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MATHEMATICS 184 (Section 921) - TERM EXAM #2 SOLUTIONS NAME: STUDENT ID NUMBER: SIGNATURE: INSTRUCTIONS: Notes, books and calculators are not permitted. Answers may receive no credit if accompanying work is not provided. When indicated, simplify answers as much as possible, and be sure to clearly indicate the final answer for each question. Please use the back of the pages if you need extra space. 1. John Dough has now decided to invest his millions in the stock market. He purchases shares in a new paperclip company, ACME Paperclips Inc., and over the next 100 days John tracks the price of ACME Paperclips shares. 2 (a) If P(t) is the price, in dollars, of a single share t days after John has purchased his shares, what are the units of P0 (t)? Units of P0 (t) = 2 dP units o f P(t) dollars are So the units of P0 (t) are or dollars per day. dt units o f t day (b) What does P0 (t) represent? P0 (t) represents the rate of increase of the share price at time t or How quickly the share price (and John’s investment) are growing t days after his purchase. (c) What is happening to the share price and consequently to John’s investment when: 1 i. P0 (t) > 0? It is growing with time. He’s making money (ie. the share price is increasing with time). 1 ii. P0 (t) < 0? It is decreasing in value (ie. the share price is decreasing). 1 iii. P0 (t) = 0? It’s value is constant with time (ie. the share price is not changing). Page 2 (d) At the end of the 100 days, John graphs the share price over time and gets the following graph. Using this graph, fill in the following table with the signs (+, − or 0) of P(t), P0 (t), and P00 (t) at the indicated points on the graph. 1.8 P(t) - Share Price (dollars per share) 6 C 1.6 1.4 1.2 A 1 B 0.8 0.6 D 0.4 0.2 0 0 20 40 60 t (days since purchase) 80 100 Point Sign of P(t) Sign of P0 (t) Sign of P00 (t) A + + − B + 0 + C + 0 − D + − + Page 3 7 2. (a) Using the limit definition of the derivative (no shortcuts) differentiate the function (Simplify your answer) 4x − 1 f (x) = 3x + 7 4x − 1 3x + 7 f (x + h) − f (x) f 0 (x) = lim h→0 h f (x) = = lim 4(x+h)−1 3(x+h)+7 − (4x−1) (3x+7) h 1 (4x + 4h − 1) (3x + 7) (4x − 1) (3x + 3h + 7) lim · · − · h→0 h (3(x + h) + 7) (3x + 7) (3x + 7) (3(x + h) + 7) 12x2 + 12xh − 3x + 28x + 28h − 7 − (12x2 + 12xh + 28x − 3x − 3h − 7) lim h→0 h(3x + 7)(3(x + h) + 7) 31h lim h→0 h(3x + 7)(3(x + h) + 7) 31 lim h→0 (3x + 7)(3(x + h) + 7) 31 (3x + 7)2 h→0 = = = = f 0 (x) = Page 4 4 (b) Using the quotient rule, differentiate the same function (Simplify your answer) 4x − 1 f (x) = . 3x + 7 u du dx v dv dx = 4x − 1 = 4 = 3x + 7 = 3 so f 0 (x) = = = = du dv ·v−u· dx dx v2 4 · (3x + 7) − (4x − 1) · (3) (3x + 7)2 12x + 28 − (12x − 3) (3x + 7)2 31 (3x + 7)2 Page 5 10 3. Find all the points on the curve y = ex (x2 − 3) at which the tangent line is horizontal. This means we must find the points on this curve at which the derivative is equal to zero: By the product rule with f (x) = ex f 0 (x) = ex g(x) = (x2 − 3) g0 (x) = 2x y0 = f 0 (x)g(x) + f (x)g0 (x) y0 = ex (x2 − 3) + ex (2x) So we set this equal to zero and solve for x. 0 0 0 0 = = = = ex (x2 − 3) + ex (2x) ex (x2 + 2x − 3) x2 + 2x − 3 (x + 3)(x − 1) (Note that ex is never zero) So the tangent line is horizontal when x = 1 and x = −3 x = 1 ⇒ y = e1 (12 − 3) = −2e x = −3 ⇒ y = e−3 ((−3)2 − 3) = 6e−3 So (1, −2e) and (−3, 6e−3 ) are the coordinates at which the tangent line is horizontal. Page 6 4. Differentiate the following functions: 2 (a) f (x) = 7x7 + 5x3 − 2x + 8 f 0 (x) = 7(7x6 ) + 5(3x2 ) − 2(1) + 0 f 0 (x) = 49x6 + 15x2 − 2 2 2 (b) y = 3ex+7 + xe − π y0 = 3(e7 )(ex ) + e2 xe y0 = 3ex+7 + e2 x 3 2 −1 −0 e2 −1 (c) y = (x7 − x6 + x3 − 2x + 3)(2x4 + 6x3 − 3x2 + 1) This is product rule: y0 = f 0 (x)g(x) + f (x)g0 (x) Here: f (x) = x7 − x6 + x3 − 2x + 3 ⇒ f 0 (x) = 7x6 − 6x5 + 3x2 − 2 g(x) = 2x4 + 6x3 − 3x2 + 1 ⇒ g0 (x) = 8x3 + 18x2 − 6x So, y0 = f 0 (x)g(x) + f (x)g0 (x) y0 = (7x6 − 6x5 + 3x2 − 2)(2x4 + 6x3 − 3x2 + 1) + (x7 − x6 + x3 − 2x + 3)(8x3 + 18x2 − 6x) Page 7 4 (d) g(t) = t−5 et (t 2 +2) This is quotient rule: ( uv )0 = Here: du dv dt v−u dt 2 v du =1 dt dv v = et (t 2 + 2) ⇒ = et (t 2 + 2) + et (2t) dt u = t −5 ⇒ (product rule) So, 1 · et (t 2 + 2) + (t − 5)[et (t 2 + 2) + et (2t)] (et (t 2 + 2))2 t 3 − 2t 2 − 8t − 8 g0 (t) = et (t 2 + 2)2 g0 (t) = Page 8 (if simplified)