Find unit tangent vector at indicated point

[olivia333]

Homework Statement

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

r(t) = e^(19t)costi + e^(19t)sintj + e^(19t) kT(pi/2) = <___i+___j+___k>

Homework Equations

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

The Attempt at a Solution

Answers:
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!

 

[Dick]

The overall sign is wrong on the i component.

 

[olivia333]

The overall sign is wrong on the i component.

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.

 

[Dick]

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.

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.

 

[olivia333]

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.

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.

 

[Dick]

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.

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.