Length of a curve

**spaz84** · August 28th 2011, 04:12 AM

Question is: Find the length of the curve x = t-sint, y = 1-cost for 0<=t<=2pi

I've started the question using the formula $\int_{0}^{2\pi}\sqrt{{(\frac{dx}{dt}})^2 + (\frac{dy}{dt})^2}} dt$

differentiated and subbed back into the formula: $\int_{0}^{2\pi}\sqrt{{(1-cost)^2 + (sint)^2}} dt$

expanded simplified using trig identity: $\int_{0}^{2\pi}\sqrt{{2-2cost}} dt$

did some u substitution. u = 2-2cost du/dt=2sint du=2sint dt

$\int_{0}^{2\pi}\sqrt{{u}} dt$

this is where im getting stuck. I'm trying to change to dt to a du and i get:

$\int_{0}^{2\pi}\frac{\sqrt{u}}{sint} du$

looking at wolframalpha I should be getting
integrate sqrt(2-2cost) dt - Wolfram|Alpha

$\int_{0}^{2\pi}- \frac{1}{\sqrt{4-u}} du$

can anyone help me see how they completed this step please?
Interestingly the final solution to the question is zero(i think), is there any insight that would mean you wouldnt need to complete the integration but come to the solution earlier.. ?

again, any help would be great(man latex code takes forever to write your first time) thanks

**alexmahone** · August 28th 2011, 04:32 AM

$\int_{0}^{2\pi}\sqrt{2-2cost} dt=\int_{0}^{2\pi}\sqrt{2(1-cost)}=\int_{0}^{2\pi}\sqrt{2*2sin^2(t/2)}=\int_{0}^{2\pi}2sin(t/2)$

**Plato** · August 28th 2011, 04:34 AM

Originally Posted by spaz84

Question is: Find the length of the curve x = t-sint, y = 1-cost for 0<=t<=2pi
I've started the question using the formula $\int_{0}^{2\pi}\sqrt{{(\frac{dx}{dt}})^2 + (\frac{dy}{dt})^2}} dt$
differentiated and subbed back into the formula: $\int_{0}^{2\pi}\sqrt{{(1-cost)^2 + (sint)^2}} dt$
expanded simplified using trig identity: $\int_{0}^{2\pi}\sqrt{{2-2cost}} dt$

I would write it as $\int_0^{2\pi } {\sqrt {2 - 2\cos (t)} dt} = \sqrt 2 \int_0^{2\pi }{\sqrt {1 - \cos (t)} dt}$
Then look this Be sure to click "Show Steps".

**skeeter** · August 28th 2011, 04:40 AM

Originally Posted by alexmahone

$\int_{0}^{2\pi}\sqrt{2-2cost} dt=\int_{0}^{2\pi}\sqrt{2(1-cost)}=\int_{0}^{2\pi}\sqrt{2*2sin^2t}=\int_{0}^{2 \pi}2sint$

correction ...

$2(1 - \cos{t}) = \frac{4(1 - \cos{t})}{2} = 4\sin^2\left(\frac{t}{2}\right)$

$\int_0^{2\pi} 2\sin\left(\frac{t}{2}\right) \, dt$

**spaz84** · August 28th 2011, 04:46 AM

Originally Posted by alexmahone

$\int_{0}^{2\pi}\sqrt{2-2cost} dt=\int_{0}^{2\pi}\sqrt{2(1-cost)}=\int_{0}^{2\pi}\sqrt{2*2sin^2(t/2)}=\int_{0}^{2\pi}2sin(t/2)$

Outstanding, thank u. It looks like u are using a trig identity I am not familiar with, would u mind clarifying how u went from step 2 to 3

**alexmahone** · August 28th 2011, 04:50 AM

Originally Posted by spaz84

Outstanding, thank u. It looks like u are using a trig identity I am not familiar with, would u mind clarifying how u went from step 2 to 3

$cost=1-2sin^2(t/2)$

$1-cost=2sin^2(t/2)$

**mr fantastic** · August 28th 2011, 04:51 AM

Originally Posted by spaz84

Outstanding, thank u. It looks like u are using a trig identity I am not familiar with, would u mind clarifying how u went from step 2 to 3

You're expected to recognise that 1 - cos(t) = 2sin^2(t/2) follows from the double angle formula cos(2A) = 1 - 2 sin^2(A). Let A = t/2 ....

**spaz84** · August 28th 2011, 02:33 PM

Originally Posted by mr fantastic

You're expected to recognise that 1 - cos(t) = 2sin^2(t/2) follows from the double angle formula cos(2A) = 1 - 2 sin^2(A). Let A = t/2 ....

brilliant, thank you to everyone for their time and willingness to help...

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