Thread: Tangent plane to a surface

Hello all,
I am studying for a calc three test and came across a problem I got stuck on the problem is as follows.

given the function g(x,y,z)= x^2 ln y+y z find an equation for the tangent plane to the surface g(x,y,z) =2 at the point (2,1,2)

I started of by finding the gradient of g(x,y,z) and found that to be (2xlny,x^2/y +z, y)
I en evaluated this a the point (2,1,2) and got the gradient vector to be (0,6,1) however the g(x,y,z) = 2 is throwing me off any help guidance would be greatly appreciated.


Thanks,
Stephen

"g(x,y,z)= x^2 ln(y)+ yz" itself does not define a surface. It is a function which has a given value at each point in three dimenisional space. In order to define a surface, you have to be able to, at least theoretically, reduce to two variables (because a surface is two dimensional). Being given that g(x,y,z)= 2, or x^2ln(y)+ yz= 2 does that. We could, for example, solve for z: z= (2- x^2ln(y))/y, reducing to x and y. Also note that you are asked to find the tangent plane at (2, 1, 2). g(2, 1, 2)= 2^2ln(1)+ 1(2)= 0+ 2= 2.

Now, you know that the vector 6j+ k is normal to the tangent plane and (2, 1, 2) is a point in that plane. What is the equation of the tangent plane?

So if I am understanding you correctly the answer is just 6y+z=8

Yes, that is correct.