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February 17th, 2015, 08:29 PM   #1
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Finding the rate of change at a point in the direction of a vector

Hi, I am confused on how to find the rate of change at a point in the direction of a vector,
I know that when you are finding the rate of change at a point to another point, you take the change in distance for those two points, but I am not sure how to do that for this problem...

What is the rate of change of f(x,y)=2xy+y2 at the point (3,2) in the direction v=−i+3j?
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February 18th, 2015, 03:49 AM   #2
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This is a directional derivative. We are surfing along a line looking at how $f(x,y)$ changes along that line.

In this case our line is $<3,2>+t<-1,3> = <3-t, 2+3t>$ where $t$ varies as we move along the line. And at $t=0$ we are at $<3,2>$.

Thus we have the directional derivative$$f'(3,2) = \lim_{t \to 0} {f(3-t,2+3t) - f(3,2) \over t}$$
This formula leads to
$${\mathrm d \over \mathrm d t}f(x,y) = {\partial f \over \partial x}{\mathrm dx \over \mathrm d t} + {\partial f \over \partial x}{\mathrm dx \over \mathrm d t}$$
In our case $x = 3-t$ and $y=2+3t$.
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