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 2021-05-22Answered
Find the equation of the tangent plane to the graph of 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
Step 1
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