# Thread: Rate of change f(x,y,z) in direction v

1. ## Rate of change f(x,y,z) in direction v

At the point (1,2,-3) a vector v makes an angle of pi/3 radians with the gradient of the function,

$f(x,y,z) = x^2yz - 3xy^3$

Find the rate of change of f(x,y,z) in direction v.

Okay so if I can find the components of the vector v then I simply have to dot the gradient of the function evaluated at the point (1,2,-3) with a unit vector in the direction of v.

Now I've just got to figure out how I can get the components of v.

The dot product would give me on equation, the cross product another, but where do I get the last equation?

I guess because I specified it has to be a unit vector in the direction of v my last equation would be that the magnitude of the vector must be equal to 1.

Is this the only way to solve this problem?

EDIT: I'm stuck with this one, cross product didn't work out so well. Any ideas?

2. Note that the vector v is not well-defined because there is a whole range of vectors that make an angle of pi/3 with a given 3D vector . On the other hand, the dot product is just the product of the lengths times the cosine of the angle between the two vectors. So you don't need to know v exactly.

3. Originally Posted by emakarov
Note that the vector v is not well-defined because there is a whole range of vectors that make an angle of pi/3 with a given 3D vector . On the other hand, the dot product is just the product of the lengths times the cosine of the angle between the two vectors. So you don't need to know v exactly.
But I do need v in order to find the rate of change in that direction.

4. I don't think you do. I admit that it has been some time since I studied this, so I'll give a quote from Wikipedia.
If the hill height function H is differentiable, then the gradient of H dotted with a unit vector gives the slope of the hill in the direction of the vector. More precisely, when H is differentiable, the dot product of the gradient of H with a given unit vector is equal to the directional derivative of H in the direction of that unit vector.
Also,
The directional derivative of a multivariate differentiable function along a given vector V at a given point P intuitively represents the instantaneous rate of change of the function, moving through P, in the direction of V.
Edit: So the rate of change in the direction of v depends only on the angle of v with the gradient vector, not on v itself.

5. In other words, just multiply the gradient vector's length |∇f(1,2,-3)| with $\cos (\frac{\pi}{3})$