Unit tangent, unit normal, unit binormal, curvature

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Homework Help Overview

The discussion revolves around finding the unit tangent, normal, and binormal vectors, as well as the curvature of a specific curve defined by the parametric equations x = 4t, y = -3t^2, z = -4t^3 at t = 1. Participants are exploring the mathematical concepts related to vector calculus and curvature.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of the unit tangent vector and the challenges in deriving the unit normal vector. There is an emphasis on the need to evaluate these vectors at a specific parameter value, t = 1. Some participants express confusion about the application of the quotient rule and the implications of substituting t = 1 in the calculations.

Discussion Status

The discussion is ongoing, with participants sharing their attempts and questioning the validity of their approaches. There is recognition of the complexity of the problem, and some guidance has been offered regarding the use of the quotient rule for differentiation. Multiple interpretations of the problem are being explored, particularly concerning the evaluation of derivatives.

Contextual Notes

Participants note the difficulty in managing the calculations without computational assistance, highlighting the complexity of the expressions involved, particularly for |dT/dt|. There is an acknowledgment of the potential for errors in manual calculations and the need for careful consideration of the parameterization.

s3a
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Homework Statement


Question:
"Find the unit tangent, normal and binormal vectors T, N, B, and the curvature of the curve
x = 4t, y = -3t^2, z = -4t^3 at t = 1."

Answer:
T = 0.285714285714286 i - 0.428571428571429 j - 0.857142857142857 k
N = -0.75644794981871 i + 0.448265451744421 - 0.476282042478447 k
B = 0.588348405414552 i + 0.784464540552736 j - 0.196116135138184
ϰ = 0.0445978383113072


Homework Equations


N = dT/dt / |dT/dt|


The Attempt at a Solution


I tried to use the equation from the "Relevant equations" part above. I know there are alternative ways but I want to figure out what I am doing wrong for this method.

I (successfully) get the unit tangent vector to be:
T = (4 i - 6t j - 12t^2 k)/sqrt(4^2 + 6^2 * t^2 + 12^2 * t^4)
T = 2/7 i - 3/7 * t j - 6/7 * t^2 k

I (unsuccessfully) get the unit normal vector to be:
N = (3/7 i - 12/7*t k)/sqrt( (3/7)^2 + (12/7)^2 * t^2)

What am I doing wrong?

Any input would be greatly appreciated!
Thanks in advance!
 
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s3a said:

Homework Statement


Question:
"Find the unit tangent, normal and binormal vectors T, N, B, and the curvature of the curve
x = 4t, y = -3t^2, z = -4t^3 at t = 1."

Answer:
T = 0.285714285714286 i - 0.428571428571429 j - 0.857142857142857 k
N = -0.75644794981871 i + 0.448265451744421 - 0.476282042478447 k
B = 0.588348405414552 i + 0.784464540552736 j - 0.196116135138184
ϰ = 0.0445978383113072


Homework Equations


N = dT/dt / |dT/dt|


The Attempt at a Solution


I tried to use the equation from the "Relevant equations" part above. I know there are alternative ways but I want to figure out what I am doing wrong for this method.

I (successfully) get the unit tangent vector to be:
T = (4 i - 6t j - 12t^2 k)/sqrt(4^2 + 6^2 * t^2 + 12^2 * t^4)
The above is really T(t), the tangent vector for an arbitrary value of the parameter t.

The problem asks for the unit tangent vector and unit normal vector at t = 1. IOW, it's looking for T(1), N(1), and B(1).
s3a said:
T = 2/7 i - 3/7 * t j - 6/7 * t^2 k

I (unsuccessfully) get the unit normal vector to be:
N = (3/7 i - 12/7*t k)/sqrt( (3/7)^2 + (12/7)^2 * t^2)

What am I doing wrong?
See above. I didn't check your work, so if you have errors, I wasn't looking for them. Again, you want T(1), N(1), and B(1) - the unit vectors at a particular value of t.
s3a said:
Any input would be greatly appreciated!
Thanks in advance!
 
When you went from T = (4 i - 6t j - 12t^2 k)/sqrt(4^2 + 6^2 * t^2 + 12^2 * t^4) to T = 2/7 i - 3/7 * t j - 6/7 * t^2 k you put t=1 in the denominator. That's means you can't use the second expression to find dT/dt. You eliminated some of the t dependence. You need to use the first and use the quotient rule. BTW this is quite a messy problem.
 
Sorry, I was doing a lot of stuff in my head so the t = 1 went on and off.

I get |dT/dt| to be sqrt(144t^4 + 36t^2 + 16) and it doesn't seem that I can get rid of the square root.

So, I am assuming it's hard to do it this way and that I shouldn't do it this way assuming it is possible. Is it possible though? (I am not asking for it computed that way but I would just like to know if it is possible to get past that step without using a computer or something of the sort.)
 
s3a said:
Sorry, I was doing a lot of stuff in my head so the t = 1 went on and off.

I get |dT/dt| to be sqrt(144t^4 + 36t^2 + 16) and it doesn't seem that I can get rid of the square root.

So, I am assuming it's hard to do it this way and that I shouldn't do it this way assuming it is possible. Is it possible though? (I am not asking for it computed that way but I would just like to know if it is possible to get past that step without using a computer or something of the sort.)

I get something a LOT messier for |dT/dt|. I'm using a computer for this one and would feel sorry for someone who wasn't. It's pretty bad.
 

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