Get the solution to "The following questions are about the function f(x,y)=x2e2xy...."

Kyran Hudson

Answered question

2021-10-23

The following questions are about the function

f (x, y) = x^{2} e^{2 x y}

.
Find an equation of the tangent plane of f at the point (2,0, f(2,0))=(2,0,4).

Tasneem Almond

Skilled2021-10-24Added 91 answers

Step 1
If f(x,y) is a function of two variables x and y, then

f_{x} (x, y) = \frac{\partial f}{\partial x}

f_{y} (x, y) = \frac{\partial f}{\partial y}

The equation of the tangent plane of function f at the point

(x_{0}, y_{0})

z = f (x_{0}, y_{0}) + f_{x} (x_{0}, y_{0}) (x - x_{0}) + f_{y} (x_{0}, y_{0}) (y - y_{0})

Step 2
The value of the given function at (2,0) is

f (2, 0) = {(2)}^{2} e^{2 (2) (0)}

= 4 e^{0}

=4(1)
=4
The equation of the tangent plane of function f at the point (2,0) is

z = f (2, 0) + f_{x} (2, 0) (x - 2) + f_{y} (2, 0) (y - 0)

z=4+4(x-2)+16(y)
z=4+4x-8+16y
z=4x+16y-4

Jeffrey Jordon

Expert2022-08-15Added 2605 answers

Answer is given below (on video)

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