Question

If x = a\left\\{ {\cos \theta + \log \tan \left( {\dfrac{\theta }{2}} \right)} \right\\} and $y = a\sin \theta$ , then $\dfrac{{dy}}{{dx}}$ is equal to
A) $\cot \theta$
B) $\tan \theta$
C) $\sin \theta$
D) $\cos \theta$

Answer 1

We can find the derivative of x with respect to $\theta$. It can be found by finding the derivative of the terms inside the bracket and by using chain rule. Then we can find the derivative of y with respect to $\theta$. Then we can divide them to get the required solution.

Complete step by step solution:
We have x = a\left\\{ {\cos \theta + \log \tan \left( {\dfrac{\theta }{2}} \right)} \right\\} and $y = a\sin \theta$
We can find the derivative of x with respect to $\theta$ .
\Rightarrow \dfrac{{dx}}{{d\theta }} = \dfrac{d}{{d\theta }}\left( {a\left\\{ {\cos \theta + \log \tan \left( {\dfrac{\theta }{2}} \right)} \right\\}} \right)
We know that, $\dfrac{d}{{dx}}af\left( x \right) = a\dfrac{d}{{dx}}f\left( x \right)$ . So, we get
\Rightarrow \dfrac{{dx}}{{d\theta }} = a\dfrac{d}{{d\theta }}\left\\{ {\cos \theta + \log \tan \left( {\dfrac{\theta }{2}} \right)} \right\\}
We know that, $\dfrac{d}{{dx}}\left[ {f\left( x \right) + g\left( x \right)} \right] = \dfrac{d}{{dx}}f\left( x \right) + \dfrac{d}{{dx}}g\left( x \right)$ . So, we get
\Rightarrow \dfrac{{dx}}{{d\theta }} = a\left\\{ {\dfrac{d}{{d\theta }}\cos \theta + \dfrac{d}{{d\theta }}\log \tan \left( {\dfrac{\theta }{2}} \right)} \right\\}
We know that $\dfrac{d}{{d\theta }}\cos \theta = - \sin \theta$ . On applying this, we get
\Rightarrow \dfrac{{dx}}{{d\theta }} = a\left\\{ { - \sin \theta + \dfrac{d}{{d\theta }}\log \tan \left( {\dfrac{\theta }{2}} \right)} \right\\}
We know that $\dfrac{d}{{dx}}\log x = \dfrac{1}{x}$ and on applying chain rule, we get
\Rightarrow \dfrac{{dx}}{{d\theta }} = a\left\\{ { - \sin \theta + \dfrac{1}{{\tan \left( {\dfrac{\theta }{2}} \right)}} \times \dfrac{d}{{d\theta }}\tan \left( {\dfrac{\theta }{2}} \right)} \right\\}
We know that $\dfrac{d}{{dx}}\tan \left( x \right) = {\sec ^2}x$ and on applying chain rule, we get,
\Rightarrow \dfrac{{dx}}{{d\theta }} = a\left\\{ { - \sin \theta + \dfrac{1}{{\tan \left( {\dfrac{\theta }{2}} \right)}} \times {{\sec }^2}\left( {\dfrac{\theta }{2}} \right) \times \dfrac{d}{{d\theta }}\left( {\dfrac{\theta }{2}} \right)} \right\\}
We know that, $\dfrac{d}{{dx}}af\left( x \right) = a\dfrac{d}{{dx}}f\left( x \right)$ . So, we get
\Rightarrow \dfrac{{dx}}{{d\theta }} = a\left\\{ { - \sin \theta + \dfrac{1}{{\tan \left( {\dfrac{\theta }{2}} \right)}} \times {{\sec }^2}\left( {\dfrac{\theta }{2}} \right) \times \dfrac{1}{2}\dfrac{d}{{d\theta }}\left( \theta \right)} \right\\}
We know that $\dfrac{d}{{dx}}x = 1$ . On using this, we get
\Rightarrow \dfrac{{dx}}{{d\theta }} = a\left\\{ { - \sin \theta + \dfrac{1}{{\tan \left( {\dfrac{\theta }{2}} \right)}} \times {{\sec }^2}\left( {\dfrac{\theta }{2}} \right) \times \dfrac{1}{2}} \right\\}
We know that $\tan x = \dfrac{{\sin x}}{{\cos x}}$ and $\sec x = \dfrac{1}{{\cos x}}$ . On applying this relation, we get
\Rightarrow \dfrac{{dx}}{{d\theta }} = a\left\\{ { - \sin \theta + \dfrac{{\cos \left( {\dfrac{\theta }{2}} \right)}}{{\sin \left( {\dfrac{\theta }{2}} \right)}} \times \dfrac{1}{{{{\cos }^2}\left( {\dfrac{\theta }{2}} \right)}} \times \dfrac{1}{2}} \right\\}
On cancelling the common terms and rearranging, we get
\Rightarrow \dfrac{{dx}}{{d\theta }} = a\left\\{ { - \sin \theta + \dfrac{1}{{2\sin \left( {\dfrac{\theta }{2}} \right)\cos \left( {\dfrac{\theta }{2}} \right)}}} \right\\}
We know that $2\sin A\cos A = \sin 2A$ . So, the denominator will become
\Rightarrow \dfrac{{dx}}{{d\theta }} = a\left\\{ { - \sin \theta + \dfrac{1}{{\sin \theta }}} \right\\}
On taking the LCM, we get
\Rightarrow \dfrac{{dx}}{{d\theta }} = a\left\\{ {\dfrac{{ - {{\sin }^2}\theta + 1}}{{\sin \theta }}} \right\\}
We know that $1 - {\sin ^2}A = {\cos ^2}A$ . So, the numerator will become
\Rightarrow \dfrac{{dx}}{{d\theta }} = a\left\\{ {\dfrac{{{{\cos }^2}\theta }}{{\sin \theta }}} \right\\} … (1)
Now let’s consider $y = a\sin \theta$
We can find the derivative of x with respect to $\theta$ .
$\Rightarrow \dfrac{{dy}}{{d\theta }} = \dfrac{d}{{d\theta }}a\sin \theta$
We know that, $\dfrac{d}{{dx}}af\left( x \right) = a\dfrac{d}{{dx}}f\left( x \right)$ . So, we get
$\Rightarrow \dfrac{{dy}}{{d\theta }} = a\dfrac{d}{{d\theta }}\sin \theta$
We know that $\dfrac{d}{{d\theta }}\sin \theta = \cos \theta$ . So, the derivative will become
$\Rightarrow \dfrac{{dy}}{{d\theta }} = a\cos \theta$ … (2)
Now we need to find the value of $\dfrac{{dy}}{{dx}}$ . For that we can divide (2) with (1)
\Rightarrow \dfrac{{\dfrac{{dy}}{{d\theta }}}}{{\dfrac{{dx}}{{d\theta }}}} = \dfrac{{a\cos \theta }}{{a\left\\{ {\dfrac{{{{\cos }^2}\theta }}{{\sin \theta }}} \right\\}}}
We can cancel the common terms in the numerator and denominator of LHS. So, we get
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{a\cos \theta }}{{a\left\\{ {\dfrac{{{{\cos }^2}\theta }}{{\sin \theta }}} \right\\}}}
Now we can simplify the RHS.
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{a\cos \theta \sin \theta }}{{a{{\cos }^2}\theta }}$
On cancelling the common terms, we get
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\sin \theta }}{{\cos \theta }}$
We know that $\tan x = \dfrac{{\sin x}}{{\cos x}}$ . So, we get
$\Rightarrow \dfrac{{dy}}{{dx}} = \tan \theta$
Therefore, the required solution is $\tan \theta$

So, the correct answer is option B.

Note:
This method of finding the derivative is known as derivatives using parametric form. This method can be used to find the derivative of the curve which can be expressed in parametric form. We must take care of the order when dividing the derivatives of x and y. If we take the reverse order, we will get the derivative of x with respect to y. While taking the derivatives, we must apply the chain rule.