Explanation Given that the point is changed to ( 40.0 , 50.0 , 5.55 ) and the slope of the function, z = y 2 327 − x 2 763 . Here, the dependent variable is z and the independent variables are x and y . Differentiate z partially with respect to x in order to obtain the slope of the tangent line parallel to xz -plane as follows. ∂ z ∂ x = ∂ ∂ x ( y 2 327 − x 2 763 ) = ∂ ∂ x ( y 2 327 ) + ∂ ∂ x ( − x 2 763 ) = 0 − 1 763 ( 2 x ) = − 2 x 763 Substitute x = 40 in the derivative obtained, thus the slope is computed as follows. ∂ z ∂ x | x = 40.0 = − 2 ( 40.0 ) 763 = − 80 763 ≈ − 0

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Chapter 29.3, Problem 2E

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The slopes of the tangent lines that are parallel to the xz- and yz-planes when the point is changed to (40.0,50.0,5.55).

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Chapter 29.3, Problem 1E

Chapter 29.3, Problem 3E

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Chapter 29 Solutions

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The slopes of the tangent lines that are parallel to the xz- and yz- planes when the point is changed to ( 40.0 , 50.0 , 5.55 ) .

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The slopes of the tangent lines that are parallel to the xz- and yz-planes when the point is changed to (40.0,50.0,5.55).

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Chapter 29 Solutions