Finding the Unit Tangent Vector for a Given Curve

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Homework Help Overview

The discussion revolves around finding the unit tangent vector for a given curve defined by the vector function r(t) = < -1t^(2)+2, -3e^(5t), -5sin(-4t > at the point t=0. Participants are exploring the differentiation of the vector function and the subsequent calculations needed to determine the unit tangent vector.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants attempt to differentiate the vector function and calculate the unit tangent vector, with some expressing confusion about the application of the chain rule in their derivatives. Others question their own calculations and seek clarification on specific derivative steps.

Discussion Status

Some participants have provided guidance on the correct application of the chain rule and the simplification of expressions. There are multiple interpretations of the derivative calculations, and while some participants have arrived at what they believe to be correct answers, there is no explicit consensus on the final outcomes.

Contextual Notes

Participants are working under the constraints of homework assignments, which may limit the information they can share or the methods they can use. There is also a mention of rounding issues affecting the perceived correctness of answers.

the7joker7
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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...
 
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the7joker7 said:

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 et is et but the derivative of e5t is NOT e5t. 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!
 
Last edited by a moderator:
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:
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.
 
the7joker7 said:
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, -2et], 3sin(-t)> is indeed <-2t, -2et, -3 cos(t)> and at t= 0, that is <0, -2, -3> which has length [itex]\sqrt{13}[/itex] or approximately 3.605. What, exactly, was the original problem?
 
Let `bar r(t) = < -1t^(2)+4, -2e^(1t), 3sin(-1t) >`

Find the unit tangent vector `bar T(t)` at the point `t=0`
 
The the correct answer is [itex]<0, -2/\sqrt{13}, -3/\sqrt{13}>[/itex]. What makes you say that "isn't right"?
 
Nevermind, it was the right answer, it just didn't want me to round.

Thanks.
 

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