x 2 y = 1 2. Problem 2. Compute the Taylor series (at the base point 0) for the function 1 (1 x) 3.

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1 MATH FINAL EXAM - SOME REVIEW PROBLEMS WITH SOLUTIONS 8.0 Calculus, Fall 207 Professor: Jared Speck Problem. Consider the following curve in the plane: x 2 y = 2. Let a be a number. The portion of the curve with x a is revolved around the x axis to generate a solid of revolution. Show that the surface area of the solid does not become infinite when a. Problem 2. Compute the Taylor series (at the base point 0) for the function ( x) 3. Problem 3. Compute the Taylor series (at the base point 0) for the function f(x) = arcsin x. In particular, compute the first four non-zero terms in the series and illustrate the general pattern for the coefficients. Problem 4. Find all numbers p 0 such that the series ln ( + k ) p converges. k= Problem 5. Find all numbers p 0 such that the series k(ln k) p + converges. k=2 Problem 6. Show that if x is not a number of the form x = π 2 sec 2 x = (sin x) 2k. k=0 + mπ for some integer m, then Problem 7. Let f(x) = e x2. Compute f (208) (0), where f (208) (x) is the 208 th derivative of f(x).

2 2 MATH FINAL EXAM - SOME REVIEW PROBLEMS WITH SOLUTIONS Problem 8. Consider the following curve in the plane, described in polar form: r = ln θ. Find the area of the region in the second quadrant (x 0, y 0) that is trapped between the x axis, the y axis, and the curve. Problem 9. Compute e x + cos 2 x lim. x e x

3 MATH FINAL EXAM - SOME REVIEW PROBLEMS WITH SOLUTIONS 3 Solutions Problem. Consider the following curve in the plane: x 2 y = 2. Let a be a number. The portion of the curve with x a is revolved around the x axis to generate a solid of revolution. Show that the surface area of the solid does not become infinite when a. Solution: We first compute that y = 2 x 2, dy dx = x 3, ( ) 2 dy ds = + dx = + x dx 6 dx. The surface area of the region of interest is thin conical strip circumference conical strip slant height = 2πy ds. Hence, the surface area is a 2πy ds = a 2π 2 x 2 + x 6 dx. Now when x, we have + x = 2. Hence, the above integral is π 2 a = 2π [ x ] a = ( 2π ). a x 2 dx Now as a, the above term converges to the finite number 2π.

4 4 MATH FINAL EXAM - SOME REVIEW PROBLEMS WITH SOLUTIONS Problem 2. Compute the Taylor series (at the base point 0) for the function ( x) 3. Solution: We first note that ( x) = d dx 2 x. The standard formula for the geometric series yields that x = x k = + x + x 2 + x 3 + x 4 +. k=0 Twice differentiating the above series term-by-term and multiplying by /2, we deduce that ( x) = d 2 d dx 2 dx 2 xk = k(k )x k 2 2 k=0 The above series is the desired Taylor series. k=2 = + 3x + 6x 2 +.

5 MATH FINAL EXAM - SOME REVIEW PROBLEMS WITH SOLUTIONS 5 Problem 3. Compute the Taylor series (at the base point 0) for the function f(x) = arcsin x. In particular, compute the first four non-zero terms in the series and illustrate the general pattern for the coefficients. Solution: Our strategy is to first compute the Taylor series for d arcsin x = dx x 2 and to then integrate that series term-by-term to recover the Taylor Series for arcsin x. More precisely, we first use the fundamental theorem of calculus to deduce that arcsin x = x 0 dy y 2. We will compute the Taylor series for the integrand in two steps. In the first step, we compute y2 the Taylor series for the function Differentiating with respect to u, we deduce that u = ( u) /2. d du ( u) /2 = 2 ( u) 3/2, ( ) ( ) d 2 3 du ( 2 u) /2 = ( u) 5/2, 2 2 ( ) ( ) ( ) d du ( 3 u) /2 = ( u) 7/2, Setting u = 0 in the above formulas, we deduce that From the Taylor formula d du ( u) /2 u=0 = 2, ( ) ( ) d 2 3 du ( 2 u) /2 u=0 =, 2 2 ( ) ( d 3 3 du ( 3 u) /2 u=0 = 2 2. ) ( 5 2 ), f(u) = a 0 + a u + a 2 u 2 + a 3 u 3 +, a n = f (n) (0), n! and the above computations, it follows that u = + 2! u ! u ! u3 +.

6 6 MATH FINAL EXAM - SOME REVIEW PROBLEMS WITH SOLUTIONS We now substitute u = y 2 to deduce the Taylor series for the integrand y 2 : = + y 2 2! y ! y ! y6 +. Integrating this series term-by-term, we conclude that x dy arcsin x = y 2 0 x ( = + 0 2! y ! y ) 3! y6 + dy ( = y + 2! 3 y ! 5 y ) 3! 7 y7 + = x + 2! 3 x ! 5 x ! 7 x7 +. y=x y=0

7 MATH FINAL EXAM - SOME REVIEW PROBLEMS WITH SOLUTIONS 7 Problem 4. Find all numbers p 0 such that the series ln ( + k ) p converges. lim k k= Solution: We first claim that for large k, ln ( ) + k. To verify this claim, we use L Hôpital s p k p rule in the form to deduce that ln ( ) + ( p)k p k p + k p Therefore, by limit comparison, converges if and only if k p k= = lim k = lim k + k p =. ln ( + k ) p k= k p ( p)k p converges. By the integral comparison test, this latter series converges if and only if p >.

8 8 MATH FINAL EXAM - SOME REVIEW PROBLEMS WITH SOLUTIONS Problem 5. Find all numbers p 0 such that the series k(ln k) p + converges. Solution: We first note that k(ln k) p + k(ln k), p as k, so by limit comparison, it suffices to investigate the convergence of k(ln k). p By the integral comparison test, it suffices to investigate the convergence of x=2 k=2 k=2 dx x(ln x) p = lim M M x=2 dx x(ln x) p. To this end, we make the substitution u = ln x, du = dx, which leads to x { dx du x(ln x) = p u = p u p, if p, p ln u, if p =. { = (ln p x) p, if p, ln(ln x), if p =. Hence, we compute that Thus, M lim M x=2 { dx x(ln x) = (ln p 2) p, if p >, p limit does not exist, if p. k=2 converges if p > and diverges if 0 p. k(ln k) p +

9 MATH FINAL EXAM - SOME REVIEW PROBLEMS WITH SOLUTIONS 9 Problem 6. Show that if x is not a number of the form x = π 2 sec 2 x = (sin x) 2k. k=0 + mπ for some integer m, then Solution: As long as x is not of the form x = π +mπ for some integer m, we have that sin x <. 2 Thus, under this assumption, it follows that the geometric series (sin x) 2k = [sin 2 x] k converges to k=0 k=0 sin 2 x = cos 2 x = sec2 x.

10 0 MATH FINAL EXAM - SOME REVIEW PROBLEMS WITH SOLUTIONS Problem 7. Let f(x) = e x2. Compute where f (208) (x) is the 208 th derivative of f(x). f (208) (0), Solution: Using Taylor s formula, we simply have to expand and use the relation f(x) = a 0 + a x + a 2 x 2 + a n = f (n) (0) n! (with n = 208). To this end, we first recall the expansion for e u : e u = + u + u2 2! + u3 3! + u4 4! + We then substitute u = x 2 to deduce that e x2 = + x 2 + x4 2! + x6 3! + x8 4! + From this series expansion, it follows that a 208 = and hence 009! f (208) (0) = 208! a 208 = 208! 009! =

11 MATH FINAL EXAM - SOME REVIEW PROBLEMS WITH SOLUTIONS Problem 8. Consider the following curve in the plane, described in polar form: r = ln θ. Find the area of the region in the second quadrant (x 0, y 0) that is trapped between the x axis, the y axis, and the curve. Solution: We first note that the second quadrant corresponds to π θ π. We then recall the 2 standard formula for the area under a curve in polar coordinates: Area = 2 θ2 θ r 2 dθ = 2 π π/2 (ln θ) 2 dθ. To evaluate the integral, we first use the integration by parts relations u = ln θ, du = dθ, θ dv = ln θ dθ, v = θ ln θ θ to deduce that (ln θ) 2 dθ = u dv = uv v du = θ(ln θ) 2 θ ln θ (ln θ ) dθ = θ(ln θ) 2 2θ ln θ + 2θ. Inserting the factor of and the bounds of integration, we conclude that the area of the region of 2 interest is π [ ] θ=π (ln θ) 2 dθ = u dv = 2 π/2 2 θ(ln θ)2 θ ln θ + θ θ=π/2 = π 2 (ln π)2 π ln π π [ ( π )] 2 π ( π ) ln ln + π 2 2.

12 2 MATH FINAL EXAM - SOME REVIEW PROBLEMS WITH SOLUTIONS Problem 9. Compute e x + cos 2 x lim. x e x Solution: This is a type limit. deduce that e x + cos 2 x lim x e x If you try to use L Hôpital s rule repeatedly, then you e x 2 sin x cos x = lim x e x e x sin(2x) = lim x e x = lim x e x 2 cos(2x) e x e x + 4 sin(2x) = lim x e x =. This will go on indefinitely and hence you will not solve the problem in this fashion. To solve the problem, we avoid L Hôpital s rule altogether and instead multiply the top and bottom of the original fraction by e x. After performing some algebra, we see that we have to compute lim + x e x cos 2 x. This limit is because e x goes to 0 as x, while cos 2 x is never larger than one.