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  • Need solution for RD Sharma Maths Class 12 Chapter 10 Differentiation Excercise 10.7 Question 24

Need solution for RD Sharma Maths Class 12 Chapter 10 Differentiation Excercise 10.7 Question 24

Answers (1)

Answer:

            \frac{d y}{d x}\; {\mathrm{At}}\; \left(t=\frac{\pi}{4}\right)=\frac{b}{a}

Hint:

            Use product rule and   \frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}

Given:

            \begin{aligned} &x=a \sin 2 t(1+\cos 2 t) \\ &y=b \cos 2 t(1-\cos 2 t) \end{aligned}

Solution:

x=a \sin 2 t(1+\cos 2 t) \\

\begin{aligned} & &\frac{d x}{d t}=a\left[\sin 2 t \times \frac{d(1+\cos 2 t)}{d t}+(1+\cos 2 t) \frac{d(\sin 2 t)}{d t}\right] \end{aligned}                        [Using product rule]

=a[\sin 2 t \times(0+(-2 \sin 2 t))+(1+\cos 2 t)(2 \cos 2 t)] \\

=a\left[-2 \sin ^{2} 2 t+2 \cos 2 t+2 \cos ^{2} 2 t\right] \\

=a\left[2 \cos 2 t+2 \cos ^{2} 2 t-2 \sin ^{2} 2 t\right] \\

=a\left[2 \cos 2 t+2\left(\cos ^{2} 2 t-\sin ^{2} 2 t\right)\right] \

\begin{aligned} &\ &=2 a[\cos 2 t+\cos 4 t] \end{aligned}                                                                                    \left[\because \cos 2 \theta=\cos ^{2} \theta-\sin ^{2} \theta\right]

\frac{d x}{d t}=2 a(\cos t+\cos 4 t)                                                                                                                              (1)

y=b \cos 2 t(1-\cos 2 t) \\

\begin{aligned} & &\frac{d y}{d t}=b\left[\cos 2 t \times \frac{d(1-\cos 2 t)}{d t}+(1-\cos 2 t) \frac{d \cos 2 t}{d t}\right] \end{aligned}              [Using product rule]

=b[\cos 2 t(0+2 \sin 2 t)]+(1-\cos 2 t)(-2 \sin 2 t)                          \quad\left[\because \frac{d \cos \theta}{d \theta}=-\sin \theta\right]

=b[2 \sin 2 t \cos 2 t-2 \sin 2 t+2 \sin 2 t \cos 2 t] \\

=b[4 \sin 2 t \cos 2 t-2 \sin t 2 t] \\

=b[2(2 \sin 2 t \cos 2 t)-2 \sin 2 t] \\

\begin{aligned} & &=2 b[\sin 4 t-\sin 2 t] \end{aligned}                                                                                     [\because \sin 2 \theta=2 \sin \theta \cos \theta]

\frac{d y}{d t}=2 b(\sin 4 t-\sin 2 t) \\                                                                                                                             (2)

\begin{aligned} & &\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}} \end{aligned}

Put the values of   \frac{d y}{d t} \text { and } \frac{d x}{d t}  from the equations (2) and (1) respectively

\frac{d y}{d x}=\frac{2 b(\sin 4 t-\sin 2 t)}{2 a(\cos 2 t+\cos 4 t)} \\

\frac{d y}{d x}=\frac{b}{a} \times\left(\frac{\sin 4 t-\sin 2 t}{\cos 2 t+\cos 4 t}\right) \\

\begin{aligned} & &\text { At } t=\frac{\pi}{4} \end{aligned}

\frac{d y}{d x}=\frac{b}{a} \times \frac{\left(\sin \left(4 \times \frac{\pi}{4}\right)-\sin \left(2 \times \frac{\pi}{4}\right)\right)}{\cos \left(2 \times \frac{\pi}{4}\right)+\cos \left(4 \times \frac{\pi}{4}\right)} \\

=\frac{b}{a}\left(\frac{\sin \pi-\sin \frac{\pi}{2}}{\cos \frac{\pi}{2}+\cos \pi}\right) \\

\begin{aligned} & &\because \sin \frac{\pi}{2}=1, \cos \frac{\pi}{2}=0, \sin \pi=0, \cos \pi=-1 \end{aligned}

=\frac{b}{a}\left[\frac{0-1}{0+(-1)}\right] \\

=\frac{b}{a}\left(\frac{-1}{-1}\right) \\

\begin{aligned} & &\frac{d y}{d x}_{\mathrm{At}}\left(t=\frac{\pi}{4}\right)=\left(\frac{b}{a}\right) \end{aligned}

Hence proved

 

Posted by

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