Find an equation of the plane tangent to the following

sonkitty2h

sonkitty2h

Answered question

2021-11-12

Find an equation of the plane tangent to the following surface at the given point.
yzexz+14=0; (0,7,2)

Answer & Explanation

Ancessitere

Ancessitere

Beginner2021-11-13Added 17 answers

yzexz+14=0
f=yzexz+14
take partial derivative w.r.t. x
fx=ddx(yzexz+14)
fx=ddx(yzexz)+ddz(14)
fx=yz2ezx+0
fx=yz2ezx
put the point (0,-7,2)
a=722e20
a=28
f=yzexz+14
take partial derivative w.r.t. y
fy=ddy(yzexz+14)
fy=ddy(yzexz)+ddy(14)
fy=zezx+0
fy=zezx
put the point (0,-7,2)
b=e022
b=2
the equation of the plane is given by
a(xx0)+b(yy0)+z(zz0)=0
here we have (a,b,c)=(28,2,7), (x0,y0,z0)=(0,7,2)
28(x0)+2(y+7)7(z2)=0
28x+2y+147z+14=0
28x+2y7z+28=0

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?