Find the equation of the tangent plane to the graph of f(x,y)=8x^{2}-2xy^{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

zi2lalZ 2021-05-22 Answered
Find the equation of the tangent plane to the graph of f(x,y)=8x22xy2 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
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Expert Answer

Dora
Answered 2021-05-23 Author has 98 answers
Step 1
We first find the partial derivatives
f=8x22xy2
fx=16x2y2
fy=4xy
At (5,4)
fx=16x2y2=16(5)2(4)2=48
fy=4xy=4(5)(4)=80
So the direction vector of the normal is = (48,-80)
Step 2
So the equation of the tangent plane is:
z=(8(5)202(5)(4)2)+48(x5)+(80)(y4)
z=40+48x-240-80y+320
z=48x-80y+120
Answer: B
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Jeffrey Jordon
Answered 2021-11-24 Author has 2262 answers

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