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ISI2015-MMA-75

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

  1. $5 \sqrt{5}+1$
  2. $8(5 \sqrt{5}+1)$
  3. $5 \sqrt{5}-1$
  4. $8(5 \sqrt{5}-1)$
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Answer: D

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

 

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

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