Question

Find the tangent plane to the surface x^2+y^2+xyz=4+z^3(y-2x) at the point (1, 1, 1).

So the equation of the tangent plane is z-z0=f'x(x-x0)+f'y(y-y0).

I find the partials using implicit differentiation:
f'x=(2x)/(3yz^2-6z^2-y) @ (1,1,1) = -1/2
f'y=(2y)/(3z^2-6xz^2-x) @ (1,1,1) = -1/2

Then the equation of the plane (ie plugging into the above equation) is z+(1/2)x+(1/2)y=2
But the answer is 5x+2y+4z=11

I have two question from here:
1, Where am I going wrong in the implicit differentiation? Wolfram gives me a different result.
2, I know using this method is incorrect. Why is that? Why must I use the gradient vector to the level surface instead of implicitly deriving?

Answer 1

I'm not sure what you mean by that, the way I would solve this would be to define F(x,y,z)=x^2+y^2+xyz-z^3(y-2x)
solve for gradF and dot gradF with <x-x0, y-y0, z-z0>

Answer 2

This is calc III by the way

Answer 3

I'm not so confused about the problem itself but rather why these two methods are not equivalent

Answer 4

And also, why wolfram gives me a completely different partial derivatives when solved implicitly. I have no idea where I am going wrong

Answer 5

So where \(f(x, y, z) = (x^2+y^2+xyz) - (4+z^3(y-2x))\)
The surface you are considering is \(f^{-1}(0,0,0)\)
So the method you discussed earlier is the only reasonable one I can think of. That is where you consider the vectors orthogonal to the gradient at a given point on the level set.

Answer 6

In our textbook, they taught us how to find the equation of a tangent plane for z=f(x,y) using the formula z-z0=f'x(x-x0)+f'y(y-y0). What I'm trying to do is define my equation above as a implicit equation with z(x,y). So I use the process of implicit partial differentiation to find f'x and f'y and plug into the equation for the tangent plane. I don't know why that isn't equivalent to what you just wrote down

Answer 7

That is my real question. I understand the method you outlined but I don't get why that wouldn't be equivalent to the method I just described above

Answer 8

The way I define the equation would be:

x^2+y^2+xyz(z,y)=4+(z(x,y))^3 (y-2x)

So z is implicitly a function of (x,y)

And wondering now why my solution for the first partials of this equation are wrong according to wolfram.

Answer 9

The methods are indeed equivalent, my partial derivatives were incorrect. I found the correct partial derivatives and arrived at the right answer. Thanks for the help