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  • Need solution for RD Sharma Maths Class 12 Chapter 10 Differentiation Excercise 10.7 Question 25

Need solution for RD Sharma Maths Class 12 Chapter 10 Differentiation Excercise 10.7 Question 25

Answers (1)

Answer:

             \frac{d y}{d x}\; {\mathrm{At}}\; \left(t=\frac{\pi}{4}\right)=1

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=\cos t\left(3-2 \cos ^{2} t\right) \\ &y=\sin t\left(3-2 \sin ^{2} t\right) \end{aligned}

Solution:

x=\cos t\left(3-2 \cos ^{2} t\right)

\frac{d x}{d t}=\cos t \cdot \frac{d\left(3-2 \cos ^{2} t\right)}{d t}+\left(3-2 \cos ^{2} t\right) \frac{d \cos t}{d t}             [Using product rule]

\frac{d x}{d t}=\cos t[0-4 \cos t(-\sin t)]+\left(3-2 \cos ^{2} t\right) \times(-\sin t)               \quad\left[\because \frac{d(\cos \theta)}{d \theta}=-\sin \theta\right]

\frac{d x}{d t}=4 \cos ^{2} t \sin t-3 \sin t+2 \cos ^{2} t \sin t \\

=6 \cos ^{2} t \sin t-3 \sin t \\

\begin{aligned} &\frac{d x}{d t}=3 \sin t\left(2 \cos ^{2} t-1\right) \end{aligned}                                                                                             (1)

y=\sin t\left(3-2 \sin ^{2} t\right) \\

\begin{aligned} &\frac{d y}{d t}=\sin t \frac{\left(3-2 \sin ^{2} t\right)}{d t}+\left(3-2 \sin ^{2} t\right) \frac{d \sin t}{d t} \end{aligned}                                   [Using product rule]

=\sin t \times\left[\frac{d 3}{d t}-\frac{d\left(\sin ^{2} t\right)}{d t}\right]+\left(3-2 \sin ^{2} t\right) \cos t                                 \left[\because \frac{d \sin \theta}{d \theta}=\cos \theta\right]

=\sin t[0-4 \sin t \cos t]+3 \cos t-2 \sin ^{2} t \cos t \\

=-4 \sin ^{2} t \cos t+3 \cos t-2 \sin ^{2} t \cos t \\

=-6 \sin ^{2} t \cos t+3 \cos t \\

\begin{aligned} & &\therefore \frac{d y}{d t}=3 \cos t\left(-2 \sin ^{2} t+1\right) \end{aligned}                                                                                                     (2)

\because \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}

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

\frac{d y}{d x}=\frac{3 \cos t\left(1-\sin ^{2} t\right)}{3 \sin t\left(2 \cos ^{2} t-1\right)} \\

\begin{aligned} &\frac{d y}{d x}=\cot t \frac{\left(1-2 \sin ^{2} t\right)}{\left(2 \cos ^{2} t-1\right)} \end{aligned}                                                                                               \left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]

\frac{d y}{d x}=\cot t \frac{\left(1-2\left(1-\cos ^{2} t\right)\right)}{\left(2 \cos ^{2} t-1\right)}                                                                      \left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]  

\frac{d y}{d x}=\cot t \times \frac{\left(1-2+2 \cos ^{2} t\right)}{\left(2 \cos ^{2} t-1\right)} \\

=\cot t \times \frac{\left(2 \cos ^{2} t-1\right)}{\left(2 \cos ^{2} t-1\right)} \\

=\cot t \\  

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

\text { At } t=\frac{\pi}{4}, \frac{d y}{d x}=\cot \frac{\pi}{4}=1 \\

\begin{aligned} & &\frac{d y}{d x}\; {\mathrm{At}}\; \left(t=\frac{\pi}{4}\right)=1 \end{aligned}

 

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