# Directional derivatives and the gradient vector

1. Oct 15, 2014

### BondKing

If the unit vector u makes an angle theta with the positive x axis then we can write u = <cos theta, sin theta>

Duf(x, y) = fx(x,y) cos theta + fy(x,y) sin theta

What if I am dealing with a function with three variables (x, y, z)?

How can I find the directional derivative if I have been given an angle?

2. Oct 15, 2014

### Matterwave

In three dimensions, a unit vector will be distinguished by 2 angles. It could be a polar angle $\theta$ and an azimuthal angle $\phi$ or some other two numbers to define a direction (sometimes the designation is flipped). In spherical polar coordinates a unit vector will be $\hat{u}=(\sin\phi\cos\theta,\sin\phi\cos\phi,\sin\phi)$

3. Oct 15, 2014

### BondKing

How should I approach this question?

Find the directional derivative of f(x,y,z)=xz+y2 at the point (3,1,2) in the direction of a vector making an angle of −π/4 with ∇f(3,1,2).

I found the gradient to be <2,2,3>

4. Oct 15, 2014

### Matterwave

I don't understand what it means: "a vector making an angle of −π/4 with ∇f(3,1,2)."? In 3-D space, there are an infinity of vectors making a specific angle with respect to another vector...they form a cone.

5. Oct 15, 2014

### BondKing

Is there a way to obtain a vector (out of many)?

6. Oct 16, 2014

### deluks917

You do not have to think about angles to understand the idea of a directional derivative. Let x and v be in Rn, v be a unit vector and t in R . The directional derivative of f at a in the direction v is just the derivative of the single variable function h(t) = f(a + tv) at t=0 (that is h'(0) ).

This gives the definition of the directional derivative without discussing the gradient. And in fact some directional derivatives can exist without f having a defined gradient. Say f is differentible with respect to x but not with respect to y. However if f is differentible then the direction derivative of f at a in the direction v is grad(f)(a)*v.

7. Oct 16, 2014

### HallsofIvy

Staff Emeritus
Given function f(x, y, z), with gradient $\nabla f$, we can talk about the "directional derivative" in the direction of unit vector $\vec{v}$ as the dot product: $\nabla f\cdot \vec{v}$.

In three dimensions, a direction cannot be specified by a single angle. We need two angles such as the "$\theta$" and "$\phi$" used in spherical coordinates. Or we can use the "direction cosines", the cosines of the angles the direction makes with the three coordinate axes: $\theta_x$ is the angle a line in the given direction makes with the x-axis, $\theta_y$ the angle it makes with the y-axis, and $\theta_z$, the angle it makes with the z-axis. Of course, those three angles are not idependent. We can show that we must have $cos^2(\theta_x)+ cos^2(\theta_y)+ cos^2(\theta_z)= 1$ which means that the vector $cos(\theta_x)\vec{i}+ cos(\theta_y)\vec{j}+ cos(\theta_z)\vec{k}$ is the unit vector in that direction.

That is, the "directional derivative" of f(x, y, z) in the direction that makes angles $\theta_x$, $\theta_y$, and $\theta_z$ with the x, y, and z axes, respectively, is given by $\nabla f\cdot (cos(\theta_x)\vec{i}+ cos(\theta_y)\vec{j}+ cos(\theta_z)\vec{k})= f_x cos(\theta_x)+ f_y cos(\theta_y)+ f_z cos(\theta_z)$.