Directional Derivative and Gradient Problem


by aaron27
Tags: derivative, directional, gradient
aaron27
aaron27 is offline
#1
Nov12-12, 08:28 AM
P: 6
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.
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Vorde
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#2
Nov12-12, 01:11 PM
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P: 784
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))##
aaron27
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#3
Nov12-12, 08:07 PM
P: 6
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?

Vorde
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#4
Nov12-12, 08:53 PM
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P: 784

Directional Derivative and Gradient Problem


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.
aaron27
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#5
Nov12-12, 09:15 PM
P: 6
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?
Vorde
Vorde is offline
#6
Nov12-12, 09:32 PM
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P: 784
Yes, and the magnitude will be ##\sqrt{(-3\pi)^2+4+1}##.
aaron27
aaron27 is offline
#7
Nov14-12, 12:48 AM
P: 6
Understood, thank you very much
hkid
hkid is offline
#8
Nov14-12, 06:39 AM
P: 1
thank you... very helpful!


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