Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Please solve RD Sharma class 12 chapter Higher Order Derivatives exercise 11.1 question 13 maths textbook solution

Please solve RD Sharma class 12 chapter Higher Order Derivatives exercise 11.1 question 13 maths textbook solution

Answers (1)

Answer:

        \frac{-a}{y^{2}}

HInt:

You must know about derivative of second order

Given:

\begin{aligned} &{\text { If } x=a(\theta+\sin \theta), y=a(1+\cos \theta)} \\ &\text { Prove that } \frac{d^{2} y}{d x^{2}}=\frac{-a}{y^{2}} \end{aligned}

Solution:

\begin{aligned} &{\text { Let } x=a(\theta+\sin \theta), y=a(1+\cos \theta)} \\ \end{aligned}

        \begin{aligned} &\left.\frac{d x}{d \theta}=a[1+\cos \theta] \quad \text { ( } \frac{d}{d \theta} \theta=1, \frac{d}{d \theta} \sin \theta=\cos \theta\right) \\ &\frac{d y}{d \theta}=a(-\sin \theta) \\ &\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{-\sin \theta}{(1+\cos \theta)} \end{aligned}

        \begin{aligned} &\sin 2 \theta=2 \sin \theta \cos \theta \\ &\cos 2 \theta=2 \cos ^{2} \theta-1 \\ &\frac{d y}{d x}=\frac{-2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}} \end{aligned}

        \begin{aligned} &\frac{d y}{d x}=-\tan \frac{\theta}{2} \\ &\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d \theta}\left(\frac{d y}{d x}\right) \frac{d \theta}{d x} \\ &\frac{d}{d \theta}\left(-\tan \frac{\theta}{2}\right) \cdot \frac{1}{a(1+\cos \theta)} \end{aligned}

        \begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{-1}{2} \sec ^{2} \frac{\theta}{2} \cdot \frac{1}{a(1+\cos \theta)} \\ &=\frac{-\sec ^{2} \frac{\theta}{2}}{2 a(1+\cos \theta)} \\ &\frac{-a}{y^{2}}=\frac{-a}{a(1+\cos \theta)^{2}}=\frac{-a}{a^{2}(1+\cos \theta)(1+\cos \theta)} \end{aligned}

        \begin{aligned} &\frac{-1}{a\left(1+2 \cos ^{2} \frac{\theta}{2}-1\right)(1-\cos \theta)} \\ &=\frac{-1}{2 a \cos ^{2} \frac{\theta}{2}(1+\cos \theta)}=\frac{-\sec ^{2} \frac{\theta}{2}}{2 a(1+\cos \theta)} \end{aligned}

        \begin{aligned} &R H L:-\frac{-a}{y^{2}}=\frac{-a}{a(1+\cos \theta)^{2}}=\frac{-1}{a\left(1+2 \cos ^{2} \frac{\theta}{2}-1\right)(1-\cos \theta)} \\ &\frac{d^{2} y}{d x^{2}}=\frac{-\sec ^{2} \frac{\theta}{2}}{2 a(1+\cos \theta)}=\frac{-a}{y^{2}} \\ &\frac{d^{2} y}{d x^{2}}=\frac{-a}{y^{2}} \end{aligned}

Hence proved

Posted by

Gurleen Kaur

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Class 12 board exams 2025 to have 50% competency-based questions; no change in 10th
anam-khan

CBSE Class 12 board exams 2024 over; expected result date, trends of last 10 years
anam-khan

CBSE Class 12 accountancy board exam analysis 2024 out; paper rated easy, balanced

Latest Question