# (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.

## Answer and Explanation: 1

(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}

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