answer：Okay, so Emily is trying to decide between two houses, House A and House B, for her fresh start after a divorce. She wants to make a financially sound decision, considering both the initial cost and the long-term investment potential. I need to help her figure out which house is better by solving two sub-problems: calculating the expected value after 10 years and determining the mortgage payments over those 10 years.Starting with Sub-problem 1: calculating the expected value of both houses after 10 years. I remember that appreciation is calculated using compound interest, so the formula should be something like Future Value = Present Value * (1 + rate)^time. Let me write that down.For House A:- Price: 300,000- Appreciation rate: 3% per year- Time: 10 yearsSo, Future Value of House A = 300,000 * (1 + 0.03)^10Similarly, for House B:- Price: 250,000- Appreciation rate: 4% per year- Time: 10 yearsFuture Value of House B = 250,000 * (1 + 0.04)^10I think I can calculate these using a calculator. Let me compute each step.First, for House A:(1 + 0.03) = 1.031.03^10 is approximately... let me calculate. I know that 1.03^10 is roughly 1.3439. So, 300,000 * 1.3439 = ?300,000 * 1.3439. Let's compute 300,000 * 1 = 300,000, 300,000 * 0.3439 = 103,170. So total is 300,000 + 103,170 = 403,170. So, approximately 403,170.For House B:(1 + 0.04) = 1.041.04^10 is approximately 1.4802. So, 250,000 * 1.4802 = ?250,000 * 1 = 250,000, 250,000 * 0.4802 = 120,050. So total is 250,000 + 120,050 = 370,050. Approximately 370,050.Wait, hold on. That seems a bit low for House B. Let me double-check the exponent. 1.04^10 is indeed about 1.4802. So 250,000 * 1.4802 is 370,050. Hmm, okay, so House A appreciates more in absolute terms because it's a more expensive house, even though House B has a higher appreciation rate.So, after 10 years, House A is worth about 403,170, and House B is about 370,050. So, House A is more valuable in the future.Moving on to Sub-problem 2: calculating the monthly mortgage payments and total amount paid over 10 years.Emily is taking out a mortgage for 80% of the purchase price, with a fixed annual interest rate of 5% for 30 years. She plans to live there for 10 years, so we need to calculate the monthly payments and total paid over 10 years.First, let's find the loan amounts for each house.For House A:80% of 300,000 = 0.8 * 300,000 = 240,000For House B:80% of 250,000 = 0.8 * 250,000 = 200,000Now, the monthly mortgage payment can be calculated using the formula for a fixed-rate mortgage:M = P * [i(1 + i)^n] / [(1 + i)^n - 1]Where:- M is the monthly payment- P is the principal loan amount- i is the monthly interest rate (annual rate divided by 12)- n is the number of payments (loan term in months)But since Emily is only paying for 10 years, we need to calculate the total amount paid over 10 years, which is 120 payments. However, the loan is for 30 years, so the monthly payment is based on 360 months, but she will only make 120 payments.Wait, that might complicate things. Alternatively, maybe she is considering a 10-year mortgage? But the problem says a 30-year term. So she will have a 30-year mortgage but only make payments for 10 years before selling the house. So we need to calculate the monthly payment based on a 30-year term but sum the payments for 10 years.So, let's compute the monthly payment first for each house, then multiply by 120 to get the total paid over 10 years.Starting with House A:P = 240,000Annual interest rate = 5%, so monthly rate i = 5% / 12 = 0.0041667Number of payments for the loan term n = 360Compute M:M = 240,000 * [0.0041667*(1 + 0.0041667)^360] / [(1 + 0.0041667)^360 - 1]First, compute (1 + 0.0041667)^360. That's a big exponent, but I remember that (1 + 0.0041667)^360 is approximately e^(0.05*30) = e^1.5 ≈ 4.4817, but actually, more accurately, (1 + 0.0041667)^360 ≈ 4.4677.So, numerator: 0.0041667 * 4.4677 ≈ 0.018615Denominator: 4.4677 - 1 = 3.4677So, M ≈ 240,000 * (0.018615 / 3.4677) ≈ 240,000 * 0.005368 ≈ 240,000 * 0.005368 ≈ 1,288.32So, approximately 1,288.32 per month.Total paid over 10 years: 1,288.32 * 120 = ?1,288.32 * 100 = 128,8321,288.32 * 20 = 25,766.4Total: 128,832 + 25,766.4 = 154,598.4So, approximately 154,598.40For House B:P = 200,000Same i = 0.0041667Same n = 360Compute M:M = 200,000 * [0.0041667*(1 + 0.0041667)^360] / [(1 + 0.0041667)^360 - 1]We already know (1 + 0.0041667)^360 ≈ 4.4677Numerator: 0.0041667 * 4.4677 ≈ 0.018615Denominator: 3.4677So, M ≈ 200,000 * (0.018615 / 3.4677) ≈ 200,000 * 0.005368 ≈ 1,073.60Total paid over 10 years: 1,073.60 * 120 = ?1,073.60 * 100 = 107,3601,073.60 * 20 = 21,472Total: 107,360 + 21,472 = 128,832So, approximately 128,832Wait, let me verify these calculations because the difference seems a bit large. Alternatively, maybe I should use the exact formula or a more precise method.Alternatively, I can use the present value of an annuity formula.But perhaps it's easier to use the PMT function in Excel or a financial calculator, but since I'm doing this manually, I'll try to be precise.Alternatively, I can use the formula:M = P * (i(1 + i)^n) / ((1 + i)^n - 1)For House A:i = 0.05/12 ≈ 0.00416667n = 360Compute (1 + i)^n = (1.00416667)^360I can compute this as e^(n * ln(1.00416667)) ≈ e^(360 * 0.004158) ≈ e^(1.4969) ≈ 4.4677So, (1 + i)^n ≈ 4.4677Then, numerator: i*(1 + i)^n ≈ 0.00416667 * 4.4677 ≈ 0.018615Denominator: (1 + i)^n - 1 ≈ 4.4677 - 1 = 3.4677So, M ≈ 240,000 * (0.018615 / 3.4677) ≈ 240,000 * 0.005368 ≈ 1,288.32Yes, that seems consistent.Similarly for House B:M ≈ 200,000 * (0.018615 / 3.4677) ≈ 200,000 * 0.005368 ≈ 1,073.60So, the monthly payments are approximately 1,288.32 for House A and 1,073.60 for House B.Total paid over 10 years:House A: 1,288.32 * 120 ≈ 154,598.40House B: 1,073.60 * 120 ≈ 128,832.00So, now, to compare the total cost over 10 years, we need to consider both the initial down payment and the mortgage payments.Wait, actually, Emily is taking out a mortgage for 80% of the purchase price, so she needs to pay 20% down. So, the initial down payment is 20% of the house price.For House A:Down payment: 0.2 * 300,000 = 60,000Mortgage payments over 10 years: 154,598.40Total cost over 10 years: 60,000 + 154,598.40 = 214,598.40For House B:Down payment: 0.2 * 250,000 = 50,000Mortgage payments over 10 years: 128,832.00Total cost over 10 years: 50,000 + 128,832 = 178,832.00But wait, we also need to consider the appreciation. Because after 10 years, she sells the house, so the total cost is the initial down payment plus the mortgage payments, but she also gets the sale proceeds.So, the net gain would be the sale price minus the total amount she has paid (down payment + mortgage).Alternatively, the total cost is the down payment plus the mortgage payments, and the total return is the sale price.So, for each house, the net gain is sale price - (down payment + mortgage payments).Let me compute that.For House A:Sale price after 10 years: 403,170Total cost: 60,000 + 154,598.40 = 214,598.40Net gain: 403,170 - 214,598.40 ≈ 188,571.60For House B:Sale price after 10 years: 370,050Total cost: 50,000 + 128,832 = 178,832Net gain: 370,050 - 178,832 ≈ 191,218Wait, that's interesting. House B has a slightly higher net gain despite having a lower sale price. Because the total cost is lower.But let me verify the calculations.House A:Down payment: 60kMortgage: ~154.6kTotal cost: ~214.6kSale: ~403.17kNet: 403.17k - 214.6k ≈ 188.57kHouse B:Down payment: 50kMortgage: ~128.83kTotal cost: ~178.83kSale: ~370.05kNet: 370.05k - 178.83k ≈ 191.22kSo, House B gives a slightly higher net gain of about 191k vs. 188.5k for House A.But wait, let's think about the appreciation rates again. House A has a lower appreciation rate but a higher sale price. House B has a higher appreciation rate but a lower initial price.Alternatively, maybe we should consider the return on investment, like the net gain divided by the initial investment.For House A:Net gain: ~188.57kInitial investment: 60k down + 0 mortgage upfront? Wait, no, the initial investment is the down payment, and the mortgage is a liability.Wait, actually, the initial investment is the down payment, and the mortgage is a loan that she will pay over time. So, the return on investment would be the net gain divided by the initial down payment.So, ROI for House A: 188,571.60 / 60,000 ≈ 3.1428 or 314.28%ROI for House B: 191,218 / 50,000 ≈ 3.82436 or 382.44%So, House B has a higher ROI.Alternatively, considering the total cost, which includes both down payment and mortgage payments, the net gain is higher for House B.But another way to look at it is the cash flow. Emily is paying mortgage each month, so the total amount she has put in is the down payment plus the mortgage payments. The sale proceeds are the return.So, the net gain is sale price - (down payment + mortgage payments). As calculated, House B gives a higher net gain.Alternatively, we can compute the internal rate of return, but that might be more complex.Alternatively, think about the total amount she has invested: for House A, it's ~214.6k, and she gets back ~403.17k, so the gain is ~188.57k. For House B, she invests ~178.83k and gets back ~370.05k, so gain is ~191.22k.So, in absolute terms, House B gives a slightly higher gain.But perhaps we should also consider the opportunity cost or other factors, but based purely on these numbers, House B seems better.However, another consideration is the loan balance after 10 years. Because she is selling the house after 10 years, she might have a remaining balance on the mortgage. Wait, no, because she is selling the house, she would pay off the remaining mortgage balance from the sale proceeds.So, actually, the net gain is sale price minus (down payment + total mortgage payments + remaining loan balance). Wait, no, because the mortgage payments include paying down the principal and interest. So, the remaining balance after 10 years would be the original loan amount minus the principal paid over 10 years.Wait, this complicates things. Because the monthly mortgage payment includes both principal and interest, so over 10 years, part of the payments go to principal, reducing the loan balance.Therefore, when she sells the house, she doesn't have to pay the remaining balance because she can use the sale proceeds to pay it off. So, the net gain is sale price minus (down payment + total mortgage payments + remaining loan balance). But actually, the sale price covers the remaining loan balance, so the net gain is sale price - (down payment + total mortgage payments) - remaining loan balance + remaining loan balance? Wait, no.Wait, let me think carefully.When she sells the house, she gets the sale price. From that, she has to pay off the remaining mortgage balance. The rest is her net gain.So, the net gain is sale price - remaining mortgage balance - down payment - total mortgage payments.Wait, no. Because the down payment is already part of her initial investment, and the mortgage payments are her outflows. The sale price is her inflow, from which she pays off the remaining mortgage.So, the net gain is (sale price - remaining mortgage balance) - (down payment + total mortgage payments).But actually, the total mortgage payments include both principal and interest. So, the remaining mortgage balance after 10 years is the original loan amount minus the principal paid over 10 years.Therefore, to compute the net gain, we need to calculate:Net gain = sale price - remaining mortgage balance - down payment - total mortgage paymentsBut this seems a bit convoluted. Alternatively, another approach is:Total amount invested = down payment + total mortgage paymentsTotal amount received = sale price - remaining mortgage balanceNet gain = total amount received - total amount investedBut this might not be the standard way to calculate it. Alternatively, perhaps the correct way is:Net gain = sale price - (down payment + total mortgage payments + remaining mortgage balance)But that doesn't make sense because the sale price covers the remaining mortgage balance.Wait, perhaps it's better to think of it as:When she sells the house, she gets sale price, from which she pays off the remaining mortgage. The rest is her profit.So, profit = sale price - remaining mortgage balance - down payment - total mortgage paymentsBut that would be negative because she has already paid the mortgage payments, which include both principal and interest.Wait, maybe I'm overcomplicating.Alternatively, think of it as:Total cost = down payment + total mortgage paymentsTotal proceeds = sale price - remaining mortgage balanceNet gain = total proceeds - total costBut that would be:Net gain = (sale price - remaining mortgage balance) - (down payment + total mortgage payments)But since the total mortgage payments include paying down the principal, the remaining mortgage balance is less than the original loan.Wait, perhaps it's better to compute the remaining mortgage balance after 10 years and then calculate the net gain accordingly.Let me try that.For House A:Original loan: 240,000Monthly payment: ~1,288.32Number of payments made: 120We need to compute the remaining balance after 120 payments.This can be done using the formula:Remaining balance = P * [(1 + i)^n - (1 + i)^p] / [(1 + i)^n - 1]Where:- P = principal- i = monthly interest rate- n = total number of payments- p = number of payments madeSo, for House A:P = 240,000i = 0.0041667n = 360p = 120Remaining balance = 240,000 * [(1.0041667)^360 - (1.0041667)^120] / [(1.0041667)^360 - 1]We already know (1.0041667)^360 ≈ 4.4677Compute (1.0041667)^120:Again, using the same method, (1.0041667)^120 ≈ e^(0.05*10) = e^0.5 ≈ 1.6487, but more accurately, it's about 1.6386.So, numerator: 4.4677 - 1.6386 ≈ 2.8291Denominator: 4.4677 - 1 ≈ 3.4677So, remaining balance ≈ 240,000 * (2.8291 / 3.4677) ≈ 240,000 * 0.816 ≈ 195,840So, remaining balance ≈ 195,840Therefore, when she sells House A, she gets 403,170, from which she pays off 195,840, leaving her with 403,170 - 195,840 = 207,330Her total investment was down payment + total mortgage payments: 60,000 + 154,598.40 ≈ 214,598.40So, net gain = 207,330 - 214,598.40 ≈ -7,268.40Wait, that can't be right. She's losing money?Wait, that doesn't make sense. Maybe I made a mistake in the calculation.Wait, no, because the remaining balance is 195,840, which is subtracted from the sale price, giving her 207,330. Her total investment was 214,598.40, so she's actually at a loss of about 7,268.But that contradicts the earlier calculation where net gain was 188k. So, clearly, I'm making a mistake here.Wait, perhaps the correct way is:Total proceeds from sale: 403,170Minus remaining mortgage: 195,840Equals equity: 207,330Total investment: down payment + total mortgage payments: 60,000 + 154,598.40 = 214,598.40So, net gain: 207,330 - 214,598.40 ≈ -7,268.40That suggests a loss, which is odd because the house appreciated.Wait, perhaps the issue is that the mortgage payments include interest, so the total amount paid is more than the principal, hence the negative gain.But that doesn't seem right because the house appreciated, so she should have a gain.Wait, let's think differently. The total amount she has put into the house is the down payment plus the mortgage payments. The total amount she gets back is the sale price minus the remaining mortgage balance.So, net gain is (sale price - remaining mortgage) - (down payment + mortgage payments)Which is (403,170 - 195,840) - (60,000 + 154,598.40) = 207,330 - 214,598.40 ≈ -7,268.40So, she's at a loss. That seems counterintuitive because the house appreciated, but the mortgage interest is eating into her gains.Similarly, for House B:Remaining balance after 10 years:P = 200,000i = 0.0041667n = 360p = 120Remaining balance = 200,000 * [(1.0041667)^360 - (1.0041667)^120] / [(1.0041667)^360 - 1]We already have (1.0041667)^360 ≈ 4.4677 and (1.0041667)^120 ≈ 1.6386Numerator: 4.4677 - 1.6386 ≈ 2.8291Denominator: 3.4677Remaining balance ≈ 200,000 * (2.8291 / 3.4677) ≈ 200,000 * 0.816 ≈ 163,200Sale price: 370,050Proceeds after paying off mortgage: 370,050 - 163,200 ≈ 206,850Total investment: down payment + mortgage payments = 50,000 + 128,832 ≈ 178,832Net gain: 206,850 - 178,832 ≈ 28,018So, for House B, she makes a net gain of about 28,018, whereas for House A, she loses about 7,268.This is a significant difference. So, House B is better.But wait, this seems to contradict the earlier calculation where House B had a higher net gain when not considering the remaining mortgage balance. So, why is that?Because in the first approach, I didn't account for the remaining mortgage balance, which is a significant amount. When considering that, House A's sale price minus remaining mortgage is less than the total investment, leading to a loss, while House B still makes a small profit.Therefore, the correct way is to consider the remaining mortgage balance because she has to pay it off when selling. So, the net gain is sale price minus remaining mortgage minus total investment (down payment + mortgage payments).So, based on this, House B is better because she makes a small profit, while House A results in a loss.But wait, let me double-check the remaining balance calculations because they are crucial.For House A:Remaining balance after 10 years: ~195,840Sale price: 403,170Proceeds: 403,170 - 195,840 = 207,330Total investment: 60,000 + 154,598.40 = 214,598.40Net gain: 207,330 - 214,598.40 ≈ -7,268.40For House B:Remaining balance: ~163,200Sale price: 370,050Proceeds: 370,050 - 163,200 = 206,850Total investment: 50,000 + 128,832 = 178,832Net gain: 206,850 - 178,832 ≈ 28,018So, yes, House B is better.Alternatively, another way to look at it is the cash flow over 10 years.For House A:She pays 60k down, then 1,288.32 per month for 10 years, totaling ~154.6k.After 10 years, she sells the house for ~403.17k, but has to pay off ~195.84k mortgage, leaving her with ~207.33k.So, her net is 207.33k - (60k + 154.6k) ≈ -7.27kFor House B:Down payment: 50kMortgage: ~128.83kSale proceeds: ~370.05kRemaining mortgage: ~163.2kProceeds after mortgage: ~206.85kNet: 206.85k - (50k + 128.83k) ≈ 28.02kSo, House B is better.Therefore, despite House A having a higher sale price, the higher mortgage payments and remaining balance result in a loss, while House B results in a small profit.So, Emily should choose House B.But wait, let me check if the remaining balance calculation is correct.For House A:Using the formula:Remaining balance = P * [(1 + i)^n - (1 + i)^p] / [(1 + i)^n - 1]P = 240,000i = 0.0041667n = 360p = 120Compute (1 + i)^n = 4.4677(1 + i)^p = (1.0041667)^120 ≈ 1.6386So, numerator: 4.4677 - 1.6386 ≈ 2.8291Denominator: 4.4677 - 1 ≈ 3.4677So, 2.8291 / 3.4677 ≈ 0.816Thus, remaining balance ≈ 240,000 * 0.816 ≈ 195,840Yes, that's correct.Similarly for House B:200,000 * 0.816 ≈ 163,200Yes.So, the calculations are correct.Therefore, the conclusion is that House B is a better financial investment over the 10-year period because it results in a net gain, while House A results in a loss.

answer：First, I need to calculate the number of chairs in the original 5 rows. Since each row has 8 chairs, multiplying 5 by 8 gives 40 chairs.Next, I'll determine the number of chairs in the additional 3 rows. Each of these rows has 10 chairs, so multiplying 3 by 10 results in 30 chairs.Finally, to find the total number of chairs Maria will set up, I'll add the chairs from the original rows to the chairs from the additional rows: 40 plus 30 equals 70 chairs.

answer：Alright, so I have this problem about seismic wave propagation, and I need to figure out the time it takes for the waves to reach the sea surface and then another point 4 km away. Let me break this down step by step.Starting with part 1: The wave is traveling directly from the earthquake's origin to the sea surface above. The origin is 5 km beneath the ocean floor, and the ocean floor is another 3 km beneath the sea surface. So, the total depth from the earthquake to the sea surface is 5 + 3 = 8 km. That makes sense.The velocity of the seismic wave is given by this function: ( v(z, theta) = v_0 + alpha z cos(theta) + beta z^2 sin^2(theta) ). They've given me specific values for ( v_0 ), ( alpha ), and ( beta ). Also, for part 1, the angle ( theta ) is 0 degrees because the wave is moving vertically upwards. So, ( theta = 0 ).Let me plug in ( theta = 0 ) into the velocity equation. Cosine of 0 is 1, and sine of 0 is 0. So, the velocity simplifies to ( v(z, 0) = v_0 + alpha z times 1 + beta z^2 times 0 ). That means the velocity is just ( v(z) = v_0 + alpha z ).Given ( v_0 = 1.5 ) km/s and ( alpha = 0.02 ) s(^{-1}). So, ( v(z) = 1.5 + 0.02 z ). Hmm, okay. So the velocity increases linearly with depth. That makes sense because as you go deeper, the velocity increases.Now, I need to calculate the time it takes for the wave to travel from 8 km depth to the surface. Since the velocity is a function of depth, I can't just use a simple ( time = distance / velocity ) because the velocity isn't constant. I need to integrate the reciprocal of velocity over the path.In other words, the time ( t ) is the integral from ( z = 8 ) km to ( z = 0 ) of ( dz / v(z) ). So, ( t = int_{8}^{0} frac{dz}{1.5 + 0.02 z} ). Wait, actually, since we're moving from 8 km to 0, it's the integral from 8 to 0, but integrating from a higher limit to a lower limit is the same as negative integral from 0 to 8. But since time is positive, I can write it as ( t = int_{0}^{8} frac{dz}{1.5 + 0.02 z} ).Let me compute this integral. The integral of ( 1/(a + bz) dz ) is ( (1/b) ln(a + bz) ). So, applying that here, the integral becomes ( frac{1}{0.02} ln(1.5 + 0.02 z) ) evaluated from 0 to 8.Calculating the constants: ( 1/0.02 = 50 ). So, ( t = 50 [ln(1.5 + 0.02 times 8) - ln(1.5 + 0.02 times 0)] ).Simplify inside the logarithms: 0.02 * 8 = 0.16, so the first term is ( ln(1.5 + 0.16) = ln(1.66) ). The second term is ( ln(1.5) ).So, ( t = 50 [ln(1.66) - ln(1.5)] ). Using logarithm properties, this is ( 50 ln(1.66 / 1.5) ).Calculating ( 1.66 / 1.5 ): Let's see, 1.5 goes into 1.66 once with 0.16 remaining. 0.16 / 1.5 is approximately 0.1067. So, 1.66 / 1.5 ≈ 1.1067.So, ( t ≈ 50 ln(1.1067) ). Calculating ( ln(1.1067) ): I know that ( ln(1.1) ≈ 0.09531 ), and ( ln(1.1067) ) should be a bit more. Let me compute it more accurately.Using a calculator, ( ln(1.1067) ≈ 0.1013 ). So, ( t ≈ 50 * 0.1013 ≈ 5.065 ) seconds.Wait, that seems really fast. 8 km in about 5 seconds? Let me check my calculations.Wait, 1.5 km/s is the initial velocity. So, if the velocity were constant at 1.5 km/s, the time would be 8 / 1.5 ≈ 5.333 seconds. But since the velocity increases with depth, the wave is actually moving faster as it goes deeper, which would make the average velocity higher, so the time should be less than 5.333 seconds. So, 5.065 seconds seems reasonable.Wait, but hold on. The velocity is increasing with depth, meaning that as the wave goes deeper, it's moving faster. But since it's starting at 8 km depth, moving up to 0 km, so the velocity starts at ( v(8) = 1.5 + 0.02*8 = 1.5 + 0.16 = 1.66 ) km/s and decreases to 1.5 km/s at the surface. Wait, no, hold on. Wait, z is the depth, so as the wave moves up, z decreases. So, the velocity is 1.66 km/s at 8 km depth and decreases to 1.5 km/s at the surface.Wait, that contradicts my earlier thought. So, actually, the wave starts at 8 km depth with a higher velocity and slows down as it ascends. So, the average velocity would be less than 1.66 km/s but more than 1.5 km/s. So, the time should be more than 8 / 1.66 ≈ 4.819 seconds and less than 5.333 seconds. So, 5.065 seconds is within that range. Okay, that seems okay.But let me double-check the integral. So, ( t = int_{0}^{8} frac{dz}{1.5 + 0.02 z} ). Let me make substitution: Let u = 1.5 + 0.02 z. Then, du/dz = 0.02, so dz = du / 0.02.When z = 0, u = 1.5. When z = 8, u = 1.5 + 0.16 = 1.66.So, the integral becomes ( int_{1.5}^{1.66} frac{1}{u} * (du / 0.02) ) = ( (1 / 0.02) int_{1.5}^{1.66} frac{du}{u} ) = 50 [ln(u)] from 1.5 to 1.66 = 50 (ln(1.66) - ln(1.5)).Yes, that's correct. So, my calculation is correct. So, t ≈ 5.065 seconds. Let me write that as approximately 5.07 seconds.Wait, but let me compute it more accurately. Let's compute ln(1.66) and ln(1.5):ln(1.66): Let's use calculator approximation. ln(1.66) ≈ 0.509.ln(1.5) ≈ 0.4055.So, the difference is 0.509 - 0.4055 = 0.1035.Multiply by 50: 0.1035 * 50 = 5.175 seconds.Wait, earlier I had 5.065, but with more accurate ln values, it's 5.175. Hmm, so which is it?Wait, perhaps I approximated ln(1.1067) as 0.1013, but actually, 1.66 / 1.5 is approximately 1.106666..., so ln(1.106666) is approximately 0.1013. So, 50 * 0.1013 ≈ 5.065.But when I compute ln(1.66) - ln(1.5) directly, it's 0.509 - 0.4055 = 0.1035, which is 5.175.Wait, so which is correct? Let me compute ln(1.66) and ln(1.5) more precisely.Using calculator:ln(1.66): Let me compute it step by step.We know that ln(1.6) ≈ 0.4700, ln(1.65) ≈ 0.5033, ln(1.66) is a bit higher.Compute ln(1.66):Using Taylor series around 1.65:Let me take x = 1.65, h = 0.01.ln(1.65 + 0.01) ≈ ln(1.65) + (0.01)/1.65 - (0.01)^2/(2*(1.65)^2) + ...ln(1.65) ≈ 0.5033First term: 0.01 / 1.65 ≈ 0.00606Second term: (0.0001)/(2*(2.7225)) ≈ 0.0001 / 5.445 ≈ 0.00001836So, ln(1.66) ≈ 0.5033 + 0.00606 - 0.00001836 ≈ 0.50934.Similarly, ln(1.5):We know ln(1.5) ≈ 0.4055.So, ln(1.66) - ln(1.5) ≈ 0.50934 - 0.4055 ≈ 0.10384.Multiply by 50: 0.10384 * 50 ≈ 5.192 seconds.So, approximately 5.19 seconds.Wait, but earlier I had 5.065 when I did ln(1.1067). Wait, 1.66 / 1.5 is 1.106666..., so ln(1.106666) is approximately 0.1013, which when multiplied by 50 gives 5.065. But when I compute ln(1.66) - ln(1.5) directly, I get approximately 0.10384, which is 5.192.So, which is correct? It seems that the direct computation is more accurate because when I compute the difference of logs, it's the same as the log of the ratio, so both should give the same result. Wait, but 1.66 / 1.5 is 1.106666..., so ln(1.106666) should equal ln(1.66) - ln(1.5). So, perhaps my initial approximation of ln(1.106666) was too low.Let me compute ln(1.106666) more accurately.We can use the Taylor series for ln(1+x) around x=0.106666.Wait, ln(1.106666) = ln(1 + 0.106666). The Taylor series is x - x^2/2 + x^3/3 - x^4/4 + ...So, x = 0.106666.Compute up to, say, x^4 term.First term: 0.106666Second term: -(0.106666)^2 / 2 ≈ -0.011378Third term: +(0.106666)^3 / 3 ≈ +0.001213Fourth term: -(0.106666)^4 / 4 ≈ -0.000129So, adding these up:0.106666 - 0.011378 = 0.0952880.095288 + 0.001213 = 0.0965010.096501 - 0.000129 = 0.096372So, ln(1.106666) ≈ 0.096372.Wait, but earlier, when I computed ln(1.66) - ln(1.5) ≈ 0.10384, which is higher than 0.096372. That can't be. There must be a mistake here.Wait, no, actually, ln(1.106666) is equal to ln(1.66) - ln(1.5). So, if ln(1.66) - ln(1.5) ≈ 0.10384, then ln(1.106666) should also be ≈ 0.10384. But my Taylor series gave me 0.096372, which is less. So, perhaps the Taylor series isn't converging quickly enough or I made a miscalculation.Alternatively, maybe I should use a calculator for more accurate values.Using a calculator:ln(1.66) ≈ 0.509327ln(1.5) ≈ 0.405465Difference: 0.509327 - 0.405465 ≈ 0.103862So, 0.103862 * 50 ≈ 5.1931 seconds.Therefore, the more accurate value is approximately 5.19 seconds.So, perhaps my initial approximation was too low because I didn't compute the Taylor series far enough. So, the correct time is approximately 5.19 seconds.But let me check with another method. Maybe using a substitution in the integral.Alternatively, perhaps I can use the formula for the integral of 1/(a + bz) dz, which is (1/b) ln(a + bz) + C.So, as I did earlier, the integral from 0 to 8 is 50 [ln(1.66) - ln(1.5)] ≈ 50 * 0.103862 ≈ 5.1931 seconds.So, rounding to three decimal places, 5.193 seconds. Let's say approximately 5.19 seconds.Wait, but in the initial substitution, I thought of z as depth, so when integrating from 0 to 8, but actually, the wave is moving from 8 km to 0 km, so the integral should be from 8 to 0, which would be negative, but since time is positive, we take the absolute value.But regardless, the calculation remains the same because integrating from 0 to 8 gives the same positive value as integrating from 8 to 0 and taking the negative.So, I think 5.19 seconds is the correct answer for part 1.Moving on to part 2: Now, the wave travels from the earthquake origin to a point 4 km horizontally from the epicenter on the sea surface. So, the origin is at (0, -8 km), and the destination is at (4 km, 0). So, the straight-line distance is the hypotenuse of a right triangle with legs 4 km and 8 km.Calculating the straight-line distance: ( sqrt{4^2 + 8^2} = sqrt{16 + 64} = sqrt{80} ≈ 8.944 km ).But the wave is traveling through an anisotropic medium, so the velocity depends on both depth z and the angle θ that the wavefront makes with the vertical.Given the velocity function: ( v(z, theta) = 1.5 + 0.02 z cos(theta) + 0.005 z^2 sin^2(theta) ).Since the wave is traveling along a straight line, the angle θ is constant along the path. So, θ is the angle between the wave path and the vertical. In this case, the wave is moving from (0, -8) to (4, 0), so the angle θ can be found using trigonometry.In the triangle, the horizontal leg is 4 km, and the vertical leg is 8 km. So, tan(theta) = opposite / adjacent = 4 / 8 = 0.5. So, theta = arctan(0.5) ≈ 26.565 degrees.So, θ ≈ 26.565 degrees. Let me convert that to radians for calculations, but maybe I can keep it in degrees for now.So, cos(theta) ≈ cos(26.565°) ≈ 0.8944, and sin(theta) ≈ sin(26.565°) ≈ 0.4472.So, cos(theta) ≈ 0.8944, sin(theta) ≈ 0.4472.So, the velocity function becomes ( v(z) = 1.5 + 0.02 z * 0.8944 + 0.005 z^2 * (0.4472)^2 ).Calculating the coefficients:0.02 * 0.8944 ≈ 0.017888(0.4472)^2 ≈ 0.2, so 0.005 * 0.2 = 0.001.So, the velocity function simplifies to ( v(z) = 1.5 + 0.017888 z + 0.001 z^2 ).So, ( v(z) = 1.5 + 0.017888 z + 0.001 z^2 ).Now, to find the travel time, we need to integrate the reciprocal of velocity along the path. However, since the wave is moving along a straight line, we can parameterize the path in terms of z.Wait, but how is z changing along the path? Since the wave is moving from depth 8 km to 0 km while moving horizontally 4 km, the depth z decreases from 8 km to 0 km as the wave travels.But in terms of the path, we can express z as a function of the distance traveled along the path. Let me denote s as the arc length along the wave path, from 0 to S, where S = sqrt(4^2 + 8^2) ≈ 8.944 km.But actually, since the wave is moving straight, we can express z as a function of the vertical position. Wait, maybe it's easier to parameterize in terms of z.Wait, let me think. The wave is moving along a straight line, so for each z, the horizontal position x is proportional. Since the total horizontal distance is 4 km when z = 0, and x = 0 when z = 8 km.So, the relationship between x and z is linear: x = (4 / 8) * (8 - z) = 0.5 * (8 - z). So, x = 4 - 0.5 z.But maybe that's complicating things. Alternatively, since the wave is moving along a straight line, the angle θ is constant, so we can express the differential time element as dt = ds / v(z, θ), where ds is the differential arc length.But since θ is constant, ds = dz / cos(theta). Because in the vertical direction, dz = ds * cos(theta), so ds = dz / cos(theta).Therefore, the total time t is the integral from z = 8 km to z = 0 of [dz / (v(z, θ) * cos(theta))].So, t = ∫ (dz / (v(z) * cos(theta))) from 8 to 0.But since we're integrating from 8 to 0, it's the same as ∫ from 0 to 8 of [dz / (v(z) * cos(theta))].So, let me write that:t = (1 / cos(theta)) ∫_{0}^{8} [dz / v(z)]Because cos(theta) is a constant for this path.We already have v(z) = 1.5 + 0.017888 z + 0.001 z^2.So, t = (1 / 0.8944) ∫_{0}^{8} [dz / (1.5 + 0.017888 z + 0.001 z^2)].So, first, compute the integral ∫ [dz / (1.5 + 0.017888 z + 0.001 z^2)] from 0 to 8.This integral looks a bit more complicated because the denominator is a quadratic in z. Let me write the denominator as:0.001 z^2 + 0.017888 z + 1.5.Let me factor out 0.001 to make it easier:0.001 (z^2 + 17.888 z + 1500).Wait, 0.017888 / 0.001 = 17.888, and 1.5 / 0.001 = 1500.So, the integral becomes:∫ dz / [0.001 (z^2 + 17.888 z + 1500)] = (1 / 0.001) ∫ dz / (z^2 + 17.888 z + 1500).Which is 1000 ∫ dz / (z^2 + 17.888 z + 1500).Now, to integrate 1/(z^2 + bz + c) dz, we can complete the square.Let me complete the square for the quadratic in the denominator:z^2 + 17.888 z + 1500.The coefficient of z is 17.888, so half of that is 8.944, and squaring that gives approximately 80.So, z^2 + 17.888 z + 80 - 80 + 1500 = (z + 8.944)^2 + (1500 - 80) = (z + 8.944)^2 + 1420.So, the integral becomes:1000 ∫ dz / [(z + 8.944)^2 + (sqrt(1420))^2].Compute sqrt(1420): sqrt(1420) ≈ 37.68.So, the integral is 1000 ∫ dz / [(z + 8.944)^2 + (37.68)^2].The integral of 1/(u^2 + a^2) du is (1/a) arctan(u/a) + C.So, let me make substitution u = z + 8.944, du = dz.Limits: when z = 0, u = 8.944; when z = 8, u = 16.944.So, the integral becomes:1000 [ (1 / 37.68) arctan(u / 37.68) ] evaluated from u = 8.944 to u = 16.944.So, compute:1000 * (1 / 37.68) [ arctan(16.944 / 37.68) - arctan(8.944 / 37.68) ].Calculate the arguments:16.944 / 37.68 ≈ 0.44958.944 / 37.68 ≈ 0.2374Now, compute arctan(0.4495) and arctan(0.2374).Using calculator approximations:arctan(0.4495) ≈ 24.2 degrees ≈ 0.422 radiansarctan(0.2374) ≈ 13.3 degrees ≈ 0.232 radiansSo, the difference is approximately 0.422 - 0.232 = 0.190 radians.So, the integral is approximately:1000 * (1 / 37.68) * 0.190 ≈ 1000 * 0.00504 * 0.190 ≈ 1000 * 0.0009576 ≈ 0.9576 seconds.Wait, that seems very low. Let me check my calculations.Wait, 1000 * (1 / 37.68) ≈ 26.54.Then, 26.54 * 0.190 ≈ 5.043 seconds.Ah, yes, I made a mistake in the calculation earlier. Let me correct that.So, 1000 * (1 / 37.68) ≈ 26.54.Then, 26.54 * (0.422 - 0.232) ≈ 26.54 * 0.190 ≈ 5.043 seconds.So, the integral ∫ [dz / v(z)] ≈ 5.043 seconds.But remember, this is multiplied by 1 / cos(theta), which is approximately 1 / 0.8944 ≈ 1.118.So, total time t ≈ 5.043 * 1.118 ≈ 5.64 seconds.Wait, that seems a bit high. Let me verify the steps again.First, the integral ∫ [dz / (1.5 + 0.017888 z + 0.001 z^2)] from 0 to 8.We rewrote it as 1000 ∫ dz / (z^2 + 17.888 z + 1500).Completed the square: (z + 8.944)^2 + 1420.So, integral becomes 1000 ∫ dz / [(z + 8.944)^2 + (37.68)^2].Then, substitution u = z + 8.944, limits from 8.944 to 16.944.Integral is 1000 * (1 / 37.68) [ arctan(u / 37.68) ] from 8.944 to 16.944.Compute arctan(16.944 / 37.68) ≈ arctan(0.4495) ≈ 0.422 radians.arctan(8.944 / 37.68) ≈ arctan(0.2374) ≈ 0.232 radians.Difference: 0.422 - 0.232 = 0.190 radians.So, integral ≈ 1000 * (1 / 37.68) * 0.190 ≈ 1000 * 0.00504 * 0.190 ≈ 1000 * 0.0009576 ≈ 0.9576? Wait, no, that's not correct.Wait, 1000 * (1 / 37.68) is approximately 26.54.Then, 26.54 * 0.190 ≈ 5.043.Yes, that's correct. So, the integral ∫ [dz / v(z)] ≈ 5.043 seconds.Then, multiply by 1 / cos(theta) ≈ 1.118.So, t ≈ 5.043 * 1.118 ≈ 5.64 seconds.But wait, in part 1, the time was approximately 5.19 seconds for the vertical path, and here, for a longer path (8.944 km vs 8 km), the time is only slightly longer, 5.64 seconds. That seems plausible because the wave is moving faster in some parts due to the velocity function.But let me check if I did the integral correctly.Alternatively, maybe I should use numerical integration for the integral ∫ [dz / (1.5 + 0.017888 z + 0.001 z^2)] from 0 to 8.Let me approximate it using the trapezoidal rule or Simpson's rule.First, let's compute the function f(z) = 1 / (1.5 + 0.017888 z + 0.001 z^2) at several points between z=0 and z=8.Let me choose z = 0, 2, 4, 6, 8.Compute f(z):At z=0: f(0) = 1 / 1.5 ≈ 0.6667At z=2: f(2) = 1 / (1.5 + 0.035776 + 0.004) = 1 / (1.539776) ≈ 0.6500At z=4: f(4) = 1 / (1.5 + 0.071552 + 0.016) = 1 / (1.587552) ≈ 0.6300At z=6: f(6) = 1 / (1.5 + 0.107328 + 0.036) = 1 / (1.643328) ≈ 0.6080At z=8: f(8) = 1 / (1.5 + 0.143104 + 0.064) = 1 / (1.707104) ≈ 0.5856Now, using Simpson's rule for n=4 intervals (5 points):Simpson's rule formula: Δz/3 [f(z0) + 4f(z1) + 2f(z2) + 4f(z3) + f(z4)]Δz = 2 km.So,Integral ≈ (2/3) [0.6667 + 4*0.6500 + 2*0.6300 + 4*0.6080 + 0.5856]Compute each term:0.6667 + 4*0.6500 = 0.6667 + 2.6 = 3.26672*0.6300 = 1.264*0.6080 = 2.4320.5856Adding all together: 3.2667 + 1.26 + 2.432 + 0.5856 ≈ 7.5443Multiply by (2/3): 7.5443 * (2/3) ≈ 5.0295 seconds.So, Simpson's rule gives approximately 5.03 seconds for the integral ∫ [dz / v(z)].Then, multiply by 1 / cos(theta) ≈ 1.118:t ≈ 5.03 * 1.118 ≈ 5.62 seconds.This is close to the previous result of 5.64 seconds, so it seems consistent.Therefore, the total travel time is approximately 5.62 seconds.But let me check if this makes sense. The vertical path took about 5.19 seconds, and this path is longer (8.944 km vs 8 km), but the wave is moving faster in some parts because the velocity function includes a term that depends on z and theta.Wait, in the vertical case, the velocity was increasing with depth, but in this case, the velocity is a function of both z and theta. So, the velocity is higher because of the angle term, which might make the wave move faster overall, but the path is longer.So, the time is slightly longer than the vertical case, which makes sense.Alternatively, let me compute the average velocity for this path.Total distance along the path: S ≈ 8.944 km.Total time: ≈5.62 seconds.Average velocity: 8.944 / 5.62 ≈ 1.59 km/s.Which is higher than the vertical case's average velocity of 8 / 5.19 ≈ 1.54 km/s.So, that makes sense because the wave is moving at a higher velocity due to the angle term.Therefore, the total travel time is approximately 5.62 seconds.But let me see if I can get a more accurate value.Alternatively, perhaps I can use a better approximation for the integral.But given that both the substitution method and Simpson's rule give around 5.03 seconds for the integral, and then multiplied by 1.118 gives around 5.62 seconds, I think that's a reasonable estimate.So, summarizing:1. Time for vertical path: approximately 5.19 seconds.2. Time for 4 km horizontal path: approximately 5.62 seconds.But let me write the answers with more precise decimals.For part 1, the integral gave us approximately 5.1931 seconds, so 5.19 seconds.For part 2, using Simpson's rule, we got 5.03 seconds for the integral, then multiplied by 1.118 gives approximately 5.62 seconds.Alternatively, using the substitution method, we had 5.043 * 1.118 ≈ 5.64 seconds.So, perhaps the answer is approximately 5.63 seconds.But let me compute it more precisely.Compute 5.043 * 1.118:5.043 * 1 = 5.0435.043 * 0.1 = 0.50435.043 * 0.01 = 0.050435.043 * 0.008 = 0.040344Adding up: 5.043 + 0.5043 = 5.54735.5473 + 0.05043 = 5.59775.5977 + 0.040344 ≈ 5.638 seconds.So, approximately 5.64 seconds.Alternatively, using Simpson's rule, 5.03 * 1.118 ≈ 5.62 seconds.So, the answer is approximately 5.63 seconds.Therefore, the total travel time for part 2 is approximately 5.63 seconds.But let me check if I made any mistakes in the substitution method.Wait, in the substitution method, I had:t = (1 / cos(theta)) * integral.But wait, the integral was ∫ [dz / v(z)] from 0 to 8, which we computed as approximately 5.043 seconds.Then, t = 5.043 / cos(theta) ≈ 5.043 / 0.8944 ≈ 5.64 seconds.Yes, that's correct.Alternatively, if I use the more precise integral value from Simpson's rule, which was 5.03 seconds, then t ≈ 5.03 / 0.8944 ≈ 5.62 seconds.So, depending on the method, it's either 5.62 or 5.64 seconds. I think 5.63 seconds is a reasonable approximation.Therefore, the answers are:1. Approximately 5.19 seconds.2. Approximately 5.63 seconds.

answer：First, I need to determine the original force that each wing can withstand, which is 5,000 pounds.Next, I'll calculate the additional force resistance provided by the supports, which is 20% of 5,000 pounds. This equals 1,000 pounds.Adding this additional resistance to the original force, the wing can now withstand 6,000 pounds.Finally, after testing, it's found that the wing can withstand an extra 1,000 pounds beyond the safety requirement. Therefore, the total force each wing can withstand after adding the supports is 7,000 pounds.

answer：Alright, so I have this problem where Dr. Lee is analyzing the geopolitical stability between the US and Korea using a partial differential equation. The equation given is:[ frac{partial S}{partial t} = k_1 frac{partial^2 E}{partial x^2} - k_2 frac{partial M}{partial x} ]And I need to solve two sub-problems. Let me start with Sub-problem 1.**Sub-problem 1:**They tell me that economic cooperation ( E ) is a Gaussian function:[ E(x,t) = A e^{-frac{(x - vt)^2}{2sigma^2}} ]Where ( A ), ( v ), and ( sigma ) are constants. I need to find the second partial derivative of ( E ) with respect to ( x ) and substitute it back into the PDE.Okay, so first, let's compute ( frac{partial^2 E}{partial x^2} ). Since ( E ) is a function of ( x ) and ( t ), but in this case, it's specifically given as a function of ( x - vt ), which suggests that it's a traveling Gaussian wave moving with velocity ( v ).Let me denote ( u = x - vt ). Then, ( E ) can be written as:[ E(u) = A e^{-frac{u^2}{2sigma^2}} ]So, ( E ) is a function of ( u ), which itself is a function of ( x ) and ( t ). Therefore, when taking derivatives with respect to ( x ), I need to use the chain rule.First, compute the first derivative ( frac{partial E}{partial x} ):[ frac{partial E}{partial x} = frac{dE}{du} cdot frac{partial u}{partial x} ]Since ( u = x - vt ), ( frac{partial u}{partial x} = 1 ). So,[ frac{partial E}{partial x} = frac{dE}{du} = A cdot e^{-frac{u^2}{2sigma^2}} cdot left( -frac{2u}{2sigma^2} right) = -frac{A u}{sigma^2} e^{-frac{u^2}{2sigma^2}} ]Simplify that:[ frac{partial E}{partial x} = -frac{A (x - vt)}{sigma^2} e^{-frac{(x - vt)^2}{2sigma^2}} ]Now, let's compute the second derivative ( frac{partial^2 E}{partial x^2} ). Again, we'll take the derivative of ( frac{partial E}{partial x} ) with respect to ( x ):[ frac{partial^2 E}{partial x^2} = frac{d}{dx} left( -frac{A (x - vt)}{sigma^2} e^{-frac{(x - vt)^2}{2sigma^2}} right) ]Let me denote ( w = x - vt ) to make it easier. Then, the expression becomes:[ frac{partial^2 E}{partial x^2} = frac{d}{dw} left( -frac{A w}{sigma^2} e^{-frac{w^2}{2sigma^2}} right) cdot frac{dw}{dx} ]But ( frac{dw}{dx} = 1 ), so we can ignore that for now and just compute the derivative with respect to ( w ):Let me compute the derivative:Let ( f(w) = -frac{A w}{sigma^2} e^{-frac{w^2}{2sigma^2}} )So, ( f'(w) = -frac{A}{sigma^2} cdot left[ e^{-frac{w^2}{2sigma^2}} + w cdot e^{-frac{w^2}{2sigma^2}} cdot left( -frac{2w}{2sigma^2} right) right] )Simplify step by step:First, the derivative of ( w ) is 1, so:[ f'(w) = -frac{A}{sigma^2} cdot left[ e^{-frac{w^2}{2sigma^2}} + w cdot left( -frac{w}{sigma^2} right) e^{-frac{w^2}{2sigma^2}} right] ]Simplify inside the brackets:[ e^{-frac{w^2}{2sigma^2}} - frac{w^2}{sigma^2} e^{-frac{w^2}{2sigma^2}} = e^{-frac{w^2}{2sigma^2}} left( 1 - frac{w^2}{sigma^2} right) ]So, putting it back:[ f'(w) = -frac{A}{sigma^2} cdot e^{-frac{w^2}{2sigma^2}} left( 1 - frac{w^2}{sigma^2} right) ]Therefore, substituting back ( w = x - vt ):[ frac{partial^2 E}{partial x^2} = -frac{A}{sigma^2} e^{-frac{(x - vt)^2}{2sigma^2}} left( 1 - frac{(x - vt)^2}{sigma^2} right) ]Alternatively, we can write this as:[ frac{partial^2 E}{partial x^2} = -frac{A}{sigma^2} left( 1 - frac{(x - vt)^2}{sigma^2} right) e^{-frac{(x - vt)^2}{2sigma^2}} ]So that's the second derivative of ( E ) with respect to ( x ). Now, we need to substitute this back into the original PDE:[ frac{partial S}{partial t} = k_1 frac{partial^2 E}{partial x^2} - k_2 frac{partial M}{partial x} ]But wait, in Sub-problem 1, they only ask to substitute ( frac{partial^2 E}{partial x^2} ) back into the original PDE. So, I think that's all for Sub-problem 1. So, substituting:[ frac{partial S}{partial t} = k_1 left[ -frac{A}{sigma^2} left( 1 - frac{(x - vt)^2}{sigma^2} right) e^{-frac{(x - vt)^2}{2sigma^2}} right] - k_2 frac{partial M}{partial x} ]But wait, in Sub-problem 1, are we supposed to substitute only the second derivative of E, or also handle the term with M? Let me check the problem statement."Sub-problem 1: Assuming ( E ) is a Gaussian function... find the expression for ( frac{partial^2 E}{partial x^2} ) and substitute it back into the original PDE."So, yes, substitute ( frac{partial^2 E}{partial x^2} ) into the PDE. So, the equation becomes:[ frac{partial S}{partial t} = k_1 left[ -frac{A}{sigma^2} left( 1 - frac{(x - vt)^2}{sigma^2} right) e^{-frac{(x - vt)^2}{2sigma^2}} right] - k_2 frac{partial M}{partial x} ]But since in Sub-problem 1, they don't specify anything about ( M ), maybe we just leave it as is. So, perhaps the answer for Sub-problem 1 is just the expression for ( frac{partial^2 E}{partial x^2} ), which I found earlier.Wait, the problem says "find the expression for ( frac{partial^2 E}{partial x^2} ) and substitute it back into the original PDE." So, that would mean that the PDE becomes:[ frac{partial S}{partial t} = k_1 cdot text{[expression]} - k_2 frac{partial M}{partial x} ]But since in Sub-problem 1, we don't have information about ( M ), maybe we just leave it as an expression with ( frac{partial M}{partial x} ). So, perhaps the answer is just the expression for ( frac{partial^2 E}{partial x^2} ), which is:[ frac{partial^2 E}{partial x^2} = -frac{A}{sigma^2} left( 1 - frac{(x - vt)^2}{sigma^2} right) e^{-frac{(x - vt)^2}{2sigma^2}} ]But the problem says to substitute it back into the PDE, so maybe I need to write the entire PDE with this substituted. So, the PDE becomes:[ frac{partial S}{partial t} = -k_1 frac{A}{sigma^2} left( 1 - frac{(x - vt)^2}{sigma^2} right) e^{-frac{(x - vt)^2}{2sigma^2}} - k_2 frac{partial M}{partial x} ]But since in Sub-problem 1, we don't have information about ( M ), perhaps that's as far as we can go. So, maybe the answer is just the expression for ( frac{partial^2 E}{partial x^2} ).Wait, let me check the problem again:"Sub-problem 1: Assuming ( E ) is a Gaussian function... find the expression for ( frac{partial^2 E}{partial x^2} ) and substitute it back into the original PDE."So, yes, substitute it into the PDE, so the resulting equation is:[ frac{partial S}{partial t} = k_1 cdot text{[expression]} - k_2 frac{partial M}{partial x} ]But since in Sub-problem 1, we don't have ( M ) defined, perhaps we just leave it as an expression with ( frac{partial M}{partial x} ). So, the substituted PDE is:[ frac{partial S}{partial t} = -k_1 frac{A}{sigma^2} left( 1 - frac{(x - vt)^2}{sigma^2} right) e^{-frac{(x - vt)^2}{2sigma^2}} - k_2 frac{partial M}{partial x} ]But since we don't have ( M ) yet, maybe that's all we can do for Sub-problem 1. So, perhaps the answer is just the expression for ( frac{partial^2 E}{partial x^2} ), which is:[ frac{partial^2 E}{partial x^2} = -frac{A}{sigma^2} left( 1 - frac{(x - vt)^2}{sigma^2} right) e^{-frac{(x - vt)^2}{2sigma^2}} ]But the problem says to substitute it back into the PDE, so maybe I need to write the entire PDE with this substituted. So, the PDE becomes:[ frac{partial S}{partial t} = -k_1 frac{A}{sigma^2} left( 1 - frac{(x - vt)^2}{sigma^2} right) e^{-frac{(x - vt)^2}{2sigma^2}} - k_2 frac{partial M}{partial x} ]But since in Sub-problem 1, we don't have information about ( M ), perhaps that's as far as we can go. So, maybe the answer is just the expression for ( frac{partial^2 E}{partial x^2} ).Wait, perhaps I should just compute the second derivative and present that as the answer for Sub-problem 1, since the substitution into the PDE is straightforward once we have that.So, to recap, I found:[ frac{partial^2 E}{partial x^2} = -frac{A}{sigma^2} left( 1 - frac{(x - vt)^2}{sigma^2} right) e^{-frac{(x - vt)^2}{2sigma^2}} ]So, that's the expression for the second derivative. Substituting this into the PDE gives:[ frac{partial S}{partial t} = -k_1 frac{A}{sigma^2} left( 1 - frac{(x - vt)^2}{sigma^2} right) e^{-frac{(x - vt)^2}{2sigma^2}} - k_2 frac{partial M}{partial x} ]But since Sub-problem 1 doesn't involve ( M ), maybe the answer is just the expression for ( frac{partial^2 E}{partial x^2} ).I think that's it for Sub-problem 1. Now, moving on to Sub-problem 2.**Sub-problem 2:**Here, military tension ( M ) is modeled as ( M(x) = B sin(Cx) ), where ( B ) and ( C ) are constants. I need to find ( frac{partial M}{partial x} ) and substitute it back into the original PDE. Then, solve the resulting equation for ( S ) as a function of ( x ) and ( t ).First, let's compute ( frac{partial M}{partial x} ). Since ( M ) is a function of ( x ) only, the derivative is straightforward.[ frac{partial M}{partial x} = frac{d}{dx} [B sin(Cx)] = B C cos(Cx) ]So, ( frac{partial M}{partial x} = B C cos(Cx) ).Now, substitute this into the original PDE:[ frac{partial S}{partial t} = k_1 frac{partial^2 E}{partial x^2} - k_2 frac{partial M}{partial x} ]From Sub-problem 1, we have ( frac{partial^2 E}{partial x^2} ) as:[ frac{partial^2 E}{partial x^2} = -frac{A}{sigma^2} left( 1 - frac{(x - vt)^2}{sigma^2} right) e^{-frac{(x - vt)^2}{2sigma^2}} ]So, substituting both terms into the PDE:[ frac{partial S}{partial t} = k_1 left[ -frac{A}{sigma^2} left( 1 - frac{(x - vt)^2}{sigma^2} right) e^{-frac{(x - vt)^2}{2sigma^2}} right] - k_2 (B C cos(Cx)) ]Simplify:[ frac{partial S}{partial t} = -frac{k_1 A}{sigma^2} left( 1 - frac{(x - vt)^2}{sigma^2} right) e^{-frac{(x - vt)^2}{2sigma^2}} - k_2 B C cos(Cx) ]So, now we have a PDE for ( S(x,t) ):[ frac{partial S}{partial t} = -frac{k_1 A}{sigma^2} left( 1 - frac{(x - vt)^2}{sigma^2} right) e^{-frac{(x - vt)^2}{2sigma^2}} - k_2 B C cos(Cx) ]This is a first-order linear PDE in ( S ). To solve it, we can integrate both sides with respect to ( t ), treating ( x ) as a constant.So, integrating both sides from ( t = 0 ) to ( t ):[ S(x,t) - S(x,0) = int_0^t left[ -frac{k_1 A}{sigma^2} left( 1 - frac{(x - vtau)^2}{sigma^2} right) e^{-frac{(x - vtau)^2}{2sigma^2}} - k_2 B C cos(Cx) right] dtau ]Assuming that ( S(x,0) ) is the initial condition, which we can denote as ( S_0(x) ). So,[ S(x,t) = S_0(x) + int_0^t left[ -frac{k_1 A}{sigma^2} left( 1 - frac{(x - vtau)^2}{sigma^2} right) e^{-frac{(x - vtau)^2}{2sigma^2}} - k_2 B C cos(Cx) right] dtau ]Let me split the integral into two parts:[ S(x,t) = S_0(x) - frac{k_1 A}{sigma^2} int_0^t left( 1 - frac{(x - vtau)^2}{sigma^2} right) e^{-frac{(x - vtau)^2}{2sigma^2}} dtau - k_2 B C int_0^t cos(Cx) dtau ]Now, let's compute each integral separately.First, the second integral:[ int_0^t cos(Cx) dtau ]Since ( cos(Cx) ) is independent of ( tau ), this integral is simply:[ cos(Cx) cdot int_0^t dtau = t cos(Cx) ]So, the second term becomes:[ -k_2 B C cdot t cos(Cx) ]Now, the first integral is more complicated:[ int_0^t left( 1 - frac{(x - vtau)^2}{sigma^2} right) e^{-frac{(x - vtau)^2}{2sigma^2}} dtau ]Let me make a substitution to simplify this integral. Let ( u = x - vtau ). Then, ( du = -v dtau ), so ( dtau = -frac{du}{v} ).When ( tau = 0 ), ( u = x ). When ( tau = t ), ( u = x - vt ).So, changing the limits and variables:[ int_{u=x}^{u=x - vt} left( 1 - frac{u^2}{sigma^2} right) e^{-frac{u^2}{2sigma^2}} cdot left( -frac{du}{v} right) ]The negative sign flips the limits:[ frac{1}{v} int_{x - vt}^{x} left( 1 - frac{u^2}{sigma^2} right) e^{-frac{u^2}{2sigma^2}} du ]So, the integral becomes:[ frac{1}{v} int_{x - vt}^{x} left( 1 - frac{u^2}{sigma^2} right) e^{-frac{u^2}{2sigma^2}} du ]Let me denote this integral as ( I ):[ I = frac{1}{v} int_{x - vt}^{x} left( 1 - frac{u^2}{sigma^2} right) e^{-frac{u^2}{2sigma^2}} du ]To compute this integral, let's split it into two parts:[ I = frac{1}{v} left[ int_{x - vt}^{x} e^{-frac{u^2}{2sigma^2}} du - frac{1}{sigma^2} int_{x - vt}^{x} u^2 e^{-frac{u^2}{2sigma^2}} du right] ]Let me compute each integral separately.First integral:[ I_1 = int e^{-frac{u^2}{2sigma^2}} du ]This is a standard Gaussian integral, which relates to the error function (erf). The integral is:[ I_1 = sigma sqrt{frac{pi}{2}} text{erf}left( frac{u}{sigma sqrt{2}} right) + C ]Second integral:[ I_2 = int u^2 e^{-frac{u^2}{2sigma^2}} du ]This integral can be computed using integration by parts or by recognizing it as related to the Gaussian function. The result is:[ I_2 = frac{sigma^3 sqrt{pi}}{2} text{erf}left( frac{u}{sigma sqrt{2}} right) + sigma^2 u e^{-frac{u^2}{2sigma^2}} + C ]Wait, let me verify that. Alternatively, we can recall that:[ int u^2 e^{-a u^2} du = frac{sqrt{pi}}{4 a^{3/2}} text{erf}(a u) + frac{u e^{-a u^2}}{2a} + C ]In our case, ( a = frac{1}{2sigma^2} ), so:[ I_2 = int u^2 e^{-frac{u^2}{2sigma^2}} du = frac{sqrt{pi}}{4 left( frac{1}{2sigma^2} right)^{3/2}} text{erf}left( sqrt{frac{1}{2sigma^2}} u right) + frac{u e^{-frac{u^2}{2sigma^2}}}{2 cdot frac{1}{2sigma^2}} + C ]Simplify:First term:[ frac{sqrt{pi}}{4 left( frac{1}{2sigma^2} right)^{3/2}} = frac{sqrt{pi}}{4} cdot (2sigma^2)^{3/2} = frac{sqrt{pi}}{4} cdot (2)^{3/2} sigma^3 = frac{sqrt{pi}}{4} cdot 2 sqrt{2} sigma^3 = frac{sqrt{pi} cdot 2 sqrt{2} sigma^3}{4} = frac{sqrt{pi} cdot sqrt{2} sigma^3}{2} ]Second term:[ frac{u e^{-frac{u^2}{2sigma^2}}}{2 cdot frac{1}{2sigma^2}} = frac{u e^{-frac{u^2}{2sigma^2}}}{frac{1}{sigma^2}} = u sigma^2 e^{-frac{u^2}{2sigma^2}} ]So, putting it together:[ I_2 = frac{sqrt{2pi} sigma^3}{2} text{erf}left( frac{u}{sigma sqrt{2}} right) + u sigma^2 e^{-frac{u^2}{2sigma^2}} + C ]Therefore, the integral ( I ) becomes:[ I = frac{1}{v} left[ sigma sqrt{frac{pi}{2}} left( text{erf}left( frac{x}{sigma sqrt{2}} right) - text{erf}left( frac{x - vt}{sigma sqrt{2}} right) right) - frac{1}{sigma^2} left( frac{sqrt{2pi} sigma^3}{2} left( text{erf}left( frac{x}{sigma sqrt{2}} right) - text{erf}left( frac{x - vt}{sigma sqrt{2}} right) right) + sigma^2 x e^{-frac{x^2}{2sigma^2}} - sigma^2 (x - vt) e^{-frac{(x - vt)^2}{2sigma^2}} right) right] ]Simplify term by term:First, the ( I_1 ) part:[ sigma sqrt{frac{pi}{2}} left( text{erf}left( frac{x}{sigma sqrt{2}} right) - text{erf}left( frac{x - vt}{sigma sqrt{2}} right) right) ]Second, the ( I_2 ) part:[ -frac{1}{sigma^2} cdot frac{sqrt{2pi} sigma^3}{2} left( text{erf}left( frac{x}{sigma sqrt{2}} right) - text{erf}left( frac{x - vt}{sigma sqrt{2}} right) right) = -frac{sqrt{2pi} sigma}{2} left( text{erf}left( frac{x}{sigma sqrt{2}} right) - text{erf}left( frac{x - vt}{sigma sqrt{2}} right) right) ]And the other terms from ( I_2 ):[ -frac{1}{sigma^2} left( sigma^2 x e^{-frac{x^2}{2sigma^2}} - sigma^2 (x - vt) e^{-frac{(x - vt)^2}{2sigma^2}} right) = -x e^{-frac{x^2}{2sigma^2}} + (x - vt) e^{-frac{(x - vt)^2}{2sigma^2}} ]Putting all together:[ I = frac{1}{v} left[ sigma sqrt{frac{pi}{2}} left( text{erf}left( frac{x}{sigma sqrt{2}} right) - text{erf}left( frac{x - vt}{sigma sqrt{2}} right) right) - frac{sqrt{2pi} sigma}{2} left( text{erf}left( frac{x}{sigma sqrt{2}} right) - text{erf}left( frac{x - vt}{sigma sqrt{2}} right) right) - x e^{-frac{x^2}{2sigma^2}} + (x - vt) e^{-frac{(x - vt)^2}{2sigma^2}} right] ]Let me factor out ( sigma sqrt{frac{pi}{2}} ) from the first two terms:[ sigma sqrt{frac{pi}{2}} left( 1 - frac{sqrt{2}}{2} right) left( text{erf}left( frac{x}{sigma sqrt{2}} right) - text{erf}left( frac{x - vt}{sigma sqrt{2}} right) right) ]Simplify ( 1 - frac{sqrt{2}}{2} ):[ 1 - frac{sqrt{2}}{2} = frac{2 - sqrt{2}}{2} ]So,[ sigma sqrt{frac{pi}{2}} cdot frac{2 - sqrt{2}}{2} left( text{erf}left( frac{x}{sigma sqrt{2}} right) - text{erf}left( frac{x - vt}{sigma sqrt{2}} right) right) ]Simplify ( sqrt{frac{pi}{2}} cdot frac{2 - sqrt{2}}{2} ):Let me compute ( sqrt{frac{pi}{2}} cdot frac{2 - sqrt{2}}{2} ):[ sqrt{frac{pi}{2}} cdot frac{2 - sqrt{2}}{2} = frac{sqrt{pi}}{sqrt{2}} cdot frac{2 - sqrt{2}}{2} = frac{sqrt{pi}}{sqrt{2}} cdot left( 1 - frac{sqrt{2}}{2} right) ]But this might not lead to a significant simplification, so perhaps we can leave it as is.So, putting it all together, the integral ( I ) is:[ I = frac{1}{v} left[ sigma sqrt{frac{pi}{2}} cdot frac{2 - sqrt{2}}{2} left( text{erf}left( frac{x}{sigma sqrt{2}} right) - text{erf}left( frac{x - vt}{sigma sqrt{2}} right) right) - x e^{-frac{x^2}{2sigma^2}} + (x - vt) e^{-frac{(x - vt)^2}{2sigma^2}} right] ]This seems quite complicated, but perhaps we can leave it in terms of the error function and exponentials.Therefore, the expression for ( S(x,t) ) becomes:[ S(x,t) = S_0(x) - frac{k_1 A}{sigma^2} cdot I - k_2 B C t cos(Cx) ]Substituting ( I ):[ S(x,t) = S_0(x) - frac{k_1 A}{sigma^2} cdot frac{1}{v} left[ sigma sqrt{frac{pi}{2}} cdot frac{2 - sqrt{2}}{2} left( text{erf}left( frac{x}{sigma sqrt{2}} right) - text{erf}left( frac{x - vt}{sigma sqrt{2}} right) right) - x e^{-frac{x^2}{2sigma^2}} + (x - vt) e^{-frac{(x - vt)^2}{2sigma^2}} right] - k_2 B C t cos(Cx) ]Simplify the constants:[ frac{k_1 A}{sigma^2 v} cdot sigma sqrt{frac{pi}{2}} cdot frac{2 - sqrt{2}}{2} = frac{k_1 A}{sigma v} cdot sqrt{frac{pi}{2}} cdot frac{2 - sqrt{2}}{2} ]Let me compute ( sqrt{frac{pi}{2}} cdot frac{2 - sqrt{2}}{2} ):[ sqrt{frac{pi}{2}} cdot frac{2 - sqrt{2}}{2} = frac{sqrt{pi}}{sqrt{2}} cdot frac{2 - sqrt{2}}{2} = frac{sqrt{pi} (2 - sqrt{2})}{2 sqrt{2}} = frac{sqrt{pi} (2 - sqrt{2})}{2 sqrt{2}} ]We can rationalize the denominator:[ frac{sqrt{pi} (2 - sqrt{2})}{2 sqrt{2}} = frac{sqrt{pi} (2 - sqrt{2}) sqrt{2}}{2 cdot 2} = frac{sqrt{pi} (2 sqrt{2} - 2)}{4} = frac{sqrt{pi} ( sqrt{2} - 1 )}{2} ]So, the term becomes:[ frac{k_1 A}{sigma v} cdot frac{sqrt{pi} ( sqrt{2} - 1 )}{2} ]Therefore, the expression for ( S(x,t) ) simplifies to:[ S(x,t) = S_0(x) - frac{k_1 A}{sigma v} cdot frac{sqrt{pi} ( sqrt{2} - 1 )}{2} left( text{erf}left( frac{x}{sigma sqrt{2}} right) - text{erf}left( frac{x - vt}{sigma sqrt{2}} right) right) + frac{k_1 A}{sigma^2 v} left( x e^{-frac{x^2}{2sigma^2}} - (x - vt) e^{-frac{(x - vt)^2}{2sigma^2}} right) - k_2 B C t cos(Cx) ]This is quite a complex expression, but it represents the solution for ( S(x,t) ) given the forms of ( E ) and ( M ).However, perhaps there's a simpler way to express this. Let me think about the integral again.Wait, another approach: since ( E(x,t) ) is a Gaussian moving with velocity ( v ), and we're integrating over ( tau ), perhaps we can change variables to ( xi = x - vtau ), which would make the integral a function of ( xi ), but I think we already did that.Alternatively, perhaps we can recognize that the integral of the Gaussian and its derivatives can be expressed in terms of the error function, which we did.Given that, I think the expression we have is as simplified as it can get without additional constraints or specific forms for ( S_0(x) ).Therefore, the solution for ( S(x,t) ) is:[ S(x,t) = S_0(x) - frac{k_1 A}{sigma^2 v} left[ sigma sqrt{frac{pi}{2}} cdot frac{2 - sqrt{2}}{2} left( text{erf}left( frac{x}{sigma sqrt{2}} right) - text{erf}left( frac{x - vt}{sigma sqrt{2}} right) right) - x e^{-frac{x^2}{2sigma^2}} + (x - vt) e^{-frac{(x - vt)^2}{2sigma^2}} right] - k_2 B C t cos(Cx) ]Alternatively, we can factor out constants:[ S(x,t) = S_0(x) - frac{k_1 A}{sigma v} cdot frac{sqrt{pi} ( sqrt{2} - 1 )}{2} left( text{erf}left( frac{x}{sigma sqrt{2}} right) - text{erf}left( frac{x - vt}{sigma sqrt{2}} right) right) + frac{k_1 A}{sigma^2 v} left( x e^{-frac{x^2}{2sigma^2}} - (x - vt) e^{-frac{(x - vt)^2}{2sigma^2}} right) - k_2 B C t cos(Cx) ]This is the expression for ( S(x,t) ) in terms of the error function and exponentials, plus the term involving ( cos(Cx) ).So, summarizing, after substituting both ( frac{partial^2 E}{partial x^2} ) and ( frac{partial M}{partial x} ) into the PDE, we integrated with respect to ( t ) to find ( S(x,t) ) as a function involving the error function and exponential terms, along with a term linear in ( t ) involving ( cos(Cx) ).Therefore, the final expression for ( S(x,t) ) is as above, assuming the initial condition ( S_0(x) ) is known.**Final Answer**Sub-problem 1:The expression for ( frac{partial^2 E}{partial x^2} ) is boxed{ -frac{A}{sigma^2} left( 1 - frac{(x - vt)^2}{sigma^2} right) e^{-frac{(x - vt)^2}{2sigma^2}} }.Sub-problem 2:The solution for ( S(x,t) ) is:[boxed{ S(x,t) = S_0(x) - frac{k_1 A}{sigma v} cdot frac{sqrt{pi} ( sqrt{2} - 1 )}{2} left( text{erf}left( frac{x}{sigma sqrt{2}} right) - text{erf}left( frac{x - vt}{sigma sqrt{2}} right) right) + frac{k_1 A}{sigma^2 v} left( x e^{-frac{x^2}{2sigma^2}} - (x - vt) e^{-frac{(x - vt)^2}{2sigma^2}} right) - k_2 B C t cos(Cx) }]

answer：First, I need to calculate the total time Alex spent watching the films. Each film has a runtime of 120 minutes, and Alex watched 5 films. So, the total watching time is 120 minutes multiplied by 5, which equals 600 minutes.Next, I'll determine the total break time Alex took. After each film, Alex took a 15-minute break. Since there are 5 films, there are 4 breaks in between them. Therefore, the total break time is 15 minutes multiplied by 4, resulting in 60 minutes.Finally, I'll add the total watching time and the total break time to find the overall time Alex spent. Adding 600 minutes and 60 minutes gives a total of 660 minutes.