251sol6 - ∂x ∂t = 4 cos 4 t and ∂y ∂t = 6 cos 6 t ....

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Assignment 6 Multivariate Calculus Math 251 (Fall 2010) A1. A bicycle travels over a surface z = f ( x, y ) where which is deFned implicitly by x 3 + y 3 + z 3 = 4 + sin( x + y + z ) The path of the bicycle is described by equations x = sin 4 t y = sin 6 t where t is time (in seconds) and distance measured in meters. When t = 0, the bicycle is at the point (0 , 0 , z 0 ) where z 0 is a number close to 1 . 709. (a) How fast is the altitude of the cyclist increasing at time t = 0? (Note: Your answer should be an expression involving z 0 , and measured in meters per second.) (b) Is the bicycle going uphill or downhill at t = 0. (Note: you will need a calculator for this part) Solution We are asked to compute ∂z ∂t at the point ( x, y, z ) = (0 , 0 , z 0 ) and t = 0 on the surface which satisFes F ( x, y, z ) = x 3 + y 3 + z 3 - sin( x + y + z ) - 4 = 0, on the parameterized curve satisfying x = sin 5 t and y = sin 6 t . By the chain rule we Fnd that ∂z ∂t = ∂z ∂x ∂x ∂t + ∂z ∂y ∂y ∂t . (1) We have that
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Unformatted text preview: ∂x ∂t = 4 cos 4 t and ∂y ∂t = 6 cos 6 t . ±or the other two quantities in (1), we use the implicit function rule. ∂z ∂x =-∂F ∂x ∂F ∂z = cos ( x + y + z )-3 x 2 3 z 2-cos ( x + y + z ) ∂z ∂y =-∂F ∂y ∂F ∂z = cos ( x + y + z )-3 y 2 3 z 2-cos ( x + y + z ) Substituting all of this into (1), we get ∂z ∂t = cos ( x + y + z )-3 x 2 3 z 2-cos ( x + y + z ) · 5 cos 5 t + cos ( x + y + z )-3 y 2 3 z 2-cos ( x + y + z ) · 6 cos 6 t. (2) To answer part (a), we substitute the values ( x, y, z ) = (0 , , z ) and t = 0 into (2) to Fnd ∂z ∂t = cos z 3 z 2-cos z 4 · cos 0 + cos z 3 z 2-cos z · 6 cos 0 = 10 cos z 3 z 2-cos z meters/sec . ±or part (b) we use a calculator to evaluate ∂z ∂t ≈ -. 248 m/s, so the bicycle is moving downhill at t = 0. a...
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This note was uploaded on 12/13/2010 for the course MATH 251 taught by Professor Unknown during the Spring '08 term at Simon Fraser.

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251sol6 - ∂x ∂t = 4 cos 4 t and ∂y ∂t = 6 cos 6 t ....

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