Problem Finding the Speed of Particle (Vector)

Hey everyone,

I am having a bit of difficulty solving this problem. It says:

A particle has position function $\displaystyle \ r(t) = (2\sqrt{2})i + (e^{2t})j + (e^{-2t})k$. What is its speed at time t?

Well I first took the derivative of the function. Then I took the magnitude to find the length which corresponds to the speed. I got my answer to be

$\displaystyle \sqrt{8+4e^{2t}+4e^{-2t}}$, is this answer correct? If so, how would I further simplify it? If it is incorrect, what am I doing wrong?

Any help and feedback is greatly appreciated, thanks.

Re: Problem Finding the Speed of Particle (Vector)

Hello, Beevo!

Quote:

A particle has position function: $\displaystyle \ r(t) \:=\: (2\sqrt{2})i + (e^{2t})j + (e^{-2t})k$.

What is its speed at time $\displaystyle t$?

Well, I first took the derivative of the function.

Then I took the magnitude to find the length which corresponds to the speed.

I got my answer to be: $\displaystyle \sqrt{8+4e^{2t}+4e^{-2t}}$.

Is this answer correct? . Yes!

If so, how would I further simplify it?

It can be simplified . . .

$\displaystyle \sqrt{4e^{2t} + 8 + 4e^{-2t}} \;=\;\sqrt{4(e^{2t} + 2 + e^{-2t})} \;=\;2\sqrt{e^{2t} + 2 + e^{-2t}}$

. . . . . . . . . . . . .$\displaystyle =\;2\sqrt{(e^t + e^{-t})^2} \;=\;2(e^t + e^{-t}) $

Re: Problem Finding the Speed of Particle (Vector)

Quote:

Originally Posted by

**Beevo** Hey everyone,

I am having a bit of difficulty solving this problem. It says:

A particle has position function $\displaystyle \ r(t) = (2\sqrt{2})i + (e^{2t})j + (e^{-2t})k$. What is its speed at time t?

Well I first took the derivative of the function. Then I took the magnitude to find the length which corresponds to the speed. I got my answer to be

$\displaystyle \sqrt{8+4e^{2t}+4e^{-2t}}$, is this answer correct? If so, how would I further simplify it? If it is incorrect, what am I doing wrong?

Any help and feedback is greatly appreciated, thanks.

Looks good to me, unless you have to use hyperbolic functions to simplify.

-Dan

PS okay I should have seen that one coming Soroban! Thanks.

Re: Problem Finding the Speed of Particle (Vector)

Thanks for the feedback guys. The simplification part just kind of threw me off.

Re: Problem Finding the Speed of Particle (Vector)

Okay, I'm confused! Where did that "8" come from? With the position $\displaystyle r(t)= 2\sqrt{2}\vec{i}+ e^{2t}\vec{j}+ e^{-2t}\vec{k}$,I get the velocity to be $\displaystyle 2e^{2t}\vec{j}- 2e^{-2t}\vec{k}$ and so the speed is $\displaystyle \sqrt{4e^{4t}+ 4e^{-4t}}= 2\sqrt{e^{4t}+ e^{-4t}}$

Re: Problem Finding the Speed of Particle (Vector)

Quote:

Originally Posted by

**HallsofIvy** Okay, I'm confused! Where did that "8" come from? With the position $\displaystyle r(t)= 2\sqrt{2}\vec{i}+ e^{2t}\vec{j}+ e^{-2t}\vec{k}$,I get the velocity to be $\displaystyle 2e^{2t}\vec{j}- 2e^{-2t}\vec{k}$ and so the speed is $\displaystyle \sqrt{4e^{4t}+ 4e^{-4t}}= 2\sqrt{e^{4t}+ e^{-4t}}$

My fault, there should be a 't' after $\displaystyle \2\sqrt{2}$