9A_quiz5soln

# 9A_quiz5soln - a =-4 Solution Here we have f a = √ 25 = 5...

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Math 9A Quiz 5 Tuesday, May 15 2007 Name: 1. The radius r and height h of a right circular cylinder are realted to the cylinder’s volume V by the formula V = πr 2 h. Solution: before we answer the parts, let’s calculate dV/dt : dV/dt = d dt ( πr 2 h ) = π ± d dt ( r 2 ) h + r 2 d dt ( h ) ² = π ± 2 r dr dt h + r 2 dh dt ² . (a) How is dV/dt related to dh/dt if r is constant? If r is constant, then dr/dt = 0, and so dV/dt = πr 2 dh dt . (b) How is dV/dt related to dr/dt if h is constant? If h is constant, then dh/dt = 0, and so dV/dt = 2 πr dr dt h. (c) How is dV/dt related to dr/dt and dh/dt if neither r nor h is constant? This is just the expression we found at the beginning: dV/dt = π ± 2 r dr dt h + r 2 dh dt ² .

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2. Find the linearization of f ( x ) = x 2 + 9 at
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Unformatted text preview: a =-4. Solution: Here we have f ( a ) = √ 25 = 5 and f ( x ) = ( √ x 2 + 9) = (( x 2 + 9) 1 / 2 ) = 1 / 2( x 2 + 9)-1 / 2 ( x 2 ) = x √ x 2 + 9 , and so f (-4) =-4 / 5. Therefore L ( x ) = 5-4 / 5( x + 4) . 3. Find the diﬀerential dy for the following functions. (a) y = x 3-3 √ x Solution: dy = d ( x 3-3 √ x ) = 3 x 2 dx-3 · 1 2 x-1 / 2 dx = 3 x 2 dx-3 2 √ x dx = 3 ± x 2-1 2 √ x ² dx. (b) y = cos( x 2 ) Solution: dy = d (cos( x 2 )) =-sin( x 2 ) d ( x 2 ) =-sin( x 2 )2 xdx....
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## This note was uploaded on 04/16/2008 for the course MATH 9A taught by Professor Apoorva during the Winter '07 term at UC Riverside.

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9A_quiz5soln - a =-4 Solution Here we have f a = √ 25 = 5...

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