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A particle moves along the intersection of the elliptic paraboloid $z = x^2 + 2y^2$ and the plane $x = 2$. At the moment when the particle is at $(2, 1, 5)$, what is the rate of change of $z$ with respect to $y$?

Attempt: This question confuses me. I thought of using the directional derivative, but since they are asking just for the rate of change with respect to $y$ only I suppose taking a simple partial derivative of $z$ with respect to $y$ is sufficient: $$z=f(x,y),\;f_y=4y$$ $$f_y(2,1,5)=4$$

So thats the rate of change of $z$ with respect to $y$ at $(2,1,5)$. However, it is suspiciously easy and I think I am doing it wrong. Help please.

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$+1$ for the great way of asking the question. –  Gigili May 25 '12 at 6:25
    
I don't understand this. $(2,1,5)$ doesn't lie on $z = x^2 + 2y^2$. –  user17762 May 25 '12 at 6:27
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I presume you mean (2, 1, 6), right? But yes, you're doing this correctly. It really is that easy. –  user22805 May 25 '12 at 6:33
    
It is from a sample exam and there is no mistake in the point coordinates. (Typo?). The extraneous material about the intersection confused me a lot. –  Koba May 25 '12 at 6:39
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1 Answer 1

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Yes, what you have done is correct. Note that the particle moves along the intersection of the elliptic paraboloid $z=x^2+2y^2$ and the plane $x=2$. Hence, it is actually moving along a parabola given by $z = 2y^2 + 4$. Hence, the rate of change of $z$ with respect to $y$ can be obtained directly by differentiating $z$ with respect to $y$.

The only thing is as David pointed out in his comment, the particle should be at $(2,1,6)$ and $(2,1,5)$.

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