# directional derivative Help

• Oct 17th 2009, 08:07 PM
zpwnchen
directional derivative Help
Quote:
i could get the directional derivative which is -12. but how to get the unit vector out of it?
• Oct 17th 2009, 08:16 PM
mr fantastic
Quote:

Originally Posted by zpwnchen
i could get the directional derivative which is -12. but how to get the unit vector out of it?

The direction of maximum rate of change is given by $\displaystyle \nabla f$ evaluated at the given point. Divide this vector by it's magnitude to get a unit vector.
• Oct 17th 2009, 09:03 PM
zpwnchen
$\displaystyle F_{u} = | \bigtriangledown F(x,y)| . |u|$ gives the max rate of change.

so i find the
$\displaystyle \bigtriangledown F(x,y) = <6x/ \sqrt{36x^2+64y^2} ,8y/ \sqrt{36x^2+64y^2}>$
$\displaystyle = <12/ \sqrt{36x^2+64y^2}, 8/ \sqrt{36x^2+64y^2}>$
and then take a magnitude of it and gives me 1. so unit vector is going to be "-12" which is not right. :{
• Oct 17th 2009, 09:32 PM
mr fantastic
Quote:

Originally Posted by zpwnchen
$\displaystyle F_{u} = | \bigtriangledown F(x,y)| . |u|$ gives the max rate of change.

so i find the
$\displaystyle \bigtriangledown F(x,y) = <6x/ \sqrt{36x^2+64y^2} ,8y/ \sqrt{36x^2+64y^2}>$
$\displaystyle = <12/ \sqrt{36x^2+64y^2}, 8/ \sqrt{36x^2+64y^2}>$
and then take a magnitude of it and gives me 1. so unit vector is going to be "-12" which is not right. :{

I don't have the faintest idea what you're doing or why you're doing it.

The given function is $\displaystyle f(x, y) = 3x^2 + 4y^2$. I have already told you that $\displaystyle \nabla f$ evaluated at the given point is the direction of maximum rate of change. And I have also told you to divide that vector by it's magnitude to get a unit vector in that direction.

I cannot see that you have done or attempted to do any of what I said.

By the way, "-12" is a scalar not a vector.
• Oct 17th 2009, 09:42 PM
zpwnchen
got it! thank you so much