Final Exam Solutions

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1 Final Exam Solutions Laurence Field Math, Section March, Name: Solutions Instructions: This exam has 8 questions for a total of points. The value of each part of each question is stated. The time allowed is two hours. Partial credit may be given for incorrect answers if your working is legible. No books, notes or calculators are allowed. Since there are no calculators, give your answers in exact form. No decimal approximation is needed. Important: For question 8, answer EITHER part 8(a) OR part 8(b). [Your score for question 8 will be the greater of your 8(a) score and your 8(b) score.] Answer all parts of questions 7. Question Points Score Total: i

2 . Compute the following integrals. x (a) x x + dx Let u = x x + = x / +, so that du = x/ dx, so x dx = du. Then x x x + dx = du u = log u + C = log(x x + ) + C. (b) tan(x) dx Let u = cos(x), so that du = sin(x) dx. Then tan(x) dx = du u = log cos(x) + C. π/ (c) sin x + cos x dx π/ Note that sin x and cos x are odd and even functions of x respectively. Therefore π/ π/ sin x + cos x dx = π/ cos x dx = [sin x] π/ =.. Consider the function f(x) = e /x. (a) Is f even, odd or neither? Since f is neither even nor odd. f( x) = e /x = /f(x) ±f(x), (b) Find all local extrema, points of inflection, and horizontal and vertical asymptotes of the graph of f. 8 As x ±, /x, so f(x), so there is a horizontal asymptote at y =. The only gap in the domain is at. As x +, /x, so f(x), so x = is a vertical asymptote. As x, /x, so f(x). For local extrema, we calculate f (x) = x e/x <, so there are no local extrema as the function decreases on both branches (, ) and (, ). Now ( f (x) = x + ) x e /x = + x x e /x,

3 so f (x) vanishes at x = /, is negative for x < / and positive otherwise. There is a point of inflection at x = /. (c) Sketch the graph of f, showing all the features you found in (b) (d) Write a formula for f (y). If y = f(x) = e /x, then log y = /x, so x = /log y. Thus f (y) = log y.. (a) Give the definition of a x, where a > and x is real but not necessarily rational. You should use the exponential and/or natural logarithm functions in your answer. a x = exp(x log a) = e x log a. (b) Let f be a continuous function on [a, b]. Let P be the partition a = x < x < x < < x n < x n = b. Let x i be the length of the ith subinterval, that is, x i = x i x i. Give the definition of U f (P ), the upper sum for f on the partition P. U f (P ) = M x + M x + + M n x n,

4 where M i = max{f(x) : x i x x i } is the maximum value of f on the ith subinterval of P. (c) Answer true or false to each statement. No justification is required. (i) Let f be continuous on [a, b]. Then for all partitions P of [a, b] we have L f (P ) b a f(x) dx U f (P ). True (by definition of the integral). (ii) If a function f is concave down, then it is decreasing. False, for example: f(x) = x. (iii) Let f be continuous on (a, b) and let c (a, b). If f (x) + as x c, then the graph of f has a vertical tangent at c. True (by definition of a vertical tangent). (iv) If f is differentiable and one-to-one on (a, b), then the inverse function f is differentiable everywhere in its domain. True (by a theorem from class).. Consider the region R bounded by the curves y = x and y = x. (a) Find the area A of the region R. The curves intersect where x = x, i.e. x =,. The area is A = (x x ) dx = [ x ] x =. (b) Find the coordinates (x, y) of the centroid of R. 6 We have and Thus x = / and y = /. xa = ya = x (x x ) dx = (x x 6 ) dx =. (c) Using Pappus theorem or otherwise, find the volume obtained by revolving R around the line x =.

5 Since the region is entirely to the right of the line, and the distance from the centroid to the line is / ( ) = /, the volume is V = π = π.. In this question, consider a hemispherical bowl of radius inches. (a) Find a formula for V (h), the volume (in cubic inches) of the water in the bowl when the bowl is 8 filled to a depth of h inches. The bowl is obtained by revolving the semicircle x + (y ) = 9, y around the y-axis. Using the disk method, V (h) = h π(9 (y ) ) dy = π [9y y ] h = π [9h h ]. (b) The bowl has a small hole in the bottom and is leaking water at a rate of cubic inch per minute. 7 How quickly is the depth of the water in the bowl decreasing when the depth is inch? By the chain rule, dv dt = dv dh dh dt = π(6h h ) dh dt. At the relevant point, = dv/dt = π(6 ) dh/dt and so dh/dt = /π, so the depth is decreasing at /π inches per minute. 6. (a) A sum of $ is kept in a savings account which earns continuously compounded interest at the rate of 6% per year. (i) What is the balance of the account after year? In the notation from class, A(t) = A() e rt, so the balance in dollars is A() = e.6. (ii) How many years from the initial deposit will it take for the balance to reach $? We want to find t satisfying = A(t) = e.6t, so t = (log )/.6 years is how long it will take. (b) Consider the function f(x) = + t dt.

6 (i) Show that f is one-to-one. What is the range of f? By the first fundamental theorem of calculus, f (x) = + x >, so f is increasing, hence one-to-one. The range of f is the domain of f, namely R. (ii) Calculate (f ) (). Since f() =, f () =, so (f ) () = f () = + =. 7. (a) Consider the function f whose graph is depicted below. 8 y y = f(x) 6 x Sketch the graph of the function g(x) = f(t) dt, showing all local extrema and points of inflection. y local maximum y = g(x) point of inflection 6 local minimum x (b) The position of a particle P moving on a circle is described by an angle θ(t) which depends on time t. Suppose that θ (t) = sin(t/a), where a > is a constant. Suppose also that θ () = a and θ() =. (i) Find a formula for the angle θ(t) at time t.

7 P θ(t) Integrating, θ (t) = a cos(t/a) + C for some constant C. Since θ () = a, C =. Integrating again, θ(t) = a sin(t/a) + D for some constant D. Since θ() =, D =. Therefore θ(t) = a sin(t/a). (ii) For which values of a does the particle P eventually visit every point on the circle? For this to happen, we need the range of θ to cover every angle on the circle, that is, a full range of π radians. Since the range of θ is [ a, a ], this happens iff a π, i.e., a π. 8. Important: For question 8, answer EITHER part 8(a) OR part 8(b). [Your score for question 8 will be the greater of your 8(a) score and your 8(b) score.] EITHER: (a) Suppose f is a function with domain ( π, π ). Suppose also that, for all x, f (x) + f(x) =. (i) Show that f (x) cos x + f(x) sin x ] =. dx f (x) cos x + f(x) sin x ] = f (x) cos x f (x) sin x + f (x) sin x + f(x) cos(x) dx = [f (x) + f(x)] cos x =. (ii) Hence prove that ] f(x) sec x = C sec x ( ) dx for some constant C. Integrating the result in (i), f (x) cos x + f(x) sin x = C 6

8 for some constant C. Multiplying by sec x, f (x) sec x + f(x) sec x tan x = C sec x. Now observe that the left-hand side is d dx[ f(x) sec x ]. (iii) By taking indefinite integrals of both sides of ( ), prove that 6 for some constant D. Integrating ( ), f(x) = C sin x + D cos x f(x) sec x = C tan x + D for some constant D. Multiplying both sides by cos x, we get the result. OR: (b) Let f be a nonnegative continuous function and a, b nonnegative constants. Suppose that, for all x, f(x) a + b f(t) dt. ( ) (i) Let g(x) = f(t) dt. Show that g (x) a + b g(x) for all x. By the first fundamental theorem of calculus, for all x where the inequality follows from ( ). g (x) = f(x) a + b g(x) (ii) Show that g(t) e bt ] = ( g (t) b g(t) ) e bt. dt g(t) e bt ] = g (t) e bt + g(t) ( be bt ) = ( g (t) b g(t) ) e bt. dt (iii) Using the results of (i) and (ii), show that for all t. g(t) e bt ] a e bt. ( ) dt By (ii) and then (i), g(t) e bt ] = ( g (t) b g(t) ) e bt (a + b g(t) b g(t)) e bt = a e bt. dt 7

9 (iv) By integrating both sides of ( ) from to x, show that for all x, g(x) e bx a b [ e bx + ]. Since the integral is order-preserving, by integrating ( ) from to x we get g(x) e bx g() a e bt dt = a b [ e bx + ]. Now observe that g() = by definition of g, and we get the result. (v) Deduce from (iv) and ( ) that, for all x, f(x) a e bx. For any x, by ( ), f(x) a + bg(x) a + be bx a b [ e bx + ], where the second inequality follows from (iv). The right-hand side simplifies as a a+a e bx = a e bx, whence the result follows. 8