ISI2015-MMA-75

Question

The length of the curve $x=t^3$, $y=3t^2$ from $t=0$ to $t=4$ is

• A. $5 \sqrt{5}+1$
• B. $8(5 \sqrt{5}+1)$
• C. $5 \sqrt{5}-1$
• D. $8(5 \sqrt{5}-1)$

Comments

https://gateoverflow.in/321802/isi2015-mma-75

---

https://math.stackexchange.com/questions/1364129/find-the-length-of-the-loop-of-the-given-curve-x-3t-t3-y-3t2

Answer 1

Answer: D

---

$\frac{dx}{dt}=3t^2$ and $\frac{dy}{dt}=6t$

 

Length of the curve = $\int_0^4\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}dt=8(5\sqrt{5}-1)$

Comments

Simplify and substitute $t^2+4=u$ and solve the intergral by change of variable method.