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A directional derivative

mark

Let

f(x,y) = x^3 - xy+y^2.
  1. Compute the gradient \nabla f.
  2. Compute D_{\vec{u}}f, where \vec{u} is the unit vector in the direction \langle 3,4 \rangle.
  3. In what direction is f increasing the fastest from the point (3,4)?
impish_wyvern

I computed the gradient to be \langle 3x^2-y,-x+2y\rangle.
That would mean that D_{\vec{u}}f = \langle (3x^2-y)(\frac{3}{5}),(-x+2y)(\frac{4}{5})\rangle.
I know that the direction where f is increasing the fastest from point (3,4) has got to be the direction of the gradient \nabla f, but I’m not entirely confident with how to show that information. I found the magnitude of the gradient to be \approx 14.368, but that just means that max rate of increase, not the direction. Right?

mark

No, that’s not quite correct. The directional derivative is

\nabla f \cdot \vec{u},

which is scalar valued, not vector valued, as you’ve written. Thus, I think the correct answer would be

D_{\vec{u}}f = (3x^2-y)\left(\frac{3}{5}\right) + (-x+2y)\left(\frac{4}{5}\right).

The things you’ve written about the gradient are correct. This missing piece is that you simply plug in the point (3,4) into the gradient to get the direction in which f is increasing the fastest from that point.

impish_wyvern

Thanks! So the direction is the vector

\langle 3\times3^2-4, -3 + 2\times4 \rangle = \langle23,5\rangle?