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2011a_x2a_sols

# 2011a_x2a_sols - Spring 2011 Math 151 Exam II Version A...

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Unformatted text preview: Spring 2011 Math 151 Exam II Version A Solutions 1. A Differentiate to find the velocity: s ′ ( t ) = 6 t 2 − 42 t + 72 , so s ′ (1) = 36 . 2. C Differentiate to find the velocity: r ′ ( t ) = ( 1 + e t , 1 + 2 t ) . At the point (1 , 0) , t + e t = 1 and t + t 2 = 0 , so t = 0 . The velocity is r ′ (0) = ( 2 , 1 ) , so the speed is | r ′ (0) | = √ 5 . 3. C If you don’t just remember the key limit, multiply by the con- jugate: lim x → 1 − cos x 2 x · 1 + cos x 1 + cos x = lim x → 1 − cos 2 x 2 x (1 + cos x ) = lim x → sin 2 x 2 x (1 + cos x ) = lim x → sin x x · sin x 2(1 + cos x ) = 1 · 0 = . 4. C Switch x and y and solve for y : x = √ y − 5 , so x 2 = y − 5 and y = x 2 + 5 . The domain of f − 1 is the range of f , which is [ , ∞ ) . 5. B dy dx = dy/dt dx/dt = 3 t 2 2 t − 4 . When t = 2 , dy dt = 12 and dx dt = 0 , so the graph has a vertical tangent at t = 2 . When t = 2 , x = 2 2 − 4(2) + 1 = − 3 , so the equation is x = − 3 ....
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2011a_x2a_sols - Spring 2011 Math 151 Exam II Version A...

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