Chapter 14, Problem 14RE

### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

Chapter
Section

### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

# Find the slope of the tangent in the x-direction to the surface z   = 5 x 4 − 3 x y 2 +   y 2  at  ( 1 ,   2 ,   −   3 ) .

To determine

To calculate: The slope of the tangent in the x-direction to the surface z=5x43xy2+y2 at the point (1,2,3).

Explanation

Given Information:

The provided surface is z=5x4âˆ’3xy2+y2.

The provided point is (1,2,âˆ’3).

Formula used:

The slope of the tangent of a function f(x,y) in the positive x-direction is given by âˆ‚fâˆ‚x.

For a function f(x,y), the partial derivative of f with respect to x is calculated by taking the derivative of f(x,y) with respect to x and keeping the other variable y constant. The partial derivative of f with respect to x is denoted by âˆ‚fâˆ‚x and the partial derivative of f with respect to y is denoted by âˆ‚fâˆ‚y.

Power of x rule for a real number n is such that, if f(x)=xn then fâ€²(x)=nxnâˆ’1.

Constant function rule for a constant c is such that, if f(x)=c then fâ€²(x)=0.

Coefficient rule for a constant c is such that, if f(x)=câ‹…u(x), where u(x) is a differentiable function of x, then fâ€²(x)=câ‹…uâ€²(x).

Calculation:

Consider the provided surface, z=5x4âˆ’3xy2+y2.

Since, z is a function of the variables x and y.

Thus, the function is z(x,y)=5x4âˆ’3xy2+y2

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