathy scale where higher scores mean more empathy.The first question is asking for the probability that a randomly selected student from the team sports group has an empathy score higher than the mean empathy score of the non-participating group. Let me break this down.First, I need to note the given data. For the team sports group, the mean empathy score is 75 with a standard deviation of 10. For the non-participating group, the mean is 82 with a standard deviation of 12. Both groups are assumed to have empathy scores that follow a normal distribution.So, the question is: What's the probability that a student from the team sports group has a score higher than 82? Because 82 is the mean of the non-participating group.To find this probability, I can use the z-score formula. The z-score tells us how many standard deviations an element is from the mean. The formula is:z = (X - μ) / σWhere X is the value we're interested in, μ is the mean, and σ is the standard deviation.In this case, X is 82, μ is 75, and σ is 10. Plugging these into the formula:z = (82 - 75) / 10 = 7 / 10 = 0.7So, the z-score is 0.7. Now, I need to find the probability that a z-score is greater than 0.7. This is the area under the standard normal curve to the right of z = 0.7.I remember that standard normal distribution tables give the area to the left of the z-score. So, if I look up z = 0.7, I can find the area to the left and subtract it from 1 to get the area to the right.Looking up z = 0.7 in the standard normal table, the cumulative probability is approximately 0.7580. Therefore, the area to the right is 1 - 0.7580 = 0.2420.So, the probability is about 24.2%.Wait, let me double-check that. If the mean is 75, and we're looking for scores above 82, which is 7 points above the mean. With a standard deviation of 10, that's 0.7 standard deviations above. Since the normal distribution is symmetric, the area beyond 0.7 standard deviations should be less than 50%, which 24.2% is. That seems reasonable.Alternatively, I can use the empirical rule, but 0.7 isn't one of the standard deviations that correspond to the 68-95-99.7 rule. So, the table method is more accurate here.Okay, so I think 24.2% is the right answer for the first part.Moving on to the second question. The activist wants to test whether there's a statistically significant difference in the mean empathy scores between the two groups. Specifically, she wants to know if the mean score of non-participating students is significantly higher than that of participating students. We need to conduct a hypothesis test at a 5% significance level, assuming equal variances.First, let me set up the hypotheses. Since she believes that sports participation might negatively impact empathy, she's hypothesizing that non-participating students have higher empathy scores. So, the alternative hypothesis is that the mean of non-participating is higher than the mean of participating.In hypothesis testing terms:Null hypothesis (H0): μ1 - μ2 ≤ 0 (Mean of non-participating - Mean of participating ≤ 0)Alternative hypothesis (H1): μ1 - μ2 > 0 (Mean of non-participating - Mean of participating > 0)But since we're testing if non-participating is significantly higher, it's a one-tailed test.Given that we're assuming equal variances, we can use the pooled variance t-test. First, we need to calculate the pooled variance.But wait, the problem says to assume equal variances, but the standard deviations are different: 10 and 12. Hmm. Maybe they mean we can assume equal variances for the test, even though the sample standard deviations are different. Or perhaps we need to check if the variances are equal? Wait, the question says to assume equal variances, so we don't need to test for that.So, moving on.We have two independent samples. Let me note the given data:Group 1: Team sports participants- Sample size (n1): Let's see, total sample is 200, but it's divided into two groups. It doesn't specify the sizes, so I might have to assume equal sample sizes? Wait, the problem doesn't specify, so perhaps it's 100 each? Because 200 divided into two groups, unless specified otherwise, it's often equal. But let me check the original problem.Wait, the problem says "a sample of 200 students, divided into two groups: those who regularly participate in team sports and those who do not participate in sports at all." It doesn't specify the sizes of each group. Hmm. That complicates things because without knowing n1 and n2, I can't compute the t-statistic.Wait, maybe I missed something. Let me go back.In the first question, it just gives the means and standard deviations for each group, but doesn't specify the sample sizes. Hmm. So, perhaps in the second question, we can assume equal sample sizes? Or maybe the sample sizes are equal? Since the total is 200, maybe 100 each? Or perhaps the sample sizes are not equal, but we don't have that information.Wait, this is a problem. Without knowing the sample sizes, I can't compute the t-test. Maybe the problem assumes equal sample sizes? Or perhaps it's a typo, and the sample sizes are given elsewhere? Let me check the original problem again.Looking back: "Suppose the mean empathy score for the group participating in team sports is 75 with a standard deviation of 10, and the mean empathy score for the non-participating group is 82 with a standard deviation of 12." So, only means and standard deviations are given, not sample sizes.Hmm. Maybe in the second question, the sample sizes are equal? Or perhaps the sample sizes are 100 each? Since the total is 200, and divided into two groups, it's likely 100 each. I think that's a reasonable assumption if not specified otherwise.So, let's assume n1 = n2 = 100.Alternatively, if the sample sizes are different, we can't proceed without that information. So, I think the problem expects us to assume equal sample sizes of 100 each.Okay, moving forward with n1 = n2 = 100.Given that, we can proceed with the pooled variance t-test.First, let's note the means:μ1 (non-participating) = 82μ2 (participating) = 75But in hypothesis testing, we use sample means, so let's denote:x̄1 = 82x̄2 = 75Sample standard deviations:s1 = 12s2 = 10Sample sizes:n1 = 100n2 = 100Since we're assuming equal variances, we can calculate the pooled variance (s_p^2):s_p^2 = [(n1 - 1)s1^2 + (n2 - 1)s2^2] / (n1 + n2 - 2)Plugging in the numbers:s_p^2 = [(99)(144) + (99)(100)] / (198)Calculate numerator:99*144 = 14,25699*100 = 9,900Total numerator = 14,256 + 9,900 = 24,156Denominator = 198So, s_p^2 = 24,156 / 198 ≈ 122Therefore, s_p = sqrt(122) ≈ 11.045Now, the standard error (SE) for the difference in means is:SE = s_p * sqrt(1/n1 + 1/n2) = 11.045 * sqrt(1/100 + 1/100) = 11.045 * sqrt(0.02) ≈ 11.045 * 0.1414 ≈ 1.56Now, the t-statistic is calculated as:t = (x̄1 - x̄2) / SE = (82 - 75) / 1.56 ≈ 7 / 1.56 ≈ 4.487Now, we need to determine the degrees of freedom. Since we're using the pooled variance, df = n1 + n2 - 2 = 100 + 100 - 2 = 198.At a 5% significance level, for a one-tailed test with df = 198, the critical t-value is approximately 1.65 (since for large degrees of freedom, it approaches the z-score of 1.645).Our calculated t-statistic is 4.487, which is much greater than 1.65. Therefore, we reject the null hypothesis.Alternatively, we can calculate the p-value. With a t-statistic of 4.487 and df = 198, the p-value is extremely small, much less than 0.05. So, we have strong evidence to reject the null hypothesis.Therefore, we can conclude that the mean empathy score of students who do not participate in sports is significantly higher than that of students who participate in team sports at the 5% significance level.Wait, let me double-check the calculations.Pooled variance: [(99*144) + (99*100)] / 198 = (14,256 + 9,900)/198 = 24,156 / 198 = 122. Correct.Standard error: sqrt(122*(1/100 + 1/100)) = sqrt(122*0.02) = sqrt(2.44) ≈ 1.56. Correct.t-statistic: (82 - 75)/1.56 ≈ 4.487. Correct.Degrees of freedom: 198. Correct.Critical t-value for one-tailed at 5% is indeed around 1.65. So, 4.487 > 1.65, so reject H0.Yes, that seems correct.Alternatively, if the sample sizes were different, the calculation would change, but since we assumed equal sample sizes, this is fine.So, summarizing:1. The probability is approximately 24.2%.2. The hypothesis test shows a statistically significant difference, with non-participating students having higher empathy scores.Final Answer1. The probability is boxed{0.242}.2. The mean empathy score of non-participating students is significantly higher, so we reject the null hypothesis. The final answer is boxed{text{Reject } H_0}.</think>

🧠:<think>Alright, so I have this problem about a department director trying to streamline bureaucratic procedures. There are two parts to the problem. Let me try to tackle them step by step.Starting with part 1: I need to derive a simplified formula for the total initial time ( T ) which is the sum from ( i = 1 ) to ( n ) of ( t_i ), where each ( t_i = frac{2}{i+1} ) hours. Then, I have to compute ( T ) when ( n = 50 ).Okay, so the total time ( T ) is given by:[T = sum_{i=1}^{n} frac{2}{i+1}]Hmm, that looks like a harmonic series but shifted by one. Let me recall that the harmonic series is ( H_n = sum_{k=1}^{n} frac{1}{k} ). So, if I adjust the indices, maybe I can express ( T ) in terms of harmonic numbers.Let me rewrite the sum:[T = 2 sum_{i=1}^{n} frac{1}{i+1} = 2 left( sum_{k=2}^{n+1} frac{1}{k} right)]Yes, that's right. By substituting ( k = i + 1 ), the sum becomes from ( k = 2 ) to ( k = n + 1 ). So, this can be expressed as:[T = 2 left( H_{n+1} - 1 right)]Because ( H_{n+1} = sum_{k=1}^{n+1} frac{1}{k} ), so subtracting 1 gives us the sum from ( k = 2 ) to ( k = n + 1 ).Therefore, the simplified formula for ( T ) is:[T = 2(H_{n+1} - 1)]Now, I need to compute ( T ) for ( n = 50 ). So, I have to calculate ( H_{51} ) and then plug it into the formula.But wait, calculating ( H_{51} ) exactly might be tedious. Maybe I can approximate it using the formula for harmonic numbers. I remember that ( H_n ) can be approximated by ( ln(n) + gamma + frac{1}{2n} - frac{1}{12n^2} ), where ( gamma ) is the Euler-Mascheroni constant, approximately 0.5772.So, for ( n = 51 ):[H_{51} approx ln(51) + 0.5772 + frac{1}{102} - frac{1}{12 times 2601}]Let me compute each term step by step.First, ( ln(51) ). I know that ( ln(50) ) is approximately 3.9120, so ( ln(51) ) should be a bit more. Let me use a calculator for precision.Calculating ( ln(51) ): approximately 3.9318.Next, adding ( gamma approx 0.5772 ):3.9318 + 0.5772 = 4.5090.Then, adding ( frac{1}{102} approx 0.0098 ):4.5090 + 0.0098 = 4.5188.Subtracting ( frac{1}{12 times 2601} ). Let's compute that:12 * 2601 = 31212, so ( frac{1}{31212} approx 0.000032 ).So, subtracting that:4.5188 - 0.000032 ≈ 4.518768.Therefore, ( H_{51} approx 4.5188 ).So, plugging back into ( T ):[T = 2(H_{51} - 1) = 2(4.5188 - 1) = 2(3.5188) = 7.0376]So, approximately 7.0376 hours.Wait, but let me check if this approximation is accurate enough. Maybe I should compute ( H_{51} ) more precisely.Alternatively, I can compute ( H_{51} ) by adding up the terms step by step, but that would take a lot of time. Alternatively, I can use a calculator or a known value.Looking up ( H_{51} ), it's approximately 4.518738517.So, using that exact value:( H_{51} approx 4.518738517 )Thus,( T = 2(4.518738517 - 1) = 2(3.518738517) = 7.037477034 )So, approximately 7.0375 hours.Therefore, for ( n = 50 ), ( T approx 7.0375 ) hours.Wait, but let me think again. The problem says ( t_i = frac{2}{i+1} ). So, for ( i = 1 ) to ( n ), each term is ( 2/(i+1) ). So, the sum is 2 times the sum from ( i=2 ) to ( n+1 ) of ( 1/i ), which is 2*(H_{n+1} - 1). So, that's correct.So, moving on to part 2.The director wants to reduce the total time by 25%, so the new total time should be 0.75T. To achieve this, redundant steps are eliminated, reducing the number of steps to ( m ) where ( m < n ). The remaining steps are the ones that take the least time.So, we need to find ( m ) such that the sum of the ( m ) smallest ( t_i ) is equal to 0.75T.But wait, the problem says "the remaining steps are the ones that take the least time". So, to minimize the total time, we should keep the steps with the smallest ( t_i ). Since ( t_i = frac{2}{i+1} ), as ( i ) increases, ( t_i ) decreases. So, the smallest ( t_i ) are the ones with the largest ( i ).Wait, that seems counterintuitive. Let me think. If ( t_i = 2/(i+1) ), then for ( i = 1 ), ( t_1 = 2/2 = 1 ); for ( i = 2 ), ( t_2 = 2/3 approx 0.6667 ); for ( i = 3 ), ( t_3 = 2/4 = 0.5 ); and so on. So, as ( i ) increases, ( t_i ) decreases. Therefore, the smallest ( t_i ) correspond to the largest ( i ).Therefore, to minimize the total time, we should keep the steps with the largest ( i ), i.e., the last ( m ) steps. So, the total time after reorganization would be the sum from ( i = n - m + 1 ) to ( n ) of ( t_i ).Wait, but the problem says "the remaining steps are the ones that take the least time". So, yes, that would be the last ( m ) steps, since they have the smallest ( t_i ).Therefore, the total time after reorganization is:[T' = sum_{i = n - m + 1}^{n} frac{2}{i + 1}]And we want this to be equal to 0.75T.So, we have:[sum_{i = n - m + 1}^{n} frac{2}{i + 1} = 0.75 times sum_{i=1}^{n} frac{2}{i + 1}]Simplify both sides by dividing by 2:[sum_{i = n - m + 1}^{n} frac{1}{i + 1} = 0.75 times sum_{i=1}^{n} frac{1}{i + 1}]Let me denote ( S = sum_{i=1}^{n} frac{1}{i + 1} = H_{n+1} - 1 ). Then, the equation becomes:[sum_{i = n - m + 1}^{n} frac{1}{i + 1} = 0.75 S]But ( sum_{i = n - m + 1}^{n} frac{1}{i + 1} = sum_{k = n - m + 2}^{n + 1} frac{1}{k} = H_{n+1} - H_{n - m + 1} )Therefore:[H_{n+1} - H_{n - m + 1} = 0.75 S = 0.75 (H_{n+1} - 1)]So, substituting back:[H_{n+1} - H_{n - m + 1} = 0.75 H_{n+1} - 0.75]Let me rearrange this equation:[H_{n+1} - 0.75 H_{n+1} = H_{n - m + 1} - 0.75]Simplify the left side:[0.25 H_{n+1} = H_{n - m + 1} - 0.75]Therefore:[H_{n - m + 1} = 0.25 H_{n+1} + 0.75]So, we have:[H_{n - m + 1} = 0.25 H_{n+1} + 0.75]Now, our goal is to find ( m ) such that this equation holds. Given that ( n = 50 ), let's plug that in.First, compute ( H_{51} ). Earlier, we found ( H_{51} approx 4.518738517 ).So,[H_{51} approx 4.518738517]Therefore,[0.25 H_{51} + 0.75 = 0.25 times 4.518738517 + 0.75 approx 1.129684629 + 0.75 = 1.879684629]So, we need ( H_{n - m + 1} approx 1.879684629 ).We need to find ( k = n - m + 1 ) such that ( H_k approx 1.879684629 ).Given that ( n = 50 ), ( k = 51 - m ).So, we need to find ( k ) where ( H_k approx 1.879684629 ).Let me recall that ( H_k ) increases as ( k ) increases, so we need to find ( k ) such that ( H_k approx 1.8797 ).Let me list some harmonic numbers:- ( H_1 = 1 )- ( H_2 = 1 + 0.5 = 1.5 )- ( H_3 = 1.5 + 0.3333 ≈ 1.8333 )- ( H_4 = 1.8333 + 0.25 ≈ 2.0833 )- ( H_5 ≈ 2.0833 + 0.2 ≈ 2.2833 )- ( H_6 ≈ 2.2833 + 0.1667 ≈ 2.45 )- ( H_7 ≈ 2.45 + 0.1429 ≈ 2.5929 )- ( H_8 ≈ 2.5929 + 0.125 ≈ 2.7179 )- ( H_9 ≈ 2.7179 + 0.1111 ≈ 2.829 )- ( H_{10} ≈ 2.829 + 0.1 ≈ 2.929 )Wait, but we need ( H_k ≈ 1.8797 ). Looking at the values:- ( H_3 ≈ 1.8333 )- ( H_4 ≈ 2.0833 )So, 1.8797 is between ( H_3 ) and ( H_4 ). Let me compute ( H_3.5 ) or use linear approximation.Alternatively, since ( H_k ) is approximately ( ln(k) + gamma + 1/(2k) ), let's use that approximation.We have:[H_k approx ln(k) + 0.5772 + frac{1}{2k}]Set this equal to 1.8797:[ln(k) + 0.5772 + frac{1}{2k} = 1.8797]Let me denote ( x = k ). Then,[ln(x) + 0.5772 + frac{1}{2x} = 1.8797]Let me rearrange:[ln(x) + frac{1}{2x} = 1.8797 - 0.5772 = 1.3025]So,[ln(x) + frac{1}{2x} = 1.3025]This is a transcendental equation, so we can solve it numerically.Let me make an initial guess. Let's try ( x = 3.5 ):Compute ( ln(3.5) + 1/(2*3.5) ≈ 1.2528 + 0.1429 ≈ 1.3957 ), which is higher than 1.3025.Try ( x = 3 ):( ln(3) + 1/(2*3) ≈ 1.0986 + 0.1667 ≈ 1.2653 ), which is lower than 1.3025.So, the solution is between 3 and 3.5.Let me try ( x = 3.2 ):( ln(3.2) ≈ 1.16315 )( 1/(2*3.2) = 1/6.4 ≈ 0.15625 )Sum ≈ 1.16315 + 0.15625 ≈ 1.3194, which is still higher than 1.3025.Try ( x = 3.1 ):( ln(3.1) ≈ 1.1314 )( 1/(2*3.1) ≈ 0.1613 )Sum ≈ 1.1314 + 0.1613 ≈ 1.2927, which is lower than 1.3025.So, between 3.1 and 3.2.Let me try ( x = 3.15 ):( ln(3.15) ≈ 1.1482 )( 1/(2*3.15) ≈ 0.1587 )Sum ≈ 1.1482 + 0.1587 ≈ 1.3069, which is slightly higher than 1.3025.So, between 3.1 and 3.15.Let me try ( x = 3.12 ):( ln(3.12) ≈ 1.1375 )( 1/(2*3.12) ≈ 0.1603 )Sum ≈ 1.1375 + 0.1603 ≈ 1.2978, which is lower than 1.3025.So, between 3.12 and 3.15.Let me try ( x = 3.13 ):( ln(3.13) ≈ 1.1412 )( 1/(2*3.13) ≈ 0.1600 )Sum ≈ 1.1412 + 0.1600 ≈ 1.3012, which is very close to 1.3025.So, ( x ≈ 3.13 ).Therefore, ( k ≈ 3.13 ). Since ( k ) must be an integer, we can check ( k = 3 ) and ( k = 4 ).But wait, ( H_3 ≈ 1.8333 ) and ( H_4 ≈ 2.0833 ). Our target is approximately 1.8797, which is between ( H_3 ) and ( H_4 ).But since ( H_k ) is only defined for integer ( k ), we can't have ( k = 3.13 ). So, perhaps the closest integer is ( k = 4 ), but ( H_4 = 2.0833 ) which is higher than 1.8797. Alternatively, maybe we need to consider that ( k ) is not an integer, but in reality, ( k ) must be an integer because it's the number of steps.Wait, but in our earlier equation, ( k = n - m + 1 = 51 - m ). So, ( k ) must be an integer because ( m ) is an integer. Therefore, we need to find the smallest integer ( k ) such that ( H_k geq 1.8797 ).Looking at the harmonic numbers:- ( H_3 ≈ 1.8333 ) (less than 1.8797)- ( H_4 ≈ 2.0833 ) (greater than 1.8797)So, the smallest integer ( k ) where ( H_k geq 1.8797 ) is ( k = 4 ).Therefore, ( k = 4 ), which means:[51 - m = 4 implies m = 51 - 4 = 47]Wait, but let me verify this.If ( k = 4 ), then ( H_4 ≈ 2.0833 ). But our target was ( H_k ≈ 1.8797 ). So, ( H_4 ) is larger than our target. Therefore, if we take ( k = 4 ), the sum ( H_{n+1} - H_{n - m + 1} = H_{51} - H_4 ≈ 4.5187 - 2.0833 ≈ 2.4354 ). Then, 2.4354 compared to 0.75T.Wait, let's compute 0.75T. Earlier, ( T ≈ 7.0375 ), so 0.75T ≈ 5.2781.But ( T' = 2.4354 times 2 = 4.8708 ), which is less than 5.2781. So, that's not enough.Wait, perhaps I made a mistake in the earlier steps.Let me recap:We have:[H_{n - m + 1} = 0.25 H_{n+1} + 0.75]With ( n = 50 ), ( H_{51} ≈ 4.5187 ), so:[H_{51 - m + 1} = 0.25 * 4.5187 + 0.75 ≈ 1.1297 + 0.75 = 1.8797]So, ( H_{52 - m} ≈ 1.8797 ). So, we need ( H_k ≈ 1.8797 ), where ( k = 52 - m ).Looking for ( k ) such that ( H_k ≈ 1.8797 ). As we saw earlier, ( H_3 ≈ 1.8333 ) and ( H_4 ≈ 2.0833 ). So, 1.8797 is between ( H_3 ) and ( H_4 ). Since ( H_k ) is increasing, the smallest ( k ) where ( H_k geq 1.8797 ) is ( k = 4 ).But then ( k = 4 ), so ( 52 - m = 4 implies m = 52 - 4 = 48 ).Wait, earlier I thought ( k = 51 - m ), but actually, ( k = n - m + 1 = 50 - m + 1 = 51 - m ). So, ( k = 51 - m ).So, if ( k = 4 ), then ( 51 - m = 4 implies m = 47 ).But when ( k = 4 ), ( H_k = 2.0833 ), which is larger than 1.8797. So, perhaps we need a larger ( k ) to get a smaller ( H_k ). Wait, no, ( H_k ) increases with ( k ). So, to get a smaller ( H_k ), we need a smaller ( k ).Wait, this is confusing. Let me think again.We have:[H_{51 - m} = 1.8797]But ( H_{51 - m} ) must be approximately 1.8797. Since ( H_3 ≈ 1.8333 ) and ( H_4 ≈ 2.0833 ), 1.8797 is between ( H_3 ) and ( H_4 ). Therefore, ( 51 - m ) must be between 3 and 4. But since ( 51 - m ) must be an integer, the closest integer ( k ) where ( H_k ) is just above 1.8797 is ( k = 4 ). Therefore, ( 51 - m = 4 implies m = 47 ).But when we compute ( T' ), it's ( 2*(H_{51} - H_4) ≈ 2*(4.5187 - 2.0833) ≈ 2*(2.4354) ≈ 4.8708 ), which is less than 0.75T ≈ 5.2781.So, 4.8708 < 5.2781, meaning that we've cut too many steps. Therefore, we need a smaller ( m ), i.e., keep more steps, so that ( T' ) is larger.Wait, but if we take ( k = 3 ), then ( H_3 ≈ 1.8333 ), which is less than 1.8797. So, ( H_{51 - m} = 1.8333 ) would give us:[H_{51 - m} = 1.8333 implies 51 - m = 3 implies m = 48]Then, ( T' = 2*(H_{51} - H_3) ≈ 2*(4.5187 - 1.8333) ≈ 2*(2.6854) ≈ 5.3708 ), which is greater than 0.75T ≈ 5.2781.So, ( m = 48 ) gives ( T' ≈ 5.3708 ), which is slightly more than 0.75T.But we need ( T' = 0.75T ≈ 5.2781 ). So, perhaps ( m = 47.5 ), but since ( m ) must be an integer, we need to check between ( m = 47 ) and ( m = 48 ).Wait, but ( m ) must be an integer, so we can't have half steps. Therefore, we need to find the smallest ( m ) such that ( T' leq 0.75T ). But in our case, ( m = 47 ) gives ( T' ≈ 4.8708 ), which is too low, and ( m = 48 ) gives ( T' ≈ 5.3708 ), which is higher than 0.75T.But 5.3708 is higher than 5.2781, so perhaps ( m = 48 ) is the answer, but it's slightly over. Alternatively, maybe we need to find a non-integer ( m ), but since ( m ) must be integer, perhaps we need to accept that we can't get exactly 0.75T, but the closest is ( m = 48 ).Alternatively, perhaps my approach is flawed. Let me think differently.Instead of trying to solve for ( k ), maybe I can compute the cumulative sum from the end and find when it reaches 0.75T.Given that ( T ≈ 7.0375 ), 0.75T ≈ 5.2781.We need to find the smallest ( m ) such that the sum of the last ( m ) terms ( t_i ) is at least 5.2781.Since the terms are decreasing, the sum of the last ( m ) terms will be the largest possible sum for ( m ) terms. So, we need to find the smallest ( m ) where the sum of the last ( m ) terms is ≥ 5.2781.Alternatively, since the sum of the last ( m ) terms is:[sum_{i = n - m + 1}^{n} frac{2}{i + 1}]We can compute this sum for different ( m ) until it reaches approximately 5.2781.Given that ( n = 50 ), let's compute the sum for ( m = 48 ):Sum from ( i = 3 ) to ( 50 ) of ( 2/(i + 1) ). Wait, no, ( i = 50 - 48 + 1 = 3 ) to 50.Wait, no, ( i = n - m + 1 = 50 - 48 + 1 = 3 ) to 50.So, the sum is ( sum_{i=3}^{50} frac{2}{i + 1} = 2 sum_{k=4}^{51} frac{1}{k} = 2 (H_{51} - H_3) ).We know ( H_{51} ≈ 4.5187 ) and ( H_3 ≈ 1.8333 ), so:( 2*(4.5187 - 1.8333) = 2*(2.6854) ≈ 5.3708 ), which is as before.Similarly, for ( m = 47 ):Sum from ( i = 4 ) to 50 of ( 2/(i + 1) = 2*(H_{51} - H_4) ≈ 2*(4.5187 - 2.0833) ≈ 2*(2.4354) ≈ 4.8708 ).So, 4.8708 < 5.2781 < 5.3708.Therefore, the sum for ( m = 48 ) is 5.3708, which is more than 0.75T, and for ( m = 47 ) it's less.Therefore, the smallest ( m ) such that the sum is at least 0.75T is ( m = 48 ).But wait, the problem says "the total time after reorganization is 0.75T". So, we need the sum to be exactly 0.75T. Since we can't have a non-integer ( m ), perhaps we need to find ( m ) such that the sum is as close as possible to 0.75T.Alternatively, maybe we can use a more precise approximation for ( H_k ).Wait, let's try to compute ( H_{51 - m} ) more accurately.We have:[H_{51 - m} = 0.25 H_{51} + 0.75 ≈ 0.25 * 4.5187 + 0.75 ≈ 1.1297 + 0.75 = 1.8797]So, we need ( H_{51 - m} ≈ 1.8797 ).Looking up exact values or using a calculator for ( H_k ):- ( H_3 = 1 + 1/2 + 1/3 ≈ 1.8333 )- ( H_4 = H_3 + 1/4 ≈ 1.8333 + 0.25 = 2.0833 )So, 1.8797 is between ( H_3 ) and ( H_4 ). Therefore, ( 51 - m ) is between 3 and 4. Since ( 51 - m ) must be an integer, the closest is 4, but as we saw, that gives a sum that's too high.Alternatively, perhaps we can use a weighted average between ( H_3 ) and ( H_4 ).Let me denote ( k = 51 - m ). We have:[H_k = 1.8797]Between ( k = 3 ) and ( k = 4 ):The difference between ( H_4 ) and ( H_3 ) is ( 2.0833 - 1.8333 = 0.25 ).We need ( H_k = 1.8797 ), which is ( 1.8797 - 1.8333 = 0.0464 ) above ( H_3 ).So, the fraction is ( 0.0464 / 0.25 ≈ 0.1856 ).Therefore, ( k ≈ 3 + 0.1856 ≈ 3.1856 ).So, ( k ≈ 3.1856 implies 51 - m ≈ 3.1856 implies m ≈ 51 - 3.1856 ≈ 47.8144 ).Since ( m ) must be an integer, we can round to ( m = 48 ).Therefore, the value of ( m ) is 48.But let me verify this.If ( m = 48 ), then ( k = 51 - 48 = 3 ). Wait, no, ( k = 51 - m = 51 - 48 = 3 ). But earlier, we saw that ( H_3 ≈ 1.8333 ), which is less than 1.8797. So, that would mean:[H_{51 - m} = H_3 ≈ 1.8333]But we need ( H_{51 - m} ≈ 1.8797 ). Therefore, ( k = 51 - m ) must be slightly larger than 3, which would mean ( m ) is slightly smaller than 48. But since ( m ) must be integer, we can't have ( m = 47.81 ). Therefore, the closest integer is ( m = 48 ), even though it results in ( H_{51 - 48} = H_3 ≈ 1.8333 ), which is less than 1.8797.But wait, if we take ( m = 48 ), the sum ( T' = 2*(H_{51} - H_3) ≈ 5.3708 ), which is more than 0.75T ≈ 5.2781.Alternatively, if we take ( m = 47 ), the sum is 4.8708, which is less than 0.75T.Therefore, there's no integer ( m ) that gives exactly 0.75T. So, the director would have to choose between ( m = 47 ) (which gives a sum less than 0.75T) or ( m = 48 ) (which gives a sum more than 0.75T). Since the director wants to reduce the total time by 25%, i.e., achieve 0.75T, and the sum for ( m = 48 ) is the closest above 0.75T, perhaps ( m = 48 ) is the answer.Alternatively, maybe the problem expects us to use the approximation where ( m ≈ 47.81 ), and round to 48.Therefore, the value of ( m ) is 48.</think>

🧠:<think>Okay, so I have this problem about modeling the growth and health of Komodo dragons at a zoo. It's split into two parts. Let me try to tackle them one by one.Starting with the first part: I need to find the general solution for the growth rate ( G(t) ) given a differential equation. The equation is:[frac{dG(t)}{dt} = k cdot frac{D(t)}{L(t) + a}]where ( D(t) = 5e^{0.1t} ) kg/month and ( L(t) = 2 + 0.3t ). The initial condition is ( G(0) = G_0 ).So, this is a first-order linear ordinary differential equation (ODE). To solve it, I think I can separate variables or use an integrating factor. Let me see.First, let's write down the equation with the given functions:[frac{dG}{dt} = k cdot frac{5e^{0.1t}}{2 + 0.3t + a}]Hmm, so it's actually a separable equation. I can write it as:[dG = k cdot frac{5e^{0.1t}}{2 + 0.3t + a} dt]So, to find ( G(t) ), I need to integrate both sides:[G(t) = G_0 + int_0^t k cdot frac{5e^{0.1tau}}{2 + 0.3tau + a} dtau]Wait, but the problem says to find the general solution. So maybe I can express it in terms of an integral, but perhaps it's better to see if the integral can be evaluated in closed form.Let me denote the denominator as ( L(t) + a = 2 + 0.3t + a ). Let me denote ( b = 2 + a ), so the denominator becomes ( b + 0.3t ).So, the integral becomes:[int frac{5e^{0.1t}}{b + 0.3t} dt]Hmm, integrating ( e^{kt} ) over a linear function. I think this might not have an elementary antiderivative. Let me check.Let me make a substitution. Let ( u = b + 0.3t ). Then, ( du = 0.3 dt ), so ( dt = frac{du}{0.3} ).Expressing ( t ) in terms of ( u ): ( t = frac{u - b}{0.3} ).So, ( e^{0.1t} = e^{0.1 cdot frac{u - b}{0.3}} = e^{frac{u - b}{3}} ).Substituting back into the integral:[int frac{5e^{frac{u - b}{3}}}{u} cdot frac{du}{0.3} = frac{5}{0.3} int frac{e^{frac{u - b}{3}}}{u} du]Simplify constants:[frac{5}{0.3} = frac{50}{3}]So, the integral becomes:[frac{50}{3} int frac{e^{frac{u - b}{3}}}{u} du]Hmm, this integral is similar to the exponential integral function, which is a special function. Specifically, the integral ( int frac{e^{kt}}{t} dt ) is related to the exponential integral ( Ei(kt) ).So, perhaps we can express the integral in terms of ( Ei ).Let me write:Let ( v = frac{u - b}{3} ), so ( u = 3v + b ), and ( du = 3 dv ).Substituting into the integral:[frac{50}{3} int frac{e^{v}}{3v + b} cdot 3 dv = 50 int frac{e^{v}}{3v + b} dv]Hmm, this is still not a standard exponential integral because of the denominator ( 3v + b ). Maybe another substitution.Let me set ( w = 3v + b ), so ( v = frac{w - b}{3} ), and ( dv = frac{dw}{3} ).Substituting:[50 int frac{e^{frac{w - b}{3}}}{w} cdot frac{dw}{3} = frac{50}{3} int frac{e^{frac{w - b}{3}}}{w} dw]Hmm, this seems to be going in circles. Maybe I need to accept that this integral doesn't have an elementary form and express it in terms of the exponential integral function.The exponential integral is defined as:[Ei(x) = - int_{-x}^{infty} frac{e^{-t}}{t} dt]But our integral is:[int frac{e^{v}}{3v + b} dv]Wait, maybe I can manipulate it to look like the exponential integral.Let me factor out constants:Let me write ( 3v + b = 3(v + frac{b}{3}) ). So,[int frac{e^{v}}{3(v + frac{b}{3})} dv = frac{1}{3} int frac{e^{v}}{v + frac{b}{3}} dv]Let me set ( z = v + frac{b}{3} ), so ( v = z - frac{b}{3} ), ( dv = dz ). Then,[frac{1}{3} int frac{e^{z - frac{b}{3}}}{z} dz = frac{e^{- frac{b}{3}}}{3} int frac{e^{z}}{z} dz = frac{e^{- frac{b}{3}}}{3} Ei(z) + C]Substituting back:( z = v + frac{b}{3} = frac{u - b}{3} + frac{b}{3} = frac{u}{3} )So,[frac{e^{- frac{b}{3}}}{3} Eileft( frac{u}{3} right) + C]But ( u = b + 0.3t ), so:[frac{e^{- frac{b}{3}}}{3} Eileft( frac{b + 0.3t}{3} right) + C]Therefore, putting it all together, the integral is:[frac{50}{3} times frac{e^{- frac{b}{3}}}{3} Eileft( frac{b + 0.3t}{3} right) + C = frac{50}{9} e^{- frac{b}{3}} Eileft( frac{b + 0.3t}{3} right) + C]But ( b = 2 + a ), so:[frac{50}{9} e^{- frac{2 + a}{3}} Eileft( frac{2 + a + 0.3t}{3} right) + C]Therefore, the general solution for ( G(t) ) is:[G(t) = G_0 + frac{50}{9} k e^{- frac{2 + a}{3}} Eileft( frac{2 + a + 0.3t}{3} right)]Wait, but I think I might have made a mistake in the constants. Let me retrace.Wait, earlier, when I substituted ( u = b + 0.3t ), then ( du = 0.3 dt ), so ( dt = du / 0.3 ). Then, ( e^{0.1t} = e^{(0.1)( (u - b)/0.3 )} = e^{(u - b)/3} ).So, the integral becomes:[int frac{5 e^{(u - b)/3}}{u} cdot frac{du}{0.3} = frac{5}{0.3} int frac{e^{(u - b)/3}}{u} du = frac{50}{3} int frac{e^{(u - b)/3}}{u} du]Then, I set ( v = (u - b)/3 ), so ( u = 3v + b ), ( du = 3 dv ). Substituting:[frac{50}{3} int frac{e^{v}}{3v + b} cdot 3 dv = 50 int frac{e^{v}}{3v + b} dv]Then, set ( w = 3v + b ), so ( v = (w - b)/3 ), ( dv = dw / 3 ). Substituting:[50 int frac{e^{(w - b)/3}}{w} cdot frac{dw}{3} = frac{50}{3} int frac{e^{(w - b)/3}}{w} dw]Which is:[frac{50}{3} e^{-b/3} int frac{e^{w/3}}{w} dw = frac{50}{3} e^{-b/3} cdot 3 Eileft( frac{w}{3} right) + C = 50 e^{-b/3} Eileft( frac{w}{3} right) + C]But ( w = 3v + b = 3 cdot frac{u - b}{3} + b = u ). Wait, that can't be right because ( w = 3v + b = (u - b) + b = u ). So, ( w = u ).Therefore, the integral becomes:[50 e^{-b/3} Eileft( frac{u}{3} right) + C]But ( u = b + 0.3t ), so:[50 e^{-b/3} Eileft( frac{b + 0.3t}{3} right) + C]Therefore, the integral is:[int frac{5e^{0.1t}}{b + 0.3t} dt = 50 e^{-b/3} Eileft( frac{b + 0.3t}{3} right) + C]So, going back to the expression for ( G(t) ):[G(t) = G_0 + k cdot left[ 50 e^{-b/3} Eileft( frac{b + 0.3t}{3} right) right]]But ( b = 2 + a ), so substituting back:[G(t) = G_0 + 50 k e^{-(2 + a)/3} Eileft( frac{2 + a + 0.3t}{3} right)]Wait, but I think I might have missed the constant factor earlier. Let me check.Wait, no, because when I did the substitution, I had:Original integral: ( int frac{5 e^{0.1t}}{b + 0.3t} dt )After substitution, it became ( 50 e^{-b/3} Eileft( frac{b + 0.3t}{3} right) + C )So, the integral is ( 50 e^{-b/3} Eileft( frac{b + 0.3t}{3} right) + C )Therefore, the solution is:[G(t) = G_0 + k cdot 50 e^{-(2 + a)/3} Eileft( frac{2 + a + 0.3t}{3} right)]Simplify constants:50 is 50, so:[G(t) = G_0 + 50 k e^{-(2 + a)/3} Eileft( frac{2 + a + 0.3t}{3} right)]Alternatively, we can write this as:[G(t) = G_0 + C e^{-(2 + a)/3} Eileft( frac{2 + a + 0.3t}{3} right)]where ( C = 50k ).But perhaps it's better to leave it in terms of the constants as given.So, summarizing, the general solution is:[G(t) = G_0 + 50 k e^{-(2 + a)/3} Eileft( frac{2 + a + 0.3t}{3} right)]I think that's as far as we can go analytically. So, that's the general solution for ( G(t) ).Now, moving on to the second part: evaluating the health index ( H(t) ) at ( t = 12 ) months.The health index is given by:[H(t) = int_0^t left( frac{2G(tau)}{L(tau)} + frac{D(tau)}{L(tau)^2} right) dtau]We need to substitute ( G(tau) ) from the first part, which is:[G(tau) = G_0 + 50 k e^{-(2 + a)/3} Eileft( frac{2 + a + 0.3tau}{3} right)]And ( L(tau) = 2 + 0.3tau ), ( D(tau) = 5 e^{0.1tau} ).So, plugging these into ( H(t) ):[H(t) = int_0^t left( frac{2 [G_0 + 50 k e^{-(2 + a)/3} Eileft( frac{2 + a + 0.3tau}{3} right)] }{2 + 0.3tau} + frac{5 e^{0.1tau}}{(2 + 0.3tau)^2} right) dtau]This integral looks quite complicated. Let's break it down into two parts:1. The first part involves ( G(tau) ), which includes the exponential integral function. Integrating this might not be straightforward.2. The second part is ( frac{5 e^{0.1tau}}{(2 + 0.3tau)^2} ), which might also be challenging to integrate.Given that both integrals might not have elementary antiderivatives, perhaps we need to evaluate them numerically at ( t = 12 ). However, since this is a theoretical problem, maybe there's a way to express ( H(t) ) in terms of the exponential integral function or other special functions.Alternatively, perhaps we can find a relationship between ( G(t) ) and ( H(t) ) to simplify the expression.Wait, let me think. From the first part, we have:[frac{dG}{dt} = k cdot frac{D(t)}{L(t) + a}]Which is:[frac{dG}{dt} = k cdot frac{5 e^{0.1t}}{2 + 0.3t + a}]So, ( frac{dG}{dt} ) is known. Maybe we can express ( G(t) ) in terms of ( frac{dG}{dt} ) and substitute into ( H(t) ).Looking at ( H(t) ):[H(t) = int_0^t left( frac{2G(tau)}{L(tau)} + frac{D(tau)}{L(tau)^2} right) dtau]Let me see if I can relate this to ( frac{dG}{dt} ).From the first equation:[frac{dG}{dt} = k cdot frac{D(t)}{L(t) + a}]So, ( D(t) = frac{L(t) + a}{k} cdot frac{dG}{dt} )Substituting ( D(t) ) into ( H(t) ):[H(t) = int_0^t left( frac{2G(tau)}{L(tau)} + frac{frac{L(tau) + a}{k} cdot frac{dG}{dtau}}{L(tau)^2} right) dtau]Simplify the second term:[frac{frac{L(tau) + a}{k} cdot frac{dG}{dtau}}{L(tau)^2} = frac{(L(tau) + a)}{k L(tau)^2} cdot frac{dG}{dtau}]So, ( H(t) ) becomes:[H(t) = int_0^t left( frac{2G(tau)}{L(tau)} + frac{(L(tau) + a)}{k L(tau)^2} cdot frac{dG}{dtau} right) dtau]Hmm, maybe integrating the second term by parts. Let me consider the integral:[int frac{(L(tau) + a)}{k L(tau)^2} cdot frac{dG}{dtau} dtau]Let me set ( u = frac{L(tau) + a}{k L(tau)^2} ) and ( dv = dG ). Then, ( du = frac{d}{dtau} left( frac{L(tau) + a}{k L(tau)^2} right) dtau ) and ( v = G(tau) ).But integrating by parts might complicate things further. Alternatively, perhaps we can find a relationship between ( H(t) ) and ( G(t) ).Wait, let me think differently. Maybe express ( H(t) ) in terms of ( G(t) ) and its derivative.Given that ( H(t) ) is an integral involving ( G(tau) ) and ( D(tau) ), and ( D(tau) ) is related to ( frac{dG}{dtau} ), perhaps we can find a differential equation for ( H(t) ).Let me compute ( frac{dH}{dt} ):[frac{dH}{dt} = frac{2G(t)}{L(t)} + frac{D(t)}{L(t)^2}]From the first part, ( frac{dG}{dt} = k cdot frac{D(t)}{L(t) + a} ), so ( D(t) = frac{L(t) + a}{k} cdot frac{dG}{dt} )Substituting into ( frac{dH}{dt} ):[frac{dH}{dt} = frac{2G(t)}{L(t)} + frac{frac{L(t) + a}{k} cdot frac{dG}{dt}}{L(t)^2} = frac{2G(t)}{L(t)} + frac{(L(t) + a)}{k L(t)^2} cdot frac{dG}{dt}]Hmm, this seems similar to what I had before. Maybe I can express this as a derivative of some combination of ( G(t) ) and ( L(t) ).Alternatively, perhaps we can find an integrating factor or see if ( H(t) ) can be expressed in terms of ( G(t) ) and ( L(t) ).Wait, let me try to manipulate the expression for ( frac{dH}{dt} ):[frac{dH}{dt} = frac{2G(t)}{L(t)} + frac{(L(t) + a)}{k L(t)^2} cdot frac{dG}{dt}]Let me factor out ( frac{1}{L(t)} ):[frac{dH}{dt} = frac{1}{L(t)} left( 2G(t) + frac{(L(t) + a)}{k L(t)} cdot frac{dG}{dt} right )]Hmm, not sure if that helps.Alternatively, let me consider the expression:[frac{d}{dt} left( frac{G(t)}{L(t)} right ) = frac{G'(t) L(t) - G(t) L'(t)}{L(t)^2}]Compute ( L'(t) ):( L(t) = 2 + 0.3t ), so ( L'(t) = 0.3 )So,[frac{d}{dt} left( frac{G(t)}{L(t)} right ) = frac{G'(t) (2 + 0.3t) - 0.3 G(t)}{(2 + 0.3t)^2}]Compare this with the expression for ( frac{dH}{dt} ):[frac{dH}{dt} = frac{2G(t)}{2 + 0.3t} + frac{(2 + 0.3t + a)}{k (2 + 0.3t)^2} G'(t)]Hmm, not immediately obvious. Maybe another approach.Alternatively, perhaps express ( H(t) ) in terms of ( G(t) ) and its integral.Given that ( H(t) ) is:[H(t) = int_0^t left( frac{2G(tau)}{L(tau)} + frac{D(tau)}{L(tau)^2} right ) dtau]And from the first part, ( D(tau) = frac{L(tau) + a}{k} cdot frac{dG}{dtau} )So,[H(t) = int_0^t left( frac{2G(tau)}{L(tau)} + frac{frac{L(tau) + a}{k} cdot frac{dG}{dtau}}{L(tau)^2} right ) dtau]Simplify the second term:[frac{L(tau) + a}{k L(tau)^2} cdot frac{dG}{dtau} = frac{L(tau) + a}{k L(tau)^2} G'(tau)]So, ( H(t) ) becomes:[H(t) = int_0^t left( frac{2G(tau)}{L(tau)} + frac{(L(tau) + a)}{k L(tau)^2} G'(tau) right ) dtau]Let me see if I can combine these terms. Let me write the integrand as:[frac{2G(tau)}{L(tau)} + frac{(L(tau) + a)}{k L(tau)^2} G'(tau)]Let me factor out ( frac{1}{L(tau)} ):[frac{1}{L(tau)} left( 2G(tau) + frac{(L(tau) + a)}{k L(tau)} G'(tau) right )]Hmm, not sure. Alternatively, perhaps consider integrating the second term by parts.Let me denote:Let ( u = frac{L(tau) + a}{k L(tau)^2} ), then ( du = frac{d}{dtau} left( frac{L(tau) + a}{k L(tau)^2} right ) dtau )And ( dv = G'(tau) dtau ), so ( v = G(tau) )Then, integrating by parts:[int u dv = uv - int v du]So,[int frac{(L(tau) + a)}{k L(tau)^2} G'(tau) dtau = frac{(L(tau) + a)}{k L(tau)^2} G(tau) - int G(tau) cdot frac{d}{dtau} left( frac{L(tau) + a}{k L(tau)^2} right ) dtau]Compute ( frac{d}{dtau} left( frac{L(tau) + a}{k L(tau)^2} right ) ):First, ( L(tau) = 2 + 0.3tau ), so ( L'(tau) = 0.3 )Let me compute the derivative:[frac{d}{dtau} left( frac{L + a}{k L^2} right ) = frac{1}{k} cdot frac{L' cdot L^2 - (L + a) cdot 2 L L'}{L^4} = frac{1}{k} cdot frac{L' L^2 - 2 L (L + a) L'}{L^4}]Simplify numerator:[L' L^2 - 2 L (L + a) L' = L' [L^2 - 2 L (L + a)] = L' [L^2 - 2 L^2 - 2 a L] = L' [- L^2 - 2 a L] = - L' L (L + 2a)]Therefore,[frac{d}{dtau} left( frac{L + a}{k L^2} right ) = frac{ - L' L (L + 2a) }{k L^4} = frac{ - L' (L + 2a) }{k L^3 }]Substituting back:[int frac{(L + a)}{k L^2} G' dtau = frac{(L + a)}{k L^2} G - int G cdot left( frac{ - L' (L + 2a) }{k L^3 } right ) dtau]Simplify:[= frac{(L + a)}{k L^2} G + frac{L'}{k} cdot frac{(L + 2a)}{L^3} int G dtau]Wait, no, the integral is:[int G cdot frac{ L' (L + 2a) }{k L^3 } dtau]So, putting it all together, the integral becomes:[int frac{(L + a)}{k L^2} G' dtau = frac{(L + a)}{k L^2} G + frac{L'}{k} cdot frac{(L + 2a)}{L^3} int G dtau]This seems to complicate things further because now we have an integral of ( G ) inside the expression.Given the complexity, perhaps it's better to accept that ( H(t) ) cannot be expressed in a simple closed form and instead focus on expressing it in terms of the exponential integral function or other special functions, or perhaps evaluate it numerically.However, since the problem asks to evaluate ( H(t) ) at ( t = 12 ), and given that the integral involves special functions, it's likely that we need to express ( H(t) ) in terms of ( G(t) ) and its integral, but without specific numerical values for ( k ), ( a ), and ( G_0 ), we can't compute a numerical answer.Wait, but maybe there's a way to express ( H(t) ) in terms of ( G(t) ) and its derivatives without integrating. Let me think.From the expression for ( frac{dH}{dt} ):[frac{dH}{dt} = frac{2G(t)}{L(t)} + frac{(L(t) + a)}{k L(t)^2} cdot frac{dG}{dt}]We can write this as:[frac{dH}{dt} = frac{2G(t)}{L(t)} + frac{(L(t) + a)}{k L(t)^2} cdot frac{dG}{dt}]But from the first part, ( frac{dG}{dt} = k cdot frac{D(t)}{L(t) + a} = k cdot frac{5 e^{0.1t}}{L(t) + a} )So, substituting ( frac{dG}{dt} ):[frac{dH}{dt} = frac{2G(t)}{L(t)} + frac{(L(t) + a)}{k L(t)^2} cdot k cdot frac{5 e^{0.1t}}{L(t) + a} = frac{2G(t)}{L(t)} + frac{5 e^{0.1t}}{L(t)^2}]Wait, that's interesting. So, the expression simplifies to:[frac{dH}{dt} = frac{2G(t)}{L(t)} + frac{5 e^{0.1t}}{L(t)^2}]But this is exactly the integrand of ( H(t) ). So, this doesn't help us directly. Hmm.Alternatively, perhaps we can express ( H(t) ) in terms of ( G(t) ) and some other function.Wait, let me consider the expression for ( H(t) ):[H(t) = int_0^t left( frac{2G(tau)}{L(tau)} + frac{D(tau)}{L(tau)^2} right ) dtau]We know ( D(tau) = 5 e^{0.1tau} ), so:[H(t) = int_0^t frac{2G(tau)}{L(tau)} dtau + int_0^t frac{5 e^{0.1tau}}{L(tau)^2} dtau]Let me denote the first integral as ( I_1 ) and the second as ( I_2 ):[I_1 = int_0^t frac{2G(tau)}{L(tau)} dtau][I_2 = int_0^t frac{5 e^{0.1tau}}{L(tau)^2} dtau]So, ( H(t) = I_1 + I_2 )We already have an expression for ( G(tau) ):[G(tau) = G_0 + 50 k e^{-(2 + a)/3} Eileft( frac{2 + a + 0.3tau}{3} right )]So, substituting into ( I_1 ):[I_1 = int_0^t frac{2 [G_0 + 50 k e^{-(2 + a)/3} Eileft( frac{2 + a + 0.3tau}{3} right ) ] }{2 + 0.3tau} dtau]This integral is still complicated because it involves the exponential integral function.Similarly, ( I_2 ) is:[I_2 = int_0^t frac{5 e^{0.1tau}}{(2 + 0.3tau)^2} dtau]This integral might be more manageable. Let me try to compute ( I_2 ).Let me make a substitution for ( I_2 ):Let ( u = 2 + 0.3tau ), so ( du = 0.3 dtau ), ( dtau = frac{du}{0.3} ). When ( tau = 0 ), ( u = 2 ). When ( tau = t ), ( u = 2 + 0.3t ).Also, ( e^{0.1tau} = e^{0.1 cdot frac{u - 2}{0.3}} = e^{frac{u - 2}{3}} )So, substituting into ( I_2 ):[I_2 = int_{2}^{2 + 0.3t} frac{5 e^{frac{u - 2}{3}}}{u^2} cdot frac{du}{0.3} = frac{5}{0.3} int_{2}^{2 + 0.3t} frac{e^{frac{u - 2}{3}}}{u^2} du = frac{50}{3} int_{2}^{2 + 0.3t} frac{e^{frac{u - 2}{3}}}{u^2} du]Let me make another substitution: let ( v = frac{u - 2}{3} ), so ( u = 3v + 2 ), ( du = 3 dv ). When ( u = 2 ), ( v = 0 ). When ( u = 2 + 0.3t ), ( v = frac{0.3t}{3} = 0.1t ).Substituting:[I_2 = frac{50}{3} int_{0}^{0.1t} frac{e^{v}}{(3v + 2)^2} cdot 3 dv = 50 int_{0}^{0.1t} frac{e^{v}}{(3v + 2)^2} dv]This integral still doesn't have an elementary antiderivative, but perhaps it can be expressed in terms of the exponential integral function or other special functions.Alternatively, we can express it as:[I_2 = 50 int_{0}^{0.1t} frac{e^{v}}{(3v + 2)^2} dv]Which might be left as is or evaluated numerically.Given that both ( I_1 ) and ( I_2 ) involve integrals that can't be expressed in elementary terms, perhaps the best approach is to leave ( H(t) ) expressed in terms of these integrals, or use numerical methods to evaluate them at ( t = 12 ).However, since the problem doesn't provide specific values for ( k ), ( a ), ( G_0 ), and asks for an evaluation at ( t = 12 ), it's likely that we need to express ( H(12) ) in terms of the exponential integral function or other special functions.Alternatively, perhaps there's a relationship between ( H(t) ) and ( G(t) ) that can simplify the expression.Wait, let me recall that ( H(t) ) is defined as:[H(t) = int_0^t left( frac{2G(tau)}{L(tau)} + frac{D(tau)}{L(tau)^2} right ) dtau]And from the first part, we have an expression for ( G(t) ). So, perhaps substituting ( G(t) ) into ( H(t) ) gives us an expression in terms of the exponential integral function.But as we saw earlier, this leads to integrals involving ( Ei ) which are difficult to evaluate.Given the complexity, perhaps the problem expects us to express ( H(t) ) in terms of ( G(t) ) and its integral, but without specific constants, it's hard to proceed.Alternatively, maybe the problem is designed such that ( H(t) ) can be expressed in terms of ( G(t) ) and ( L(t) ) without integrating, but I don't see an immediate way.Wait, perhaps differentiating ( H(t) ) and relating it to ( G(t) ) and ( L(t) ).We have:[frac{dH}{dt} = frac{2G(t)}{L(t)} + frac{D(t)}{L(t)^2}]But from the first part, ( D(t) = 5 e^{0.1t} ), so:[frac{dH}{dt} = frac{2G(t)}{L(t)} + frac{5 e^{0.1t}}{L(t)^2}]But we also have ( frac{dG}{dt} = k cdot frac{5 e^{0.1t}}{L(t) + a} )So, perhaps we can write:[frac{dH}{dt} = frac{2G(t)}{L(t)} + frac{5 e^{0.1t}}{L(t)^2} = frac{2G(t)}{L(t)} + frac{L(t) + a}{k} cdot frac{dG}{dt} cdot frac{1}{L(t)^2}]Wait, that's similar to what I had before. Not sure.Alternatively, perhaps express ( frac{dH}{dt} ) in terms of ( G(t) ) and ( frac{dG}{dt} ):[frac{dH}{dt} = frac{2G(t)}{L(t)} + frac{5 e^{0.1t}}{L(t)^2}]But since ( frac{dG}{dt} = k cdot frac{5 e^{0.1t}}{L(t) + a} ), we can solve for ( 5 e^{0.1t} ):[5 e^{0.1t} = frac{L(t) + a}{k} cdot frac{dG}{dt}]Substituting into ( frac{dH}{dt} ):[frac{dH}{dt} = frac{2G(t)}{L(t)} + frac{frac{L(t) + a}{k} cdot frac{dG}{dt}}{L(t)^2} = frac{2G(t)}{L(t)} + frac{(L(t) + a)}{k L(t)^2} cdot frac{dG}{dt}]Which is the same expression as before. So, it seems we're going in circles.Given the time constraints and the complexity, perhaps the best approach is to accept that ( H(t) ) cannot be expressed in a simple closed form and instead leave it in terms of integrals involving the exponential integral function.Therefore, the final answer for ( H(12) ) would be:[H(12) = int_0^{12} left( frac{2G(tau)}{L(tau)} + frac{5 e^{0.1tau}}{L(tau)^2} right ) dtau]Where ( G(tau) ) is given by the general solution from part 1.However, since the problem asks to evaluate ( H(t) ) at ( t = 12 ), and given that the integral involves special functions, it's likely that the answer is expected to be expressed in terms of these functions without numerical evaluation.Alternatively, perhaps the problem expects us to recognize that ( H(t) ) can be expressed in terms of ( G(t) ) and ( L(t) ) in a simplified form, but I don't see a straightforward way to do that.Given all this, I think the best approach is to present the general solution for ( G(t) ) as found in part 1 and express ( H(t) ) in terms of integrals involving ( G(t) ) and ( D(t) ), acknowledging that further simplification would require numerical methods or special functions.So, summarizing:1. The general solution for ( G(t) ) is:[G(t) = G_0 + 50 k e^{-(2 + a)/3} Eileft( frac{2 + a + 0.3t}{3} right )]2. The health index ( H(t) ) at ( t = 12 ) is:[H(12) = int_0^{12} left( frac{2G(tau)}{2 + 0.3tau} + frac{5 e^{0.1tau}}{(2 + 0.3tau)^2} right ) dtau]Which can be expressed in terms of the exponential integral function and other integrals, but without specific values for ( k ), ( a ), and ( G_0 ), we cannot simplify further.Therefore, the final answer for ( H(12) ) is expressed as the integral above.</think>

🧠:<think>Alright, so I have this problem about a system of differential equations modeling two competing species. The equations are:[begin{cases}frac{dx}{dt} = x(1 - x - a y) frac{dy}{dt} = y(1 - y - b x)end{cases}]where (a) and (b) are positive constants. I need to find the fixed points and classify their stability by looking at the Jacobian matrix. Then, for specific values (a = 2) and (b = 3), I have to construct the phase portrait and describe the behavior near each fixed point, as well as discuss how initial conditions affect the long-term behavior.Okay, let's start with part 1: finding the fixed points. Fixed points occur where both derivatives are zero. So, set (frac{dx}{dt} = 0) and (frac{dy}{dt} = 0).First, for (frac{dx}{dt} = 0), we have:(x(1 - x - a y) = 0)Similarly, for (frac{dy}{dt} = 0):(y(1 - y - b x) = 0)So, the possible solutions are when either x or y is zero, or the terms in the parentheses are zero.Case 1: x = 0If x = 0, plug into the second equation:(y(1 - y - 0) = 0), so y(1 - y) = 0. Thus, y = 0 or y = 1.So, fixed points are (0, 0) and (0, 1).Case 2: y = 0If y = 0, plug into the first equation:(x(1 - x - 0) = 0), so x(1 - x) = 0. Thus, x = 0 or x = 1.So, fixed points are (0, 0) and (1, 0).Case 3: Neither x nor y is zero. So, we have:1 - x - a y = 01 - y - b x = 0So, we have a system of linear equations:x + a y = 1b x + y = 1Let me write this as:x + a y = 1b x + y = 1We can solve for x and y. Let's write it in matrix form:[begin{pmatrix}1 & a b & 1end{pmatrix}begin{pmatrix}x yend{pmatrix}=begin{pmatrix}1 1end{pmatrix}]To solve this, we can use Cramer's rule or find the inverse of the matrix. Let's compute the determinant of the coefficient matrix:Determinant D = (1)(1) - (a)(b) = 1 - a bSo, if D ≠ 0, there is a unique solution. So,x = (1 * 1 - a * 1) / D = (1 - a) / (1 - a b)Similarly,y = (1 * 1 - b * 1) / D = (1 - b) / (1 - a b)Wait, hold on, maybe I should write it properly.Using Cramer's rule:x = |1 a; 1 1| / D = (1*1 - a*1) / D = (1 - a)/DSimilarly, y = |1 1; b 1| / D = (1*1 - 1*b)/D = (1 - b)/DSo, x = (1 - a)/(1 - a b)y = (1 - b)/(1 - a b)So, the third fixed point is at ( (1 - a)/(1 - a b), (1 - b)/(1 - a b) )But we have to make sure that D ≠ 0, so 1 - a b ≠ 0, which implies a b ≠ 1.So, if a b ≠ 1, we have this third fixed point. If a b = 1, then the determinant is zero, so either no solution or infinitely many solutions.But in our case, since the equations are x + a y = 1 and b x + y = 1, if a b = 1, the two equations are multiples of each other?Wait, let's check. If a b = 1, then the second equation becomes (1/a) x + y = 1, since b = 1/a.But the first equation is x + a y = 1.So, if a b = 1, then the two equations are:x + a y = 1(1/a) x + y = 1Multiply the second equation by a: x + a y = aBut the first equation is x + a y = 1. So, unless a = 1, these are inconsistent. So, if a b = 1, and a ≠ 1, then the system is inconsistent, so no solution. If a = 1, then b = 1, and both equations become x + y = 1, so infinitely many solutions along the line x + y = 1.But in our case, since a and b are positive constants, but not necessarily equal to 1. So, unless a = 1 and b = 1, which would lead to infinitely many fixed points along x + y = 1, but in general, if a b ≠ 1, we have a unique fixed point.So, in summary, the fixed points are:1. (0, 0)2. (1, 0)3. (0, 1)4. ( (1 - a)/(1 - a b), (1 - b)/(1 - a b) ) provided that a b ≠ 1.Wait, but actually, in the case when a b = 1, we have either no solution or infinitely many solutions. So, if a b = 1, the only fixed points are (0,0), (1,0), (0,1). If a b ≠ 1, we have an additional fixed point at ( (1 - a)/(1 - a b), (1 - b)/(1 - a b) ).So, that's the fixed points.Now, moving on to classifying their stability. For that, we need to compute the Jacobian matrix at each fixed point and analyze its eigenvalues.The Jacobian matrix J is given by:[J = begin{pmatrix}frac{partial}{partial x} frac{dx}{dt} & frac{partial}{partial y} frac{dx}{dt} frac{partial}{partial x} frac{dy}{dt} & frac{partial}{partial y} frac{dy}{dt}end{pmatrix}]Compute each partial derivative:First, (frac{partial}{partial x} frac{dx}{dt}):The derivative of x(1 - x - a y) with respect to x is (1 - x - a y) + x*(-1) = 1 - x - a y - x = 1 - 2x - a ySimilarly, (frac{partial}{partial y} frac{dx}{dt}):Derivative of x(1 - x - a y) with respect to y is x*(-a) = -a xSimilarly, (frac{partial}{partial x} frac{dy}{dt}):Derivative of y(1 - y - b x) with respect to x is y*(-b) = -b yAnd (frac{partial}{partial y} frac{dy}{dt}):Derivative of y(1 - y - b x) with respect to y is (1 - y - b x) + y*(-1) = 1 - y - b x - y = 1 - 2y - b xSo, putting it all together, the Jacobian matrix is:[J = begin{pmatrix}1 - 2x - a y & -a x -b y & 1 - 2y - b xend{pmatrix}]Now, we need to evaluate this matrix at each fixed point and find its eigenvalues to determine stability.Let's start with each fixed point.1. Fixed point (0, 0):Plug x=0, y=0 into J:[J(0,0) = begin{pmatrix}1 - 0 - 0 & -0 -0 & 1 - 0 - 0end{pmatrix} = begin{pmatrix}1 & 0 0 & 1end{pmatrix}]So, the eigenvalues are the diagonal entries, which are both 1. Since both eigenvalues are greater than 1, the fixed point (0, 0) is an unstable node.2. Fixed point (1, 0):Plug x=1, y=0 into J:First, compute each entry:1 - 2(1) - a(0) = 1 - 2 = -1- a x = -a(1) = -a- b y = -b(0) = 01 - 2(0) - b(1) = 1 - bSo, J(1,0) is:[begin{pmatrix}-1 & -a 0 & 1 - bend{pmatrix}]To find eigenvalues, solve det(J - λ I) = 0:[begin{vmatrix}-1 - λ & -a 0 & 1 - b - λend{vmatrix} = (-1 - λ)(1 - b - λ) - 0 = (-1 - λ)(1 - b - λ)]Set equal to zero:(-1 - λ)(1 - b - λ) = 0So, λ = -1 or λ = 1 - bSo, eigenvalues are -1 and (1 - b)Since a and b are positive constants, 1 - b could be positive or negative depending on b.But in general, for the fixed point (1, 0):- One eigenvalue is -1, which is negative.- The other eigenvalue is (1 - b). If b > 1, then 1 - b < 0; if b < 1, 1 - b > 0.So, the stability depends on the sign of the second eigenvalue.If both eigenvalues are negative, it's a stable node.If one eigenvalue is negative and the other is positive, it's a saddle point.So, for (1, 0):- If b > 1: both eigenvalues negative → stable node.- If b < 1: one eigenvalue negative, one positive → saddle point.- If b = 1: one eigenvalue is zero, so it's a non-hyperbolic fixed point, and we can't classify it just by eigenvalues.Similarly, for fixed point (0, 1):Plug x=0, y=1 into J:Compute each entry:1 - 2(0) - a(1) = 1 - a- a x = -a(0) = 0- b y = -b(1) = -b1 - 2(1) - b(0) = 1 - 2 = -1So, J(0,1) is:[begin{pmatrix}1 - a & 0 -b & -1end{pmatrix}]Eigenvalues are the diagonal entries since it's a diagonal matrix:λ1 = 1 - aλ2 = -1So, similar to (1, 0):- If a > 1: 1 - a < 0 → both eigenvalues negative → stable node.- If a < 1: 1 - a > 0 → one positive, one negative → saddle point.- If a = 1: one eigenvalue zero → non-hyperbolic.Now, the fourth fixed point, which exists when a b ≠ 1:( (1 - a)/(1 - a b), (1 - b)/(1 - a b) )Let me denote this point as (x*, y*), where:x* = (1 - a)/(1 - a b)y* = (1 - b)/(1 - a b)We need to compute the Jacobian at (x*, y*). Let's compute each entry.First, compute 1 - 2x* - a y*:1 - 2x* - a y* = 1 - 2*(1 - a)/(1 - a b) - a*(1 - b)/(1 - a b)Let me compute this:= [ (1 - a b) - 2(1 - a) - a(1 - b) ] / (1 - a b)Compute numerator:1 - a b - 2 + 2a - a + a bSimplify:1 - 2 + (2a - a) + (-a b + a b)= (-1) + a + 0= a - 1So, 1 - 2x* - a y* = (a - 1)/(1 - a b)Similarly, compute -a x*:= -a*(1 - a)/(1 - a b) = (-a + a^2)/(1 - a b)Compute -b y*:= -b*(1 - b)/(1 - a b) = (-b + b^2)/(1 - a b)Compute 1 - 2y* - b x*:1 - 2y* - b x* = 1 - 2*(1 - b)/(1 - a b) - b*(1 - a)/(1 - a b)Compute numerator:1 - a b - 2(1 - b) - b(1 - a)= 1 - a b - 2 + 2b - b + a bSimplify:1 - 2 + (2b - b) + (-a b + a b)= (-1) + b + 0= b - 1So, 1 - 2y* - b x* = (b - 1)/(1 - a b)Therefore, the Jacobian matrix at (x*, y*) is:[J(x*, y*) = begin{pmatrix}frac{a - 1}{1 - a b} & frac{-a + a^2}{1 - a b} frac{-b + b^2}{1 - a b} & frac{b - 1}{1 - a b}end{pmatrix}]Factor out 1/(1 - a b):[J(x*, y*) = frac{1}{1 - a b} begin{pmatrix}a - 1 & -a + a^2 -b + b^2 & b - 1end{pmatrix}]Let me denote the matrix as:[M = begin{pmatrix}a - 1 & a(a - 1) b(b - 1) & b - 1end{pmatrix}]So, J = M / (1 - a b)To find eigenvalues, we can compute the trace and determinant of J.But since J = M / (1 - a b), the eigenvalues of J are the eigenvalues of M divided by (1 - a b).Alternatively, we can compute the trace and determinant of J.Compute trace of J:Tr(J) = [ (a - 1) + (b - 1) ] / (1 - a b) = (a + b - 2)/(1 - a b)Compute determinant of J:det(J) = [ (a - 1)(b - 1) - (-a + a^2)(-b + b^2) ] / (1 - a b)^2Wait, let me compute numerator:det(M) = (a - 1)(b - 1) - [a(a - 1)][b(b - 1)]= (a - 1)(b - 1) - a b (a - 1)(b - 1)= (a - 1)(b - 1)(1 - a b)So, det(J) = det(M)/(1 - a b)^2 = [ (a - 1)(b - 1)(1 - a b) ] / (1 - a b)^2 = (a - 1)(b - 1)/(1 - a b)So, determinant of J is (a - 1)(b - 1)/(1 - a b)So, to summarize:Tr(J) = (a + b - 2)/(1 - a b)det(J) = (a - 1)(b - 1)/(1 - a b)Now, to find eigenvalues, we can use the quadratic formula:λ = [ Tr(J) ± sqrt(Tr(J)^2 - 4 det(J)) ] / 2But perhaps we can find a better way.Alternatively, note that the eigenvalues of J are the eigenvalues of M scaled by 1/(1 - a b). So, if we can find the eigenvalues of M, we can just divide them by (1 - a b) to get eigenvalues of J.Let me try that.Compute eigenvalues of M:M = [ [a - 1, a(a - 1)], [b(b - 1), b - 1] ]Let me factor out (a - 1) from the first row and (b - 1) from the second column:M = (a - 1)(b - 1) * [ [1, a], [b, 1] ] / (a - 1)(b - 1) ? Wait, no.Wait, actually, M can be written as:M = (a - 1)(b - 1) * [ [1, a/(b - 1)], [b/(a - 1), 1] ]Hmm, maybe not helpful.Alternatively, let's compute the eigenvalues of M.The characteristic equation is:| M - λ I | = 0So,| (a - 1 - λ) a(a - 1) || b(b - 1) (b - 1 - λ) | = 0Compute determinant:(a - 1 - λ)(b - 1 - λ) - a(a - 1) b(b - 1) = 0Let me expand this:= (a - 1)(b - 1) - (a - 1)λ - (b - 1)λ + λ^2 - a b (a - 1)(b - 1) = 0Wait, that's a bit messy. Maybe factor terms.Let me denote c = (a - 1), d = (b - 1). Then,The equation becomes:(c - λ)(d - λ) - a b c d = 0Expanding:c d - c λ - d λ + λ^2 - a b c d = 0Rearranged:λ^2 - (c + d) λ + c d (1 - a b) = 0So, eigenvalues are:λ = [ (c + d) ± sqrt( (c + d)^2 - 4 c d (1 - a b) ) ] / 2Substituting back c = a - 1, d = b - 1:λ = [ (a - 1 + b - 1) ± sqrt( (a + b - 2)^2 - 4 (a - 1)(b - 1)(1 - a b) ) ] / 2Simplify:λ = [ (a + b - 2) ± sqrt( (a + b - 2)^2 - 4 (a - 1)(b - 1)(1 - a b) ) ] / 2This is getting complicated, but perhaps we can factor or simplify the discriminant.Compute discriminant D:D = (a + b - 2)^2 - 4 (a - 1)(b - 1)(1 - a b)Let me expand both terms.First, (a + b - 2)^2 = a^2 + b^2 + 4 + 2ab - 4a - 4bSecond, 4 (a - 1)(b - 1)(1 - a b):First compute (a - 1)(b - 1) = ab - a - b + 1Then multiply by (1 - a b):(ab - a - b + 1)(1 - a b) = ab(1 - a b) - a(1 - a b) - b(1 - a b) + 1(1 - a b)= ab - a^2 b^2 - a + a^2 b - b + a b^2 + 1 - a bCombine like terms:ab - a^2 b^2 - a + a^2 b - b + a b^2 + 1 - a b= (-a^2 b^2) + (a^2 b + a b^2) + (ab - a b) + (-a - b) + 1= -a^2 b^2 + a^2 b + a b^2 + 0 + (-a - b) + 1So, 4*(that) = 4*(-a^2 b^2 + a^2 b + a b^2 - a - b + 1)So, D = (a^2 + b^2 + 4 + 2ab - 4a - 4b) - 4*(-a^2 b^2 + a^2 b + a b^2 - a - b + 1)= a^2 + b^2 + 4 + 2ab - 4a - 4b + 4a^2 b^2 - 4a^2 b - 4a b^2 + 4a + 4b - 4Simplify term by term:a^2 + b^2 + 4 + 2ab - 4a - 4b + 4a^2 b^2 - 4a^2 b - 4a b^2 + 4a + 4b - 4Combine like terms:a^2 + b^2 + (4 - 4) + 2ab + (-4a + 4a) + (-4b + 4b) + 4a^2 b^2 - 4a^2 b - 4a b^2Simplify:a^2 + b^2 + 2ab + 4a^2 b^2 - 4a^2 b - 4a b^2Factor where possible:Note that a^2 + 2ab + b^2 = (a + b)^2So, D = (a + b)^2 + 4a^2 b^2 - 4a^2 b - 4a b^2Factor 4a^2 b^2 - 4a^2 b - 4a b^2:= 4a^2 b^2 - 4a^2 b - 4a b^2 = 4a b^2(a - 1) - 4a^2 bWait, maybe factor 4ab:= 4ab(a b - a - b)Wait, let's see:4a^2 b^2 - 4a^2 b - 4a b^2 = 4a^2 b^2 - 4a^2 b - 4a b^2 = 4a b(a b - a - b)Wait, 4a^2 b^2 - 4a^2 b - 4a b^2 = 4a b(a b - a - b + 0). Hmm, not quite.Alternatively, factor 4ab:= 4ab(a b - a - b + 0). Hmm, not helpful.Alternatively, factor 4ab:= 4ab(a b - a - b + 1) - 4abWait, maybe not. Alternatively, let's see:Let me factor 4ab from the first two terms:= 4ab(a b - a) - 4a b^2 = 4ab(a(b - 1)) - 4a b^2 = 4a^2 b(b - 1) - 4a b^2Hmm, not helpful.Alternatively, perhaps factor 4ab:= 4ab(a b - a - b) + something? Not sure.Alternatively, perhaps factor as:= 4a^2 b^2 - 4a^2 b - 4a b^2 = 4a^2 b^2 - 4a^2 b - 4a b^2 = 4a b(a b - a - b)Wait, 4a b(a b - a - b) = 4a^2 b^2 - 4a^2 b - 4a b^2. Yes, exactly.So, D = (a + b)^2 + 4ab(a b - a - b)So, D = (a + b)^2 + 4ab(a b - a - b)Hmm, not sure if this helps, but perhaps we can write it as:D = (a + b)^2 - 4ab(a + b - a b)Wait, because 4ab(a b - a - b) = -4ab(a + b - a b)So, D = (a + b)^2 - 4ab(a + b - a b)Hmm, maybe factor further.Alternatively, perhaps factor (a + b):D = (a + b)[(a + b) - 4ab(1 - (a b)/(a + b))]Not sure. Maybe this is not the right path.Alternatively, perhaps we can consider specific values for a and b, but since we are supposed to do this generally, maybe we can proceed differently.Alternatively, note that the trace and determinant of J can be used to determine the type of fixed point.Recall that for a 2x2 system, the fixed point is:- Stable node if both eigenvalues are negative.- Unstable node if both eigenvalues are positive.- Saddle point if eigenvalues have opposite signs.- Stable spiral if eigenvalues are complex with negative real part.- Unstable spiral if eigenvalues are complex with positive real part.So, let's analyze the trace and determinant of J.Tr(J) = (a + b - 2)/(1 - a b)det(J) = (a - 1)(b - 1)/(1 - a b)Note that 1 - a b is in the denominator. So, depending on whether 1 - a b is positive or negative, the signs of Tr(J) and det(J) will change.Let me consider two cases:Case 1: 1 - a b > 0, i.e., a b < 1Case 2: 1 - a b < 0, i.e., a b > 1Case 1: a b < 1Then, 1 - a b > 0, so Tr(J) = (a + b - 2)/(positive)det(J) = (a - 1)(b - 1)/(positive)So, Tr(J) sign depends on (a + b - 2)det(J) sign depends on (a - 1)(b - 1)Case 1a: a + b - 2 > 0 and (a - 1)(b - 1) > 0Then, Tr(J) > 0, det(J) > 0So, both eigenvalues have positive real parts if they are real, or spiral outwards if complex.But since det(J) > 0 and Tr(J) > 0, it's either a stable spiral (if complex) or unstable node (if real). Wait, no:Wait, if Tr(J) > 0 and det(J) > 0, the eigenvalues have positive real parts, so it's either an unstable node or unstable spiral.But to check if eigenvalues are real or complex, we can look at the discriminant D of the characteristic equation.If D > 0, real eigenvalues; if D < 0, complex eigenvalues.But D is complicated, so perhaps we can think in terms of the parameters.Alternatively, perhaps we can note that if a b < 1, then the fixed point (x*, y*) is a stable node or spiral, depending on the trace and determinant.Wait, actually, let's think about the signs.If a b < 1, then 1 - a b > 0.If a > 1 and b > 1, then (a - 1)(b - 1) > 0, so det(J) > 0.And if a + b > 2, then Tr(J) > 0.So, in this case, Tr(J) > 0 and det(J) > 0, so eigenvalues have positive real parts, so it's an unstable node or spiral.But if a + b < 2, then Tr(J) < 0, and det(J) > 0, so eigenvalues have negative real parts, so it's a stable node or spiral.Wait, but if a + b = 2, Tr(J) = 0, which is a special case.Similarly, if a < 1 and b < 1, then (a - 1)(b - 1) > 0, so det(J) > 0.If a + b > 2, Tr(J) > 0; if a + b < 2, Tr(J) < 0.So, in summary, for a b < 1:- If a + b > 2: Tr(J) > 0, det(J) > 0 → unstable node or spiral.- If a + b < 2: Tr(J) < 0, det(J) > 0 → stable node or spiral.- If a + b = 2: Tr(J) = 0, det(J) > 0 → eigenvalues are purely imaginary, so it's a center.But in our case, a and b are positive constants, so a + b > 0, but whether it's greater than 2 or not depends on the specific values.Similarly, if a b > 1, then 1 - a b < 0.So, Tr(J) = (a + b - 2)/(negative)det(J) = (a - 1)(b - 1)/(negative)So, Tr(J) sign is opposite of (a + b - 2)det(J) sign is opposite of (a - 1)(b - 1)So, Case 2: a b > 1Case 2a: a + b - 2 > 0 → Tr(J) < 0(a - 1)(b - 1) > 0 → det(J) < 0So, Tr(J) < 0, det(J) < 0 → eigenvalues have opposite signs → saddle point.Case 2b: a + b - 2 < 0 → Tr(J) > 0(a - 1)(b - 1) > 0 → det(J) < 0So, Tr(J) > 0, det(J) < 0 → eigenvalues have opposite signs → saddle point.Case 2c: a + b - 2 = 0 → Tr(J) = 0det(J) = (a - 1)(b - 1)/(negative)If (a - 1)(b - 1) > 0, then det(J) < 0If (a - 1)(b - 1) < 0, then det(J) > 0But since a b > 1, and a + b = 2, it's possible that a and b are both greater than 1 or both less than 1, but since a b > 1, they can't both be less than 1.Wait, if a + b = 2 and a b > 1, then both a and b must be greater than 1, because if one is less than 1, the other would have to be greater than 1, but their product would be less than 1*(something). Wait, let's see.Suppose a + b = 2, a b > 1.Let me solve for a and b.Let a = 2 - b.Then, a b = (2 - b) b = 2b - b^2 > 1So, 2b - b^2 - 1 > 0 → -b^2 + 2b - 1 > 0 → -(b^2 - 2b + 1) > 0 → -(b - 1)^2 > 0But -(b - 1)^2 is always ≤ 0, so it can't be > 0. So, there is no solution where a + b = 2 and a b > 1.Therefore, when a b > 1, a + b cannot be equal to 2.So, in Case 2, a b > 1, and a + b ≠ 2, so Tr(J) and det(J) have opposite signs, leading to a saddle point.Wait, but earlier, when a b > 1, if a + b > 2, Tr(J) < 0, det(J) < 0 (if a > 1, b > 1). Wait, no:Wait, when a b > 1, 1 - a b < 0.So, det(J) = (a - 1)(b - 1)/(1 - a b)If a > 1 and b > 1, (a - 1)(b - 1) > 0, so det(J) < 0.Similarly, if a < 1 and b < 1, (a - 1)(b - 1) > 0, but since a b > 1, both a and b must be greater than 1, because if one is less than 1, the other would have to be greater than 1 to have a b > 1, but then (a - 1)(b - 1) would be negative.Wait, let's clarify:If a b > 1, possible cases:1. Both a > 1 and b > 1: Then, (a - 1)(b - 1) > 0.2. One of a or b > 1, the other < 1: But then, a b > 1 requires that the larger one is sufficiently large to compensate.But in this case, (a - 1)(b - 1) < 0.So, in Case 2: a b > 1.Subcase 1: a > 1, b > 1.Then, det(J) = (positive)/(negative) = negative.Tr(J) = (a + b - 2)/(negative). If a + b > 2, Tr(J) < 0; if a + b < 2, Tr(J) > 0.But if a > 1 and b > 1, a + b > 2 is possible, but not necessarily always.Wait, for example, a = 1.5, b = 1.5: a + b = 3 > 2.But a = 1.2, b = 1.2: a + b = 2.4 > 2.Wait, actually, if a > 1 and b > 1, then a + b > 2 only if both are greater than 1, but actually, a = 1.1, b = 1.1: a + b = 2.2 > 2.But if a = 1.5, b = 1.5: a + b = 3 > 2.Wait, actually, if a > 1 and b > 1, then a + b > 2 is always true, because a > 1 and b > 1 implies a + b > 2.Wait, no. If a = 1.1, b = 1.1: a + b = 2.2 > 2.But if a = 1.01, b = 1.01: a + b = 2.02 > 2.Wait, actually, if a > 1 and b > 1, then a + b > 2 is always true, because a > 1 and b > 1 implies a + b > 2.Wait, no, actually, if a approaches 1 from above, and b approaches 1 from above, a + b approaches 2 from above.So, a + b can be just slightly above 2.So, in Subcase 1: a > 1, b > 1, a + b > 2.Thus, Tr(J) = (a + b - 2)/(1 - a b) < 0, since denominator is negative.det(J) = (a - 1)(b - 1)/(1 - a b) < 0, since numerator positive, denominator negative.So, Tr(J) < 0, det(J) < 0.Thus, eigenvalues have opposite signs: one positive, one negative. So, it's a saddle point.Subcase 2: a b > 1, but one of a or b < 1.Say, a > 1, b < 1, but a b > 1.Then, (a - 1)(b - 1) < 0, since a - 1 > 0, b - 1 < 0.Thus, det(J) = (negative)/(negative) = positive.Tr(J) = (a + b - 2)/(1 - a b). Since a > 1, b < 1, a + b could be > or < 2.But since a b > 1, and b < 1, a must be > 1/b > 1.So, a + b: depends on how large a is.But regardless, Tr(J) = (a + b - 2)/(negative). So, if a + b > 2, Tr(J) < 0; if a + b < 2, Tr(J) > 0.But in this subcase, a b > 1, b < 1, a > 1.So, a + b can be greater or less than 2.For example, a = 3, b = 0.5: a b = 1.5 > 1, a + b = 3.5 > 2.Another example: a = 2, b = 0.6: a b = 1.2 > 1, a + b = 2.6 > 2.Another example: a = 1.5, b = 0.8: a b = 1.2 > 1, a + b = 2.3 > 2.Wait, is it possible for a + b < 2 in this subcase?Suppose a = 1.1, b = 0.91: a b = 1.001 > 1, a + b = 2.01 > 2.If a = 1.01, b = 0.99: a b ≈ 1.00, but slightly above 1, a + b = 2.00.So, it seems that in this subcase, a + b is just slightly above 2, but not necessarily always.Wait, actually, if a approaches 1 from above, and b approaches 1 from below, a + b approaches 2 from above.But can a + b be less than 2?Wait, suppose a = 1.5, b = 0.75: a b = 1.125 > 1, a + b = 2.25 > 2.Another example: a = 1.2, b = 0.85: a b = 1.02 > 1, a + b = 2.05 > 2.Wait, maybe it's not possible for a + b < 2 in this subcase.Wait, let's think algebraically.Given a b > 1, and b < 1, then a > 1/b.Since b < 1, 1/b > 1.So, a > 1/b.Thus, a + b > 1/b + b.We can analyze the function f(b) = b + 1/b for b > 0.The minimum of f(b) is at b = 1, f(1) = 2.Since b < 1, f(b) = b + 1/b > 2.Thus, a + b > 2.Therefore, in Subcase 2, a + b > 2, so Tr(J) = (a + b - 2)/(1 - a b) < 0 (since denominator negative).det(J) = (a - 1)(b - 1)/(1 - a b) = (positive)(negative)/(negative) = positive.So, Tr(J) < 0, det(J) > 0.Thus, eigenvalues have negative real parts (if real) or spiral inwards (if complex). So, it's a stable node or spiral.Wait, but earlier, in Subcase 1, we had a saddle point, and in Subcase 2, we have a stable node or spiral.So, in summary, for the fixed point (x*, y*):- If a b < 1: - If a + b > 2: Unstable node or spiral. - If a + b < 2: Stable node or spiral. - If a + b = 2: Center (neutral stability).- If a b > 1: - If a > 1, b > 1: Saddle point. - If one of a or b < 1, but a b > 1: Stable node or spiral.Wait, but in Subcase 2, when a b > 1 and one of a or b < 1, we have a stable node or spiral.This seems a bit conflicting, but perhaps it's correct.Alternatively, perhaps it's better to note that when a b > 1, the fixed point (x*, y*) is a saddle point if both a > 1 and b > 1, and a stable node otherwise.But I need to make sure.Wait, in Subcase 2, when a b > 1 and one of a or b < 1, we have Tr(J) < 0 and det(J) > 0, so eigenvalues have negative real parts, so it's a stable node or spiral.So, in that case, (x*, y*) is stable.But in Subcase 1, when a > 1, b > 1, a b > 1, we have Tr(J) < 0 and det(J) < 0, so eigenvalues have opposite signs, so it's a saddle point.So, in conclusion, for the fixed point (x*, y*):- If a b < 1: - If a + b > 2: Unstable node. - If a + b < 2: Stable node. - If a + b = 2: Center.- If a b > 1: - If a > 1 and b > 1: Saddle point. - If one of a or b < 1: Stable node.Wait, but in the case when a b > 1 and one of a or b < 1, we have a stable node.But in reality, when a b > 1 and one of a or b < 1, the fixed point (x*, y*) is stable.So, to summarize:Fixed points:1. (0, 0): Unstable node.2. (1, 0): - If b > 1: Stable node. - If b < 1: Saddle point. - If b = 1: Non-hyperbolic.3. (0, 1): - If a > 1: Stable node. - If a < 1: Saddle point. - If a = 1: Non-hyperbolic.4. (x*, y*): - If a b < 1: - If a + b > 2: Unstable node. - If a + b < 2: Stable node. - If a + b = 2: Center. - If a b > 1: - If a > 1 and b > 1: Saddle point. - If one of a or b < 1: Stable node.But this is getting quite involved. Maybe it's better to proceed to part 2, where specific values are given: a = 2, b = 3.So, let's move on to part 2.Given a = 2, b = 3.First, let's find the fixed points.From earlier, fixed points are:1. (0, 0)2. (1, 0)3. (0, 1)4. (x*, y*) = ( (1 - a)/(1 - a b), (1 - b)/(1 - a b) )Compute x* and y*:a = 2, b = 3.So, x* = (1 - 2)/(1 - 2*3) = (-1)/(1 - 6) = (-1)/(-5) = 1/5 = 0.2y* = (1 - 3)/(1 - 6) = (-2)/(-5) = 2/5 = 0.4So, fixed point is (0.2, 0.4)Now, let's classify each fixed point.First, (0, 0):As before, Jacobian is [[1, 0], [0, 1]], eigenvalues 1, 1. So, unstable node.Second, (1, 0):Compute Jacobian at (1, 0):J(1,0) = [[-1, -a], [0, 1 - b]] = [[-1, -2], [0, 1 - 3]] = [[-1, -2], [0, -2]]Eigenvalues are -1 and -2, both negative. So, stable node.Third, (0, 1):J(0,1) = [[1 - a, 0], [-b, -1]] = [[1 - 2, 0], [-3, -1]] = [[-1, 0], [-3, -1]]Eigenvalues are -1 and -1, both negative. So, stable node.Fourth, (0.2, 0.4):Compute Jacobian at (0.2, 0.4):From earlier, J(x*, y*) = M / (1 - a b), where M is:M = [[a - 1, a(a - 1)], [b(b - 1), b - 1]] = [[2 - 1, 2(2 - 1)], [3(3 - 1), 3 - 1]] = [[1, 2], [6, 2]]So, M = [[1, 2], [6, 2]]Thus, J = M / (1 - 6) = [[1, 2], [6, 2]] / (-5) = [[-1/5, -2/5], [-6/5, -2/5]]Compute trace and determinant:Tr(J) = (-1/5) + (-2/5) = -3/5det(J) = (-1/5)(-2/5) - (-2/5)(-6/5) = (2/25) - (12/25) = -10/25 = -2/5So, trace is negative, determinant is negative.Thus, eigenvalues have opposite signs: one positive, one negative. So, it's a saddle point.So, summarizing the fixed points for a = 2, b = 3:1. (0, 0): Unstable node.2. (1, 0): Stable node.3. (0, 1): Stable node.4. (0.2, 0.4): Saddle point.Now, to construct the phase portrait, we need to understand the behavior of trajectories near each fixed point and how they interact.Given that (0, 0) is an unstable node, trajectories near (0, 0) will move away from it.(1, 0) and (0, 1) are stable nodes, so trajectories near them will converge towards them.The saddle point (0.2, 0.4) will have trajectories approaching it along one direction and moving away along another.Given that a = 2, b = 3, which are both greater than 1, and a b = 6 > 1, so in this case, the fixed point (0.2, 0.4) is a saddle point.Now, to describe the behavior of the solutions near each fixed point:Near (0, 0): Since it's an unstable node, trajectories move away from (0, 0) in all directions.Near (1, 0): Trajectories spiral or move directly towards (1, 0). Since the Jacobian at (1, 0) has eigenvalues -1 and -2, both real and negative, so it's a stable node, not a spiral. So, trajectories approach (1, 0) along straight lines.Similarly, near (0, 1): The Jacobian has eigenvalues -1 and -1, so it's a stable node with trajectories approaching along straight lines.Near (0.2, 0.4): As a saddle point, trajectories approach along the stable manifold and move away along the unstable manifold.Now, how do initial conditions influence the long-term behavior?Given that we have two stable nodes at (1, 0) and (0, 1), and a saddle point at (0.2, 0.4), the phase portrait will have trajectories that either converge to (1, 0), (0, 1), or are repelled by the saddle point.The presence of the saddle point suggests that there is a separatrix (the stable manifold of the saddle point) that divides the phase plane into regions where trajectories approach either (1, 0) or (0, 1).Thus, depending on the initial conditions, the populations will either tend to (1, 0), meaning species x survives and y dies out, or tend to (0, 1), meaning species y survives and x dies out.If the initial conditions are exactly on the stable manifold of the saddle point, the trajectory will approach the saddle point, but since it's a saddle, it's unstable, so slight perturbations will lead to convergence to one of the stable nodes.Therefore, the long-term behavior is determined by whether the initial populations are above or below the separatrix defined by the stable manifold of the saddle point.In summary, for a = 2, b = 3:- The system has four fixed points: three nodes and a saddle.- The unstable node at (0, 0) repels trajectories.- The stable nodes at (1, 0) and (0, 1) attract trajectories.- The saddle point at (0.2, 0.4) acts as a separator.- Depending on initial conditions, the system will approach either (1, 0) or (0, 1), with the separatrix determining the basin of attraction for each stable node.</think>