Question

If $x=a\left(t-\sin t\right)$ and $y=a\left(1-\cos t\right)$, then $\frac{dy}{dx}=$

• A. $\tan \left(\frac{t}{2}\right)$
• B. $-\tan \left(\frac{t}{2}\right)$
• C. $\cot \left(\frac{t}{2}\right)$
• D. $-\cot \left(\frac{t}{2}\right)$

Answer 1

The correct option is C

$\cot \left(\frac{t}{2}\right)$

Explanation for the correct option.

Step 1. Find the value of $\frac{dx}{dt}$ and $\frac{dy}{dt}$.

Differentiate $x=a\left(t-\sin t\right)$ with respect to $t$.

$\begin{array}{rcl}\frac{dx}{dt}& =& \frac{d}{dt}\left(a\left(t-\sin t\right)\right)\\ & =& a\left(1-\cos t\right)\end{array}$

Differentiate $y=a\left(1-\cos t\right)$ with respect to $t$.

$\begin{array}{rcl}\frac{dy}{dt}& =& \frac{d}{dt}\left(a\left(1-\cos t\right)\right)\\ & =& a\left(0-\left(-\sin t\right)\right)\\ & =& a\sin t\end{array}$

Step 2. Find the value of $\frac{dy}{dx}$.

The term $\frac{dy}{dx}$ is given as: $\frac{dy}{dx}=\frac{dy}{dt}·\frac{dt}{dx}$.

Substitute the found values:

$\begin{array}{rcl}\frac{dy}{dx}& =& a\sin t·\frac{1}{a\left(1-\cos t\right)}\\ & =& \frac{\sin t}{1-\cos t}\\ & =& \frac{\sin \left(2\times {\displaystyle \frac{t}{2}}\right)}{1-\cos \left(2\times \frac{t}{2}\right)}\\ & =& \frac{2\sin {\displaystyle \frac{t}{2}}\cos {\displaystyle \frac{t}{2}}}{2{\sin }^2{\displaystyle \frac{t}{2}}}\left[\because sin2A=2sinAcosA;1-cos2A=2si{n}^{2}A\right]\\ & =& \frac{\cos {\displaystyle \frac{t}{2}}}{\sin {\displaystyle \frac{t}{2}}}\\ & =& \cot \left(\frac{t}{2}\right)\end{array}$

So, the value of $\frac{dy}{dx}$ is $\cot \left(\frac{t}{2}\right)$.

Hence, the correct option is C.