Math 16A, Summer 2009 Exam #2 Name: Solutions. Problem Total Score / 120. (x 2 2x + 1) + (e x + x)(2x 2)

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1 Math 16A, Summer 2009 Exam #2 Name: Solutions Each Problem is worth 10 points. You must show work to get credit. Problem Total Score / 120 Problem 1. Compute the derivatives of the following functions: f(x) = (e x + x)(x 2 2x + 1) f(x) = 3x2 + 2x 1 x + 1 Solution 1. f (x) = ( e x x) (x 2 2x + 1) + (e x + x)(2x 2) f (x) = (6x + 2)( x + 1) 1 2 x+1 (3x2 + 2x 1) x + 1 Problem 2. Let f(x) = x + e x, g(x) = x 2. Compute d dx [f(g(x))]. Let y = 1+u 1 u and u = 2x2 x + 1, and compute dx. Solution 2. We have f (x) = 1+ex 2 x+e x, g (x) = 2x. Then d [f(g(x))] = f (g(x))g (x) dx 1 + e x2 = 2 ( 2x) x 2 x2 + e = x(1 + ) e x2 x2 + e x2

2 We have dx = du du dx (1 u) + (1 + u) = (4x 1) (1 u) 2 8x 2 = (1 (2x 2 x + 1)) 2 8x 2 = ( 2x 2 + x) 2 Problem 3. Find the equation of the tangent line to the curve at the point (8, 8). Solution 3. Use implicit differentiation: Plug in x = 8, y = 8: The equation for the tangent line is x 2/3 + y 2/3 = x 1/ y 1/3 dx = 0 dx = y1/3 x 1/3 dx = 2 (8, 8) 2 = 1 y ( 8) = 1(x 8) y = x 16. Problem 4. Suppose x and y are differentiable functions of t, and are related by the equation Find when dx =.3, x = 4 and y = 18. Solution 4. Use implicit differentiation: Now plug in x = 4, y = 18 and dx =.3: y 2 5x 3 = 4. 2y 15x2 dx = 0 = dx 15x2 2y = = 2

3 Problem 5. Compute the derivatives of the following functions: Solution 5. f (x) = f(x) = 1 3x(e 3x ) f(x) = ln 4 (1 + x)e 2x (3x 2 1) 1 2x x (e3x ) + 1 3x (6xe 3x2 1 ) First simplify Now compute f(x) = 1 ln (1 + x)e2x (3x 2 1) 4 ( 1 2x ln(1 + x) + ln e 2x + ln(3x 2 1) ln(1 2x) ) = 1 4 = 1 4 ( ln(1 + x) + 2x + ln(3x 2 1) ln(1 2x) ) f (x) = 1 4 ( x x 3x ) 1 2x Problem 6. Simplify the following: 3 ln2x ln x ln 4 e ln4 3 lnx+ln(x+1). Solution 6. 3 ln2x ln x ln 4 = ln 8x 3 ln x ln 4 = ln 8x3 4x = ln 2x 2 e ln 4 3ln x+ln(x+1) = e ln4 ln x3 +ln(x+1) 4(x+1) ln = e x 3 4(x + 1) = x 3

4 Problem 7. Use logarithmic differentiation to differentiate the following functions: (1 + x)3 (3x f(x) = 2 1) f(x) = x 1+ x Solution 7. First simplify ln f(x): Now differentiate: Then Simplify: Differentiate: Then ln f(x) = 1 2 ln (1 + x)3 (3x 2 1) = 1 ( ) 3 ln(1 + x) + ln(3x 2 1) ln(). 2 f (x) = d [ln f(x)] = 1 ( 3 dx x + 6x 3x ) (1 + x)3 (3x 2 1) 1 ( x + 6x 3x ) ln f(x) = (1 + x) ln x d 1 [ln f(x)] = dx 2 x ln x + (1 + x) 1 x ( f (x) = x 1+ x 1 2 x ln x + x ) x Problem 8. Solve the following equations for x: ln(x 1)(x + 1) ln(x + 1) 2 ln 4 = x 2 2 2x 8 2 x = 0. Solution 8.

5 ln(x 1) + ln(x + 1) 2 ln(x + 1) = ln4 ln(x 1) ln(x + 1) = ln4 ln x 1 x + 1 = ln4 x 1 x + 1 = 4 x 1 = 4x = 3x 5 3 = x Set Y = 2 x, then Y 3 2Y 2 8Y = 0 Y (Y 2 2Y 8) = 0 Y (Y + 2)(Y 4) = 0 So Y is 0, 2 or 4. Since Y = 2 x > 0, Y = 4. So and therefore x = 2. 2 x = 4 Problem 9. An observer on the ground is watching an airplane flying 200 feet per second at an altitude of 300 feet, as shown in the picture below. How fast is the distance from the observer changing at a time when the airplane is 500 feet from the observer? x Plane 300 y Observer Solution 9. From the Pythagorean theorem we get the relation x = y 2.

6 Use implicit differentiation: When y = 500, 2x dx = 2y = dx x y x = = ( ) = = 400 Plug in x = 40, y = 500, dx = 200: = = 160 So the distance y from the observer to the plane is changing at a speed of 160 feet per second when the plane is 500 feet away. Problem 10. A rectangular corral of 60 square meters is to be fenced off and divided into two sections, as shown below. Suppose that the cost of the fencing for the boundary costs $5 per meter, and the dividing fence costs $2 per meter. Find the dimensions of the corral that minimize the cost of the fencing. x w Solution 10. The objective equation is the cost: 5(2x + 2w) + 2w. The constraint equation is that the area is 60 square meters: xw = 60. Solve for x: Now minimize Find critical points: x = 60 w. ( f(w) = ) w + 2w + 2w = 600 w + 12w. f (w) = 600 w = 0 w 2 = = 50 w = 50

7 Check concavity: f (w) = 600 w > 0 3 for w > 0, so the graph of f(w) is always concave up and the minimimum occurs at the critical point w = 50. So the cost is minimized when w = 50, x = Problem 11. A bookstore expects to sell copies of a certain book during the coming year at a stea rate. Suppose that each new order costs $50 to process, and that the carrying costs, based on the average number of books in inventory, is $4 per book per year. How many times per year should orders be placed to minimize inventory costs? Solution 11. Let r be the number of orders placed and x the number of books in each order. The inventory cost is the ordering costs plus the carrying costs: 50r + 4 x 2 The constraint equation is xr = Solve for x: Minimize Find critical points: Check concavity: x = r f(r) = 50r + 2x = 50r r f (r) = r 2 = 0 r 2 = = r = 20 f (r) = r 3 > 0 for r > 0, so the graph of f(r) is always concave up and the minimum occurs at r = 20. So the inventory cost is minimized when 20 orders are placed each year. Problem 12. The owner of a cafe estimates that if there are 10 tables available, the daily profit will be $20 per table. Because of overcrowding, for each additional table the daily profit per table will be reduced by $.50. How many tables should be provided to maximize the daily profit from the cafe? Solution 12. Let x be the number of tables available, x 10. The daily profit per table is so the total daily profit is 20.5(x 10), P(x) = x(20.5(x 10)) =.5x x.

8 Find critical points: P (x) = x + 25 = 0 x = 25. Since the graph P(x) is a parabola opening downward, the maximum occurs that the critical point x = 25. So the profit is maximized when 25 tables are made available.