Directional derivatives and the gradient vector

[BondKing]

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

D_uf(x, y) = f_x(x,y) cos theta + f_y(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?

 

[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 $\hat{u}=(\sin\phi\cos\theta,\sin\phi\cos\phi,\sin\phi)$

 

[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>

 

[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.

 

[BondKing]

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.

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

 

[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.

 

[HallsofIvy]

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).