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?

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

3. Originally Posted by usvn
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?
here, we have $\bold{r}(t) = \left< 2 \cos t, 3 \sin t \right>$

Now, the unit tangent vector is given by $\bold{T}(t) = \frac {\bold{r}'(t)}{|\bold{r}'(t)|}$

the curvature $\kappa$ is given by $\kappa (t) = \frac {|\bold{T}'(t)|}{|\bold{r}'(t)|}$

for the point (2,0), we need $t = 0$

for the point (0,3), we need $t = \frac {\pi}2$

4. Thanks. I knew those formulas just didn't know they where the CAS or whatever. Let me try it

5. ## Curviture...

comment removed

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?

7. Originally Posted by usvn
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?
How did you get the term in red?

8. sqrt(4sin^2(t) + 9cos^2(t)

=sqrt (4+9)

=sqrt(13)

9. No not quite

|r '(t)| = sqrt(4+5cos^2(t))

at the point (2,0) t= 0

|r '(0)| = 3

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

Now compute T ' (t)

I get then T ' (0) = -2/3 but again check my results

you then get k = 2/9 at (2,0)

10. Originally Posted by usvn
sqrt(4sin^2(t) + 9cos^2(t)

=sqrt (4+9)

=sqrt(13)
Unfortunately, that's not right. Substitute $t = 0$ which would give

$
\sqrt{4 \sin^2 0 + 9 \cos^2 0} = \sqrt{9} = 3 \ne \sqrt{13}
$
.

11. how does |r '(t)| = sqrt(4+5cos^2(t)) ??

12. 4sin^2(t) + 9cos^2(t)

= 4sin^2(t) + 4 cos^2(t) + 5cos^2(t)

= 4 + 5*cos^2(t)

13. To answer the question originally posed, "CAS" means "computer algebra system" such as MAPLE or MATHEMATICA.

14. In fact I used Mathcad for T ' (t) even though the derivative wouldn't have been impossible by hand

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)

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