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exam2practicesolutions

# exam2practicesolutions - Math 2250 Exam#2 Practice Problem...

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Math 2250 Exam #2 Practice Problem Solutions 1. Let u ( x ) = p f ( x ) and suppose f (3) = 1, f 0 (3) = 8, and f 00 (3) = - 2. What is the value of u 00 (3)? Answer: Using the Chain Rule, u 0 ( x ) = 1 2 ( f ( x )) - 1 / 2 f 0 ( x ) . Hence, using the Chain Rule and the Product Rule, u 00 ( x ) = 1 2 - 1 2 ( f ( x )) - 3 / 2 f 0 ( x ) f 0 ( x ) + ( f ( x )) - 1 / 2 f 00 ( x ) = - ( f 0 ( x )) 2 4( f ( x )) 3 / 2 + f 00 ( x ) 2 p f ( x ) . Therefore, u 00 (3) = - ( f 0 (3)) 2 4( f (3)) 3 / 2 + f 00 (3) p f (3) = - 8 2 4(1) 3 / 2 + - 2 2 1 = - 17 . 2. Let f ( x ) = 1 2 sin( x 2 ) cos( x 2 ). What is f 0 q 5 π 6 ? Answer: Using the Product and Chain Rules, f 0 ( x ) = 1 2 ( cos( x 2 ) · 2 x · cos( x 2 ) + sin( x 2 ) · 2 x · ( - sin( x 2 )) ) = x cos 2 ( x 2 ) - x sin 2 ( x 2 ) = x ( cos 2 ( x 2 ) - sin 2 ( x 2 ) ) . Therefore, f 0 r 5 π 6 ! = r 5 π 6 cos 2 5 π 6 - sin 2 5 π 6 = r 5 π 6 - 3 2 ! 2 - 1 2 2 = r 5 π 6 3 4 - 1 4 = 1 2 r 5 π 6 . 3. Suppose you’ve leaned a 10 foot ladder against a vertical wall, but haven’t properly secured the bottom of the ladder. Before you can climb onto the ladder, the bottom starts to slide away from the wall at 3 ft/sec. How fast does the top of the ladder slide down the wall when the bottom is 6 feet from the wall?

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