# Thread: Stuck on Parametric Arc Length

1. ## Stuck on Parametric Arc Length

Hey I am not sure how to finish this problem...

$\displaystyle x= e^t +e^-t$
$\displaystyle y=5-2t$
$\displaystyle 0\leq{t}\leq{5}$

$\displaystyle \frac{dx}{dt}=e^t-e^{-t}$
$\displaystyle \frac{dy}{dt}=-2$
$\displaystyle (\frac{dx}{dt})^2+(\frac{dy}{dt})^2= e^{2t}-e^{-2t}+4$
$\displaystyle A= \int_0^5 \! \sqrt{(e^{2t}-e^{-2t}+4)} \, dt$

Do I have to complete the square to simplify?
Thanks

2. $\displaystyle (e^{t}-e^{-t})^2$ does not simplify to $\displaystyle e^{2t}-e^{-2t}$

3. Originally Posted by mmattson07
Hey I am not sure how to finish this problem...

$\displaystyle x= e^t +e^-t$
$\displaystyle y=5-2t$
$\displaystyle 0\leq{t}\leq{5}$

$\displaystyle \frac{dx}{dt}=e^t-e^{-t}$
$\displaystyle \frac{dy}{dt}=-2$
$\displaystyle (\frac{dx}{dt})^2+(\frac{dy}{dt})^2= e^{2t}{\color{red}+e^{-2t}+2}$
... which factorises, to square factors that cancel the square root... and then you can substitute u = e^t

4. Originally Posted by mmattson07
Hey I am not sure how to finish this problem...

$\displaystyle x= e^t +e^-t$
$\displaystyle y=5-2t$
$\displaystyle 0\leq{t}\leq{5}$

$\displaystyle \frac{dx}{dt}=e^t-e^{-t}$
$\displaystyle \frac{dy}{dt}=-2$
$\displaystyle (\frac{dx}{dt})^2+(\frac{dy}{dt})^2= e^{2t}$$\displaystyle -2e^{t}e^{-t}$$\displaystyle +e^{-2t}+4$
$\displaystyle A= \int_0^5 \! \sqrt{(e^{2t}-e^{-2t}+4)} \, dt$

Do I have to complete the square to simplify?
Thanks
Something times its reciprocal is simply 1.

5. So $\displaystyle (\frac{dx}{dt})^2+(\frac{dy}{dt})^2=e^{2t}-2e^te^{-t}+e^{-2t}+4$

and

$\displaystyle A= \int_0^5 \! \sqrt{(e^{2t}+e^{-2t}+2)} \, dt$

I still am unsure how this simplifies...

6. Originally Posted by mmattson07
So $\displaystyle (\frac{dx}{dt})^2+(\frac{dy}{dt})^2=e^{2t}-2e^te^{-t}+e^{-2t}+4$

and

$\displaystyle A= \int_0^5 \! \sqrt{(e^{2t}+e^{-2t}+2)} \, dt$

I still am unsure how this simplifies...
$\displaystyle e^{2t} + e^{-2t} + 2 = (e^t + e^{-t})^2$.

7. Thank you...don't know how I didn't see that.