Question

How do you find the length of the curve x=3t−t3x=3t-t^3, y=3t2y=3t^2, where 0≤t≤3√0<=t<=sqrt(3) ?

Answer 1

It's simple, you just have to do this:


`L=\int\limits_(a)^(b)\sqrt{\left((dx)/(dt)\right)^2+\left((dy)/(dt)\right)^2}~dt`


`x=3t-t^3`


`(dx)/(dt)=3-3t^2`


`y=3t^2`


`(dy)/(dt)=6t`

replacing


`L=\int\limits_(0)^(√(3))√(\left(3-3t^2\right)^2+\left(6t\right)^2)~dt`


`L=\int\limits_(0)^(√(3))√(9-18t^2+9t^4+36t^2)~dt`


`L=\int\limits_(0)^(√(3))√(9+9t^4+18t^2)~dt`


`L=\int\limits_(0)^(√(3))√((3t^2+3)^2)~dt`


`L=\int\limits_(0)^(√(3))(3t^2+3)~dt`


`L=t^3+3t|_0^(√(3))`


`\boxed{\boxed{L=3√(3)+3√(3)=6√(3)}}`