0 votes 0 votes The length of the curve $x=t^3$, $y=3t^2$ from $t=0$ to $t=4$ is $5 \sqrt{5}+1$ $8(5 \sqrt{5}+1)$ $5 \sqrt{5}-1$ $8(5 \sqrt{5}-1)$ Geometry isi2015-mma curves non-gate + – Arjun asked Sep 23, 2019 recategorized Nov 17, 2019 by Lakshman Bhaiya Arjun 477 views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply `JEET commented Nov 18, 2019 reply Follow Share https://gateoverflow.in/321802/isi2015-mma-75 0 votes 0 votes `JEET commented Nov 18, 2019 reply Follow Share https://math.stackexchange.com/questions/1364129/find-the-length-of-the-loop-of-the-given-curve-x-3t-t3-y-3t2 1 votes 1 votes Please log in or register to add a comment.
0 votes 0 votes Answer: D $\frac{dx}{dt}=3t^2$ and $\frac{dy}{dt}=6t$ Length of the curve = $\int_0^4\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}dt=8(5\sqrt{5}-1)$ NastyBall answered Jun 20, 2021 NastyBall comment Share Follow See 1 comment See all 1 1 comment reply KAUNIL commented Mar 9, 2023 reply Follow Share Simplify and substitute $t^2+4=u$ and solve the intergral by change of variable method. 0 votes 0 votes Please log in or register to add a comment.