MTH164FA10-M1d_ans

1 51 9 and z 1 as before page 2 of 8 tuesday october

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Unformatted text preview: deﬁned implicitly by the equation above. Then using implicit diﬀerentiation on the equation ∂z ∂z ∂z ∂z above we can calculate that ∂ x (2, 1) = 4 and ∂ y (2, 1) = −5 so dz = ∂ x dx + ∂ y dy = 4(.1) − 5(−.1) = .9 and z = −.1 as before. Page 2 of 8 Tuesday, October 19, 2010 1st Midterm Exam MTH 164 (Multivariable Calculus) (c) Find the maximum rate of change of the height of the surface at P (2, 1, −1) above the xy plane and the direction in the xy -plane at which it occurs. Answer: dz = 4dz − 5dy = ￿4, −5￿ · ￿dx, dy ￿ = ￿￿4, −5￿￿ ￿￿dx, dy ￿￿ cos(θ) So the maximum rate of a change (for a unit vector) will be when the vector is parallel to ￿4, −5￿. The √ ￿ optimal direction is ￿ = √4,−5￿ and the rate of change is 16 + 25. u...
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This document was uploaded on 05/06/2013.

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