Practice Final Exam Solutions

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1 Important Notice: To prepare for the final exam, study past exams and practice exams, and homeworks, quizzes, and worksheets, not just this practice final. A topic not being on the practice final does not mean it won t appear on the final.. If the statement is always true, circle the printed capital T. If the statement is sometimes false, circle the printed capital F. In each case, write a careful and clear justification or a counterexample. (a) If a force of F (x) = 6x pounds is required to stretch a spring x feet beyond its rest length, then 36 ft-lbs of work is done in stretching the spring from its natural length to 6 feet beyond its rest length. Work: W = b a F (x) dx Work done stretching the spring from x = to x = 6: W = 6 6 6x dx = 3x 2 = 8 ft-lbs F (b) The trapezoid rule with n = 5 for 4 dx 2x + will be an overestimate. T Justification: The curve y = /(2x + ) is decreasing and concave up, so the trapezoid tops lie above the graph throughout. The estimate is larger than the integral. y 4 x n (.5)n (c) ln(2.5) = ( ) n. F We have n xn ln( + x) = ( ) n with radius of convergence ; the series on the right diverges for x >. Therefore when we set x =.5, the left side is ln(2.5) but the right side is divergent. Page of 2

2 x 2 (d) The improper integral (x 3 dx converges. F + 7) /3 x 2 b x 2 (x 3 dx = lim + 7) /3 b (x 3 dx + 7) /3 This limit does not exist so = lim b 2 (x3 + 7) 2/3 ( = lim b 2 (b3 + 7) 2/3 ) 2 (8)2/3 b = x 2 (x 3 dx does not converge. + 7) /3 (e) The tangent line to the parametric curve (x, y) = (t /t, 4 + t 2 ) at the point where t = has equation y = x + 5. Slope dy dx = dy/dt dx/dt = 2t + /t 2 dy dx = 2 2 = when t = The tangent line is y = x + b and passes through (x, y) = (, 5) T The line is y = x For each multiple choice question, circle the correct answer. There is only one correct choice for each answer. (a) A cylindrical tank with a radius of meter and a height of 8 meters is half full. Letting y = correspond to the top of the tank, the density of water be kg/m 3, and g be the acceleration due to gravity in m/sec 2, the work required to pump the water out of the tank is (i) πg 8 4 y dy (ii) πg 8 y dy (iii) πg 4 y dy (iv) 6πg The y-values for the water in the tank are 4 y 8. Partition [4, 8] into n parts. For k =,..., n, V k = πr 2 y = π y weight = π y g k-th slice is lifted y k meters (since y = is the top of the tank). n n W k gπ y k y W k gπ y k y W = πg 8 4 k= k= y dy, which is (i). It is also πg 4 (8 y) dy. 8 4 y dy Page 2 of 2

3 (b) The Taylor series at x = for sin x is (i) ( ) n x2n+ (ii) ( ) n x2n n! (2n)! n= n= The Taylor series at x = a is n= Let f(x) = sin x and a =. Then f() = f (4) () = sin = f () = f (5) () = cos = f () = f (6) () = sin = f (3) () = f (7) () = cos = (iii) ( ) n x2n (2n )! f (n) (a) (x a) n. n! The Taylor series for sin x at is x 3! x3 + 5! x5 7! x = is (iv). (It is not (iii), which is off by a minus sign.) (iv) ( ) n x2n+ (2n + )! n= ( ) n x2n+ (2n + )!, which n= (c) A parametric curve tracing out the circle once clockwise for t π starting at (, ) is (i) (cos t, sin t) traces out the top half of the circle counter-clockwise starting at (, ). (ii) (cos t, sin t) traces out the bottom half of the circle clockwise starting at (, ). (iii) (cos(2t), sin(2t)) traces out the complete circle counter-clockwise starting at (, ). (iv) (cos(2t), sin(2t)) traces out the complete circle clockwise starting at (, ). The answer is (iv). 3. This question is about the curve y = tan x. (a) Write a definite integral that gives the arc length of curve y = tan x from x = to x = π/4. π/4 π/4 Since dy/dx = sec 2 x, the arc length is + (dy/dx) 2 dx = + sec 4 x dx. (b) Write a definite integral that gives the area of the surface formed by revolving the curve y = tan x from x = to x = π/4 around the x-axis. The surface area is π/4 2πy + (dy/dx) 2 dx = π/4 2π tan x + sec 4 x dx 4. Use the error bound formulas on the last page to determine an n such that the Trapezoid rule with n subintervals approximates dx to within.. ex Page 3 of 2

4 By the last page, error f(x) = e x. K(b a)3 2n 2, where f (x) K for all x in [a, b] with a =, b =, f(x) = e x = f (x) = e x = f (x) = e x Since f (x) is decreasing, on [, ] we have f (x) = e x f () =, so we can use K =. Thus error K(b a)3 2n 2 = (3 ) 2n 2?. = Solve for n: n The least n we can choose by this method is n =. 5. (a) Obtain the Taylor series for + x at x = from the geometric series for x, writing your final answer in summation notation. x = + x + x2 + x 3 + = x n for x <. Replacing x with x, n= + x = x + x2 x 3 + = ( ) n x n for x <. (b) Use your result from part (a) and integration to write down the Taylor series at x = for ln( + x) and then find the radius of convergence of that series. Integrating termwise, + x dx = dx x dx + x 2 dx x 3 dx +..., so ln ( + x) + C = x x2 2 + x3 3 x4 To find C, let x = : ln() + C = = C =. Thus ln ( + x) = x x2 2 + x3 3 x4 4 + = = n= n xn ( ) n xn ( ) The radius of convergence comes from the ratio test: when a n = ( ) n x n /n, a n+ /a n = (n/(n + )) x x as n, so the radius of convergence is. 6. (a) Find the 3rd-order Taylor polynomial centered at 4 for T 3 (x) = f(4) + f (4)(x 4) + f (4) 2! x. n. n. (x 4) 2 + f (4) (x 4) 3 3! Page 4 of 2

5 f(x) = x = x /2 f (x) = 2 x 3/2 = 2x 3/2 f (x) = 3 4 x 5/2 = 3 4x 5/2 f (x) = 5 8 x 7/2 = 5 8x 7/2 f(4) = 2 f (4) = 2(2 3 ) = 6 f (4) = 3 4(2 5 ) = 3 28 f (4) = f (4) = 5 8(2 7 ) f (4) = Thus T 3 (x) = 2 3 (x 4) (x 4) (x 4)3. (b) Use Taylor s inequality (see the last page) to give an upper bound for the error in approximating by the polynomial in part (a) at x = We need an M for which f (4) (x) M when x =.. Then error M ! f (4) (x) = 5 6 x 9/2 = 5, which is a decreasing function for x > 6x9/2 For x , f (4) (x) f (4) 5 (3.99) =.3. We can use 6(3.99) 9/2 M =.3. error = (The exact error is T 3 (3.99) / ! , so the upper bound on the error is quite sharp.) 7. Use the error bound for alternating series to determine how many terms of the series need to be added to estimate the full series with error <.. ( ) n n 2 + The series is alternating (signs are alternating) and the magnitude of the terms /(n 2 + ) is decreasing to since n 2 + is increasing to. Therefore the error in approximating the series using the nth partial sum is at most the magnitude of the first omitted term. So if we use the nth partial sum then the error is < (n + ) 2 +. We seek n such that (n + ) 2 + <. =. Since (n + ) 2 + > (n + ) 2 > 999 n + > 3.6 n > 3.6. We can use n 3. Thus the sum of the first 3 terms approximates the full series with the desired error. 8. Use the Integral Test to show converges if p > and diverges if < p <. np Let f(x) = x p = x p. This is continuous and positive. Also Page 5 of 2

6 f (x) = px p = p < so f(x) decreases and xp+ lim f(x) = lim x x x p =. Suppose p >. dx = lim xp b p = p. Thus Suppose < p <. dx = lim xp b So b converges if p >. np b x p x p dx = lim b p x p x p dx = lim b p diverges for < p <. np b b b p = lim b p p = lim b ( p)b p + b p = lim b p = since p >. p 9. Determine which of the following series converges conditionally, converges absolutely or diverges. Specify which convergence test you use and show how it leads to the answer. (a) n=2 Compare ( ) n ln n Test for absolute convergence first. ( )n ln n = ln n to n using the Comparison Test. ln n < n = ln n > n To show ln n < n for n 2, we look at f(x) = x ln x. Then f (x) = /x > for x >, so f(x) increasing for x >. Thus x 2 = f(x) f(2) = 2 ln 2 >, so x > ln x for x 2. n=2 n diverges = ln n diverges n=2 (by the Comparison Test) Now use the Alternating Series Test since the series is not absolutely convergent. Set b n = / ln n. Two ways to show b n is decreasing: () the denominators ln n are increasing or (2) consider f(x) = / ln x for x > and its first derivative: f (x) = (ln x) 2 x <. Page 6 of 2

7 That is negative, so f(x) is decreasing for x >. Since b n as n, the series converges by the Alternating Series Test but is not absolutely convergent. It is conditionally convergent. (b). lim n n 2 n n 2 n = lim n + 5/n 2 = 2. The series diverges by the Divergence Test. (c) ( ) n n 5 a n+ a n = n! (n + ) 5 (n + )! n 5 = Apply the Ratio Test to test for absolute convergence. (n + )5 n 5 n! (n + )! = ( n + n! ( a n+ lim = lim + ) 5 n a n n n n + = <. n ) 5 ( n + = + ) 5 n n + ( ) n n 5 Thus n! converges absolutely. (d) n= values) 5 2 n + 5n + 3 Apply the Comparison Test (terms all positive, so no need for absolute 5 < 2 n + 5n + 3 < 5 2 n and 5 converges since this is a geometric series, so the series 2n n= 5 2 n converges absolutely using the Comparison Test. (Note: The Ratio Test and + 5n + 3 Limit Comparison Test could be used also.) Page 7 of 2

8 . Below are graphs of r = 3 sin θ and r = + sin θ. y x (a) Determine both polar and rectangular coordinates for all points where the curves cross in the first and second quadrants (not including the origin). Finding crossing points: 3 sin θ = + sin θ 2 sin θ = sin θ = 2. Thus θ = π/6 and 5π/6 are choices for the angle at crossing points, and here r = 3 sin θ = 3/2. Polar coordinates of crossing points: (r, θ) = (3/2, π/6), (3/2, 5π/6) Rectangular coordinates: (x, y) = (r cos θ, r sin θ) = (3 3/4, 3/4), ( 3 3/4, 3/4) (b) Set up, but do not evaluate, an integral for the area of the region inside r = 3 sin θ and outside r = + sin θ. The rays π/6 θ 5π/6 sweep out the region. To see which curve corresponds to which equation, write r = 3 sin θ in rectangular coordinates: r 2 = 3r sin θ is the same as x 2 + y 2 = 3y, so x 2 + (y 2 (3/2)y + 9/4) = 9/4, so x 2 + (y 3/2) 2 = (3/2) 2. That is a circle centered at (, 3/2) with radius 3. The other curve is r = + sin θ. The area inside r = 3 sin θ and outside r = + sin θ is A = 5π/6 π/6 2 ( (3 sin θ) 2 ( + sin θ) 2) dθ = 5π/6 ( (3 sin θ) 2 ( + sin θ) 2) dθ. 2 π/6. Solve for y exactly: (a) dy dx = sin x y 2 with y() = 3. Page 8 of 2

9 . Write as y 2 dy = sin x dx 2. y 2 dy = sin x dx 3. y 3 3 = cos x + C 4. Apply initial condition y() = 3: 9 = + C = C =. 5. Thus y3 3 = cos x = y = (3 3 cos x)/3. (b) dy = y cos x + xy with y() = 3. dx. 2. dy dx = y(cos x + x) = dy = (cos x + x) dx y y dy = (cos x + x) dx 3. ln y = sin x + x2 2 + C = y = esin x+x2 /2+C = e C e sin x+x2 /2 4. y = ±e C e sin x+x2 /2 = Ke sin x+x2 /2, where K = ±e C. Apply initial condition y() = 3: 3 = Ke sin +2 /2 = K Solution is y = 3e sin x+x2 /2. 2. Find the orthogonal trajectories of the family of curves y = kx 4, where k is an arbitrary (nonzero) constant. For points (x, y) on the curve y = kx 4, we have k = y/x 4 and dy dx = 4kx3 = 4 y x 4 x3 = 4y x. A curve orthogonal to this family must have derivative (slope) equal to the negative reciprocal of 4y/x, hence an orthogonal curve to the family has Now we solve this equation. 4y dy = x dx dy dx = x 4y. Page 9 of 2

10 4y dy = x dx 2y 2 = 2 x2 + C for any C. Thus the families y = kx 4 and 2 x2 + 2y 2 = C are orthogonal trajectories of each other. 3. A tank contains 3 L of water with 3 kg of salt dissolved in it. Brine that contains 5 kg of salt per liter enters the tank at a rate of 4 L/min. The solution in the tank is kept well mixed and is drained at a rate of 4 L/min. Use differential equations to determine how much salt remains in the tank after 3 minutes. Let y(t) = amount of salt (in kg) after t minutes, so y() = 3 kg. Then dy/dt = inflow rate of salt outflow rate of salt. The volume remains constant since the rate of flow in is the same as the rate of the flow out, namely 4 L/min, so the concentration (in kg/l) of salt in the tank after t minutes is y(t)/3. inflow rate = 5 kg L 4 L min outflow rate = y(t) 3 kg L 4 L = 2 kg min. Our differential equation becomes dy dt dy dt 6 4y = 3 4 dy 6 4y = = dy 6 4y = 3 dt 3 dt min = 4y(t) kg 3 min = 2 4y(t) 3 with initial condition y() = 3. t 4t ln 6 4y = + C = ln 6 4y = 3 3 4C = 6 4y = e 4C e 4t/3 6 4y = ±e 4C e 4t/3 = Ke 4t/3 = y(t) = (6 Ke 4t/3 )/4 = 5 (K/4)e 4t/3 From the initial condition y() = 3 we have 3 = 5 K/4, so K/4 = 5 3 = 47. Thus the solution to the differential equation is y(t) = 5 47e 4t/3. y(3) = 5 47e 4(3)/ kg 4. Compute e 2x sin(x) dx. Show all work. We integrate by parts, starting with u = e 2x and dv = sin x dx. Then du = 2e 2x dx and v = cos x, so e 2x sin x dx = u dv = uv v du = e 2x cos x 2 e 2x cos x dx. Page of 2

11 Doing integration by parts again with u = e 2x and dv = cos x dx, so du = 2e 2x dx and v = sin x, e 2x cos x dx = u dv = uv v du = e 2x sin x + 2 e 2x sin x dx, so so 5 Therefore ( ) e 2x sin x dx = e 2x cos x 2 e 2x sin x + 2 e 2x sin x dx = e 2x cos x 2e 2x sin x 4 e 2x sin x dx, e 2x sin x dx = e 2x cos x 2e 2x sin x. Therefore e 2x sin x dx = e 2x cos x 5 2e 2x sin x 5 b = cos x 2 sin x 5e2x 5e 2x. e 2x sin x dx = lim e 2x sin x dx b = lim cos x 2 sin x b b 5e2x 5e 2x = lim cos b 2 sin b b 5e2b 5e 2b + 5 = 5 since (cos b)/e 2b /e 2b and (sin b)/e 2b /e 2b as b. dx 5. Compute x 2, where a. Your answer will depend on a. Show all work. ax First we decompose x 2 ax = into partial fractions: x(x a) Multiply both sides by x(x a) to get x(x a) = A x + B x a. = A(x a) + Bx. Set x = a and x = to get = ab and = aa, so A = /a and B = /a. Thus Integrating both sides, dx dx x(x a) = a x + a x(x a) = ax + a(x a). dx x a = a ln x + ln x a + C. a Page of 2

12 Error Bound Formulas Trapezoid Rule and Error Bound: Let a = x < x < x 2 < < x n < x n = b with x i x i = x = b a b n for all i. The nth approximation T n to f(x) dx using the Trapezoid a rule is T n = (f(x ) + 2f(x ) + 2f(x 2 ) + + 2f(x n 2 ) + 2f(x n ) + f(x n )) x 2 and the error in approximating the integral by T n can be bounded as: b K(b a) f(x) dx T n ( x) 2 K(b a)3 = 2 2n 2, a where K is any upper bound on f (x) over [a, b]: f (x) K for a x b. Taylor s Inequality: Let T n (x) be the nth-order Taylor polynomial for f(x) at x = a. If f (n+) (x) M for x a d then x a n+ T n (x) f(x) M (n + )! for x a d. Page 2 of 2