Find the equation of the tangent plane to the graph of $f(x,y)=8{x}^{2}-2x{y}^{2}$ at the point (5,4).

A)z=23x-15y+42

B)z=48x-80y+120

C)0=48x-80y+120

D)0=23x-15y+42

A)z=23x-15y+42

B)z=48x-80y+120

C)0=48x-80y+120

D)0=23x-15y+42

zi2lalZ
2021-05-22
Answered

Find the equation of the tangent plane to the graph of $f(x,y)=8{x}^{2}-2x{y}^{2}$ at the point (5,4).

A)z=23x-15y+42

B)z=48x-80y+120

C)0=48x-80y+120

D)0=23x-15y+42

A)z=23x-15y+42

B)z=48x-80y+120

C)0=48x-80y+120

D)0=23x-15y+42

You can still ask an expert for help

Dora

Answered 2021-05-23
Author has **98** answers

Step 1

We first find the partial derivatives

$f=8{x}^{2}-2x{y}^{2}$

$f}_{x}=16x-2{y}^{2$

${f}_{y}=-4xy$

At (5,4)

${f}_{x}=16x-2{y}^{2}=16\left(5\right)-2{\left(4\right)}^{2}=48$

${f}_{y}=-4xy=-4\left(5\right)\left(4\right)=-80$

So the direction vector of the normal is = (48,-80)

Step 2

So the equation of the tangent plane is:

$z=\left(8{\left(5\right)}^{2}02\left(5\right){\left(4\right)}^{2}\right)+48(x-5)+(-80)(y-4)$

z=40+48x-240-80y+320

z=48x-80y+120

Answer: B

We first find the partial derivatives

At (5,4)

So the direction vector of the normal is = (48,-80)

Step 2

So the equation of the tangent plane is:

z=40+48x-240-80y+320

z=48x-80y+120

Answer: B

Jeffrey Jordon

Answered 2021-11-24
Author has **2262** answers

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