Computing The Arclength Function

[withthemotive]

Homework Statement

Consider the curve r = (e^−2 t cos(3 t), e^−2 t sin(3 t), e^−2 t) .

Compute the arclength function s(t) : (with initial point t=0 ).

The Attempt at a Solution

r'(t) = <-2e^-2t*cos(3t) + e^-2t*-3sin(3t), -2e^-2t*sin(3t) + e^-2t*cos(3t), -2e^-2t>

Then what, do I find the length of that derivative?
Then take the integral of 0 to t?
I dunno.

 

[Unco]

Hi Withthemotive,

r'(t) = <-2e^-2t*cos(3t) + e^-2t*-3sin(3t), -2e^-2t*sin(3t) + e^-2t*cos(3t), -2e^-2t>

A minor point: the e^(-2t)*cos(3t) term in your second component is missing a factor of 3. Also, be sure to use plenty of parentheses to remove any ambiguity in the future!

Then what, do I find the length of that derivative?
Then take the integral of 0 to t?

That's right, but your derivative will need to be with respect to a dummy variable, say u:
$$s(t) = \int_0^t ||\text{r}'(u)|| \, \text{du}$$.