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given the function: $$z=f(x,y)=e^{-x^2-2y^2}$$

I'd like to find a point where if I were to place a ball, it would roll towards the direction $(2,1,a)$ .

Also, at which point could I place the ball so it wouldn't move at all?

I assume I need to calculate the gradient, and then check at which point the maximum slope vector is $(-2,-1,-a)$, i.e the opposite of where the ball would roll.

And for the second question I assume the gradient is $(0,0,0)$.

Am I correct? Thank you.

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  • $\begingroup$ the gradient of $f$ has only two coordinates. $\endgroup$ – Exodd Jun 24 '16 at 17:21
  • $\begingroup$ well, I thought the answer might be that the value of "a" doesn't matter. $\endgroup$ – Tom Dawn Jun 24 '16 at 17:45
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For your second question, you are correct that the only location with zero gradient is the origin. The gradient is

$$[-2x e^{(-x^2-2 y^2)}, -4y e^{(-x^2-2 y^2)}]$$

which is only zero at $$(x,y)=(0,0)$$

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  • $\begingroup$ Thank you. Please tell me if you have an idea with the first question :) $\endgroup$ – Tom Dawn Jun 24 '16 at 18:35

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