# Directional Derivative and Gradient Problem

1. Nov 12, 2012

### aaron27

Suppose that an object is moving in a space V, so that its position at time t is
given by r=(x,y,z)= (3sin πt, t^2, 1+t)

How to find the direction of the vector along which the cat is moving
at t = 1?

I have no idea where to find out the direction of the vector along which the object is moving.

2. Nov 12, 2012

### Vorde

It seems like you wouldn't even need to use the gradient or a directional derivative to do this. If you took AB/BC Calc you might remember how to differentiate a parametric function of one variable, you can do that and then you'll get that velocity vector, from there the direction should fall straight out (directly if all you need is a direction vector, and with a little bit of trigonometry if you want angles).

If you need help, remember that $\frac{d}{dt}r(t)= (x(t),y(t),(z(t)) \Rightarrow (\frac{d}{dt}x(t),\frac{d}{dt}y(t),\frac{d}{dt}z(t))$

Last edited: Nov 12, 2012
3. Nov 12, 2012

### aaron27

By differentiating them with respect to t, I should get the velocity vector (x,y,z) = (3 π cos π t, 2t , 1). Then substitute t = 1, (x,y,z) = (-3π, 2, 1).
This is my velocity vector. So how I get the director of the vector when t=1?

4. Nov 12, 2012

### Vorde

As I said, it depends on what you mean by direction. When my professor asks for a direction, he means a direction vector; a.k.a. a unit vector in the direction of the velocity vector. Remember this is accomplished by dividing all three components by the magnitude (length) of the vector, which you find through the pythagorean theorem.

If you want actual spherical angle directions, you'll have to work through the trig.

5. Nov 12, 2012

### aaron27

Sorry, my mistake.
So if I am looking for direction vector, I just get this (-3π, 2, 1)/(magnitude of this vector (-3π, 2, 1)) where (-3π, 2, 1)=the velocity vector?

6. Nov 12, 2012

### Vorde

Yes, and the magnitude will be $\sqrt{(-3\pi)^2+4+1}$.

7. Nov 14, 2012

### aaron27

Understood, thank you very much

8. Nov 14, 2012