# Proving a particle is traveling at constant speed

• Mar 6th 2011, 01:31 AM
iva
Proving a particle is traveling at constant speed
Hi there,

I have a particle traveling on a curve defined by (x,y) = r(t) where r(t) = (cos t, sin t, t)

I know this is a helix with the image being a circle, and I need to prove that the particle travels at a constant speed. I know grad r(t) will give me the tangent vector at a point ie (-sin t, cos t, 1). I thought that since grad gives rate of change, then a magnitude of 0 would mean constant speed. So trying this:

grad (cos t, sin t, t) = (-sin t, cos t, 1) and || (-sin t, cos t, 1)|| = root(2)..

meaning its not a constant speed right? But i know i'm wrong because my problem asks me to prove the speed is constant. CAn anyone advise please?

Thanks!
• Mar 6th 2011, 01:36 AM
Prove It
Isn't $\displaystyle \displaystyle \sqrt{2}$ a constant?
• Mar 6th 2011, 03:43 AM
iva
Ok, but doesn't that represent a rate of change, so shouldn't that value be 0 to indicate that its constant? What kind of value would then be non-constant speed?

Thanks :)
• Mar 6th 2011, 04:40 AM
Prove It
Quote:

Originally Posted by iva
Ok, but doesn't that represent a rate of change, so shouldn't that value be 0 to indicate that its constant? What kind of value would then be non-constant speed?

Thanks :)

If you get a speed of 0 then that means the particle is not moving...

To have a non-constant speed, then you would have speed as a function of $\displaystyle \displaystyle t$.
• Mar 6th 2011, 06:21 AM
Soroban
Hello, iva!

Quote:

A particle travels on a curve defined by: $\displaystyle (x,y) = r(t)$, where: .$\displaystyle r(t) \:=\: (\cos t, \sin t, t)$

I know this is a helix with the image being a circle,
and I need to prove that the particle travels at a constant speed.

I know. $\displaystyle r'(t)$ gives me the tangent vector at a point, i.e. $\displaystyle (\text{-}\sin t, \cos t, 1).$

I thought that since $\displaystyle r'(t)$ gives rate of change,
then a magnitude of 0 would mean constant speed. . no!

$\displaystyle \,r'(t)$ is the velocity of the particle.
. . It provides the direction and the speed of the particle.

The speed is the magnitude of the velocity: .$\displaystyle s \:=\:|r'(t)|$
. . Hence: .$\displaystyle s \;=\;\sqrt{(\text{-}\sin t)^2 + (\cos t)^2 + 1^2} \;=\;\sqrt{2}$

At any time $\displaystyle \,t$, the speed is $\displaystyle \sqrt{2}$ units/second.
. . The speed is constant.