Multivariable calculus question

Part b)

Question

Multivariable calculus question

Part b)

anonymous · Answer

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anonymous · Answer

For part a) I got,

$$z=z_{o}+\frac{ 2 }{ \sqrt{4x_{0}} }(x-x_{0})+\frac{ 1 }{ 2\sqrt{y_{0}} }(y-y_{0})+\frac{ 1 }{ 2\sqrt{z_{0}} }(z-z_{0})$$

anonymous · Answer

Stuck on part b) ...

anonymous · Answer

What does it mean for a constant to be independent of a point?

anonymous · Answer

It doesn't matter what the input of x,y,z are, you're going to get the same numbers? Or...

anonymous · Answer

I suppose it applies to all tangent planes.

anonymous · Answer

Yeah. I guess so. Not really sure I show it's a constant though...

anonymous · Answer

anonymous · Answer

Are you sure your tangent plane is correct?

anonymous · Answer

You have $z=f(x,y,z)$ instead of $z=f(x,y)$

anonymous · Answer

Yeah i'm pretty sure that's correct.

anonymous · Answer

I'm pretty sure you shouldn't get a recursive function.

anonymous · Answer

anonymous · Answer

According to that you should have $z= L(x,y)$ and NOT $z=L(x,y,z)$.

anonymous · Answer

Alright no problem. I just solve for z then.

anonymous · Answer

So we could just eliminate the z component here and it should be okay.

anonymous · Answer

That constant vanishes. We just put a negative sign on front of the components.

anonymous · Answer

Actualyl nvm. Now we have to square both sides too...

anonymous · Answer

Let me do it out again.

anonymous · Answer

$$
z-z_0=f_x(x-x_0)+f_y(y-y_0)
$$

anonymous · Answer

Right...

anonymous · Answer

In this case $$
f(x,y)=z = (\sqrt{c}-\sqrt{x}-\sqrt{y})^2
$$

anonymous · Answer

Yep yep. I don't think I actually need to square it out since I just find the partial derivatives.

anonymous · Answer

There should a 4x in the square root. Not a x.

anonymous · Answer

Well now I get:

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