C Roettger, Fall 17 Math 165 Final Exam worksheet solutions Problem 1 Use the Fundamental Theorem of Calculus to compute f(4), where x f(t) dt = x cos(πx). Solution. From the FTC, the derivative of the left-hand side equals f(x). So f(x) = cos(πx) πx sin(πx). Substitute x = 4 to get f(4) = 1. Problem 2 Find the area of the region enclosed by the curves y = x 4 and 8x y =. Solution. First, solve the second equation for y: y = 8x. Then find where the curves intersect: x 4 = 8x gives x = or x = 2. A rough sketch tells you that the straight line is on top and y = x 4 is below that in the interval [, 2]. So the area is A = 2 8x x 4 dx = ] 2 [4x 2 x5 = 48 5 5. Problem 3 You observe an elevator pour corn into a conical pile for storage. The diameter of the base is always 3 times the height of the pile. If the volume of the pile is increasing at a rate of 6 cubic feet per minute, how fast is the height changing when the pile is 4 feet high? 1
Note: V = 1 3 πr2 h. Solution. Make a sketch! We write r or the radius (not diameter!) of the base and h for the height of the pile. These are both functions of time t, and it is given that for all t ( always ) so the volume equals at all times Now differentiate both sides: 2r(t) = 3h(t) V (t) = 3 4 πh(t)3. V (t) = 9πh(t)2 h (t). 4 Solve for h (t) and plug in V (t) = 6 (is constant, that was given) and h(t) = 4. Or plug in first, then solve for h (t). At this particular time when h = 4, 6 = 36πh (t) which gives h (t) = 1/(6π). Units for this are feet per minute. Problem 4 Find the linear approximation to f(x) = 1 + sin 2x at x = π/2. Solution. First, f (x) = 2 cos 2x, so m = f (π/2) = 2. The equation of the tangent line reads generally y = m(x a) + f(a) and here, with m = 2 and a = π/2, ( y = 2 x π ) + 1. 2 This is the exact same thing as the linearization of f(x). 2
Problem 5 Find the absolute max and absolute min for f(x) = 12x 1/3 3x 4/3 for 1 x 2. (Hint: f(2) 7.559. No credit for solution by graph from a calculator.) Solution. We compute the derivative f (x) = 4x 2/3 4x 1/3 = 4x 2/3 (1 x). Solving f (x) = gives x = 1 only (and the derivative does not exist at x = ). So we need to evaluate f(x) at the endpoints x = 1, 2 and the critical points x =, 1. x 1 1 2 f(x) 15 9 7.559 and pick the largest and smallest values of f(x). The largest value is 9, so at x = 1 is the absolute maximum. The smallest value is 15, so at x = 1 is the absolute minimum. Problem 6 If 12 cm 2 of material is available to make a rectangular box with a square 3
base and an open top, find the largest possible volume of the box. To receive full credit, you must justify that the volume you obtain is indeed a maximum. Solution. Note the units square centimeters, units for area! this is one clue that the material is supposed to make the walls (sides and bottom) of the box. So if the bottom of the box has dimensions x by x by y, the area of the bottom and the four sides is A = x 2 + 4xy. The volume is V = x 2 y. We use A = 12 to eliminate y, y = 12 x2. 4x Then we substitute this into the equation for the volume, V = V (x) = 1 4 x(12 x2 ) = 3x 1 4 x3. So we have written the volume as function of only one variable. domain, we have x and y, therefore For the 12 x 2 so x 12. The domain is then [, 12]. Solve V (x) =, 3 3 4 x2 = which gives x = ± 4 = ±2. Discard the negative answer because it is not in the domain and check that 2 is in it (yes, it is). Then you can either make a table like in the previous problem, with the endpoints and the critical point: x 2 12 V 4 which demonstrates that at x = 2, the volume has indeed the maximum value. Alternatively, you could do a First Derivative Test: the sign pattern of V (x) 4
is + which also demonstrates that the maximum volume occurs at x = 2. You still have to compute V (2) = 4 here. Problem 7 Determine the value of a > so that the area under the curve y = x a x 2a for x 1 is maximal. Verify using first or second derivative test that this is a maximum. (You may assume that the function y = x a x 2a for x 1 and any a > ). Solution. The area under the curve is F (a) = = 1 [ x x a x 2a a+1 dx = a + 1 x2a+1 2a + 1 1 a + 1 1 2a + 1. ] 1 We need the derivative of F (a) (differentiate with respect to a, there is no x in F (a)!). F (a) = 1 (a + 1) 2 + 2 (2a + 1) 2 = (2a + 1)2 2(a + 1) 2 (a + 1) 2 (2a + 1) 2. Then we need to solve F (a) =. This can only happen if the numerator equals, which gives 2a 2 + 1 = after expanding and simplifying. So a = 1/ 2 is the only positive solution. The sign pattern of F (a) is + on the domain (, ), so at a = 1/ 2 the area is indeed maximal. The second derivative test: compute F (a) = 2 (a + 1) 8 3 (2a + 1). 3 Then verify F (a) < for a = 1/ 2. This would have only told you that a = 1/ 2 is a local maximum in my book, that is not good enough. One could put in some extra brainpower F (a) is a continuous function, and there is only one local minimum at x =. That additional information, together 5
with the fact that we have only one local maximum, would guarantee that it is an absolute maximum. But if any of these additional facts were not true, I could make up an example where this local maximum is not an absolute maximum (exercise for you: sketch examples like that!). In summary: just use the First Derivative Test. 6