Question

What is the arclength of `(t^2lnt,t-lnt)` on `t in [1,2]`?

Answer 1

`f(t)=(t^2lnt,t-lnt)`

`f'(t)=(2tlnt+t,t-1/t)`

For a parametric function `f(t)=(x,y)`, the arc length for `tin[a,b]` is given by:

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

Thus, we have an arc length of:

`color(blue)(s=int_1^2sqrt((2tlnt+t)^2+(t-1/t)^2)dtapprox2.8914`

Find the approximation using a calculator or a website like WolframAlpha.