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I'm working on a practice problem for my Calculus 3 course. It gives the function: $z=x^2+y^2$, and asks to graph the contours for $c=1,2,3$. Than asks to calculate the gradient at point $(2,1)$ and graph the result.

I'm fine with the first part. I'm letting z=c and solving for y, then graphing the result.It's the gradient portion I'm having issues with.

The gradient I came up with is: $\nabla(x^2+y^2)=\langle 2x,2y\rangle$, at $(2,1)$, = (4,2)

I'm not exactly sure how to graph this. Am I supposed to graph a line from $(2,1)$ to $(4,2)$ or what? Any help/explanation would be very much appreciated!

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I would draw the vector $(4,2)$ starting at the point $(2,1)$ (ie, a line between $(2,1)$ and $(6,3)$). – copper.hat Mar 4 '13 at 6:53
How did you get (6,3)? – PseudoPsyche Mar 4 '13 at 6:53
No, a line connecting $(2,1)$ to $(6,3) = (2,1) + (4,2)$. – copper.hat Mar 4 '13 at 6:54
For the contours you just let $z=c$ and solve for $y$ and graph the resulting function – MITjanitor Mar 4 '13 at 6:54
@copper.hat, thank you! – PseudoPsyche Mar 4 '13 at 7:02
up vote 3 down vote accepted

Gradients are drawn from the point that they're taken at. This shows where gradients are taken from, and allows gradients to be perpendicular to level curves. Since the gradient was taken at the point $(2,1)$, the vector $\langle 4,2 \rangle$ should be drawn from $(2,1)$ pointing to the point $(6,3)$ because $(2,1) + (4,2) = (6,3)$.

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Ok cool thanks! I'm just confused about one thing though. For this problem, the point (2,1) is undefined in the same graph as my contours, c=1,2,3, and the provided coordinate grid on the practice sheet with the problem only has coordinates: {x,-4,4} and {y,-4,4} Does this mean that I'll just have the point (2,1) floating off away from my contours, and it will just extend off of the provided coordinate plane in the direction of (6,3)? – PseudoPsyche Mar 4 '13 at 7:18
Yes. It won't be a point on your level curves. – Andrew Salmon Mar 4 '13 at 7:24
Alright, awesome! Thank you so much! – PseudoPsyche Mar 4 '13 at 7:54

The gradient is a vectorfield, i.e. a vector attached to every point of you space. The most clear way to draw it is to draw an arrow of length (4,2) starting from the point (2,1).

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