Question

For the curve $y = 3\sin \theta \cos \theta , x = {e^\theta }\sin \theta$, $0 \leqslant \theta \leqslant \pi$, the tangent is parallel to x-axis when $\theta$ is
A. $\dfrac{\pi }{4}$
B. $\dfrac{\pi }{2}$
C. $\dfrac{{3\pi }}{4}$
D. $\dfrac{\pi }{6}$

Answer 1

We differentiate both y and x with respect to $\theta$ and write the value of $\dfrac{{dy}}{{d\theta }}$ and $\dfrac{{dx}}{{d\theta }}$. Then divide the differentiation of y by differentiation of x to get the value of $\dfrac{{dy}}{{dx}}$ which gives us the tangent. Then we equate the value to zero as the tangent is parallel to x-axis and a line parallel to x has y coordinate as constant so the differentiation of the line parallel to x-axis is equal to zero.

Complete step-by-step answer:
We have curves $y = 3\sin \theta \cos \theta$ and $x = {e^\theta }\sin \theta$.
First we will differentiate y with respect to $\theta$.
$\Rightarrow \dfrac{{dy}}{{d\theta }} = \dfrac{d}{{d\theta }}(3\sin \theta \cos \theta )$
Taking out the constant term we get
$\Rightarrow \dfrac{{dy}}{{d\theta }} = 3 \times \dfrac{d}{{d\theta }}(\sin \theta \cos \theta )$
Now using product rule of differentiation i.e. $\dfrac{d}{{dx}}(m \times n) = m \times \dfrac{{dn}}{{dx}} + n \times \dfrac{{dm}}{{dx}}$. Substitute the values of $m = \sin \theta ,n = \cos \theta$.
$\Rightarrow \dfrac{{dy}}{{d\theta }} = 3\left[ {\sin \theta \times \dfrac{d}{{d\theta }}(\cos \theta ) + \cos \theta \times \dfrac{d}{{d\theta }}(\sin \theta )} \right]$
Substituting the values of $\dfrac{d}{{d\theta }}\sin \theta = \cos \theta ,\dfrac{d}{{d\theta }}\cos \theta = - \sin \theta$ we get

$$\Rightarrow \dfrac{{dy}}{{d\theta }} = 3\left[ {\sin \theta \times ( - \sin \theta ) + \cos \theta \times (\cos \theta )} \right] \\\ \Rightarrow \dfrac{{dy}}{{d\theta }} = 3\left[ {{{\cos }^2}\theta - {{\sin }^2}\theta } \right] \\\$$

Using the formula ${a^2} - {b^2} = (a - b)(a + b)$where $a = \cos \theta ,b = \sin \theta$, we can write
$\Rightarrow \dfrac{{dy}}{{d\theta }} = 3\left[ {(\cos \theta - \sin \theta )(\cos \theta + \sin \theta )} \right]$ … (1)
Now we will differentiate x with respect to $\theta$.
$\Rightarrow \dfrac{{dx}}{{d\theta }} = \dfrac{d}{{d\theta }}({e^\theta }\sin \theta )$
Now using product rule of differentiation i.e. $\dfrac{d}{{dx}}(m \times n) = m \times \dfrac{{dn}}{{dx}} + n \times \dfrac{{dm}}{{dx}}$. Substitute the values of $m = {e^\theta },n = \sin \theta$.
$\Rightarrow \dfrac{{dx}}{{d\theta }} = \left[ {{e^\theta } \times \dfrac{d}{{d\theta }}(\sin \theta ) + \sin \theta \times \dfrac{d}{{d\theta }}({e^\theta })} \right]$
Substituting the values of $\dfrac{d}{{d\theta }}\sin \theta = \cos \theta ,\dfrac{d}{{d\theta }}{e^\theta } = {e^\theta }$ we get
$\Rightarrow \dfrac{{dx}}{{d\theta }} = \left[ {{e^\theta }(\cos \theta ) + \sin \theta \times ({e^\theta })} \right]$
$\Rightarrow \dfrac{{dx}}{{d\theta }} = \left[ {{e^\theta }(\cos \theta + \sin \theta )} \right]$ … (2)
Now we know that $\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{d\theta }} \times \dfrac{{d\theta }}{{dx}}$. Substituting the values from (1) and (2) .
$\dfrac{{dy}}{{dx}} = 3\left[ {(\cos \theta - \sin \theta )(\cos \theta + \sin \theta )} \right] \times \dfrac{1}{{[{e^\theta }(\cos \theta + \sin \theta )]}}$
Cancel out same terms from numerator and denominator we get

$$\dfrac{{dy}}{{dx}} = 3\left[ {(\cos \theta - \sin \theta )} \right] \times \dfrac{1}{{[{e^\theta }]}} \\\ \dfrac{{dy}}{{dx}} = \dfrac{{3(\cos \theta - \sin \theta )}}{{{e^\theta }}} \\\$$

Now we equate the slope of the tangent to zero.
$\Rightarrow \dfrac{{3(\cos \theta - \sin \theta )}}{{{e^\theta }}} = 0$
Cross multiplying the denominator of LHs to numerator of RHS we get

$$\Rightarrow 3(\cos \theta - \sin \theta ) = 0 \\\ \Rightarrow \cos \theta - \sin \theta = 0 \\\$$

Shifting the value of sin to opposite side of the equation we get
$\Rightarrow \cos \theta = \sin \theta$
Dividing both sides by $\cos \theta$ we get

$$\Rightarrow \dfrac{{\cos \theta }}{{\cos \theta }} = \dfrac{{\sin \theta }}{{\cos \theta }} \\\ \Rightarrow 1 = \tan \theta \\\$$

Since, we know the value of $\tan \theta = 1$ when $\theta = \dfrac{\pi }{4}$.

So, the correct answer is “Option A”.

Note: Students mostly make mistake of writing the fraction as $\dfrac{y}{x}$ and then try to differentiate which is wrong because both x and y are in terms of $\theta$ so we differentiate them with respect to $\theta$.