Solutions to Exam 1
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Solutions to Exam 1 1. 2. 1 − cos x (1 − cos x)(1 + cos x) 1 − (cos x)2 = lim = lim 2 2 2 x→0 x→0 x→0 x (1 + cos x) x x (1 + cos x) 2 sin x 1 (sin x)2 1 1 = lim = lim 2 lim =1· = x→0 x x→0 x (1 + cos x) x→0 1 + cos x 2 2 lim 3. √ 2 y = ex/ x sin √ 2 √ √ dy x 1 x 1 = √ ex/ 2 sin √ + √ ex/ 2 cos √ dx 2 2 2 2 2 √ √ x x x x 1 d y 1 1 √ 1 √ = ex/ 2 sin √ + ex/ 2 cos √ + ex/ 2 cos √ − ex/ 2 sin √ 2 dx 2 2 2 2 2 2 2 2 √ x x/ 2 =e cos √ 2 3 √ √ d y x x 1 1 = √ ex/ 2 cos √ − √ ex/ 2 sin √ 3 dx 2 2 2 2 4 √ √ x 1 x x x d y 1 1 √ 1 √ = ex/ 2 cos √ − ex/ 2 sin √ − ex/ 2 sin √ − ex/ 2 cos √ 4 dx 2 2 2 2 2 2 2 2 √ x x/ 2 = −e sin √ 2 1 4. Let y be the height of the ball and x be the distance from the wall to the shadow. Then y = −16t2 + 64 and 64 y = x − 20 x xy = 64(x − 20) dx dy dx y+x = 64 dt dt dt After 1 second, y = −16 + 64 = 48, x = 80, and 2 dy dx = −32t = −32, so = −160 ft/s. dt dt
