$ x^4 + y^4 + z^4 = x^2 + y^2 +z^2$
How can I find the gradient vector of $f(x,y,z)$ at $(1,1,1)$?
Can I rewrite the equation by just moving everything to one side? So: $x^4 - x^2 + y^4 -y^2 +z^4 -z^2 = 0$ Then find the partial derivatives?
I get <2,2,2> as a result.
$\endgroup$ 42 Answers
$\begingroup$The condition
$$ x^4 + y^4 + z^4 = x^2 + y^2 +z^2$$
defines a $0$-level surface of a function
$$f(x,y,z)=C(x^4 - x^2 + y^4 -y^2 +z^4 -z^2)$$
where $C$ is an arbitrary non-zero constant.
The gradient vector at $(1,1,1)$ is then
$$(C,C,C).$$
$\endgroup$ 2 $\begingroup$Your answer is correct. I'm assuming you got a gradient $$\nabla f=\langle 4x^3-2x,4y^3-2y,4z^3-2z\rangle$$
and plugged in the point $(1,1,1)$, which is correct.
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