Question

If $x = a\left( {t - \sin t} \right)$ and $y = a\left( {1 - \cos t} \right)$, then $\dfrac{{dy}}{{dx}}$ equals:
(A) $\tan \left( {\dfrac{t}{2}} \right)$
(B) $ - \tan \left( {\dfrac{t}{2}} \right)$
(C) $\cot \left( {\dfrac{t}{2}} \right)$
(D) $ - \cot \left( {\dfrac{t}{2}} \right)$

Answer 1

Hint: In the given problem, we are required to differentiate a parametric function with respect to x.
For differentiating parametric functions, we first find derivatives of x and y with respect to p separately and then divide both of them to find the differential $\dfrac{{dy}}{{dx}}$. Also, we will use the chain rule of differentiation in order to solve the problem. We must remember the derivatives of sine and cosine functions to get to the final answer.

Complete step-by-step answer:
Now, we have, $x = a\left( {t - \sin t} \right)$ and $y = a\left( {1 - \cos t} \right)$.
So, to find the derivative of the parametric function, we first differentiate x and y separately with respect to the parameter, t.
Hence, differentiating x with respect to p, we get,
$ \Rightarrow \dfrac{{dx}}{{dt}} = \dfrac{d}{{dx}}\left[ {a\left( {t - \sin t} \right)} \right]$
Taking constant out of differentiation, we get,
$ \Rightarrow \dfrac{{dx}}{{dt}} = a\dfrac{d}{{dx}}\left( {t - \sin t} \right)$
We know that the derivative of the sine function $\sin x$ with respect to x is $\cos x$. Also, we know the power rule of differentiation \[\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}\].So, we get,
$ \Rightarrow \dfrac{{dx}}{{dt}} = a\left( {1 - \cos t} \right) - - - - \left( 1 \right)$
Now, we differentiate y with respect to t. So, we get,
\[ \Rightarrow \dfrac{{dy}}{{dt}} = \dfrac{d}{{dt}}\left[ {a\left( {1 - \cos t} \right)} \right]\]
Taking constant out of differentiation, we get,
\[ \Rightarrow \dfrac{{dy}}{{dt}} = a\dfrac{d}{{dt}}\left( {1 - \cos t} \right)\]
We know that the derivative of a constant function with respect to x is zero. Also, the derivative of $\cos x$ with respect to x is $ - \sin x$. So, we get,
\[ \Rightarrow \dfrac{{dy}}{{dt}} = a\sin t - - - - \left( 2 \right)\]
Now, we divide the equation $\left( 2 \right)$ by the equation $\left( 1 \right)$. So, we get,
\[ \Rightarrow \dfrac{{\left( {\dfrac{{dy}}{{dt}}} \right)}}{{\left( {\dfrac{{dx}}{{dt}}} \right)}} = \dfrac{{a\left( {1 - \cos t} \right)}}{{a\sin t}}\]
Cancelling the common factors in numerator and denominator, we get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\left( {1 - \cos t} \right)}}{{\sin t}}\]
Now, we have to simplify the expression using the trigonometric formula $\cos 2x = 1 - 2{\sin ^2}x$ and $\sin 2x = 2\sin x\cos x$. So, we get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\left( {1 - \left( {1 - 2{{\sin }^2}\dfrac{t}{2}} \right)} \right)}}{{2\sin \dfrac{t}{2}\cos \dfrac{t}{2}}}\]
Opening the brackets, we get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{1 - 1 + 2{{\sin }^2}\dfrac{t}{2}}}{{2\sin \dfrac{t}{2}\cos \dfrac{t}{2}}}\]
Cancelling like terms with opposite signs,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{2{{\sin }^2}\dfrac{t}{2}}}{{2\sin \dfrac{t}{2}\cos \dfrac{t}{2}}}\]
Cancelling the common factors in numerator and denominator, we get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\sin \dfrac{t}{2}}}{{\cos \dfrac{t}{2}}}\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \tan \left( {\dfrac{t}{2}} \right)\]
Hence, the correct answer is the option (A).
So, the correct answer is “Option A”.

Note: We must remember this method to find the derivatives of the parametric function. We must have a good grip over the trigonometric formulae such as $\tan x = \dfrac{{\sin x}}{{\cos x}}$ and the double angle formula for cosine in order to get to the final answer. Take care of the calculations.