Question

What is the arc length of the curve given by `x = t^2` and `y= 2t^2 = 1`, for `1<t<3`?

Answer 1

The arc length is `8sqrt5`

Explanation:

For this problem, we will use this formula:

`int_1^3sqrt((dx/(dt))^2 + (dy/(dt))^2) dt`

First, let's find the derivatives of `x` and `y` with respect to `t`:

`dx/(dt) = d/(dt)(t^2) = 2t`

`dy/(dt) = d/(dt)(2t^2) = 4t`

Now we can plug these into the original formula:

`int_1^3sqrt((2t)^2 + (4t)^2) dt`

`int_1^3sqrt(4t^2+16t^2)dt`

`int_1^3sqrt(20t^2)dt`

`int_1^3sqrt(20) tdt`

`int_1^3 2sqrt5tdt`

`sqrt5 int_1^3 2tdt`

`sqrt5 [t^2]_1^3`

`sqrt5 (3^2 - 1^2)`

`8sqrt5`

Final Answer