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

# 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.

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

• Chapter 12 - Derivative as a Rate Measure -Ex 12.2

## Derivative as a Rate Measurer Excercise: 12CSQB

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

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

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)

-$-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)

$2cm^{2}/\sec$

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)

-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
$\inline P^{2}=x^{2}+y^{2}$ …….(1)
=64+36
=100
P=10
Differentiate (1) w.r.t t we get
\inline \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)

$-\frac{31}{18}$
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 $\inline =\frac{d}{d t}\left(\frac{x}{y}\right)=\frac{\frac{y d x}{d t}-x \frac{d y}{d t}}{y^{2}}$
$\inline =-31 / 36 \quad, \text { since } \frac{d x}{d t}=-5 \text { and } \frac{d y}{d t}=4$

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

$24cm^{2}/\sec$
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)

$60cm^{3}/\sec$
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)

$\frac{\sqrt{3}}{5}cm/\sec$
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
\inline \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)

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 = $=24cm^{2}/\sec$
1. 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)

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)

$3 \frac{d x}{d t}+2 \frac{d y}{d t}=0$
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)

13km
Hint:
Here we use the Pythagoras theorem
Given:
Here x= 5km
y=12km
Solution:
Let B
$\inline B^{2}=x^{2}+y^{2}$
=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)

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)

$\frac{15}{13}$
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)

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

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