Question

What is the arclength of `f(t) = (te^(2t)-e^t-3t,-2t^2)` on `t in [1,3]`?

Answer 1

`L = int_1^3 sqrt([(2t+1)e^(2t)-e^t-3]^2 +16t^2) dt`
After setting the above Integral I used an online Integration Calculator for the numerical approximation. No closed integral exist, but the numerical approximation of L is: `L = 1179.856725422904`

Explanation:

Given `f(t) = (te^(2t)-e^t-3t,-2t^2)` on `t in [1,3]`
Find the Arclength, L:
`L= int_a^b sqrt((f'(t))^2 + (g'(t))^2) dt`

`f(t) = te^(2t)-e^t-3t`
`f'(t) = (2t+1)e^(2t)-e^t-3`
`(f'(t))^2 = [(2t+1)e^(2t)-e^t-3]^2`

`g(t) =-2t^2`
`g'(t) = -4t`
`(g'(t))^2 = 16t^2`

`L = int_1^3 sqrt([(2t+1)e^(2t)-e^t-3]^2 +16t^2) dt`
No closed integral, but numerically you can Approximate, L:
`L = 1179.856725422904`