RD Sharma Class 12 Exercise 12.1 Derivative as a rate measure Solutions Maths

RD Sharma Class 12 Exercise 12.1 Derivative as a rate measure Solutions Maths - Download PDF Free Online

Edited By Lovekush kumar saini | Updated on Jan 24, 2022 12:24 PM IST

The RD Sharma Class 12 solution of Derivative as a rate measure Exercise 12.1 reference material is a must for the class 12 students. This helps them in understanding the concepts given in the 12th chapter. The usage of different formulae to find the measurement of various shapes becomes easy when the students are clear with the concept.

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Derivative as a rate measure Exercise: 12.1
RD Sharma Class 12 Solutions Chapter 12 Derivative as a rate measure - Other Exercise
Benefits that the students gain by using the RD Sharma Class 12th Exercise 12.1 book:

Derivative as a rate measure Exercise: 12.1

Derivative as a rate measure exercise 12.1 question 1

Answer:
$4\pi r+2\pi h$
Hint:
To find the rate of change of the total surface area of the cylinder, we need to differentiate it with respect to the radius
Total surface area of cylinder, $A=2\pi r(r+h)$
Given:
$radius\; (r)\; and \: height\; (h)\; o\! f\; cylinder.$
Solution:
Here we have to find rate of change of the total surface area of cylinder. here radius= $r$ , height= $h$
so, total surface area of cylinder.
$A=2\pi r(r+h)$
$\therefore A= 2\pi r^{2}+2\pi rh\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ....(i)$
Let’s differentiate equation (i) with respect to radius
$\therefore \frac{d A}{d r}=\frac{d}{d r}\left(2 \pi r^{2}+2 \pi r h\right)$
$\frac{d A}{d r}=\frac{d}{d r}\left(2 \pi r^{2}\right)+\frac{d}{dr}(2\pi rh)$
$\therefore \frac{d A}{d r}=4\pi r+2\pi h$ $[\because \frac{d (x^{n})}{d x}=nx^{n-1}]$
Which is the rate of change of total surface area with respect to radius.
Note:
Here we have taken rate with respect to radius because radius was varying.

Derivative as a rate measure exercise 12.1 question 2

Answer:
$\frac{\pi D^{2}}{2}, D\; \; is\; the\; diameter\; o\! f\; sphere.$
Hint:
To find rate of change of volume we have to find the differentiation of volume of sphere with respect to diameter
$V\! olume\; o\! f\; spher\! e, V=\frac{4}{3}\pi R^{3}$
Here, $R$ = Radius,
Relation between Radius $R$ and diameter $D$ is $D=2R$
$\therefore R=\frac{D}{2}$
$\therefore V\! olume\; o\! f spher\! e, V=\frac{4}{3}\pi \left ( \frac{D}{2} \right )^{3} = \frac{4}{3}\pi \frac{D^{3}}{8}$
$\therefore V=\frac{\pi D^{3}}{6}$
Given:
Volume of a Sphere
Solution:
Here we have,
$D$ = Diameter of sphere
$V$ = Volume of sphere
Volume of sphere in terms of diameter, $\therefore V=\frac{\pi D^{3}}{6}$

Let’s differentiate equation (i) with respect to diameter,

$\therefore \frac{d V}{d D}=\frac{d}{d D}\left ( \frac{\pi D^{3}}{6} \right )$

$\therefore \frac{d V}{d D}= \frac{\pi D^{2}}{2}$ $\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]$
This is the rate of change of the volume of sphere with respect to diameter.
Note:
Here answer is with respect to diameter but with respect to radius, rate of change of volume of sphere is
$\therefore \frac{d V}{d R}= 4\pi R^{2}$

Derivative as a rate measure exercise 12.1 question 3

Answer: 1cm
Hint:
Here we have to differentiate volume of sphere with respect to surface area of sphere
$V\! olume\; o\! f\; spher\! e, V=\frac{4}{3}\pi r^{2}$
$\therefore S\! ur\! f\! ace\; area\; o\! f\; spher\! e, A=4\pi r^{2}$
Given:
Radius of sphere, $r=2cm$
Solution:
Here we have,
Radius, $r=2cm$
Area, $A=4\pi r^{2}$
Let’s differentiate it with respect to area ( $A$ )
$\Rightarrow \frac{d(A)}{dA}=4(2)\pi r\frac{dr}{dA}$ $\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]$
$\Rightarrow 1=8\pi r\frac{dr}{dA}$
$\therefore \frac{dr}{dA}=\frac{1}{8\pi r}\; \; \; \; \; \; \; \; \; \; \; ......(i)$ $V\! olume, V=\frac{4}{3}\pi r^{3}$

Let’s differentiate with respect to Area (A)

$\Rightarrow \frac{d V}{d A}=4\pi r^{2}\frac{dr}{d A}$

$\Rightarrow \frac{d V}{d A}=\frac{4\pi r^{2}}{8\pi r}$ {From the Equation (i)}

$\therefore \frac{d V}{d A}=\frac{r}{2}$
To putting value of $r=2cm$
$\therefore \frac{d V}{d A}=1cm$
Note:
We cannot put Area directly in formula of volume,
$V=\frac{4}{3}\pi r^{2}$
$S\! o, we\; can\; not\; write,V=\frac{A.r}{3}$

Derivative as a rate measure exercise 12.1 question 4

Answer: 3cm
Hint:
We have to differentiate area of circular disc ( $A$ ) with respect to its circumference ( $L$ )
Area of circular disc, $A=\pi r^{2}$
Circumference of circular disc, $L=2\pi r$
Given:
Radius of a circular disc, $r=3 cm$
Solution:
Here we have,
Radius of disc, $r=3 cm$
Circumference of disc, $L=2\pi r$
Let’s differentiate with the circumference with respect to radius $r$
$\frac{dL}{dr}=2\pi$
$\therefore \frac{dr}{dL}=\frac{1}{2\pi }\: \; \; \; \; \; \; \; \; .....(i)$ $\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]$
Area of circular disc, $A=\pi r^{2}$ .
Let’s differentiate with the area with respect to $L$
$\therefore \frac{d A}{d L}=2 \pi r\frac{dr}{d L}$
$\therefore \frac{d A}{d L}=\frac{2 \pi r}{2 \pi }$ {From eq.(i)}
$\therefore \frac{d A}{d L}=r$
Putting value of $r=3cm$ $r = 3cm$
Note:
Here we cannot write directly,
$A=\frac{L}{2}r$ It's wrong

Derivative as a rate measure exercise 12.1 question 5

Answer:

$\frac{2}{3}\pi rh$
Hint:
Here we have to differentiate volume of cone ( $V$ ) with respect to radius of base ( $r$ )
$V\! olume\; o\! f\; cone, V=\frac{1}{3}\pi r^{2}h$
Where $h$ is height of cone.
Given:
Volume of cone.
Solution:
Here we have,
$r$ = Radius of base of cone
$h$ = Height of cone
$V\! olume\; o\! f\; cone, V=\frac{1}{3}\pi r^{2}h$
Let’s differentiate $V$ with respect to $r$ ,
$\therefore \frac{dV}{dr}=\frac{2}{3}\pi rh$ $\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]$
Note:
Here height ( $h$ ) is constant so we don’t have to differentiate it.

Derivative as a rate measure exercise 12.1 question 6

Answer:
$10\pi cm^{2}/cm$
Hint:
Here we have to differentiate Area of circle ( $A$ ) with respect to its radius ( $r$ )
Area of circle, $A=\pi r^{2}$
Given:
Radius of circle, $r=5cm$
Solution:
Here we have,
$r$ = Radius of circle = $5cm$
Area of circle, $A=\pi r^{2}$
Let’s differentiate $A$ with respect to radius
$\therefore \frac{d A}{d r}=\frac{d}{d r}\left(\pi r^{2}\right)$
$=2\pi r$ $\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]$
Put, $r=5cm$
$\therefore \frac{d A}{d r}=10\pi cm^{2}/cm$
Note:
Here unit of rate is cm²/cm we can’t write unit cm because it shows rate means
Area ( $A$ ) varying with respect to radius ( $r$ )

Derivative as a rate measure exercise 12.1 question 7

Answer:
$4\pi r^{2},16\pi \; cm^{3}/cm$
Hint:
Here the volume of ball is the volume of sphere ( $V$ )
$The\; volume\; o\! f\; the\; ball = the\; volume\; o\! f\; spher\! e \; V=\frac{4}{3}\pi r^{3}$
And we have to differentiate $V$ with respect to radius ( $r$ )
Given:
Radius of ball $=r=2cm$
Solution:
Here we have,
Radius of ball $=r=2cm$
$Volume\; o\! f\; ball = V=\frac{4}{3}\pi r^{3}$
Let’s differentiate $V$ with respect to $r$
$\therefore \frac{d V}{d r}=\frac{d}{d r}\left(\frac{4}{3}\pi r^{3}\right)$ $\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]$
$\therefore \frac{d V}{d r}=4\pi r^{2}\; \; \; \; \; \; \; \; \; \; \; \; .....(i)$
Equation ( $i$ ) is rate of change of volume of ball with respect to radius.
Let's put. $r=2 cm$
$\therefore \frac{d V}{d r}=4\pi 2^{2}$
$\therefore \frac{d V}{d r}=16\pi \; cm^{3}/cm$
Note:
It we take $r=2cm$
The correct answer would be
$\frac{dV}{dr}=16\pi \; cm^{3}/cm$
And we can’t write the unit of given rate cm² because it represents volume ( $V$ ) varying with respect to radius ( $r$ )

Derivative as a rate measure exercise 12.1 question 8

Answer: Rs 20.967
Hint:
To find marginal cost at given point, we have to differentiate the function with respect to its variable and put the value of variable in formula.
Given:
$c(x)=0.007x^{3}-0.003x^{2}+15x+4000$
Solution:
Here we have,
Marginal cost
$c(x)=0.007x^{3}-0.003x^{2}+15x+4000$
Let’s differentiate the c( $x$ ) with respect to $x$
$\therefore \frac{dc(x)}{dx}=c'(x)$
$=3\times 0.007x^{2}-2\times 0.003x+15$ $\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]$
Let's put $x=17$ units
$\therefore c'(17)=3\times 0.007(17)^{2}-2\times 0.003\times 17+15$
$=6.069-0.102+15$ $\therefore c'(17)=20.967 Rs$

So $20.967 Rs$ is the marginal cost when 17 units are produced.

Derivative as a rate measure exercise 12.1 question 9

Answer: Rs 208
Hint:
To find marginal revenue we have to differentiate total revenue R( $x$ ) with respect to units $x$
Given:
$R(x)=13x^{2}+26x+15$
Solution:
Here we have,
Units, $x=7$
$Total\; r\! evenue \; R(x)=13x^{2}+26x+15$
Let’s differentiate R( $x$ ) with respect to $x$
$\therefore \frac{d R(x)}{d x}=R^{\prime}(x)=\frac{d}{d x}\left(13 x^{2}+26 x+15\right)$
$\therefore R'(x)=26x+26$ $\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]$
Let's put $x=7$
$\therefore R'(7)=26(7)+26$
$\therefore R'(7)=208$
So, the marginal revenue for $x=7$ is 208.

Derivative as a rate measure exercise 12.1 question 10

Answer:
$M\! R=Rs66$ . It indicates the extra money spent when number of employees increase from 5 to 6
Given:
$R(x)=3x^{2}+36x+5$
Solution:
Here we have,
Total revenue
$R(x)=3x^{2}+36x+5$
To find marginal revenue $M\! R$ let’s differentiate R( $x$ ) with respect to $x$
$\therefore \frac{d}{dx}R(x)=M\! R=6x+36$ $\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]$
To find $M\! R$
Put $x=5$ in $M\! R$
$\therefore M\! R=6(5)+36$
$\therefore M\! R=66\; Rs$
Here we can write
$M\! R=\lim _{h \rightarrow 1} \frac{R(x+h)-R(x)}{h}=66$
So, we can say thet in first there were 5 employees but after they become 6 employees.
Note:
Here we have taken
$M\! R=\lim _{h \rightarrow 1} \frac{R(x+h)-R(x)}{h}$
$N\! ot \; \lim _{h \rightarrow 0}\; because\; h\; represents$

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

RD Sharma Class 12 Solutions Chapter 12 Derivative as a rate measure - Other Exercise

Number of employees which will always be integer and greater than zero.

The 12th chapter for class 12 mathematics, Derivatives as a Derivative as a rate measure has a specific solution book names Class12 RD Sharma Chapter 12 Exercise 12.1 Solution. The RD Sharma reference books for this exercise can be found at the Career 360 website. These RD Sharma Class 12th Exercise 12.1 books act as reference sources for the students as well as the tutors.

As the exercise 12.1 is the first exercise of the 12th chapter, it contains basic formulae which were taught in the previous academic year. This exercise helps students in attaining more marks easily, therefore, it must be learnt well. This is where the RD Sharma Class 12th Exercise 12.1 lends a helping hand.

The RD Sharma Class 12 solutions chapter 12 exercise 12.1 contains many shortcuts that arrive at the solution quickly. Tabulation of all the formulae makes the students grasp it effortlessly. In the RD Sharma Class 12th Exercise 12 .1, there are 10 questions solved. Some of the topics that these sums revolve around are rate of chance of TSA of cylinder, volume of the cylinder, application-based questions related to Derivative as a rate measurer.

Benefits that the students gain by using the RD Sharma Class 12th Exercise 12.1 book:

It is a fact that the teachers cannot be around the students all the time to clarify their doubts. At such circumstances, the RD Sharma Class 12 Exercise12.1 lends a helping hand.
The additional questions and sums in the RD Sharma Class 12th Exercise 12.1 are updated according to the latest syllabus.
It is easy for everyone to access and download the PDF material from the Career 360 website. There are no complicated steps for it.
Numerous solved sums and practice questions are available in these books.
The ultimate goal of attaining more marks is easy to achieve when the sums are practiced as given in this set of books.

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Frequently Asked Question (FAQs)

1. Why do most of the students rely on the RD Sharma books to prepare for the public exams?

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Online websites should be referred to as recent updates would take place. Yet, websites like the Career 360 provide the option for the students to download the resource material for convenience.

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4. How many questions are asked in class mathematics Exercise 12.1?

There are 10 questions asked in mathematics Exercise 12.1. The accurate solutions for these questions are available at the RD Sharma Class 12 Chapter 12 Exercise 12.1.

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