Unit tangent, unit normal, unit binormal, curvature

I'm gettingT' = (4 i - 6t j - 12t^2 k)/sqrt(16 + 36t^2 + 144t^4)And then |dT/dt| to be sqrt(16 + 36t^2 + 144t^4)/sqrt(16 + 36t^2 + 144t^4) = 1.
  • #1
s3a
818
8

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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  • #2
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!
 
  • #3
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.
 
  • #4
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.)
 
  • #5
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.
 

What is a unit tangent vector?

A unit tangent vector is a vector that is tangent to a curve at a specific point and has a magnitude of 1. It represents the direction in which the curve is moving at that point.

What is a unit normal vector?

A unit normal vector is a vector that is perpendicular to the tangent vector at a specific point on a curve. It represents the direction in which the curve is curving at that point.

What is a unit binormal vector?

A unit binormal vector is a vector that is perpendicular to both the tangent vector and the normal vector at a specific point on a curve. It represents the direction of the twisting or torsion of the curve at that point.

What is curvature?

Curvature is a measure of how much a curve deviates from a straight line. It is defined as the rate of change of the unit tangent vector with respect to arc length. In other words, it measures the amount of curvature per unit length along the curve.

What is the relationship between unit tangent, unit normal, and unit binormal vectors?

Unit tangent, unit normal, and unit binormal vectors are all mutually perpendicular and form a right-handed coordinate system. Their magnitudes are all equal to 1, hence the term "unit". Together, they provide a complete description of the orientation and curvature of a curve at a specific point.

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