# HELP! If the curve `r(t)=e^t(vecicost+vecjsint)` from t=0 to t=piIf that is the curve (above), find the length of the curve and find the parameters (define the new parameters) of the curve using...

HELP! If the curve `r(t)=e^t(vecicost+vecjsint)` from t=0 to t=pi

If that is the curve (above), find the length of the curve and find the parameters (define the new parameters) of the curve using the arc length measured from the point (1,0) when t is increasing.

I'm really confused about the parameters part!?! Thanks in advance!

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### 1 Answer

Let's just run through quickly and determine the arc length. To find the arc length of a parameterized curve, we simply take the following integral (see link below for explanation):

`s = int_a^b sqrt(((dx)/(dt))^2 + ((dy)/(dt))^2)dt`

In our case, the bounds will be 0 and pi, and we will substitute the square of the derivatives of our functions:

`s = int_0^pi sqrt(e^(2t)sin^2t + e^2tcos^2t) dt`

Keep in mind, we can factor out `e^(2t)` and use our trigonometric identities (`sin^2t + cos^2t = 1`) to simplify and solve the integral:

`s = int_0^pi sqrt(e^2t) dt=int_0^pi e^t dt = e^pi-1`

So, our arc length is `e^pi - 1`.

To make a new set of parameters, let's recognize that if we had evaluated the above integral from 0 to t and using `tau` as a dummy variable in the integral, we would have gotten the following for our arc length:

`s = e^t - 1`

We simply have to get a transform from t to the arc length, s, to allow us to use s as our new parameter:

So, let's solve for t with respect to s:

`t = ln(s+1)`

Now, we can simply substitute t for the above function of s to get our new equation based on the new parameter (arc length):

`r(s) = e^(ln(s+1))(vecicos(ln(s+1)) + vecj sin(ln(s+1)))`

`r(s) = (s+1) (vecicos(ln(s+1)) + vecjsin(ln(s+1)))`

And there you have it. A reparameterized curve based on a new parameter, arc length.

I hope that helps! I read it as our needing to make a new function based on the arc length instead of t. If that was not what was asked, then clarify, and I'll help you figure it out!

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