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 12 Derivative as a Rate Measurer exercise multiple choise question 9 maths textbook solution

Please solve RD Sharma class 12 chapter 12 Derivative as a Rate Measurer exercise multiple choise question 9 maths textbook solution

Answers (1)

Answer:

\left(3, \frac{16}{3}\right) \text { and }\left(3,-\frac{16}{3}\right)

Hint:

Here, we use the basic concept of algebra       

Given:

16x^{2}+9y^{2}=400

Solution: Let

E(x,y)=16x^{2}+9y^{2}=400

Solving for y we get

y=\pm \frac{\sqrt{400-16 x^{2}}}{3} \quad\quad \quad \quad .....(i)

Given that

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

we have to calculate

→ Differentiating (i) with respect to t,we get

\begin{aligned} \frac{d y}{d t} &=\pm \frac{1}{3} \times \frac{1}{2 \sqrt{400-46 x^{2}}} \times-32 \times \frac{d x}{d t} \\ &=\mp \frac{16 x}{3 \sqrt{400-16 x^{2}}} \frac{d x}{d t} \end{aligned}

→ Substituting values

We get

\begin{aligned} 16 x=\pm 3 \sqrt{400-16 x^{2}}\\ \end{aligned}

Squaring the equation, we get

\begin{aligned} &256 x^{2}=9\left(400-16 x^{2}\right) \end{aligned}

Solving the equation we get

x^{2}=9 \\ x=\pm 3

→ Substituting in (i), we get

\begin{aligned} &y=\pm \frac{\sqrt{400-16 \times 9}}{3}=\pm \frac{16}{3}\\ &\text { So, }\\ &\left(3, \frac{16}{3}\right) \text { and }\left(3,-\frac{16}{3}\right) \end{aligned}

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

Board Exam Time Table 2025 Live: UP Board exams from February 24; CBSE, BSEB date sheet soon
anam-khan

CBSE Date Sheet 2025 Live: Class 10, 12 board exam dates soon; syllabus, sample papers
anam-khan

Extra time in CBSE exams for students with type 1 diabetes: Kerala SHRC seeks report on plea

Latest Question