final_2009_sol

# Here we use that d dx sin x cos x 1 sin x cos x sin x

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Here we use that d dx (sin x cos x + 1) = (sin x ) 0 cos x + sin x (cos x ) 0 = cos 2 x - sin 2 x . (b) [6 points] Find d dx sec(sin(2 x + 1)). Solution: Using the chain rule: d dx sec(sin(2 x + 1)) = sec(sin(2 x + 1)) tan(sin(2 x + 1)) d dx sin(2 x + 1) = sec(sin(2 x + 1)) tan(sin(2 x + 1)) cos(2 x + 1) d dx 2 x + 1 = sec(sin(2 x + 1)) tan(sin(2 x + 1)) cos(2 x + 1) · 2 . 5. (a) [6 points] Find the linearization of f ( x ) = x 1 / 2 at x = 9. Solution: f 0 ( x ) = 1 / 2 x - 1 / 2 , so f 0 (9) = 1 / 6. Thus L ( x ) = f (9)+ f 0 (9)( x - 9) = 3 + 1 / 6( x - 9). Page 2

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Math 171: Final Exam (b) [3 points] Use your answer to part (a) to approximate 10. Simplify your answer. Solution: We use the fact that 10 1 / 2 = f (10) L (10). Evaluating, L (10) = 3 + 1 / 6(10 - 9) = 3 + 1 / 6 = 19 / 6. 6. You are flying a kite. Assume that the string is a straight line. (a) [3 points] If the angle of the string to level ground is π/ 3 radians and the kite string has been let out to 300ft, how high is the kite? Solution: In both parts of the problem we have the following picture: θ h 300 Here θ is in radians and h is in feet. The relationship between θ and h is sin( θ ) = h/ 300 (because sin is opposite/hypotenuse) or h = 300 sin( θ ). When θ = π/ 3, we have h = 300 sin( π/ 3) = 300 3 2 = 150 3ft. (b) [8 points] If the length of the string and the angle of the string to level ground are as above and the angle is changing at - 0 . 1 radians a minute, what is the rate of change of the kite’s height? Solution: Differentiating both sides of h = 300 sin( θ ) with respect to t (in minutes) we have dh dt = 300 cos( θ ) dt . At the instant we are interested in, θ = π/ 3 and dt = - 0 . 1, so the height is changing at dh dt = 300 cos( π/ 3)( - 0 . 1) = 300 1 2 ( - 0 . 1) = - 15ft/min. 7. [8 points] Find all solutions to x + | x + 2 | + | x - 3 | = 7. Solution: We can break the problem into three cases.

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