RD Sharma Class 12 Exercise 12 CSBQ Derivative as a rate measure Solutions Maths

RD Sharma Class 12 Exercise 12 CSBQ Derivative as a rate measure Solutions Maths - Download PDF Free Online

Edited By Lovekush kumar saini | Updated on Jan 20, 2022 07:21 PM IST

RD Sharma class 12th exercise CSBQ is a covered NCERT solution used by several students in India. The class 12 maths syllabus is vast and requires a lot of practice to master. However, students may rarely find the time to clear their doubts during school hours. For this purpose, the RD Sharma class 12 chapter 12 exercise CSBQ can be useful to students. The 12th chapter of the NCERT maths book has the topic Derivative as a Rate Measure. It is a pretty concise chapter but complicated nonetheless. It covers concepts like deriving rate measurer, theorems regarding the rate measurer and many sums related to it. The CSBQ exercise will have 15 questions from a mixture of all these concepts.

JEE Main 2025: Sample Papers | Mock Tests | PYQs | Study Plan 100 Days

JEE Main 2025: Maths Formulas | Study Materials

JEE Main 2025: Syllabus | Preparation Guide | High Scoring Topics

This Story also Contains

RD Sharma Class 12 Solutions Chapter 12 CSQB Derivative as a Rate Measure - Other Exercise
Derivative as a Rate Measurer Excercise: 12CSQB
RD Sharma Chapter wise Solutions

Also Read - RD Sharma Solution for Class 9 to 12 Maths

Chapter 12 - Derivative as a Rate Measure -Ex 12.2

RD Sharma Class 12 Solutions Chapter 12 CSQB Derivative as a Rate Measure - Other Exercise

Derivative as a Rate Measurer Excercise: 12CSQB

Derivative as a Rate Measurer exercise case study base question, question 1(i)

Answer:

$\frac{d x}{d t}=-5 \mathrm{~cm} / \mathrm{min} \frac{d y}{d t}=4 \mathrm{~cm} / \mathrm{min}$

Hint:

Here, we use the concept of volume.

Given:

Length decreases at the rate of 5cm and with increase at the rate of 4cm/sec

Solution:

So, let x be length of rectangular sheet and y be width of rectangular sheet

$\frac{d x}{d t}=-5$ Because length decrease so value must be negative

and $\frac{d x}{d t}=4$ because width increase. So value must be positive

Derivative as a Rate Measure exercise case study based question, question 1(ii)

Answer:

- $-2cm/\sec$ $-2cm/\sec$

Hint:

Here, we use the formula of perimeter (P) of a rectangle

Given: $x=8cm$ and $y=6cm$

Solution:

$\begin{aligned} &p=2(x+y) \\ &\frac{d p}{d t}=2\left(\frac{d x}{d t}+\frac{d y}{d t}\right)= \\ &=2(-5+4) \quad, a s\left(\frac{d x}{d t}=-5\right) \text { and }\left(\frac{d y}{d t}=4\right) \\ &=2(-1) \\ &=-2 \mathrm{~cm} / \mathrm{sec} \end{aligned}$

Hence, the perimeter is decreases at the rate of $2cm/\sec$

Derivative as a Rate Measure exercise case study based question, question 1(iii)

Answer:

$2cm^{2}/\sec$

Hint:

Here, we use the formula of area

Given:

x=8cm

Y=6cm

Solution:

$A=x\times y$

$\frac{dA}{dt}=\frac{ydx}{dt}+x\frac{dy}{dt}$

$=-5\times 6+8\times 4$

$=32-30$

$=2cm^{2}/\sec$

Derivative as a Rate Measure exercise case study based question, question 1(iv)

Answer:
-1.6 cm/sec
Hint:
Here, we use the basic concept of diagonal plate
Given:
Here x=8cm and y=6cm
Solution:
x=8cm and y=6cm
According to Pythagoras
$P^{2}=x^{2}+y^{2}$ …….(1)
=64+36
=100
P=10
Differentiate (1) w.r.t t we get
$\begin{aligned} &2 \times 10 \frac{d p}{d t}=2 \times 8 \times(-5)+2 \times 6 \times 4 \quad, \text { since } \frac{d x}{d t}=-5 \text { and } \frac{d y}{d t}=4 \\ &\frac{d p}{d t}=-1.6 \mathrm{~cm} / \mathrm{sec} \end{aligned}$

Derivative as a Rate Measure exercise case study based question, question 1(v)

Answer:
$-\frac{31}{18}$
Hint:
Here, we use basic concepts of area
Given:
x=8 y=6
Solution:
x=8 y=6
So, according to Pythagoras
P=10
So, change of ratio of length and breadth $=\frac{d}{d t}\left(\frac{x}{y}\right)=\frac{\frac{y d x}{d t}-x \frac{d y}{d t}}{y^{2}}$
$=-31 / 36 \quad, \text { since } \frac{d x}{d t}=-5 \text { and } \frac{d y}{d t}=4$

Derivative as a Rate Measure exercise case study based question, question 2(i)

Answer:
$24cm^{2}/\sec$
Hint:
Here we use concept of cube
Given:
Length= 10 cm
Total surface area of cube $\left ( A \right )=6a^{2}$
Differentiate it
$\begin{array}{r} \frac{d A}{d t}=12 a \frac{d a}{d t} \\ \end{array}$
$\begin{array}{r} =12 \times 10 \times 0.2 \quad \text { since } \frac{d a}{d t}=0.2 \\ \end{array}$
$\begin{array}{r} =24 \mathrm{~cm}^{2} / \mathrm{sec} \end{array}$

Derivative as a Rate Measurer exercise case study base question, question 2(ii)

Answer:
$60cm^{3}/\sec$
Hint:
Here, we use the basic concept
Given:
Theedge Increases at rate of $\frac{da}{dt}=0.2cm/\sec$
Length of edge of cube =10cm
Solution:
we have $\frac{da}{dt}=0.2cm/\sec$
Volume of cube = $a^{3}$
Differentiate it we get
$\frac{d v}{d t}=3 a^{2} \frac{d a}{d t}=3 \times 10 \times 10 \times 0.2=60 \mathrm{~cm}^{3} / \mathrm{sec}$

Derivative as a Rate Measurer exercise case study base question, question 2(iii)

Answer:
$\frac{\sqrt{3}}{5}cm/\sec$
Hint:
Here, we use the formula of diagonal
Given:
$\frac{da}{dt}=0.2$
Solution:
Weknow diagonal of cube $\left ( p \right )=\sqrt{3}a$
Differentiate it we get
$\begin{aligned} &\frac{d p}{d t}=\sqrt{3} \frac{d a}{d t}=\sqrt{3} \times 0.2=\frac{2 \sqrt{3}}{10} \\ &=\frac{\sqrt{3}}{5} \mathrm{~cm} / \mathrm{sec} \end{aligned}$

Derivative as a Rate Measurer exercise case study base question, question 2(iv)

Answer:
2.5cm
Hint:
Here we use the value of cube’s surface area and volume
Given:
We Know that
Volume = $=60cm^{3}/s$
And surface area = $=24cm^{2}/\sec$

So, rate of increase of volume with respect to surface area $\frac{60}{24}=2.5cm$

Derivative as a Rate Measurer exercise case study base question, question 2(v)

Answer:
4 cm
Hint:
$V=a^{3}$
Given:
Solution:
Volume of the cube,
$\begin{aligned} &V=a^{3} \\ &\frac{d v}{d t}=3 a^{2} \cdot \frac{d a}{d t} \\ &\text { at } \frac{d a}{d t}=0.2 \mathrm{~cm} / \mathrm{sec} \\ &\frac{d v}{d t}=3 a^{2} \times 0.2-(1) \end{aligned}$
Surface Area of cube,
$\begin{aligned} &s=6 a^{2} \\ &\frac{d s}{d t}=12 a \cdot \frac{d a}{d t} \\ &\text { At } \frac{d a}{d t}=0.2 \mathrm{~cm} / \mathrm{sec} \\ &\frac{d s}{d t}=12 a \times 0.2-(2) \end{aligned}$
From equation (1) & (2),
$\frac{d v}{d t}= \frac{d s}{d t}$
$\begin{aligned} &\text { at } \frac{d a}{d t}=0.2 \mathrm{~cm} / \mathrm{sec} \\ &3 a^{2} \times 0.2=12 a \times 0.2 \\ &3 a^{2}-12 a=0 \\ &3 a(a-4)=0 \\ &a=0, \\ &a=4 \end{aligned}$
Side can’t be zero
$\therefore a=4cm$

Derivative as a Rate Measure exercise case study based question, question 3(i)

Answer:
$3 \frac{d x}{d t}+2 \frac{d y}{d t}=0$
Hint:
Here, we use basic concept of area
Given:
Aarushi’seast side speed is 10km/hr. and mall to Mira’s house speed is 15km/hr.
Solution:
Here
$\begin{aligned} &15 \frac{d x}{d t}=-10 \frac{d y}{d t} \\ &\text { So, } 3 \frac{d x}{d t}=-2 \frac{d y}{d t} \\ &\text { So, } 3 \frac{d x}{d t}+2 \frac{d y}{d t}=0 \end{aligned}$

Derivative as a Rate Measure exercise case study based question, question 3(iii)

Answer:
13km
Hint:
Here we use the Pythagoras theorem
Given:
Here x= 5km
y=12km
Solution:
Let B
$B^{2}=x^{2}+y^{2}$
=25 +144
=169
B=13 km
So, the distance between Aarushi and Mira is 13km

Derivative as a Rate Measure exercise case study based question, question 3(iv)

Answer:
10km/hr.
Hint:
Here we use the concept of speed and area
Given:
Here x=5km
y=12km
So, let B =13km (From Pythagoras)
Solution:
$\frac{da}{dt}=13km/hr$
So, the speed of Aarushi and Mira is 10km/hr.

Derivative as a Rate Measure exercise case study based question, question 3(v)

Answer:
$\frac{15}{13}$
Hint:
Here we use basic concepts of Pythagoras
Given:
x=5km
y=12km
So, B=13km
Solution:
$\theta$ , Be the angle formed by mall, Aarushi and Mira.
$\begin{aligned} &\tan \theta=\frac{y}{x} \\ &\Rightarrow \sec ^{2} \theta \frac{d \theta}{d t}=\frac{x \frac{d y}{d x}-y \frac{d x}{d t}}{x^{2}} \\ &\Rightarrow \frac{13^{2}}{5^{2}} \frac{d \theta}{d t}=\frac{5 \times 15+10 \times 12}{5^{2}} \\ &\begin{aligned} \Rightarrow \frac{d \theta}{d t} &=\frac{195}{165} \\ &=\frac{15}{13} \end{aligned} \end{aligned}$

Derivative as a Rate Measure exercise case study based question, question 2(i)

Students who want to know more about the class 12 RD Sharma chapter 12 exercise CSBQ solution and benefits of using RD Sharma solutions can check out the points mentioned below:-

The RD Sharma class 12 solutions Derivative as a Rate Measure ex CSBQ is important for last minute practice as it answers some complex questions.
The RD Sharma class 12th exercise CSBQ will refer to the latest NCERT syllabus and corresponds to all changes made by CBSE board.
Students who have used RD Sharma class 12th exercise CSBQ blindly trust the book due to its reliability.
The answers in the RD Sharma class 12 solutions Derivative as a Rate Measure ex CSBQ are vital in self-study. All answers can be used to check performance at home.
RD Sharma class 12th exercise CSBQ will help identify weak points. Students can then concentrate on these areas first.
Homework questions can be easily solved by seeking help from RD Sharma class 12th exercise CSBQ. Even school teachers use them to check answers.
Career360 provides students with a free and downloaded pdf of RD Sharma class 12 chapter 12 exercise CSBQ. It is economical to use.
Not every free solution books are worth, the content in the RD Sharma books are provided by mathematical experts, making it worth to be used.

RD Sharma Chapter wise Solutions

Frequently Asked Questions (FAQs)

1. Best site to download the class 12 RD Sharma chapter 12 exercise CSBQ solution pdf?

The best site that provides free copies of class 12 RD Sharma chapter 12 exercise CSBQ solution is the Career360 website. Navigate to the website and get your free copy now.

2. Are there extra benefits of using RD Sharma solutions?

It is well known that using RD Sharma Solutions to practice at home will help students find common questions in their board exams. Many ex-students have confirmed this fact.

3. Is RD Sharma class 12 chapter 12 exercise CSBQ ideal for easy home practice?

Students should definitely give RD Sharma class 12 chapter 12 exercise CSBQ a try as the book is recommended for home practice. It can be used to make mock tests at home and even to stay up to date with lessons.

4. Where do I purchase RD Sharma solutions pdf?

Students don't have to surf the Web extensively to purchase a RD Sharma pdf. They can instead download them for free at Career360.

5. In what prices are the RD Sharma books available on the internet?

The RD Sharma books are available even for free at the Career 360 website. Therefore, the students can make use of this opportunity to get them for free.