# Directional Derivative

• Oct 27th 2009, 01:17 AM
latavee
Directional Derivative
I've been at this problem for a while!
Find the directional derivative of the given function at the point (3,-1) in the direction theta=(pi)/4.

f(x,y)=4x+xy^2-5y

fx=4+y^2
fy=2xy-5

applying ponts:
fx=4+(-1)^2=5
fy=2(3)(-1)-5=-11

<5, -11>

But how would I find my unit vector in the direction of theta=pi/4?
• Oct 27th 2009, 03:13 AM
HallsofIvy
Quote:

Originally Posted by latavee
I've been at this problem for a while!
Find the directional derivative of the given function at the point (3,-1) in the direction theta=(pi)/4.

f(x,y)=4x+xy^2-5y

fx=4+y^2
fy=2xy-5

applying ponts:
fx=4+(-1)^2=5
fy=2(3)(-1)-5=-11

<5, -11>

But how would I find my unit vector in the direction of theta=pi/4?

The unit vector in direction $\displaystyle \theta$ is $\displaystyle \left<cos(\theta), sin(\theta)\right>$.

The directional derivative of f in the direction of unit vector $\displaystyle \vec{v}$ is $\displaystyle \nabla f\cdot\vec{v}$.