Arc length of cycloid

[rir0302]

I need to find the arc length of one arch of the cycloid ( Rt - Rsin(t), R -Rcos(t) ) with R = 4.
I've been looking at this example for reference.

But how do I find the "one arch" which I assume is the bounds for the integral in the formula?

 

[Deleted member 4993]

I need to find the arc length of one arch of the cycloid ( Rt - Rsin(t), R -Rcos(t) ) with R = 4.
I've been looking at this example for reference.
View attachment 15031
But how do I find the "one arch" which I assume is the bounds for the integral in the formula?

Can you plot the function (use WA)? How far does one arch extend?

 

[MarkFL]

I think for one arch you could use a lower limit of \(\alpha\) and an upper limit of \(\alpha+2\pi\). I would choose to let \(\alpha=0\).

 

[rir0302]

Simplifying the integrand I got:

4√(2sint-2cost+3)

But I don't know how to continue solving this. I tried plugging it into an online calculator to check, but there was an error. Did I do something wrong?

 

[Deleted member 4993]

Simplifying the integrand I got:

4√(2sint-2cost+3)

But I don't know how to continue solving this. I tried plugging it into an online calculator to check, but there was an error. Did I do something wrong?

Do not know!

You have not shown us the steps to "your" answer.

 

[rir0302]

√[(4-4cost)^2+(4+4sint)^2]
4√[(cost-1)^2+(sint+1)^2]
4√[(cost)^2+(sint)^2+2sint-2cost+2]
(cost)^2+(sint)^2 = 1 so
4√(2sint-2cost+3)

 

[Dr.Peterson]

√[(4-4cost)^2+(4+4sint)^2]
4√[(cost-1)^2+(sint+1)^2]
4√[(cost)^2+(sint)^2+2sint-2cost+2]
(cost)^2+(sint)^2 = 1 so
4√(2sint-2cost+3)

This is wrong at the start, as dy/dt is not 4 + 4sin(t), but just 4sin(t)!