Review sheet for midterm 2 solutions

Transcription

1 Review sheet for midterm solutions ) Solve the following differential equations: (a) dy dx = y 5 with y() = x Identify the equation as linear. m dy dx my x = 5m To use the product rule: dm dx = m x dm m = dx x ln m = ln x + C take C=0 m = x By the product rule: d[my] dx = 5m d[y/x] = 5 dx x y = 5 ln x + C x y = 5x ln x + Cx Using the initial condition: = y() = 5 ln + C = C C = Therefore y = 5x ln x + x

2 (b) dy dx = y x with y(0) = 0 Identify the equation as linear. m dy dx my = x m To use the product rule: dm dx = m dm m = dx ln m = x + C take C=0 m = e x By the product rule: d[my] dx = x m d[e x y] = x e x dx e x y = x e x dx Using the initial condition: 0 = y(0) = + C ( = x e x + ( = x e x xe x + ) xe x dx = x e x + xe x + e x + C y = x + x + + Ce x C = Therefore y = x + x + e x ) e x dx

3 (c) (x + ) dy = y x with y(0) = 0 dx Identify the equation as linear. m dy dx m y x + = m x x + To use the product rule: dm dx = m x + dm m = dx x + By the product rule: d[my] dx = x x + m d[y/(x + )] = x ln m = ln x + + C take C=0 m = x + dx (x + ) y x + = x (x + ) dx x + = (x + ) dx = x + (x + ) dx = ln x + x + + C y = (x + ) ln x + + C(x + ) Using the initial condition: 0 = y(0) = ln + C = + C C = Therefore y = (x + ) ln x + + (x + ) = (x + ) ln x + + x 3

4 (d) csc(x) dy = y + with y(0) = 0 dx Identify as separable dy y + = sin(x)dx ln y + = cos(x) + C Check for the missed solution: y = y = + e cos(x)+c Using the intial condition: 0 = y(0) = + e +C C = ln() + = Therefore: y = + e cos(x) ) Estimate y(3) for the following functions using step sizes h =, (a) dy dx = x xy y where y() = For h = n x n = x n + h y n = y n + f(x n, y n )h n x n = x n + y n = y n + x n x n y n yn 0 + = = 4 We find y(3) 4 Now, for h = n x n = x n + h y n = y n + f(x n, y n )h n x n = x n + y n = y n + (x n x n y n yn ) ( ) = We find y(3) 4

5 (b) dy dx = sin(πy) where y( ) = For h = n x n = x n + h y n = y n + f(x n, y n )h n x n = x n + y n = y n + sin(πy n ) sin( π ) = sin( 3π ) = 3 + sin( π ) = We find y(3) Now, for h = n x n = x n + h y n = y n + f(x n, y n )h n x n = x n + y n = y n + sin(πy n ) 0 + sin( π ) = sin( 5π ) = 9 We find y(3) 9 5

6 3) (a) A colony of ants reproduces at a rate proportional to its size. Suppose you start an ant farm with only one male ant and one female ant. One year later you count your ants and find that you have 00 of them. If ants are immortal, how many ants will you have in 5 years? Now, you realize ants are, in fact, not immortal. You do some research and keep track of your ants and find that on your farm ants die at a rate of e kt where K is the same proportianality constant from the reproduction rate. With this added information, how many ants will there actually be in 5 years? Make the following variable declarations: P = population of ants, units in ants t = time, units in years Because of the phrase reproduces at a rate proportional to its size we know dp = kp for some constant k. In addition, we have two dt initial conditions: P (0) = and P () = 00. This is separable, and solved as follows: dp P = kdt ln P = kt + C P = e kt+c = P (0) = e 0+C so C = ln() P = e kt+ln() = e kt 00 = P () = e k so k = ln(50) P = e ln(50)t = (50) t So it follows after 5 years P (5) = (50) 5 is a lot of ants. (Technically, we missed the solution P = 0, but this does not satisfy our initial conditions) BUT as the second part of the question points out we forgot to account for ants dying. We lose 0 ants every year, so our differential equation changes: 6

7 dp dt = In Out = kp ekt The inital condtions stay the same, so we must solve this new differential equation. This is a linear differential equation. m dp dt kmp = ekt m To use the product rule: dm dt = km dm m = kdt ln m = kt + C choose C=0 m = e kt Using the product rule d[mp ] = e kt m dt d[e kt P ] = dt e kt P = t + C P = e kt (C t) Using initial conditions: = P (0) = e 0 (C 0) C = 00 = P () = e k ( ) = e k k = ln(00) P = e ln(00)t ( t) = (00) t ( t) So we then have after five years P (5) = (00) 5 ( 5) = 3(00) 5. Stop. This should raise a lot of red flags, because now we have a negative number of ants. Clearly this doesn t make sense, so we need to check to see if all the ants die off. The equation we got is only valid if there are actually ants to reproduce, i.e. only when it is positive. Checking our solution: P = (00) t ( t) 0 t 0 by (00) t 0 always t 7

8 And in fact, at t = we find P = 0. This means all the ants died off, so there is no more room for the population to grow or shrink. So we need to adjust our solution to the following: P = { 00 t ( t) when 0 t 0 when t So we actually find there are no ants after 5 years i.e. P (5) = 0. 8

9 (b) You are making lemonade, but you first make a few pitchers of lemonade that is 50% lemon juice. You decide this is too much. So, you take a 000mL bucket and fill it with water. You then pour your original lemonade into the bucket at a rate of 0mL/s and drain the well mixed lemonade at the same rate. Find the concentration of lemon juice in the bucket as a function of time. You decide you want lemonade that is 0% lemon juice, how long do you have to wait? First we must declare our variables. Note this is a mixing problem: Q = concetration of lemon juice, units juice/ml L = amount of lemon juice, units juice = 000Q t = time, units seconds We start with a bucket of just water, so L(0) = 0. Our rate in is the same as our rate out at 0mL/s. We get the following differential equation: dl dt = In Out = (0.50juice/mL)(0mL/s) (Qjuice/mL)(0mL/s) = 5 0Q = 5 L 00 Noting our units match up. This is both separable and linear, I will present the separable solution method: dl 000 L = dt 00 ln 000 L = t 00 + C 000 L = e C t/ L = ±e C t/00 L = ±e C t/ Using the inital condition 0 = L(0) = ±e C = ±e C must use -, so C = ln(000) Therefore L = e ln(000) t/ = 000e t/ = 000( e t/00 ) 9

10 So we then have Q = Q(t) = 0.0 = L = ( 000 e t/00 ). We need to solve for 0. 0 = ( e t/00 ) 5 = e t/00 e t/00 = 4 5 t ( 4 ) 00 = ln 5 t = 00 ln ( ) 4 5 Algebra time! Does it make sense that our answer t is positive or negative? Is 00 ln ( 4 5) positive? 0

11 (c) You buy a house for $0,000. You cannot afford this because you are a college student, so you take out a loan for the money. The loan has a yearly interest rate of 0%. You decide to pay $000 a year to pay off the loan. Write the amount of money you still owe, M, as a function of time. Are your payments large enough that you will eventually completely pay off the loan, and if so when will you finish paying off the loan? Answer the same questions if you had paid $000 per year, or $5000 per year. We know we can apply the following: dm = In Out. The money dt IN is the money being added to your debt by the interest. The money Out is the money being taken out of the debt by your payments. We then have the separable and linear differentiale equation: dm dt = 0 M 000 m dm dt m 0 M = 000m To use the product rule: dm dt = m 0 dm m = dt 0 ln m = t + C choose C=0 0 m = e t/0 Using the product rule d[mm] = 000m dt d[e t/0 M] = 000e t/0 dt e t/0 M = 0000e t/0 + C M = Ce t/0 Using the inital debt 0000 = M(0) = Ce = C Therefore M = e t/0 Will we eventually pay off the debt? and if so when? Both questions

12 can be answered by setting M = 0 and solving for t: 0 = M = e t/0 = e t/0 t = 0 ln() So the answer is yes. For payments of $5000, you will similarly( find ) the answer is yes where M = e t/0 and t = 0 ln. For payments of $000 you find M = 0000 remains constant, so you end up never repaying your debt. 5 4

13 dt (d) Recall Newton s cooling law: = k(t T dt r ) where T r is the temperature of the container and T is the temperature of the object in question. Suppose you put a 00 C cup of soup in the refrigerator, which is set to 0 C. You want to know when the soup will reach 0 C so you can eat it. Suppose you check the soup minute after you put it in the refrigerator and find the temperature is now 50 C. How long do you have to wait from the moment you put the soup in the refrigerator before you can eat the soup? dt = k(t 0) dt dt T = kdt ln T = kt + C T = e kt+c Using initial conditions: 00 = T (0) = e C C = ln(00) 50 = T () = e k+ln(00) = 00e k 3 4 = ek ( 3 k = ln 4) ( ) therefore T = 00e ln 3 ( t 4 3 t = 00 4) ( 3 t 0 = T = 00 4) ) t ( 3 0 = 4 ( ( 3 ln = ln t 0) ( 4) ) ln 0 t = ( ) = ln 3 4 ln 0 ln 3 ln 4 3

14 4) Find the following Taylor polynomials and Taylor Series using the appropriate method. (Recall cosh(x) = ex +e x ) (a) T 3 x = + (ln )(x ) + (ln ) (x ) + (ln )3 6 (x ) 3 (b) (c) (d) sin(x) T x = T x sin(x) = ( + x + x + + x n + o(x n ))(x x3 3! + + ( )n x n+ + o(x n+ )) (n + )! = x x3 3! + + ( )n x n+ + o(x n+ ) (n + )! + x x4 3! + + ( )n x n+ + o(x n+ ) (n + )! + x 3 x5 3! + + ( )n x n+3 + o(x n+3 ) (n + )! + + ( )n x n++n + o(x n++n ) (n + )! = c 0 + c x + c x + + c n x n + o(x n ) Where c n = 3! + + ( )k (k+)! for n = k + and n = k + T 9 3x = T9 ( + u) / for u=-3x ( ) ( ) ( ) ( / / / / = + u + u ( ) ( ) ( / / / = 3 x ) u 9 ) x + + ( 3) 9 ( / 9 T + t t = ( + t )( + t + t 4 + t t n + o(t n )) = + t + t 4 + t t n + o(t n ) t + t 4 + t 6 + t t n + t n+ + o(t n+ ) = + t + t t n + o(t n ) 4 ) x 9

15 (e) T e 9 t = e 9 T e t (f) T 4 cosh(x) = + x + x4 4! = e 9 e 9 t + 4e9 t + + ( )n e 9 t n + o(t n ) n! 5) (a) Find T 3 {x 5 ln( + x)}. What is the maximum possible error, (x 5 ln( + x)) T 3 {x 5 ln( + x)}, over the interval x <? T 3 {x 5 ln( + x)} = T 3 ln( + x) = x x + x 3 only affects the fifth term. because the x5 (x 5 ln( + x)) T 3 {x 5 ln( + x)} = R 3 {x 5 ln( + x)} = f (4) (ξ)x 4 4! = 0ξ 6( + ξ) 4 x 4 4 < 0ξ 6( + ξ) 4 by x < 4 4 We now have to maximize 0ξ 6( + ξ) 4 over ξ <. Critical points are solutions to 0 + 4( + ξ) 5 = 0, i.e. ( + ξ) 5 = /5: solving this equation for ξ, we obtain ξ = ( /5) /5.3, so there s no critical points with ξ <. we just have to check the end points, so we find: ( 0 6( + ) 4 = So it follows that: 5 3) < 60 < 56 6( + ) 4 = 60 6() 4 = = 56 R 3 {x 5 ln( + x)} < 0ξ 6( + ξ) <

16 (b) Estimate 0 ex3 dx Using the first order Taylor Polynomial of e x, and find the maximum error in this estimation. The integral is approximately Now to find the error, noting R e t = eξ t 0 e x3 dx 0 0 T e t = + t T e t = + x 3 + x 3 dx = + 4 = x 3 dx = by e ξ increasing and the Comparison Theorem = = = 3 4 e x3 + x 3 dx R e t dx for t = x 3 e ξ x 6 dx ex 6 dx 6) Determine if the following series converge or diverge. Be sure to justify your answer. (a) cos(n) n! By the comparison theorem we have that cos(n)/n! /n! so that cos(n) n! n! For the ratio test on the latter sum, we see that /(n + )! /n! = n! (n + )! = n + = 0 < 0 3x 6 dx 6

17 Which shows the series converges, and so by the comparison theorem so does the original. n (b) n n= We use the it comparison test to compare this series to n= n, which diverges by the integral test. n /n / n = = = = n n n n n n n (c) (d) So by the it comparison test, the two series behave the same and both diverge. e πn We have that e πn e πn. We also see that e πn = (e π ) n is a geometric series with r = e π <, and so converges. Then by the comparison test so does the series in question. (n + )(n ) (n + ) 3 We do the it comparison test to which diverges by the n+ integral test. (n + )(n )/(n + ) 3 /(n + ) = (n + )(n ) (n + ) n + 3n = n + n + + 3/n /n = + /n + /n = 7

18 So by the it comparison test both series behave the same, and so both diverge. n (e) n We apply the ratio test (n + )/ n+ n + = n/ n n = + n = < Which show the series converges. cos(πn) (f) ln(n) n= We will apply the alternating series test. Notice that cos(nπ) = ( ) n, so we can rewrite our series as n= ( ) n ln(n) We have that > 0 by n >. We also have that ln(n+) > ln(n) ln(n) by being an increasing function, so it follows that <. ln(n+) ln(n) And lastly, we have = 0. This implies the series satisfies ln(n) the conditions of the alternating series test, and so must converge. n(n + )(n + )! (g) n!(n + 3) n= Consider that n(n + )(n + )! n!(n + 3) = n(n + )(n + ) n + 3 n 3 + 3n + n = n + 3 n + 3n + = + 3/n = 0 So by the divergence test, this series diverges. 8

19 (h) n n 3 + sin(n) We will use the it comparison test to compare to the series n= n=, which converges by the integral test. n n/(n 3 + sin(n)) n 3 = /n n 3 + sin(n) = + sin(n) n 3 = (i) Which implies the two series must behave the same, and converge. arctan(n) Notice that arctan(n) = π 0 (j) so by the divergence test, the sum diverges. n! We apply the ratio test / (n + )! / n! which show the series converges. n! = (n + )! = = = 0 < n! (n + )! n + 7) Determine the radius of convergence for the following series (a) ( ) n n! x n 9

20 We apply the ratio test ( ) n+ x n+ /(n + )! ( ) n x n /n! = = 0 < x n + (b) So the series converges for all values of x. The radiua of convergence is infinite. n n + xn n= We apply the ratio test n+ = x (n + ) n(n + ) n+ xn+ n n+ xn n + 4n + 4 = x n + n + 4/n + 4/n = x + /n = x (c) So the series converges for all x < and diverges for x >. The radius of convergence is. As a point of interest, when x = ±, the divergence test implies that the series diverges. n xn 3n We apply the ratio test (n + )x n+ /3 n nx n /3 n = x (n + ) 3n = x 3 + 3n = x 3 So the series converges for all x < 3 and diverges for x > 3. The radius of convergence is 3. As a point of interest, when x = ±3, the divergence test implies that the series diverges. 0

21 x n (d) ln(n) n= We apply the ratio test x n+ / ln(n + ) x n / ln(n) (e) ln(n + ) = x ln(n) /(n + ) = x /n = x n n + = x + /n = x By applying L Hopital s Rule. So the series converges for all x < and diverges for x >. The radius of convergence is. As a point of interest, what happens when x = ±? For x = the series converges by problem 6(f). When x =, consider that ln(n) < n n= diverges by the integral n also diverges by the comparison theorem. so that ln(n) > n. We know that test, so n= ln(n) n(3n ) x n n + We apply the ratio test (n+)(3n+) x n+ n+3 n(3n ) x n+ n (n + )(n + )(3n + ) = x n(n + 3)(3n ) 3n 3 + 6n + n + 4 = x 3n 3 + 8n 3n 3 + 6/n + /n + 4/n 3 = x 3 + 8/n 3/n = x So the series converges for all x < and diverges for x >. The radius of convergence is. As a point of interest, when x = ±, the divergence test implies that the series diverges.

22 (f) n!x n We apply the ratio test (n + )!x n+ n!x n = x (n + ) { 0 x = 0 = x 0 So the series converges for x = 0 and diverges for x > 0. The radius of convergence is 0.