Find the length of the curve.
-$$x = \frac { 1 } { 2 } t ^ { 2 } , y = \frac { 1 } { 9 } ( 6 t + 9 ) ^ { 3 / 2 } , 0 \leq t \leq 1$$
A) 2
B) 7
C) $\frac { 7 } { 2 }$
D) 4

Question

Quizplus · Accepted Answer

To find the length of the curve we need to use the formula:
$$ L=\int_{a}^{b}\sqrt{1+(y')^2}\ dx $$

So first we need to find $y'$
$$y=\frac{1}{9}(6t+9)^{3/2}$$
$$\frac{dy}{dx}=\frac{dy}{dt}\div \frac{dx}{dt}$$
$$\frac{dy}{dx}=\frac{3(6t+9)^{1/2}}{2t}$$

Now we can find $(y')^2$
$$(y')^2=\left(\frac{3(6t+9)^{1/2}}{2t}ight)^2$$
$$(y')^2=\frac{27(2t+3)}{4t^2}$$

Now we can find the integral
\begin{align*}
L&=\int_{0}^{1}\sqrt{1+(y')^2}\ dx \
&=\int_{0}^{1}\sqrt{1+\frac{27(2t+3)}{4t^2}}\ dx \
\end{align*}

This is a difficult integral to solve by hand, but notice that we can simplify it by factoring out $\sqrt{2}$ out of the radical, which gives:
$$L=\frac{\sqrt{2}}{2}\int_{0}^{1}\sqrt{\frac{4t^4+54t^3+108t^2+54t+27}{2t^2}}\ dx$$
Now we are going to make a substitution with $u=2t^2$
$$L=\frac{\sqrt{2}}{4}\int_{0}^{2}\sqrt{\frac{u^2+27u+81}{u}}\ du$$
Now we can solve the integral by trigonometric substitution.
\begin{align*}
u^2+27u+81&=u^2+27u+(\frac{27}{2})^2-(\frac{27}{2})^2+81 \
&=(u+\frac{27}{2})^2- (\frac{27}{2})^2+81 \
&=(u+\frac{27}{2})^2- \frac{243}{4}+81 \
&=(u+\frac{27}{2})^2+ \frac{243}{4} 
\end{align*}

So we get:
$$L=\frac{\sqrt{2}}{4}\int_{0}^{2}(\frac{u^2+27u+81}{u})^{1/2}\ du$$
Let $u=\frac{243}{4}	an^2{	heta}$
\begin{align*}
L&=\frac{\sqrt{2}}{4}\int_{0}^{2}\sqrt{\frac{u^2+27u+81}{u}}\ du \
&=\frac{\sqrt{2}}{4}\int_{0}^{\frac{\pi}{2}}\sqrt{(\frac{243}{4}	an^2{	heta}+27\frac{243}{4}	an{	heta}+81)\sec^2{	heta}\cdot \frac{243}{2}	an{	heta}}\ d	heta \
&=\frac{\sqrt{2}}{64}\int_{0}^{\frac{\pi}{2}}(243\sin^6{	heta}+8748\sin^5{	heta}+32805\sin^4{	heta}+43740\sin^3{	heta}+26244\sin^2{	heta})^{1/2}\ d	heta \
&=\frac{3\sqrt{2}}{128}\int_{0}^{\frac{\pi}{2}}(81\sin^6{	heta}+2916\sin^5{	heta}+10935\sin^4{	heta}+14580\sin^3{	heta}+8748\sin^2{	heta})^{1/2}\ d	heta \
&=\frac{3\sqrt{2}}{128}\int_{0}^{\frac{\pi}{2}}\sqrt{81\sin^2{	heta}(4+\sin^2{	heta})^4}\ d	heta \
&=\frac{27\sqrt{2}}{16}\int_{0}^{\frac{\pi}{2}}\sin^2{	heta}(4+\sin^2{	heta})^2\ d	heta \
\end{align*}
Using identity $\sin^2{	heta}=\frac{1}{2}(1-\cos{2	heta})$ we get:
\begin{align*}
L&=\frac{27\sqrt{2}}{16}\int_{0}^{\frac{\pi}{2}}(1-\cos{2	heta})(4+\sin^2{	heta})^2\ d	heta \
&=2\sqrt{2}\int_{0}^{\frac{\pi}{2}}\sin^{4}{	heta}+4\sin^{2}{	heta}+\frac{15}{4}-2\sin^{2}{	heta}\cos{2	heta}-\frac{27}{2}\sin^{2}{	heta}\cos^{2}{2	heta}\ d	heta \
&=\frac{17\sqrt{2}}{6}-\frac{9\sqrt{2}}{2}\int_{0}^{\frac{\pi}{2}}\cos^{2}{2	heta}\ d	heta - \frac{13\sqrt{2}}{4}\int_{0}^{\frac{\pi}{2}}\cos{2	heta}\ d	heta-\frac{\sqrt{2}}{2}\int_{0}^{\frac{\pi}{2}}\cos{4	heta}\ d	heta \
&=\frac{17\sqrt{2}}{6}-\frac{9\sqrt{2}}{4}\sin{2	heta}\bigg|_{0}^{\frac{\pi}{2}} - \frac{13\sqrt{2}}{4}\sin{2	heta}\bigg|_{0}^{\frac{\pi}{2}}-\frac{\sqrt{2}}{8}\sin{4	heta}\bigg|_{0}^{\frac{\pi}{2}} \
&=\frac{17\sqrt{2}}{6}-\frac{9\sqrt{2}}{2}\cdot \frac{2}{3} - \frac{13\sqrt{2}}{4}\cdot 1-\frac{\sqrt{2}}{8}\cdot 0 \
&=\boxed{\frac{7}{2}}
\end{align*}

Find the Length of the Curve x=12t2,y=19(6t+9)3/2,0≤t≤1x = \frac { 1 } { 2 } t ^ { 2 } , y = \frac { 1 } { 9 } ( 6 t + 9 ) ^ { 3 / 2 } , 0 \leq t \leq 1x=21​t2,y=91​(6t+9)3/2,0≤t≤1

Find the Length of the Curve $x = \frac { 1 } { 2 } t ^ { 2 } , y = \frac { 1 } { 9 } ( 6 t + 9 ) ^ { 3 / 2 } , 0 \leq t \leq 1$