Find the length of the curve.
-$$x = t ^ { 3 } , y = 2 t ^ { 2 } , 0 \leq t \leq 1$$
A) $\frac { 122 } { 3 }$
B) $\frac { 1 } { 27 }$
C) $\frac { 61 } { 27 }$
D) $\frac { 125 } { 27 }$

Question

Quizplus · Accepted Answer

Using the formula for arc length, we have 
$$\int_{0}^{1} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt = \int_{0}^{1} \sqrt{9t^4 + 16t^2} dt = \int_{0}^{1} t^2\sqrt{9+16t^2} dt$$
Make a substitution $u=9+16t^2$. Then, $du = 32t dt$, and we can rewrite our integral as 
$$\frac{1}{32}\int_{9}^{25} \sqrt{u}\, du = \frac{1}{32}\cdot\frac{2}{3}u^{\frac{3}{2}}\Bigg|_{9}^{25} = \frac{1}{48}(125-27) = \frac{61}{27}$$

Find the Length of the Curve x=t3,y=2t2,0≤t≤1x = t ^ { 3 } , y = 2 t ^ { 2 } , 0 \leq t \leq 1x=t3,y=2t2,0≤t≤1

Find the Length of the Curve $x = t ^ { 3 } , y = 2 t ^ { 2 } , 0 \leq t \leq 1$