MATH141: Calculus II Exam #4 review solutions 7/20/2017 Page 1

Transcription

1 MATH4: Calculus II Exam #4 review solutions 7/0/07 Page. The limaçon r = + sin θ came up on Quiz. Find the area inside the loop of it. Solution. The loop is the section of the graph in between its two zeroes, which occur at π and. We want to find the area between those two angles. This is a bit of a mess, so put your gloves on. It will help to keep in mind the following trig values for cosine, cos = π and cos = +. We ll need some for sine too, but which ones exactly will become clear later. At its heart this question is asking us to calculate the integral ( + sin θ) dθ. There s no magical way to make this go any faster; we ll pull out the, square it, and start chipping away. ( + sin θ) dθ = ( + 4 sin θ + 4 sin θ ) dθ = dθ + sin θ dθ + sin θ dθ So we have three integrals to do. I ll do them one at a time. The first one is a warm-up. dθ = ( π ) = π. Next, the integral of sin θ is cos θ, and at the beginning of the solution I wrote down the relevant values of cosine we d need. π ( ) sin θ dθ = cos θ = =. Now for the hard one. We ll use the identity sin cos θ θ = to improve the integrand; this cancels the coefficient out front. Then we have to do the two pieces: sin θ dθ = = = 4π = 4π ( cos θ ) dθ θ. sin θ π sin π + cos θ dθ sin 4π

2 MATH4: Calculus II Exam #4 review solutions 7/0/07 Page These values of sine are quite big. π is π, so if we deduct that from our inputs we re down to 0π = 5π = π and π = π. The values of sine at these two places are both, since the two angles are in the fourth and first quadrants respectively. + 4 = 4π = 4π +. Adding the three contributions, we arrive at π + 4π + = π. Stunningly, this matches what WolframAlpha tells me.. Write down a geometric series with first term that converges to 7. Solution. A geometric series looks like n=k an. Since we want the first term to be, we should start with n = 0. When a geometric series converges, the value is a n = first term common ratio = a. We want this to equal seven, so we should solve = 7. Doing so gives a =. So the a 7 series we re seeking is ( ) n. 7. Write the number + i in polar form, then calculate its three cube roots. Solution. This number has absolute value 8 and angle π. (You can see what the 4 angle is by the fact that it lies on the line y = x, which makes a forty-five degree angle to the positive x-axis.) This means that + i = 8 e i π 4. The cube roots will have modulus that is the cube root of the starting modulus; that s 8 =. The angles of the cube roots are those angles which, when multiplied by three, become π. One such 4 angle is π, and the others differ from that by increments of π since they re evenly spaced around the circle. That means the other arguments are 9π and. The three cube roots are 8 e i π, 8 e i 9π, and 8 e i.

3 MATH4: Calculus II Exam #4 review solutions 7/0/07 Page 4. Find the Taylor series expansion centered at x = 0 for each of the following. (a) x + x Solution.. (Start from x.) Write the function in the form ( ) x ( x) so that it looks like a modification of. We get x x + x = x ( x) n = ( ) n x n+. However you want to write this is fine, as long as that s what it equals. (b) xe x. (This is the derivative of e x, so start there and take the derivative.) Solution. Okay, to be fair there are multiple ways to do this. But I wanted you to start with e x = (x ) n = x n = + x! + x4! + x! +. and then take the derivative. You could just as well multiply by x. If you take the derivative, the first term, which was just a, will drop out, so the summation begins at n = instead. However you want to write it, the answer is n= nx n = x n+ = x + 4x! + x5! +. Maybe it would be nice to clean up n = (n )! in the first expression. Eh, too late now. 5. Use the Lagrange remainder formula to decide how good of an estimate you ll get if you use + ( )! as an estimate for e. How many terms are needed to get the error to less than? 00 Solution. The remainder formula (which I will give you) is f (n+) (x) [error in stopping after exponent n] max 0 x a (n + )! a n+. Apply this with f = e x, a =. The derivatives of e x are all ±e x, which in absolute value will always be +e x. This function is decreasing, so the maximum will occur at

4 MATH4: Calculus II Exam #4 review solutions 7/0/07 Page 4 the beginning, i.e., at x = 0. That maximum is e 0 =, so that s what we ll use for the numerator. Since we cut off after the degree term, we get [error] ( ) =! 8 = 48. ( If this process is repeated with n = 4 instead we ll get ) 4, 4! which has denominator times 4. I don t know exactly what that is off the top of my head, but it s definitely bigger than. So we need up through the exponent term to get error less than Calculate the interval of convergence for xn. Solution. Ratio test: (n + )! x n+ lim = lim (n + ) x = x lim (n + ) =. n x n n n If x 0, the calculation shows that the limit we do in the ratio test will explode to infinity. That s not known for being less than one, so apparently it diverges for all nonzero x. When x = 0 on the other hand all the terms with x in them will die, so we just get + lots of zeroes = which is a number. So the sum only converges when x = 0; this means the radius of convergence is R = 0 and the interval of convergence is {0} (a single point). You can also write this as [0, 0]. 7. Write the following series in summation notation. What is its radius of convergence? x + 4! x + 8 5! x + 7! x4 + 9! x5 +. Solution. Let s have n represent the exponent on x in each term. The summation will then be from n = to infinity. The numerators are powers of two, and the exponent on said power matches the exponent on x. So we re at n n= (??)! xn at this point. The factorials are consecutive odd numbers; at each step they go up by two, so the way to write this will be with a linear function that has slope two. ( Over one, up two is what it means to have slope two.) So it s (n +?)!, and messing around with small values of n shows that the question mark should be. The series is n= n (n )! xn. Having done this, it s time for the ratio test to find the radius of convergence: n+ x n+ lim n (n + )! (n )! (n )! = x lim n x n n (n + )! = x lim n (n)(n + ) = 0. This limit is kind of the opposite of the previous one. Regardless of what we choose for x, the limit will work out to zero. The ratio test is always happy about this, since zero is less than one. So the series converges for all x; the radius of convergence is infinity.

5 MATH4: Calculus II Exam #4 review solutions 7/0/07 Page 5 8. Solve the initial value problem dy dx = xy, with initial condition y() = 4. Solution. This equation is separable, since we can write it as dy = x dx. Integrating y both sides gives us = x + C. If we take care of the initial condition at this point, y we ll plug in x = and y = 4. That means = + C, or C = 9. If we then negate 4 4 both sides and take the reciprocal, we get the answer y = x Suppose f is a function whose Taylor series centered at x = begins Explain why f has a maximum at x =. f = (x ) + 4(x ) +. Solution. The series I gave you is supposed to be equal to f() + f ()(x ) + f ()! (x ) + f ()! (x ) +, which means if we equate coefficients we find that f() =, f () = 0, f () = 4, and f () = 4. The fact that f () = 0 is suspicious; it means there s likely a max or a min at x =. To determine which, we can look at the second derivative: because f () = 4, the function is concave down. That means that f comes to a peak and has a maximum. 0. True and false questions!? ( π ) ( ( π )) T F (a) + sin + sin + converges to sin ( ). π ( π ) ( ( π )) T F (b) + tan + tan + converges to tan ( ). π Solution. (a) is true and (b) is false. It is true that an =, but this a formula only works if the series converges, and that happens if and only if a <. Since sin( π) = and that s less than, we re safe. But tan( π) = which is larger than, so that series diverges to infinity.

6 MATH4: Calculus II Exam #4 review solutions 7/0/07 Page T F (c) T F (d) 5 (x ) dx =. Solution. False. Similar to part (b), this integral is actually improper since the integrand has an asymptote at x = which is in the middle of our range of integration. If you do out the improper integral, you ll find that it diverges. What s written here can t possibly be correct in any case, because the integral of a positive function can t be a negative number. n= ( ) n cos n n converges, but not absolutely. Solution. False, since it does converge absolutely. With absolute values we have cos n n=, which is less than or equal to n n= (owing to the fact that n 0 cos n for all n). We ve just compared to a p-series with p =, so the larger series converges, meaning the smaller one does as well. T F (e) The range of arcsin(x) is [, ]. Solution. False, that s the domain of arcsine. (It s the range of sin x.) The range is [ pi T F (f) Suppose a n (x ) n is a power series which converges when x = 5. When n= x =, the series converges. Solution. True. The first fact tells us the radius of convergence is at least. Since x = is only one unit away from the center of the series (namely, ), is inside the radius of convergence.. Decide the method needed to do calculate the following integrals. The options are integration by parts, trig sub, partial fractions, and u-substitution. If you d integrate by parts, say what u and dv are. If you d do a trig substitution, say what substitution it is. If you d use partial fractions, write out the template you d use for breaking down the integrand. If you d do a u-substitution, actually do the substitution and get an integral that only involves u, but don t go any further than that. x + (a) x 4x dx Solution. Do partial fractions. The denominator factors as x(x 4), so the template is x + x 4x = A x + B x 4.

7 MATH4: Calculus II Exam #4 review solutions 7/0/07 Page 7 (b) sin x cos x dx Solution. Break off two sines, then do a u-substitution with u = cos x, du = sin x dx. You get sin x cos x dx = sin x( cos x) cos x dx = ( u )u du. (c) x sin(x) dx Solution. Integrate by parts with u = x and dv = sin(x) dx. (This one actually needs two rounds, but whatever, this is how you start.) x + 9 (d) dx x Solution. Trig sub here, with x = tan θ.