Unit tangent vector of r(t) = (e^t)(cos t ) i + (e^t)(sin t

In summary: Remember, r'(t) = (e^t)(sin t - cos t)i + (e^t)(cos t + sin t)j + (e^t)k. In summary, the unit tangent vector for the given vector valued function is [1/sqrt(3)][(cos t - sin t)i + (sin t + cos t)j + k]. One must factor out the e^t term and take more care with the algebra to calculate the magnitude correctly.
  • #1
chetzread
801
1

Homework Statement


Find the unit tangent vector T(t) for vector valued function r(t) = (e^t)(cos t ) i + (e^t)(sin t ) j + (e^t) k

Homework Equations

The Attempt at a Solution


i gt stucked here ...
, the ans is [1/ sqrt (3) ] [ (cos t -sin t ) i + (sin t + cos t ) j +k) [/B]
 

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  • #2
chetzread said:

Homework Statement


Find the unit tangent vector T(t) for vector valued function r(t) = (e^t)(cos t ) i + (e^t)(sin t ) j + (e^t) k

Homework Equations

The Attempt at a Solution


i gt stucked here ...
, the ans is [1/ sqrt (3) ] [ (cos t -sin t ) i + (sin t + cos t ) j +k) [/B]

You got ##r'(t)## okay, but you didn't calculate its magnitude correctly. I would factor out the ##e^t## term first. Then you need to take more care with your algebra.
 
  • #3
PeroK said:
You got ##r'(t)## okay, but you didn't calculate its magnitude correctly. I would factor out the ##e^t## term first. Then you need to take more care with your algebra.
My ans is [1/ sqrt (2) ] [ (cos t -sin t ) i + (sin t + cos t ) j +k)
But not
[1/ sqrt (3) ] [ (cos t -sin t ) i + (sin t + cos t )
j +k)


Is the given ans wrong ?
 
  • #4
chetzread said:
My ans is [1/ sqrt (2) ] [ (cos t -sin t ) i + (sin t + cos t ) j +k)
But not
[1/ sqrt (3) ] [ (cos t -sin t ) i + (sin t + cos t )
j +k)


Is the given ans wrong ?
No, your answer is wrong. I get the same answer that you showed in post #1.
Please show your work for |r'(t)|.
 

Related to Unit tangent vector of r(t) = (e^t)(cos t ) i + (e^t)(sin t

1. What is the unit tangent vector?

The 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 and rate of change of the curve at that point.

2. How do you find the unit tangent vector of a given curve?

To find the unit tangent vector, you first need to find the derivative of the curve. Then, divide the derivative by its magnitude to get a vector with a magnitude of 1. This vector is the unit tangent vector.

3. Can the unit tangent vector change at different points on a curve?

Yes, the unit tangent vector can change at different points on a curve. This is because the direction and rate of change of the curve can vary at different points, resulting in a different unit tangent vector.

4. How is the unit tangent vector related to the velocity vector?

The unit tangent vector and the velocity vector are closely related. The unit tangent vector is the normalized version of the velocity vector, which means it has a magnitude of 1. They both represent the direction and rate of change of the curve at a specific point.

5. Can the unit tangent vector be used to find the curvature of a curve?

Yes, the unit tangent vector can be used to find the curvature of a curve. The curvature is equal to the magnitude of the derivative of the unit tangent vector with respect to the parameter t.

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