Finding the Unit Tangent Vector for a Given Curve

[the7joker7]

Homework Statement

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

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

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...

 

[HallsofIvy]

Homework Statement

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

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

The Attempt at a Solution

Attempt:

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

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

Here is you first two errors-you didn't use the chain rule. The derivative of e^t is e^t but the derivative of e^5t is NOT e^5t. Use the chain rule. Same for sin(4t).

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)

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

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...

You are differentiating incorrectly. Use the chain rule!

 

[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?

 

[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.

 

[HallsofIvy]

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.

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

The derivative of <-t²+ 4, -2e^t], 3sin(-t)> is indeed <-2t, -2e^t, -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?

 

[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`

 

[HallsofIvy]

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

 

[the7joker7]

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

Thanks.