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Chapter 3 58

# Chapter 3 58 - 208 CHAPTER 3 DIFFERENTIATION RULES 4...

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Unformatted text preview: 208 CHAPTER 3 DIFFERENTIATION RULES 4. Substituting the values of h and 3 in Problem 3 into P(:c) = —:—I:w3 + \$3332 gives us P(:c) 2 aa:3 + b352, where a m 4.937 x 10-5 and b m 4.78 X 10-3. 7 0 64.52 APPLIED PROJECT Building a Better Roller Coaster 1. (a) f(\$)=a\$2+bm+c 2 f'(\$):2ax+b. The origin is at P : f(0) : 0 => c = 0 The slope of the ascent is 0.8: f'(0) = 0.8 => b = 0.8 The slope of the drop is ~16: f'(100) = ‘1.6 => 200a + b 2 —1.6 (b) b = 0.8, so 200a —l— b : —1.6 => 200a -I— 0.8 = —1.6 :> 200a = —2.4 => a 2 —% = #0012. Thus, f(m) = A0.012x2 + 0.8m. (c) Since L1 passes through the origin with slope 0.8, it has equation y = 0.8m. The horizontal distance between P and Q is 100, so the y-coordinate at Q is f(100) = —0.012(1OO)2 + 0.8(100) = —40. Since L2 passes through the point (100, -40) and has slope —1.6. it has equation y + 40 = ~1.6(\$ # 100) or y : —1.6x + 120. (d) The difference in elevation between P (0, 0) and Q(100, —40) is 0 — (—40) : 40 feet. 2. (a) Interval Function First Derivative Second Derivative —l (—007 0) L1 (as) : 0.8m L'1(:c) : 0.8 L'1’(:c) : O [0.10) g(a:) : [€333 + lm2 + mm + n g’(w) : 1&ka + 2lw + m g"(a:) = 6km -— 2l [10, 90] q(a:) = (1:32 + bx + c q’(9:) = 2am + b q"(m) : 2a (90,100] h(:1:) : p363 + qm2 + rm + s h’(\$) = 3pm2 + 2qan + r h"(:c) : 6px —— 2q (100, 00) L2(a:) : —1t6m + 120 L’2(m) : -1.6 L’2’ (51:) = 0 There are 4 values of :L' (0, 10, 90, and 100) for which we must make sure the function values are equal, the ﬁrst ...
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