# Unit Tangent Vector Problem

1. Apr 27, 2008

### the7joker7

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

Let bar r(t) = < -1t^(2)+2, -3e^(5t), -5sin(-4t) >

Find the unit tangent vector bar T(t) at the point t=0

3. The attempt at a solution

Attempt:

r(t) = -1t^2 + 2, -3e^5t, -5sin(-4t)

v(t) = -2t, -3e^5t, -5cos(-4t)*-4

T(t) = (-2t - 3e^(5t) - 5cos(-4t)*-4)/sqrt(-2t^2 - (3e(5t))^2 - (5cos(-4t)*-4)^2)

=

(0 - 4 + 20)/sqrt(0 - 9 - 400)

T(0) = <0, -4/3, 20/20.2237>

Actual Solution:

bar T(t) = < 0,-0.6,0.8 >

Sorry about all these questions but these unit tangent vectors have got me stumped...

2. Apr 27, 2008

### HallsofIvy

Staff Emeritus
Here is you first two errors-you didn't use the chain rule. The derivative of et is et but the derivative of e5t is NOT e5t. Use the chain rule. Same for sin(4t).

Once you have found the derivative, v(t), I would recommend setting t= 0 immediately, then find its length.

You are differentiating incorrectly. Use the chain rule!

Last edited: Apr 27, 2008
3. Apr 27, 2008

### the7joker7

Okay, I'm forgetting some of my basic derivative stuff, but...

(-3ln(e)*e^(5t)*5),

Is that right for the second?

And the third should be...

-5*cos(-4*t)*(-4)

Correct?

Last edited: Apr 27, 2008
4. Apr 27, 2008

### the7joker7

Well, that produced the right answer, so I tried to copy it for a similar problem...

bar r(t) = < -1t^(2)+4, -2e^(1t), 3sin(-1t) >

v(t) = (-2t), (-2ln(e)*e^(t)),( 3cos(-t)*-1)>

v(0) = <0 - 2 - 3>

mag = 3.61

It isn't right though. :/ dammit.

5. Apr 27, 2008

### HallsofIvy

Staff Emeritus
First. ln(e)= 1. that should simplify your work!

The derivative of <-t2+ 4, -2et], 3sin(-t)> is indeed <-2t, -2et, -3 cos(t)> and at t= 0, that is <0, -2, -3> which has length $\sqrt{13}$ or approximately 3.605. What, exactly, was the original problem?

6. Apr 27, 2008

### the7joker7

Let bar r(t) = < -1t^(2)+4, -2e^(1t), 3sin(-1t) >

Find the unit tangent vector bar T(t) at the point t=0

7. Apr 27, 2008

### HallsofIvy

Staff Emeritus
The the correct answer is $<0, -2/\sqrt{13}, -3/\sqrt{13}>$. What makes you say that "isn't right"?

8. Apr 27, 2008

### the7joker7

Nevermind, it was the right answer, it just didn't want me to round.

Thanks.