Question

If $x=a\left( \cos \theta +\theta \sin \theta \right)$ and $y=a\left( \sin \theta -\theta \cos \theta \right)$ , then $\dfrac{dy}{dx}$ is equal to.
(a) $\cos \theta $
(b) $\tan \theta $
(c) $\sec \theta $
(d) $\csc \theta $

Answer 1

Hint:For solving this question we will use some standard results differentiation of $y=\sin x$ , $y=\cos x$ and then apply the product rule of differentiation to differentiate the given term with respect to $\theta $ and get the expression of $\dfrac{dy}{d\theta }\And \dfrac{dx}{d\theta }$ as a function of $\theta $ . After that, we will use the formula $\dfrac{dy}{dx}=\dfrac{dy}{d\theta }\times \dfrac{1}{\left( \dfrac{dx}{d\theta } \right)}$ and solve further to get the value of $\dfrac{dy}{dx}$ easily.

Complete step-by-step answer:
Given:
It is given that, if $x=a\left( \cos \theta +\theta \sin \theta \right)$ and $y=a\left( \sin \theta -\theta \cos \theta \right)$ , then we have to find the value of $\dfrac{dy}{dx}$ .
Now, before we proceed we should know the following formulas and concepts of trigonometry and differential calculus:
1. If $y=f\left( x \right)\cdot g\left( x \right)$ , then $\dfrac{dy}{dx}=\dfrac{d\left( f\left( x \right)\cdot g\left( x \right) \right)}{dx}={f}'\left( x \right)\cdot g\left( x \right)+f\left( x \right)\cdot {g}'\left( x \right)$ . This is also known as the product rule of differentiation.
2. If $y=\sin x$ , then $\dfrac{dy}{dx}=\dfrac{d\left( \sin x \right)}{dx}=\cos x$ .
3. If $y=\cos x$ , then $\dfrac{dy}{dx}=\dfrac{d\left( \cos x \right)}{dx}=-\sin x$ .
4. If $x=f\left( \theta \right)$ and $y=g\left( \theta \right)$ , then $\dfrac{dy}{dx}=\dfrac{dy}{d\theta }\times \dfrac{1}{\left( \dfrac{dx}{d\theta } \right)}=\dfrac{{g}'\left( \theta \right)}{{f}'\left( \theta \right)}$ . This is also called differentiation of a parametric function.
Now, we will use the above-mentioned formulas and concepts to differentiate $x=a\left( \cos \theta +\theta \sin \theta \right)$ and $y=a\left( \sin \theta -\theta \cos \theta \right)$ separately with respect to $\theta $ .
Calculation of $\dfrac{dy}{d\theta }$ :
Now, we will differentiate $y=a\left( \sin \theta -\theta \cos \theta \right)$ with respect to $\theta $ . Then,
$\begin{align}
  & y=a\left( \sin \theta -\theta \cos \theta \right) \\
 & \Rightarrow \dfrac{dy}{d\theta }=\dfrac{d\left( a\left( \sin \theta -\theta \cos \theta \right) \right)}{d\theta } \\
\end{align}$
Now, as $a$ is a constant term so, we can write $\dfrac{d\left( a\left( \sin \theta -\theta \cos \theta \right) \right)}{d\theta }=a\dfrac{d\left( \sin \theta -\theta \cos \theta \right)}{d\theta }$ in the above equation. Then,
$\begin{align}
  & \dfrac{dy}{d\theta }=\dfrac{d\left( a\left( \sin \theta -\theta \cos \theta \right) \right)}{d\theta } \\
 & \Rightarrow \dfrac{dy}{d\theta }=a\dfrac{d\left( \sin \theta -\theta \cos \theta \right)}{d\theta } \\
 & \Rightarrow \dfrac{dy}{d\theta }=a\left( \dfrac{d\left( \sin \theta \right)}{d\theta }-\dfrac{d\left( \theta \cos \theta \right)}{d\theta } \right) \\
\end{align}$
Now, we will use the formula $\dfrac{d\left( \sin x \right)}{dx}=\cos x$ to write $\dfrac{d\left( \sin \theta \right)}{d\theta }=\cos \theta $ and product rule of differential calculus to write $\dfrac{d\left( \theta \cos \theta \right)}{d\theta }=\cos \theta \dfrac{d\theta }{d\theta }+\theta \dfrac{d\left( \cos \theta \right)}{d\theta }$ in the above equation. Then,
$\begin{align}
  & \dfrac{dy}{d\theta }=a\left( \dfrac{d\left( \sin \theta \right)}{d\theta }-\dfrac{d\left( \theta \cos \theta \right)}{d\theta } \right) \\
 & \Rightarrow \dfrac{dy}{d\theta }=a\left( \cos \theta -\left( \cos \theta \dfrac{d\theta }{d\theta }+\theta \dfrac{d\left( \cos \theta \right)}{d\theta } \right) \right) \\
\end{align}$
Now, we will write $\dfrac{d\theta }{d\theta }=1$ and use the formula $\dfrac{d\left( \cos x \right)}{dx}=-\sin x$ to write $\dfrac{d\left( \cos \theta \right)}{d\theta }=-\sin \theta $ in the above equation. Then,
$\begin{align}
  & \dfrac{dy}{d\theta }=a\left( \cos \theta -\left( \cos \theta \dfrac{d\theta }{d\theta }+\theta \dfrac{d\left( \cos \theta \right)}{d\theta } \right) \right) \\
 & \Rightarrow \dfrac{dy}{d\theta }=a\left( \cos \theta -\left( \cos \theta -\theta \sin \theta \right) \right) \\
 & \Rightarrow \dfrac{dy}{d\theta }=a\left( \cos \theta -\cos \theta +\theta \sin \theta \right) \\
 & \Rightarrow \dfrac{dy}{d\theta }=a\theta \sin \theta ..............\left( 1 \right) \\
\end{align}$
Calculation of $\dfrac{dx}{d\theta }$ :
Now, we will differentiate $x=a\left( \cos \theta +\theta \sin \theta \right)$ with respect to $\theta $ . Then,
$\begin{align}
  & x=a\left( \cos \theta +\theta \sin \theta \right) \\
 & \Rightarrow \dfrac{dx}{d\theta }=\dfrac{d\left( a\left( \cos \theta +\theta \sin \theta \right) \right)}{d\theta } \\
\end{align}$
Now, as $a$ is a constant term so, we can write $\dfrac{d\left( a\left( \cos \theta +\theta \sin \theta \right) \right)}{d\theta }=a\dfrac{d\left( \cos \theta +\theta \sin \theta \right)}{d\theta }$ in the above equation. Then,
$\begin{align}
  & \dfrac{dx}{d\theta }=\dfrac{d\left( a\left( \cos \theta +\theta \sin \theta \right) \right)}{d\theta } \\
 & \Rightarrow \dfrac{dx}{d\theta }=a\dfrac{d\left( \cos \theta +\theta \sin \theta \right)}{d\theta } \\
 & \Rightarrow \dfrac{dx}{d\theta }=a\left( \dfrac{d\left( \cos \theta \right)}{d\theta }+\dfrac{d\left( \theta \sin \theta \right)}{d\theta } \right) \\
\end{align}$
Now, we will use the formula $\dfrac{d\left( \cos x \right)}{dx}=-\sin x$ to write \[\dfrac{d\left( \cos \theta \right)}{d\theta }=-\sin \theta \] and product rule of differential calculus to write $\dfrac{d\left( \theta \sin \theta \right)}{d\theta }=\sin \theta \dfrac{d\theta }{d\theta }+\theta \dfrac{d\left( \sin \theta \right)}{d\theta }$ in the above equation. Then,
$\begin{align}
  & \dfrac{dx}{d\theta }=a\left( \dfrac{d\left( \cos \theta \right)}{d\theta }+\dfrac{d\left( \theta \sin \theta \right)}{d\theta } \right) \\
 & \Rightarrow \dfrac{dx}{d\theta }=a\left( -\sin \theta +\left( \sin \theta \dfrac{d\theta }{d\theta }+\theta \dfrac{d\left( \sin \theta \right)}{d\theta } \right) \right) \\
\end{align}$
Now, we will write $\dfrac{d\theta }{d\theta }=1$ and use the formula $\dfrac{d\left( \sin x \right)}{dx}=\cos x$ to write $\dfrac{d\left( \sin \theta \right)}{d\theta }=\cos \theta $ in the above equation. Then,
$\begin{align}
  & \dfrac{dx}{d\theta }=a\left( -\sin \theta +\left( \sin \theta \dfrac{d\theta }{d\theta }+\theta \dfrac{d\left( \sin \theta \right)}{d\theta } \right) \right) \\
 & \Rightarrow \dfrac{dx}{d\theta }=a\left( -\sin \theta +\left( \sin \theta +\theta \cos \theta \right) \right) \\
 & \Rightarrow \dfrac{dx}{d\theta }=a\left( -\sin \theta +\sin \theta +\theta \cos \theta \right) \\
 & \Rightarrow \dfrac{dx}{d\theta }=a\theta \cos \theta ..............\left( 2 \right) \\
\end{align}$
Now, we will use the formula for the differentiation of a parametric function to write $\dfrac{dy}{dx}=\dfrac{dy}{d\theta }\times \dfrac{1}{\left( \dfrac{dx}{d\theta } \right)}$ . So, we will divide the equation (1) by equation (2). Then,
$\begin{align}
  & \dfrac{dy}{dx}=\dfrac{dy}{d\theta }\times \dfrac{1}{\left( \dfrac{dx}{d\theta } \right)} \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{a\theta \sin \theta }{a\theta \cos \theta } \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{\sin \theta }{\cos \theta } \\
\end{align}$
Now, as we know that, $\dfrac{\sin \theta }{\cos \theta }=\tan \theta $ . Then,
$\begin{align}
  & \dfrac{dy}{dx}=\dfrac{\sin \theta }{\cos \theta } \\
 & \Rightarrow \dfrac{dy}{dx}=\tan \theta \\
\end{align}$
Now, from the above result, we conclude that if $x=a\left( \cos \theta +\theta \sin \theta \right)$ and $y=a\left( \sin \theta -\theta \cos \theta \right)$ , then $\dfrac{dy}{dx}=\tan \theta $ .
Hence, (b) will be the correct option.

Note: Here, the student should first understand what is asked in the question and then proceed in the right direction to get the correct result quickly. And we should not try to eliminate $\theta $ from the given equations as in every option $\dfrac{dy}{dx}$ is a function of $\theta $ so, we should apply the formula for the differentiation for the parametric function accurately, and use every formula and product rule of differentiation to solve further. Moreover, we should avoid making calculation mistakes while solving to get the correct result.