the formula for curvature is |dT/dt|/|dr/dt| where T is the unit tangent

To see an explanation see Unit Tangent Vectors/Unit Normal Vectors

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- April 28th 2009, 09:53 PM #1
## Curviture

x = 2cos(t) and y = 3sin(t), where the parameter t satisfies 0 < t < 2pi.

This is only part of the problem but i'm having trouble with it, i need to find the curvature for this parametric equation at points (2,0) and (0,3). I was told to use CAS. Can anyone explain what CAS is and how to do this?

- April 28th 2009, 10:03 PM #2
the formula for curvature is |dT/dt|/|dr/dt| where T is the unit tangent

To see an explanation see Unit Tangent Vectors/Unit Normal Vectors

- April 28th 2009, 10:04 PM #3

- April 28th 2009, 10:35 PM #4

- May 8th 2009, 03:43 PM #5

- May 13th 2009, 09:06 AM #6
So so far i have

r'(t)=(-2sint,3cost)

r''(t)=(-2cost,-3sint)

|r'(t)| = sqrt(13)

T(t)=(-2sint/sqrt(13),3cost/sqrt(13))

T'(t)=(-2cost/sqrt(13),-3sint/sqrt(13))

|T'(t)| = sqrt(4/13 + 9/13) = 1

K=1/sqrt(13)

But then i have nowhere to plug in the 2 given coordinantes, where did i go wrong?

- May 13th 2009, 09:22 AM #7

- May 13th 2009, 09:26 AM #8

- May 13th 2009, 09:31 AM #9

- May 13th 2009, 09:34 AM #10

- May 13th 2009, 10:03 AM #11

- May 13th 2009, 10:07 AM #12

- May 13th 2009, 11:02 AM #13

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- May 13th 2009, 11:04 AM #14

- May 13th 2009, 11:21 AM #15
...hmm i'm not very proficient with maple. Trying to get |T'(t)| i get

T'(t) = (-2cost/sqrt(4+5cos^2(t)),-3sint/sqrt(4+5cos^2(t)))

|T'(0)| = (sqrt(4sin^2(t)/sqrt(4+5cos^2(t)) + 9cos^2(t)/sqrt(4+5cos^2(t)) = sqrt(4/3+5/3) = sqrt(3)??

am i doing the algebra wrong?

kappa = |T'(t)|/|r'(t)|

so kappa = sqrt(3)/3 (wrong)