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. \[\dfrac{\pi }{6}\]
Verified146.1k+ views
Hint: 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 \[\theta \].
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 \[\theta \].