Question

What is the arclength of `f(t) = (t+sqrt(lnt),t-sqrtlnt))` on `t in [1,e]`?

Answer 1

The arc length between `t=a` and `t=b` on the parametric curve `y=f(t),x=g(t)` is given by
`int_a^bsqrt((dy/dt)^2+(dx/dt)^2)\ dt`

So, from the question, we have `t` from `1` to `e` and `y=t-sqrt(ln(t)),x=t+sqrt(ln(t))`. Then, `dy/dt=1-1/(2tsqrt(ln(t)))` and `dx/dt=1+1/(2tsqrt(ln(t)))`.

Thus, we need to find
`int_1^esqrt((1-1/(2tsqrt(ln(t))))^2+(1+1/(2tsqrt(ln(t))))^2)\ dx`
`=int_1^esqrt(2+1/(2t^2ln(t)))\ dx`

Using numerical approximation methods, we find that the integral is approximately equal to `3`.