Math 181 - Practice Exam 2 - solutions Problem 1 A population of dinosaurs is modeled by P (t) = 0.3t 2 + 0.1t + 10 for times t in the interval [ 5, 0]. a) Find the rate of change of this population at t = 4. b) Find the equation of the tangent line of P (t) at t = 4 and sketch it, together with the graph of P, over the interval [ 4, 0]. Your graph should include the values of P at 4, 2, 0. Solution. a) The rate of change at time t is (use Rules, not limits, if limits are not explicitly required) P (t) = 0.6t + 0.1, so the rate of change at t = 4 is P ( 4) = 2.5. b) The tangent line there has slope 2.5, and the population equals so the tangent line has equation P ( 4) = 4.8 y = 2.5(t + 4) + 4.8. Further values are P ( 2) = 8.6 and P (0) = 10. Graph of the function from Problem 1
Graph of the function from Problem 2, with local min/max and tangent line Problem 2 Consider the function f(x) = x 3 9x 2 42x + 5. a) Find the derivative of f using Rules. Use only one rule per line, and mention which rule you are using. b) Find all local maxima and minima of f. c) Tell where the function f is increasing and where it is decreasing. d) Find the equation of the tangent line to the graph of f at t = 3. Solution. For a), f (x) = (x 3 9x 2 42x + 5) = (x 3 ) (9x 2 ) (42x) + (5) (Sum/Difference Rule) = (x 3 ) 9(x 2 ) 42x + 5 1 (Constant Factor Rule) = 3x 2 18x 42 + 0 (Power Rule) = 3(x 2 6x 14) (Algebra - factor out 3). It is possible to do this in slightly different ways, eg use the Constant Rule to get (5) = 0 straight away but since that Rule is different from the Constant Multiple Rule, it would have taken me an extra line. On the other hand, I can always do the t-rule and the fact 1 = 0 in the same line as the
derivatives of t 3 and t 2, because they are special cases of the Power Rule. The last step was not needed for a) at all, it s just useful for part b). b) There are two points where f (x) = 0. Using the the quadratic formula, they are x 1/2 = 6 ± 6 2 + 56 = 3 ± 23. 2 To find out which of these is a local maximum or local minimum, we will do part c) first. c) We draw up the sign pattern of the derivative f (x), with the consequences for the increasing/decreasing of f(x). x 3 23 3 + 23 sign of f (x) + + f(x) is... increasing decreasing increasing Back to b): According to the sign pattern, 3 23 is a relative maximum and 3 + 23 is a relative minimum. d) The slope is f (3) = 69 (using part a)). And f(3) = 175, therefore the tangent line has equation y = 69(x 3) 175. Problem 3 Find the equation for the tangent line to y = 3x 2 4x at x = 2. Use a limit computation, no rules for differentiation. Solution. The derivative of f(x) at an arbitrary point x = a is f (a) = 3(a + h) 2 4(a + h) (3a 2 4a) lim h 0 h = 3(a 2 + 2ah + h 2 ) 4a 4h 3a 2 + 4a lim h 0 h = 6ah + 3h 2 4h lim h 0 h = lim 6a + 3h 4 = 6a 4. h 0 We substitute a = 2 to get f ( 2) = 16 (note also how the answer 6a 4 agrees with the answer that we could have found using the Rules!). Since
to x = 2 belongs the value y = 12 + 8 = 20, the tangent line there has equation y = 16(x + 2) + 20. Problem 4 A farmer wants to fence in a rectangular field using strong fence on two opposite sides, costing $ 5 per foot, and cheap fence on the two other sides, costing $ 4 per foot. a) What is the minimum cost to fence in 4000 square feet in this manner? b) What if the farmer can use an existing 60-foot wall instead of strong fence? Solution. a) Write a for the length of one side with strong fence, b for the length of one side with cheap fence. So the area is A = ab = 4000, leading to b = 4000 a. Then consider the cost C of a field with these dimensions. We need 2a feet of strong fence and 2b of the cheap, costing 10a and 8b (dollars), respectively. The total cost is C = 10a + 8b = 10a + a = 10a + a 1. so we have written C = C(a) as a function of a! Now we want to minimize C(a). Differentiate C(a) using the Power Rule with exponent 1! Then we need to solve C (a) = 0. dc da = 10 + 32, 000( 1)a 2 = 10. a 2 0 = 10 a 2 (clear denominator a 2 ) 0 = 10a 2 (add and divide by 10) a 2 = = 3, 200 10 a = 3200 56.57 (in feet). The negative value for a is discarded, and we get a minimum cost of C( 3200) = 10 3200 + 32000/ 3200 1131.4
(in dollars). Yes, this really is the minimum and not a maximum, because the sign pattern of the derivative is +, so the function C(a) is first decreasing, then increasing. b) For the cost, we need to look at two different scenarios. If we use all of the existing wall (meaning a 60), the setup changes to C = 5a + 5(a 60) + 8b = 10a 300 + which we differentiate to get the same derivative C (a) = 10 a 2 a as before! In particular, C (a) > 0 for a > 3200, so C(a) is increasing for a > 3200. This means for a 60, the lowest-cost fence is actually built with a = 60. The second scenario is that we don t use all of the existing wall (so a < 60). As the cost, we get C = 5a + 8b = 5a + a with C (a) = 5 /a 2 equal to zero for a = 6400 = 80 only. So inside the interval (0, 60), always C (a) < 0. This means the lowest-cost fence is built using a = 60 and that is the final answer.
Problem 5 Find the following derivatives, using Quick Rules. No need to simplify. y = x 4 πx 2 + 5x 3 y = (x 3 5x + 1) 8 y = sin 4 x y = 4 x2 4 Solution. d dx (x4 πx 2 + 5x 3) = 4x 3 2πx + 5. d ( (x 3 5x + 1) 8) dx = 8(x 3 5x + 1) 7 (3x 2 5) (PCR with u(x) = x 3 5x + 1). d dx (sin4 x) = 4 sin 3 x cos x (PCR with u(x) = sin x) ( ) d 4 = d ( 4(x 2 4) 1/2) dx x2 4 dx ( = 4 1 ) (x 2 4) 3/2 (2x) (PCR with u(x) = x 2 4) 2 4x = (x 2 4). 3/2 Problem 6 A cylindrical barrel is to be built from sheet metal. Assume that the surface area of the barrel is A = 2πr 2 + 2πhr where h is its height and r the radius of the base. The volume is V = πr 2 h. If you want to build a barrel with a volume of 432π cubic feet, what are the dimensions that use the least amount of material? Solution. From the volume equation, we get h = V πr 2 = 432 r 2. Using this to express A as a function of r alone, A = 2πr 2 + 2 432π r = 2πr 2 + 864πr 1.
The amount of material we need is proportional to A, so we are seeking to minimize A. We differentiate A(r) using the PCR. A (r) = 4πr 864π r 2. To solve A (r) = 0, we multiply both sides by r 2 and add 864π. Then we divide both sides by 4π to get r 3 = 864 4 = 216. Taking cube roots on both sides gives r = 6.The sign pattern of V (r) is + which means r = 6 really gives the minimum (it follows that h = 12, from the volume equation). Problem 7 A rocket-powered racecar is driving with constant acceleration a = 50 meters per second squared. If the car starts with velocity 0 at time t = 0, how long does it take to achieve a speed of 70 miles per hour? How far does the car travel during that time? Please include units in your answers (use 1h = 3600s and 1mi = 1600 meters). Solution. First, write s(t) for the position of the racecar at time t. Whenever the acceleration is constant, the position is a quadratic polynomial So the velocity is s(t) = At 2 + Bt + C. (1) s (t) = 2At + B. (2) Substituting the given s (0) = 0, we get from this equation B = 0. Next, differentiating again in Equation (2), the acceleration is s (t) = 2A (3) which we know equals 50. Therefore, A = 50/2 = 25. Finally, we choose the origin of the coordinate system so that at time t = 0, s = 0 which gives us C = 0 in Equation (1). Now that we know all constants A, B, C. To answer the questions, we have to translate 70 mph into metric units: 70mi h = 70 1600m 3600s 31.1 m s.
So we need to find the time when 31.1 = s (t) = 50t which gives t 0.62 (in seconds). During that time, the racecar has traveled from 0 to s(0.62) = 9.67 (in meters).