Hw14-4 - 14.4#2. (a) To differentiate ‘directly,’ we...

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Unformatted text preview: 14.4#2. (a) To differentiate ‘directly,’ we begin with the substitution w = x 2 + y 2 = (cos t + sin t ) 2 + (cos t- sin t ) 2 = 2 cos 2 t + 2 sin 2 t = 2 , which is constant, yielding dw/dt = 0. To use the chain rule, we first solve for the partial derivatives: ∂w ∂x = 2 x, ∂w ∂y = 2 y, dx dt =- sin t + cos t, dy dt =- sin t- cos t and evaluate dw dt = ∂w ∂x dx dt + ∂w ∂y dy dt = 2(cos t + sin t )(- sin t + cos t ) + 2(cos t- sin t )(- sin t- cos t ) = 0 . (b) The constant function f ( t ) = 0 applied at the point t = 0 gives an answer of 0. 14.4#6. (a) As in problem 4, we do the direct version first: w = z- sin xy = e t- 1- sin( t ln t ) , which by the rules of single-variable differentiation (including the Chain Rule!) gives dw dt = e t- 1- (1 + ln t ) cos( t ln t ) . Now we try the strategy with the partials: ∂w ∂x =- y cos xy, ∂w ∂y =- x cos xy, ∂w ∂z = 1 , dx dt = 1 , dy dt = 1 /t, dz dt = e t- 1 ....
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This note was uploaded on 06/11/2011 for the course MATH 1920 at Cornell.

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Hw14-4 - 14.4#2. (a) To differentiate ‘directly,’ we...

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