# Tangential Component of Acceleration

1. Jun 19, 2011

### scoobiedoober

1. The problem statement, all variables and given/known data
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:

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

2. Relevant equations

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

3. The attempt at a solution

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

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

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

2. Jun 19, 2011

### 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

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

3. Jun 19, 2011

### 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?

4. Jun 19, 2011

### LCKurtz

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

5. Jun 19, 2011

### scoobiedoober

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

6. Jun 19, 2011

### Char. Limit

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

7. Jun 19, 2011

### LCKurtz

You mean = 2 sinh(t).

8. Jun 19, 2011

### Char. Limit

Ooops Of course. 2 sinh(t).