SOLUTIONS FOR PRACTICE FINAL EXAM ANDREW J. BLUMBERG. Solutions () Short answer questions: (a) State the mean value theorem. Proof. The mean value theorem says that if f is continuous on (a, b) and differentiable on [a, b], then there exists a point c [a, b] such that f (c) = f(b) f(a) b a. (b) Define continuity; give an example of a function which is not continuous at 3. Proof. A function f is continuous at a point p if x p f(x) = f(p). The function x 3 is not continuous at 3. (c) Explain why a function which is differentiable is continuous. Proof. Roughly speaking, this is because you can locally approximate the function by a line and calculate the it using the line. (d) Explain the first derivative test. Proof. The first derivative test essentially looks at whether f changes sign around the critical point p. The idea is that at a local minimum, f should be decreasing (or not increasing) to the left of the point and f should be increasing (or not decreasing) to the right of the point. () A bowl is formed by rotating the curve y = x 3 around the y-axis, with upper rim a circle of radius 8. What is the volume of the bowl? Proof. We can do this a variety of ways. The upper rim has a radius when x = 8 or y = 8 3 = 5. Doing this using disks, we solve to find x = y 3 and then integrate 5 π(y 3 ) dy = π 5 y 3 dy = 3π 5 y 5 3 5 = 3π 5 (5) 5 3 = 3π 5 85. Alternatively, we can do this using cylindrical shells and then we find 8 π(x)(5 x 3 )dx = π 8 5x x 4 dx = π(56x x5 5 ) 8 = π(56(8 ) 85 5 ) = π(85 ()(85 ) ) = 3π 5 5 85. Date: December 8,.
ANDREW J. BLUMBERG (3) Find the equation of the line tangent to the graph of the curve specified by x y 4 + e x sin(x)y = at (, ). Proof. Using implicit differentiation, we find that x (4y 3 y ) + xy 4 + e x (sin(x)y) + e x (sin(x)y + cos(x)y) =. Solving for y, we have or y (4x y 3 + e x sin(x)) = xy 4 e x sin(x)y e x cos(x)y, y = xy4 e x sin(x)y e x cos(x)y 4x y 3 + e x. sin(x) At (, ), we find that the derivative has the form, which means that we can t write an expression for the tangent line in this way. (And would have to use other techniques to analyze the tangent behavior at (, ).) (4) (a) Use linear approximation to compute (6.) 4. Proof. We consider the linear approximation to the function x 4 at the point 6. Since d dx x 4 = 4 x 3 4, the linear approximation at 6 can be computed as l(x) = + (x 6). (4)(8) Plugging in 6., we find that (6.) 4 + 3. (b) Use the linear approximation to + x to compute.. Proof. We consider the linear approximation to + x at. Since d + x = dx ( + x), the linear approximation at can be computed as l(x) = + (x). Plugging in., we find that. +.. (5) Sketch the curve x 3 + 3x x + by finding the local maxima, minima, regions of convexity, etc.
SOLUTIONS FOR PRACTICE FINAL EXAM 3 Proof. We begin by finding the derivatives. Observe that f (x) = 6x + 6x = 6(x + x ) = 6(x + )(x ). and so there are critical points at x = and x =. We also have f (x) = 6(x + ). Plugging in x =, we find that f () = 8 >, and so the function is concave up at x = and thus has a local minimum at x =. Plugging in x =, we find that f ( ) = 8 and so has a local maximum at x =. Clearly f is concave up when x > and concave down when f < ; there is an inflection point at f =. When x, we see that f(x) also, and when x, f(x) (since the highest term is cubic). (6) A foot tree is falling, with the sun directly overhead. When the tree makes an angle of 3 with the horizontal, its shadow is lengthening at 5 feet per second. How fast is the angle changing? Proof. Since the sun is directly overhead (and far away), the relevant picture is a triangle with hypotenuse length (the tree), base side (the shadow) length s, and opposite side length w. We know that cos(θ) =, and so differentiating we find that s sin(θ) dθ dt = ds dt. We re interesting in the point where ds = 5, and θ = 3. Plugging in, we find or dt dθ dt = 4, dθ dt =. (7) Do the following optimization problems: (a) You have two boards of length x, and you want to enclose the maximum area by forming a triangle from the two boards and a long brick wall. What s the length of the wall side when you enclose the maximum area? Proof. The relevant drawing involves an isoceles triangle with two sides of length x and a base of length b. Dropping the height from the tip formed by the two x sides, we have A = bh. Denote the angle made by the wall and each of the sides x by θ. We then have b cos(θ) = x, or b = x cos(θ).
4 ANDREW J. BLUMBERG Similarly, we have or Therefore, sin(θ) = h x, h = x sin(θ). A = x sin(θ) cos(θ). Differentiating (with respect to θ), we find da dθ = x ( sin θ + cos θ) = x (cos θ sin θ). This has a critical point when θ = π 4, at which angle the wall length is x. (b) A line through (, ) cuts off a triangle in the first quadrant (x, y ). For what line will the triangle have minimum area? Proof. Omitted. (8) Find the following derivatives (using any method you like): (a) d x dx e t dt Proof. Using the fundamental theorem of calculus and the chain rule, we find that d x e t dt = xe x4. dx Differentiating again, we get (e x4 ) + (x)(e x4 )( 4x 3 ). (b) (c) d dx ln(sin x + x 3 ) Proof. Using the chain rule, we find that the derivative is ( ) sin ( sin x cos x + 3x ). x + x 3 d ln x dx x Proof. Using the quotient rule, we find that the derivative is (x )(x ) x ln x x 4.
SOLUTIONS FOR PRACTICE FINAL EXAM 5 (d) d (x 3 + x ) dx (x + ) 3 Proof. Omitted (but use logarithmic differentiation). (9) Use the intermediate value theorem to find a root of the polynomial x 3 x + x. Use a derivative argument to count the total number of roots for this polynomial. Proof. At x =, the polynomial has value. At x =, the polynomial has value 8 4 + = 5. Since polynomials are continuous, the intermediate value theorem tells us that there is a root in the interval [, ]. Taking the derivatives, we find p (x) = 3x x +. This is a quadratic and so can have at most zeros, which implies by the mean value theorem that p can have at most three roots. () Evaluate the following integrals: (a) sin x cos 3 xdx Proof. Since the exponent on cos x is odd, we rewrite this expression as sin x(cos x) cos xdx = sin x( sin x) cos xdx. (b) Next, substituting in u = sin x, we have du = cos(x)dx, and so the integral becomes u ( u )du = u u 4 du = u3 3 u5 5 = sin3 x sin5 x. 3 5 5 x ln xdx Proof. Integrating by parts, we set u = ln x and dv = x dx. Then we have du = x3 xdx and v = 3. Therefore the indefinite integral in question becomes x 3 ln x x 3 3 3x dx = x3 ln x x3 3 9. Evaluating at the its in question, we have ( 53 ln 5 3 53 53 ) ( 9 9 ) = 53 ln 5 3.
6 ANDREW J. BLUMBERG (c) dx x + 3x + Proof. Factoring the denominator, we have x +3x+ = (x+)(x+). We now use partial fractions, obtaining (x + )(x + ) = A x + + B x +. Multiplying through by the denominator, we have = A(x + ) + B(x + ). Plugging in x =, we find that B =. Plugging in x =, we find that A =. Therefore, the integral in question becomes dx x + + dx x + = ( ln(x + ) + ln(x + )) = ( ln(3) + ln()) ( ln() + ) = ln(3) + ln(). (d) dx (x + ) Proof. We use trigonometric substitution, with u = tan(θ). Then du = sec (θ)dθ and (using the fact that tan (θ) + = sec (θ)), we have b b sec θ dθ = cos θdθ. a Here a = tan () = and b = tan () = π 4. double-angle formulas, we have that cos (θ) = Thus, the integral becomes π 4 cos θ + dθ = π 8 + a + cos θ. θ (sin π 4 ) = π 8 + 4. Finally, using the (e) 4x (x )(x ) dx Proof. We use partial fractions, and write: 4x (x )(x ) = 4x (x ) (x + ) = A (x + ) + B (x ) + C (x ).
SOLUTIONS FOR PRACTICE FINAL EXAM 7 Multiplying by the denominator, we get the equation 4x = A(x ) + B(x + )(x ) + C(x + ) = A(x x + ) + B(x ) + C(x + ) = (A + B)x + ( A + C)x + (A B + C). Plugging in x =, we find that C = 4 or C =. Plugging in x =, we find that 4 = 4A or A =. Then since (A B + C) =, we deduce that ( B + ) = or B =. Therefore, the integral in question becomes dx x + + dx x + dx = ln(x + ) + ln(x ) + (x ) 3(x ). () Find the following its: (a) (b) (c) (d) x x e x Proof. Using L Hopital s rule, we find x x e x = x x e x = x e x =. x sin x cos x. Proof. Using L Hopital s rule, we find x sin x cos x = sin x cos x = cos x =. x sin x x x ln(5 + x) ln(5). x Proof. Recognizing that this is the it which defines the derivative of ln(x) at x = 5, we see that the it is 5. x 4 (x 4) x + 4x 3. Proof. Factoring the denominator, we find x 4 (x 4) x + 4x 3 = x 4 (x 4) (x + 8)(x 4) = x 4 x + 8 =. (e) x sin x x
8 ANDREW J. BLUMBERG Proof. Using L Hopital s rule, we find sin x x x = cos x =. x E-mail address: blumberg@math.utexas.edu