# Homework Help: Find unit tangent vector at indicated point

1. Feb 15, 2012

### olivia333

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

Find the unit tangent vector at the indicated point of the vector function

r(t) = e(19t)costi + e(19t)sintj + e(19t) k

T(pi/2) = <___i+___j+___k>

2. Relevant equations

r'(t) / |r'(t)|

3. The attempt at a solution

19e(19*∏/2)(cos(∏/2)-sin(∏/2)) / ((19e(19*∏/2)sin(∏/2))2+(19e(19*∏/2)cos(∏/2))2+(19e(19*∏/2))2)^.5 i

19e(19*∏/2)(cos(∏/2)+sin(∏/2)) / ((19e(19*∏/2)sin(∏/2))2+(19e(19*∏/2)cos(∏/2))2+(19e(19*∏/2))2)^.5 j (correct)

19e(19*∏/2) / ((19e(19*∏/2)sin(∏/2))2+(19e(19*∏/2)cos(∏/2))2+(19e(19*∏/2))2)^.5 k (correct)

What's wrong with i?

Thanks!

Last edited: Feb 15, 2012
2. Feb 15, 2012

### Dick

The overall sign is wrong on the i component.

3. Feb 15, 2012

### olivia333

Sorry, I didn't mean to put that there. I'm using a program online though and it's still wrong.

Annoying thing I actually put in:

19e^(19*pi/2)(cos(pi/2)-sin(pi/2))/((19e^(19*pi/2)sin(pi/2))^2+(19e^(19*pi/2)cos(pi/2))^2+(19e^(19*pi/2))^2)^.5

The parenthesis are correct because it shows us a much neater version of what we put in with our computer, and everything is as it should look.

4. Feb 15, 2012

### Dick

Ok, there is more than the sign wrong. The numerator should be e^(19*pi/2)*(19*cos(pi/2)-sin(pi/2)). Both terms don't have a factor of 19 in them. Check your product rule. There is a similar problem with the j component, but it didn't get caught because cos(pi/2)=0. BTW, you could simplify these expressions a LOT.

Last edited: Feb 15, 2012
5. Feb 15, 2012

### olivia333

Thank you so much for your help. I would simplify, but I can't go farther than this on tests because I don't have enough time (this long annoying answer counts as full credit) and I'd like to do it like I'll do it on the test.

6. Feb 15, 2012

### Dick

Sure and you are welcome. You can do it anyway you'll get full credit. But at least setting cos(pi/2)=0 and sin(pi/2)=1 will also save you time writing the answer down on a test.