# (a) Find the equation of the plane tangent to the graph of f(x,y) = x^{2} e^{xy}

## Question:

(a) Find the equation of the plane tangent to the graph of {eq}f(x,y) = x^{2} e^{xy} {/eq} at {eq}(1,0) . {/eq}

(b) Find the linear approximation of {eq}f(x,y) {/eq} for {eq}(x,y) {/eq} near {eq}(1,0) . {/eq}

(c) Find the differential of f at the point (1,0) .

## Tangent Plane:

The equation of the tangent plane to the surface {eq}f(x,y) {/eq} at point {eq}(x_0,y_0) {/eq}

is found arresting the Taylor series of the function at the first order terms, i.e.

{eq}\displaystyle L(x,y) = f(x_0,y_0) + f_x(x_0,y_0) (x-x_0) + f_y(x_0,y_0) (y-y_0) {/eq}

where {eq}f_x, \; f_y {/eq} are the first partial derivatives of the function.

(a) The equation of the plane tangent to the graph of

{eq}\displaystyle f(x,y) = x^{2} e^{xy} {/eq}

at point (1,0) is given by

{eq}\displaystyle f(1,0) = 1^{2} e^{1(0)} = 1 \\ \displaystyle f_x(x,y) = 2xe^{xy}+e^{xy}yx^2 \rightarrow f_x(1,0) = 2 \\ \displaystyle f_y(x,y) = e^{xy}x^3 \rightarrow f_y(1,0) = 1 \\ \displaystyle L(x,y) = f(1,0) + f_x(1,0) (x-1) + f_y(1,0) y = 1 + 2(x-1) + y = 2x+y-1 {/eq}

(b) The linear approximation of {eq}f(x,y) {/eq} for {eq}(x,y) {/eq} near {eq}(1,0) {/eq}

is equal to the equation of the tangent plane found at point (a).

(c) The differential of f at the point (1,0) is equal to

{eq}\displaystyle df =f_x(1,0) dx + f_y(1,0) dy = 2 dx + dy {/eq}

Linearization of Functions

from

Chapter 10 / Lesson 1
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Over the river and through the woods to Grandmother's house we go ... Are we there yet? In this lesson, apply linearization to estimate when we will finally get to Grandma's house!