# Thread: Find the length of curve

1. ## Find the length of curve

The equations x = t + sin t, y = 1 – cos t, 0≤t≤2pi, define a curve C parametrically from the origin O to A(2pi, 0), which is rotated completely about Ox to form a surface S.

Find the length of the curve from O to A.

As I did so far,

dx/dt = 1 + cos t and dy/dt = sin t

Curve length OA = $\int_{0}^{2\pi} [(1+cos t)^2 + sin^2t]^ \frac {1}{2}dt$
$=\int_{0}^{2\pi} (2+2cos t)^\frac{1}{2}dt$
$=\int_{0}^{2\pi}2 cos \frac{t}{2}dt$

My answer is 0. Where I did wrong?

2. Originally Posted by geton
The equations x = t + sin t, y = 1 – cos t, 0≤t≤2pi, define a curve C parametrically from the origin O to A(2pi, 0), which is rotated completely about Ox to form a surface S.

Find the length of the curve from O to A.

As I did so far,

dx/dt = 1 + cos t and dy/dt = sin t

Curve length OA = $\int_{0}^{2\pi} [(1+cos t)^2 + sin^2t]^ \frac {1}{2}dt$
$=\int_{0}^{2\pi} (2+2cos t)^\frac{1}{2}dt$
$=\int_{0}^{2\pi}2 cos \frac{t}{2}dt$

My answer is 0. Where I did wrong?

$AL=\int_a^{b}\sqrt{\left(\frac{dx}{dt}\right)^2+\l eft(\frac{dy}{dt}\right)^2}dt$

So seeing thta

$\frac{dx}{dt}=1+\cos(t)$

and

$\frac{dy}{dt}=\sin(t)$

We have

$\int_0^{2\pi}\sqrt{1+\cos^2(t)+2\cos(t)+
\sin^2(t)}dt=\sqrt{2}\int_0^{2\pi}\sqrt{1+1\cos(t) }dt$

Which is nasty...but

I just realized

$\cos\left(\frac{x}{2}\right)=\pm\sqrt{\frac{1+\cos (x)}{2}}$

So we see that

$\sqrt{2}\cos\left(\frac{x}{2}\right)=\sqrt{1+\cos( x)}$

after your sub we get 8

how did you get 0?

3. Originally Posted by geton
The equations x = t + sin t, y = 1 – cos t, 0≤t≤2pi, define a curve C parametrically from the origin O to A(2pi, 0), which is rotated completely about Ox to form a surface S.

Find the length of the curve from O to A.

As I did so far,

dx/dt = 1 + cos t and dy/dt = sin t

Curve length OA = $\int_{0}^{2\pi} [(1+cos t)^2 + sin^2t]^ \frac {1}{2}dt$
$=\int_{0}^{2\pi} (2+2cos t)^\frac{1}{2}dt$
$=\int_{0}^{2\pi}2 cos \frac{t}{2}dt$

My answer is 0. Where I did wrong?

Length is positive. So the length is the magnitude of the area enclosed by $2 \cos \left( \frac{t}{2} \right)$ and the t-axis between t = 0 and t = 2 pi. How would you find that area ......? You'd draw a graph of $2 \cos \left( \frac{t}{2} \right)$ versus t and break it up, right?:

$\int_{0}^{\pi}2 \cos \left( \frac{t}{2}\right) \, dt {\color{red}-} \int_{\pi}^{2 \pi}2 \cos \left( \frac{t}{2}\right) \, dt$ since the signed area is negative from pi to 2pi ......

Capisce?

4. Originally Posted by geton
Curve length OA = $\int_{0}^{2\pi} [(1+cos t)^2 + sin^2t]^ \frac {1}{2}dt$
$=\int_{0}^{2\pi} (2+2cos t)^\frac{1}{2}dt$
$=\int_{0}^{2\pi}2 cos \frac{t}{2}dt$
$= 2*2 [sin \frac{t}{2}]_{0}^{2\pi}$
$=4(sin\pi - sin 0)$

Here is my problem. $sin\pi = 0$
Oh! We both made a stupid mistake

There is a sign change! With the plus or minus, it isn't plus over 0 to 2pi, it changes signs at pi

EDIT: Mr. F beat me to it

5. Originally Posted by mr fantastic

$\int_{0}^{\pi}2 \cos \left( \frac{t}{2}\right) \, dt {\color{red}-} \int_{\pi}^{\pi}2 \cos \left( \frac{t}{2}\right) \, dt$ since the signed area is negative from pi to 2pi ......
But isn't

$

\int_{0}^{\pi}2 \cos \left( \frac{t}{2}\right) \, dt - \int_{\pi}^{2\pi}2 \cos \left( \frac{t}{2}\right) \, dt
$
?

6. Originally Posted by geton
But isn't

$

\int_{0}^{\pi}2 \cos \left( \frac{t}{2}\right) \, dt - \int_{\pi}^{2\pi}2 \cos \left( \frac{t}{2}\right) \, dt
$
?
Clearly I made an obvious typo. Fixed.