gradient of a function 3 variables

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I just don't get how it work at all. I am asked to find the gradient of a function and then find it for a set of coordinate.

$$f(x,y,z) = 4x^3yz^2+2xz^3+xyz$$

Find $\nabla f$ and then $\nabla f$ at $(1,-1,-1)$.

I added all the partial derivatives together then replaced $(x,y,z)$ by $(1,-1,-1)$ and got $3$ as an answer but it doesn't feel right at all.

How am I supposed to proceed?

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2 Answers

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The gradient is a vector:

$$\nabla f = (\partial f/\partial x, \partial f/\partial y, \partial f/\partial z)$$

(this particular way of writing it is specific to $\mathbb{R}^3$)

So in this case, $$\nabla f(x,y,z) = (12x^2yz^2+2z^3+yz,4x^3z^2+0+xz,8x^3yz+6xz^2+xy)$$

And so the gradient at $(1,-1,-1)$ is given by $$\nabla f(1,-1,-1) = (-13,3,13)$$

The sum of these components is $3$, as you observed, but the value of the gradient is a vector, not the sum of these components.

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I added all the partial derivatives together then replaced (x,y,z) by (1,−1,−1) and got 3 as an answer but it doesn't feel right at all.

It isn't. A scalar value is not a valid answer. The gradient is a vector whose components are the partial derivatives of the trivariate function.

$$\nabla f(x,y,z) = \dfrac{\partial f}{\partial x}\vec \imath + \dfrac{\partial f}{\partial y}\vec \jmath + \frac{\partial f}{\partial z}\vec k $$

Before adding anything, multiply the partials by the unit vectors; then substitute. The result will be a 3-vector.

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