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  • Provide Solution for RD Sharma Class 12 Chapter 10 Differentiation Exercise 10.7 Question 4

Provide Solution for RD Sharma Class 12 Chapter 10 Differentiation Exercise 10.7 Question 4

Answers (1)

Answer:

            \frac{dy}{dx}=\cot \theta

Hint:

            Use product rule

Given:

            x=a e^{\theta}(\sin \theta-\cos \theta)

            \begin{aligned}&y=a e^{\theta}(\sin \theta+\cos \theta) \end{aligned}

Solution:

x=a e^{\theta}(\sin \theta-\cos \theta)

\begin{aligned} & \\ &\frac{d x}{d \theta}=a e^{\theta} \frac{d(\sin \theta-\cos \theta)}{d \theta}+(\sin \theta-\cos \theta) \frac{d\left(a e^{\theta}\right)}{d \theta} \end{aligned}                                        [Use product rule]

\begin{aligned} &\frac{d x}{d \theta}=a e^{\theta}(\cos \theta+\sin \theta)+(\sin \theta-\cos \theta) a \cdot \frac{d e^{\theta}}{d \theta} \\ &\frac{d x}{d \theta}=a e^{\theta}(\cos \theta+\sin \theta)+(\sin \theta-\cos \theta) \cdot a \cdot e^{\theta} \\ &\frac{d x}{d \theta}=a e^{\theta}(\cos \theta+\sin \theta+\sin \theta-\cos \theta) \end{aligned}

\begin{aligned} &\frac{d x}{d \theta}=2 a e^{\theta} \sin \theta \\ &y=a e^{\theta}(\sin \theta+\cos \theta) \end{aligned}

\frac{d y}{d \theta}=a e^{\theta} \cdot \frac{d(\sin \theta+\cos \theta)}{d \theta}+(\sin \theta+\cos \theta) \cdot \frac{d\left(a e^{\theta}\right)}{d \theta}                         [Using product rule]

\begin{aligned} &=a e^{\theta}(\cos \theta-\sin \theta)+(\sin \theta+\cos \theta) a \cdot \frac{d e^{\theta}}{d \theta} \\ &=a e^{\theta}(\cos \theta-\sin \theta)+(\sin \theta+\cos \theta) a e^{\theta} \\ &=a e^{\theta}(\cos \theta-\sin \theta+\sin \theta+\cos \theta) \\ &\frac{d y}{d \theta}=2 a e^{\theta} \cos \theta \end{aligned}

So,

\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{2 a e^{\theta} \cos \theta}{2 a e^{\theta} \sin \theta}=\frac{\cos \theta}{\sin \theta}=\cot \theta

 

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