# Thread: Find the Length

1. ## Find the Length

$\displaystyle x=\frac{(2t+3)^\frac{3}{2}}{3}$, $\displaystyle y=t+\frac{t^2}{2}$, $\displaystyle 0\leq t\leq 3$ They want me to find the Length

right now ive gotten to $\displaystyle L=\int^3_0 \sqrt{t^2+5t+4} dt$ I dont know if this is even right, and I dont know how to take this integral either :-(

2. Originally Posted by jobebob
$\displaystyle x=\frac{(2t+3)^\frac{3}{2}}{3}$, $\displaystyle y=t+\frac{t^2}{2}$, $\displaystyle 0\leq t\leq 3$ They want me to find the Length

right now ive gotten to $\displaystyle L=\int^3_0 \sqrt{t^2+5t+4} dt$ I dont know if this is even right, and I dont know how to take this integral either :-(
You need:

$\displaystyle L=\int_0^3 \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy} {dt}\right)^2}~dt$

$\displaystyle \frac{dx}{dt}=\frac{1}{2}\sqrt{2t+3}$

$\displaystyle \frac{dy}{dt}=1+t$

$\displaystyle L=\frac{1}{2}~\int_0^3 \sqrt{4t^2+10t+7}~dt$

RonL

3. so $\displaystyle L=\frac{1}{2}(\frac{1}{\sqrt{4}}sinh^{-1} \frac{8t+10}{\sqrt{10}})$ from t=0 to t=3?

4. Originally Posted by jobebob
so $\displaystyle L=\frac{1}{2}(\frac{1}{\sqrt{4}}sinh^{-1} \frac{8t+10}{\sqrt{10}})$ from t=0 to t=3?
Not sure what you mean?

RonL

5. Originally Posted by CaptainBlack
Not sure what you mean?

RonL
Yea im not really sure either, but wouldn't that be the integral of $\displaystyle \sqrt{4t^2+10t+7}$