Question

If $x=a(t+\sin t),y=a(1-\cos t)$ if $\frac{dx}{dy}=\cot p$ then $p=?$

• A. $t$
• B. $2t$
• C. $\frac{t}{2}$
• D. $-t^2$

Answer 1

The correct option is C $\frac{t}{2}$

Given

$x=a(t+\sin t)⟹\frac{dx}{dt}=a(1+\cos t)=2a{\cos }^2\frac{t}{2}$

$y=a(1-\cos t)⟹\frac{dy}{dt}=a\sin t=2a\sin \frac{t}{2}\cos \frac{t}{2}$

$\frac{dx}{dy}=\frac{\frac{dx}{dt}}{\frac{dy}{dt}}=\frac{2a{\cos }^2\frac{t}{2}}{2a\sin \frac{t}{2}\cos \frac{t}{2}}=\cot \frac{t}{2}=\cot p$

$⟹p=\frac{t}{2}$