Tangential Component of Acceleration

[scoobiedoober]

Homework Statement

I was following through the notes I took in my Calc class, and either I wrote down the formulas incorrectly, my professor did this problem incorrectly, or I wrote the answer down incorrectly. I'm just going to redo it here, so could someone give me a thumbs up or thumbs down if I did it correctly? Thanks.

Problem:

Find the tangential component of the acceleration vector of the given partial motion vector:

$$r(t)=\langle e^{-t},\sqrt{2}t,e^t\rangle$$

Homework Equations

$$a_T=\frac{d}{dt}|v|$$

The Attempt at a Solution

$$v(t)=\frac{dr(t)}{dt}=\langle -e^{-t},\sqrt{2},e^t \rangle$$

$$|v|=\sqrt{e^{-2t}+2+e^{2t}}$$
$$|v|=\sqrt{(e^{-t}+e^t)^2}$$
$$|v|=e^{-t}+e^t$$


$$a_T=\frac{d}{dt}|v|=-e^{-t}+e^t$$

 

[LCKurtz]

No, that is not how you calculate it. You need to calculate the unit tangent vector T which you get by dividing the velocity vector by its length. Then you get the acceleration vector A by differentiating your velocity vector. Then you get the tangential component of the acceleration by calculating

$$A_T = \vec A\cdot T$$

 

[scoobiedoober]

Okay then I'll do it that way:


(from before)
$$v(t)=\langle -e^{-t},\sqrt{2},e^t \rangle$$
$$|v|=e^{-t}+e^t$$

(new)
$$T(t)=\frac{v(t)}{|v|}=\langle \frac{-e^{-t}}{e^{-t}+e^t}, \frac{\sqrt{2}}{e^{-t}+e^t}, \frac{e^t}{e^{-t}+e^t}\rangle$$

$$a(t)=\frac{d}{dt}v(t)=\langle e^{-t},0,e^t \rangle$$

$$a_T=a(t) \centerdot T(t) = e^{-t}\frac{-e^{-t}}{e^{-t}+e^t} + 0 + e^t\frac{e^t}{e^{-t}+e^t}=\frac{e^{-2t}-e^{2t}}{e^{-t}+e^t}=\frac{(e^{t}-e^{-t})(e^{t}+e^{-t})}{e^{-t}+e^t}=-e^{-t}+e^t$$

So apparently I CAN do it the original way...and it was a bit easier too. So am I safe to assume I have done it correctly?

 

[LCKurtz]

So apparently I CAN do it the original way...and it was a bit easier too. So am I safe to assume I have done it correctly?

Yes, I made a mistake in my calculation and had a different answer, sorry for the misstep.

 

[scoobiedoober]

Yes, I made a mistake in my calculation and had a different answer, sorry for the misstep.

No worries, even my professor with a PHD flubbed it up! Thank the lord for technology :p

 

[Char. Limit]

One note: e^(t) - e^(-t) = sinh(t), if you want to make your answer look more elegant.

 

[LCKurtz]

One note: e^(t) - e^(-t) = sinh(t), if you want to make your answer look more elegant.

You mean = 2 sinh(t).

 

[Char. Limit]

You mean = 2 sinh(t).

Ooops Of course. 2 sinh(t).