Length curve formed by one arch of the cycloid: x=4(t-sin(t)), y=4(1-cos(t))

[FritoTaco]

Hi, I'm looking at this problem from my book and can't understand how to get to step 4 , did the use the \(\displaystyle \sin^2\theta+\cos^2\theta=1\) identity to replace the \(\displaystyle \cos^2t+16\sin^2t\)?

Question: Find the length of the curve formed by one arch of the cycloid: x=4(t-sin(t)), y=4(1-cos(t)).

Formula: \(\displaystyle ds=\sqrt{\left(\dfrac{dx}{dy} \right)^2+\left(\dfrac{dy}{dt} \right)^2}dt\)

1. \(\displaystyle \dfrac{dx}{dt}=4(1-\cos t)\), \(\displaystyle \dfrac{dy}{dt}4\sin t\)

2. \(\displaystyle ds=\sqrt{\left[4(1-\cos t) \right]^2+\left(4\sin t \right)^2}dt\)

3. \(\displaystyle ds=\sqrt{\left[16(1-2\cos t+\cos ^2 t)\right]+\left(16\sin^2t \right)}dt\)

4. \(\displaystyle ds=\sqrt{16-32\cos t +16} dt\)

5. \(\displaystyle ds=\sqrt{32-32\cos t}dt\)

Evaluted using integration afterwards..

 

[ksdhart2]

Okay, so in step 3 they have:

\(\displaystyle ds=\sqrt{\left[16(1-2\cos t+\cos ^2 t)\right]+\left(16\sin^2t \right)}dt\)

Expanding everything out gives:

\(\displaystyle ds=\sqrt{16 - 32 \cos t + 16 \cos ^2 t + 16 \sin^2t }dt\)

And then, yes, the identity of \(\displaystyle sin^2(t) + cos^2(t) = 1\) was used to show that \(\displaystyle 16sin^2(t) + 16cos^2(t) = 16[sin^2(t)+cos^2(t)] = 16\), and the rest falls into place.

 

[FritoTaco]

Okay, so in step 3 they have:

\(\displaystyle ds=\sqrt{\left[16(1-2\cos t+\cos ^2 t)\right]+\left(16\sin^2t \right)}dt\)

Expanding everything out gives:

\(\displaystyle ds=\sqrt{16 - 32 \cos t + 16 \cos ^2 t + 16 \sin^2t }dt\)

And then, yes, the identity of \(\displaystyle sin^2(t) + cos^2(t) = 1\) was used to show that \(\displaystyle 16sin^2(t) + 16cos^2(t) = 16[sin^2(t)+cos^2(t)] = 16\), and the rest falls into place.

Oh, you need to factor the 16 out. Okay thank you, I was lost for a minute.