I have a problem where I am to create a function in terms of $x$ and $y$ and compute the gradient at the point $(1,1)$. I computed the gradient but in order to evaluate it at the given point do I just plug the point in to the gradient so I get back a vector with two components or do I calculate the length of that vector to get down to just a scalar?
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$\begingroup$The gradient is a vector-valued function. You evaluate at the point and you obtain a vector.
For example, if $f = x^2 + y^2$, then $\nabla f : \mathbb{R}^2 \to \mathbb{R}^2$ is given by
$$(\nabla f)(x, y) = \left(\frac{\partial f}{\partial x}(x, y), \frac{\partial f}{\partial y}(x, y)\right) = (2x, 2y).$$
So at the point $(1, 1)$ you obtain the gradient vector $(\nabla f)(1, 1) = (2, 2)$.
$\endgroup$ 1 $\begingroup$Let $\nabla f(x,y)=\left(\frac{\partial f}{\partial x}(x,y),\frac{\partial f}{\partial x}(x,y)\right)$ be the gradient of $f$, then:
- Gradient at point $(1,1)$:
$$\nabla f(1,1)=\left(\frac{\partial f}{\partial x}(1,1),\frac{\partial f}{\partial x}(1,1)\right)$$
- Length of gradient at point $(x,y)$:
$$\|\nabla f(x,y)\|=\left(\left(\frac{\partial f}{\partial x}(x,y)\right)^2+\left(\frac{\partial f}{\partial x}(x,y)\right)^2\right)^{1/2}$$
- Length of gradient at point $(1,1)$
$$\|\nabla f(1,1)\|=\left(\left(\frac{\partial f}{\partial x}(1,1)\right)^2+\left(\frac{\partial f}{\partial x}(1,1)\right)^2\right)^{1/2}$$
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