Calculating the Length of a Parametric Curve with Integration

[phospho]

THe parametric equations of the curve C are:

x = a(t-sin(t)), y = a(1-cos(t))

where 0 <= t <= 2pi

Find, by using integration, the length of C.

$\dfrac{dx}{dt} = a (1-\cos t)$
$\dfrac{dy}{dt} = a\sin t$

$\left(\dfrac{dx}{dt}\right)^2 + \left(\dfrac{dy}{dt} \right)^2 = a^2 (1 - 2\cos t + cos^2t + sin^2 t) = 2a^2(1 - \cos t)$

length of curve = $S = \displaystyle\int_0^{2\pi} \sqrt{2a^2(1-\cos t)}\ dt = -2 \sqrt{2} a \left[\sqrt{1 + \cos t} \right]_0^{2\pi}$

evaluating this I get 0... any help?

 

[Dick]

THe parametric equations of the curve C are:

x = a(1-sin(t)), y = a(1-cos(t))

where 0 <= t <= 2pi

Find, by using integration, the length of C.

$\dfrac{dx}{dt} = a (1-\cos t)$
$\dfrac{dy}{dt} = a\sin t$

$\left(\dfrac{dx}{dt}\right)^2 + \left(\dfrac{dy}{dt} \right)^2 = a^2 (1 - 2\cos t + cos^2t + sin^2 t) = 2a^2(1 - \cos t)$

length of curve = $S = \displaystyle\int_0^{2\pi} \sqrt{2a^2(1-\cos t)}\ dt = -2 \sqrt{2} a \left[\sqrt{1 + \cos t} \right]_0^{2\pi}$

evaluating this I get 0... any help?

Looks to me like dx/dt is wrong. Check it.

 

[phospho]

Looks to me like dx/dt is wrong. Check it.

Sorry, was a mistype by me of what x is.

 

[Dick]

Sorry, was a mistype by me of what x is.

Ok, that's better. The answer to a question like this is usually to break the range 0 to 2pi up into smaller parts and add the differences. But how did you get that antiderivative? That doesn't look right at all.

 

[phospho]

Ok, that's better. The answer to a question like this is usually to break the range 0 to 2pi up into smaller parts and add the differences. But how did you get that antiderivative? That doesn't look right at all.

well I took out the constants 2a^2, that's how I got √2a

I then done: $\int \sqrt{1-\cos t}\ dt$ using the substitution of u = cos(t), getting $-2\sqrt{1+cos(t)}$

checking, the derivative of $-2\sqrt{1+cos(t)}$ is $\dfrac{sin(t)}{\sqrt{1+cos(t)}} = \dfrac{\sqrt{1-cos^2t}}{\sqrt{1+cos(t)}} = \sqrt{1-cos(t)}$

 

[Dick]

well I took out the constants 2a^2, that's how I got √2a

I then done: $\int \sqrt{1-\cos t}\ dt$ using the substitution of u = cos(t), getting $-2\sqrt{1+cos(t)}$

checking, the derivative of $-2\sqrt{1+cos(t)}$ is $\dfrac{sin(t)}{\sqrt{1+cos(t)}} = \dfrac{\sqrt{1-cos^2t}}{\sqrt{1+cos(t)}} = \sqrt{1-cos(t)}$

Oh, yeah, right. I'm glad you spelled that out. What's going on is that $\sqrt{1-cos^2t}$ isn't equal to $sin(t)$, it's equal to $|sin(t)|$. You'll need to change the sign of the antiderivative depending on the sign of $sin(t)$. Split the integral range up appropriately.

 

[phospho]

Oh, yeah, right. I'm glad you spelled that out. What's going on is that $\sqrt{1-cos^2t}$ isn't equal to $sin(t)$, it's equal to $|sin(t)|$. You'll need to change the sign of the antiderivative depending on the sign of $sin(t)$. Split the integral range up appropriately.

I'm not quite sure what you mean?:

$-2 \sqrt{2} a \left[\sqrt{1 + \cos t} \right]_0^{\pi} -2 \sqrt{2} a \left[\sqrt{1 + \cos t} \right]_{\pi}^{2\pi}$

Do you mean that? (I'm not sure of the answer, so can't check.)

 

[Dick]

I'm not quite sure what you mean?:

$-2 \sqrt{2} a \left[\sqrt{1 + \cos t} \right]_0^{\pi} -2 \sqrt{2} a \left[\sqrt{1 + \cos t} \right]_{\pi}^{2\pi}$

Do you mean that? (I'm not sure of the answer, so can't check.)

For pi to 2pi the sine is negative. So sqrt(1-cos(t)^2)=(-sin(t)). You need to add an extra minus to the antiderivative in that range.
$-2 \sqrt{2} a \left[\sqrt{1 + \cos t} \right]_0^{\pi} +2 \sqrt{2} a \left[\sqrt{1 + \cos t} \right]_{\pi}^{2\pi}$

 

[phospho]

For pi to 2pi the sine is negative. So sqrt(1-cos(t)^2)=(-sin(t)). You need to add an extra minus to the antiderivative in that range.
$-2 \sqrt{2} a \left[\sqrt{1 + \cos t} \right]_0^{\pi} +2 \sqrt{2} a \left[\sqrt{1 + \cos t} \right]_{\pi}^{2\pi}$

oh I see, thanks!

 

[Dick]

oh I see, thanks!

You're welcome. The way you were doing it before you were getting a negative contribution to arclength from [pi,2pi] cancelling the positive part [0,pi]. You know that can't happen. arclength is positive everywhere.

 

[phospho]

You're welcome. The way you were doing it before you were getting a negative contribution to arclength from [pi,2pi] cancelling the positive part [0,pi]. You know that can't happen. arclength is positive everywhere.

Sorry to bump this up but wolframalpha managed to get the integral of ∫√(1-cos(t))dt = -2cot(t/2)√(1-cos(t)) + C, could anyone explain which substitution they used or whatever? I've tried (and I am still trying) to figure out how they've done it.