Question

The length of one arch of the cycloid $x=a(t-\sin t),y=a(1-\cos t)$

• A. $2a$
• B. $4a$
• C. $6a$
• D. $8a$

Answer 1

Answer: (D)

$8a$

As a point moves from one end $O$ to the other end of its first arch, the parameter $t$ increases from $0$ to $2\pi$
Also $\frac{dx}{dt}=a(1-\cos t),\frac{dy}{dt}=a\sin t$
$\therefore$ Length of an arch$={\int }_0^{2\pi }\sqrt{[{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2]}dx$
$={\int }_0^{2\pi }\sqrt{{\left[a(1-\cos t)\right]}^2+{\left(a\sin t\right)}^2}dx$
$=a{\int }_0^{2\pi }\sqrt{1+{\cos }^{2}t-2\cos t+{\sin }^{2}t}dx$
$=a{\int }_0^{2\pi }\sqrt{1+({\cos }^{2}t+{\sin }^{2}t-2\cos t)}dx$
$=a{\int }_0^{2\pi }\sqrt{2-2\cos t}dx$
$=a{\int }_0^{2\pi }\sqrt{2(1-\cos t)}$
$=2a{\int }_0^{2\pi }\sin \left(\frac{t}{2}\right)dt$
$=2a{|-\frac{\cos \frac{t}{2}}{\frac{1}{2}}|}_0^{2\pi }$
$=4a[(-\cos \pi )-(-\cos 0)]$
$=4a[-(-1)-(-1)]$
$=4a(1+1)$
$=8a$