Question

What is the arc length of `r(t)=(te^(t^2),t^2e^t,1/t)` on `tin [1,ln2]`?

Answer 1

Arc Length `~~ −2.42533 \ \` (5dp)

The arc length is negative due to the lower bound `1` being greater than the upper bound of `ln2`

Explanation:

We have a parametric vector function, given by:

`bb ul r(t) = << te^(t^2), t^2e^t, 1/t >>`

In order to calculate the arc-length we will require the vector derivative, which we can compute using the product rule:

`bb ul r'(t) = << (t)(2te^(t^2)) + (1)(e^(t^2)) , (t^2)(e^t) + (2t)(e^t) , -1/t^2 >>`
`\ \ \ \ \ \ \ \ = << 2t^2e^(t^2) + e^(t^2) , t^2e^t + 2te^t , -1/t^2 >>`

Then we compute the magnitude of the derivative vector:

`| bb ul r'(t) | = sqrt( (2t^2e^(t^2) + e^(t^2))^2 + (t^2e^t + 2te^t)^2 + (-1/t^2)^2) )`

`" " = sqrt( e^(2 t) t^4 + 1/t^4 + 4 e^(2 t) t^3 + 4 e^(2 t) t^2 + 4 e^(2 t^2) t^2 + e^(2 t^2) + 4 e^(2 t^2) t^4 )`

Then we can compute the arc-length using:

`L = int_(1)^(ln2) \ | bb ul r'(t) | \ dt`
`\ \ = int_(1)^(ln2) \ sqrt( e^(2 t) t^4 + 1/t^4 + 4 e^(2 t) t^3 + 4 e^(2 t) t^2 + 4 e^(2 t^2) t^2 + e^(2 t^2) + 4 e^(2 t^2) t^4 ) \ dt`

It is unlikely we can compute this integral using analytical technique, so instead using Numerical Methods we obtain an approximation:

`L ~~ −2.42533 \ \` (5dp)

The arc length is negative due to the lower bound `1` being greater than the upper bound of `ln2`