Thread: Calculus question

I was wondering if I did the following problem right:

Write the equation of the plane tangent to the given surface at the given point P.

xyz+ 2x^3 - y^2 + z^2 = 10; P(2, -1, -1)

So what I did was find the gradient of the above and got:

<yz + 4x, xz - 2y, xy - 2z>

Substituting P in I got, a normal vector of <9, 0, 0>

This led to a plane equation of:
9(x - 2) = 0

Originally Posted by meebo0129

I was wondering if I did the following problem right:

Write the equation of the plane tangent to the given surface at the given point P.

xyz+ 2x^3 - y^2 + z^2 = 10; P(2, -1, -1)

So what I did was find the gradient of the above and got:

<yz + 4x, xz - 2y, xy - 2z>

Substituting P in I got, a normal vector of <9, 0, 0>

This led to a plane equation of:
9(x - 2) = 0

Try again, you got the gradient all wrong.

Okay...here's my second attempt:

(x-2)(yz-2x) + (y+1)(xz-2y) + (z+1)(xy-2z) = 0

(x-2)(1-4)+(y+1)(-2+2)+(z+1)(-2+2)=0

Then, I simplified this to:
-3(x-2)=0
x-2 = 0

x = 2 <- equation of plane

Would this be correct?

You have been told: "the gradient is all wrong."

Oh, you know what? I accidentally wrote 2x^3 for the second term instead of 2x^2, which is what it was supposed to be. If this is the case, then is my original response correct?

O.K. I have corrected the above.
But note the your "z" component should be with a +.

Originally Posted by meebo0129

Write the equation of the plane tangent to the given surface at the given point P.

xyz+ 2x^3 - y^2 + z^2 = 10; P(2, -1, -1)

I was wondering if P(2,-1,-1) is on the given surface.