Changing least rapidly?

**MathSucker** · November 21st 2009, 05:20 AM

At (2,1), I have to find the direction where $f(x,y)=x^2+3xy-y^2$ is increasing fastest.

So I did:

$\triangledown f(2,1)= \langle 2(2)+3, 3(2)-2 \rangle = \langle 7,4 \rangle$

Next I have to find the rate of growth in this direction, so I did:

$||\triangledown f(2,1)|| = \sqrt{7^2+4^2} = \sqrt{65}$

Finally I have to find the direction a this point where f is changing "least rapidly." Not sure what to do here...

**HallsofIvy** · November 21st 2009, 07:16 AM

Originally Posted by MathSucker

At (2,1), I have to find the direction where $f(x,y)=x^2+3xy-y^2$ is increasing fastest.

So I did:

$\triangledown f(2,1)= \langle 2(2)+3, 3(2)-2 \rangle = \langle 7,4 \rangle$

Next I have to find the rate of growth in this direction, so I did:

$||\triangledown f(2,1)|| = \sqrt{7^2+4^2} = \sqrt{65}$

Finally I have to find the direction a this point where f is changing "least rapidly." Not sure what to do here...

What do you mean by "changing least rapidly"? A function increases most rapidly in the direction of the gradient, which you calculated, decreases most rapidly in the opposite direction, and has 0 rate of change at right angles to the gradient. Does "has 0 rate of change" qualify as "changes least rapidly"?

**MathSucker** · November 21st 2009, 07:25 AM

I'm not quite sure what it means, hence the name of the thread. I assume the magnitude of when it is decreasing fastest is the same as when it is increasing fastest. I was thinking "zero because it isn't moving sideways" but that seems to easy. Or maybe that's the point. I dunno.

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