Question

# Find $\dfrac{dy}{dx}$ , if $y=12\left( 1-\cos t \right),x=10\left( t-\sin t \right),-\dfrac{\pi }{2} < t < \dfrac{\pi }{2}$ .

Hint: In order to solve this problem, we need to know the chain rule. The chain rule is given by $\dfrac{dy}{dx}=\dfrac{dy}{dt}\times \dfrac{dt}{dx}$ . Also, in order to simplify the equation, we need to know some formulas. They are given by $\sin 2x=2\sin x.\cos x$, $1-\cos 2x=2{{\sin }^{2}}x$ and $\cot x=\dfrac{\cos x}{\sin x}$ .

As we can see that the $y$ is the function of $t$ . and $x$ is also the function of $t$ .
Hence, we cannot find the value $\dfrac{dy}{dx}$ directly by differentiating $y$ .
To solve this, we need to use the chain rule.
The chain rule says that
$\dfrac{dy}{dx}=\dfrac{dy}{dt}\times \dfrac{dt}{dx}...........(i)$
Therefore, we now need to find the values of $\dfrac{dy}{dt}$ and $\dfrac{dt}{dx}$ separately.
Differentiating $y=12\left( 1-\cos t \right)$ we get,
$\dfrac{dy}{dt}=\dfrac{d}{dt}\left( 12\left( 1-\cos t \right) \right)$
Solving this further we get,
\begin{align} & \dfrac{dy}{dt}=\dfrac{d}{dt}\left( 12-12\cos t \right) \\ & =\left( 0-12\left( -\sin t \right) \right) \\ & =12\sin t..........................(ii) \end{align}
Similarly differentiating $x=10\left( t-\sin t \right)$ , we get,
\begin{align} & \dfrac{dx}{dt}=\dfrac{d}{dt}\left( 10\left( t-\sin t \right) \right) \\ & \\ \end{align}
Solving this further we get,
\begin{align} & \dfrac{dx}{dt}=\dfrac{d}{dt}\left( 10\left( t-\sin t \right) \right) \\ & =\dfrac{d}{dt}\left( 10t-10\sin t \right) \\ & =10-10\cos t.....................(iii) \end{align}
We need to find the value of $\dfrac{dt}{dx}$ , taking the inverse of equation (iii) we get,
$\dfrac{dt}{dx}=\dfrac{1}{10-10\cos t}............(iv)$
Substituting the values of equation (ii) and (iv) in equation (i) we get,
$\dfrac{dy}{dx}=12\sin t\times \dfrac{1}{10-10\cos t}$
Solving this we get,
\begin{align} & \dfrac{dy}{dx}=\dfrac{12\sin t}{10\left( 1-\cos t \right)} \\ & =\dfrac{6\sin t}{5\left( 1-\cos t \right)} \end{align}
We can use the formulas $\sin 2x=2\sin x.\cos x$ and $1-\cos 2x=2{{\sin }^{2}}x$ .
Using the formula, we get,
$\dfrac{dy}{dx}=\dfrac{6\left( 2\sin \dfrac{t}{2}.\cos \dfrac{t}{2} \right)}{5\left( 2{{\sin }^{2}}\dfrac{t}{2} \right)}$
Solving this further we get,
\begin{align} & \dfrac{dy}{dx}=\dfrac{6\left( \cos \dfrac{t}{2} \right)}{5\left( \sin \dfrac{t}{2} \right)} \\ & =\dfrac{6}{5}\cot \dfrac{t}{2} \end{align}

Hence, the value of $\dfrac{dy}{dx}=\dfrac{6}{5}\cot \dfrac{t}{2}$.

Note: We can also solve this dividing equation (ii) by (iii). The answer will remain the same. We can also solve this by another approach. We can solve this by finding the value of $t$ in terms of $x$ and substituting in the equation of $y$ . Our aim will be to eliminate the value of $t$ . But we need to be careful as this method can turn out to be extremely complicated.