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

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

Edited By Lovekush kumar saini | Updated on Jan 20, 2022 07:14 PM IST

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

Derivative as a Rate Measurer Excercise: 12 VSA

Derivative as a rate measure exercise Very short answer type question 1

Answer: 9 Units/unit time.
Hint: Here we use the concept of rate of change of bodies and quantities
Given: $s=t^{3}-6 t^{2}+9 t+8$
Solution: Here,
$s=t^{3}-6 t^{2}+9 t+8$
Let’s differentiate along time(t)
$\frac{d s}{d t}=3 t^{2}-12 t+9$
Initial velocity= velocity at (t=0)

$\left.\frac{d s}{d t} \text { (at } \mathrm{t}=0\right)=9 \text { units/unit time }$
So, the initial velocity of the particle is 9 units/unit time.

Derivative as a rate measure exercise Very short answer type question 2

Answer: $3\; cm^{2}/sec$
Hint: Here we all know the volume of sphere is $\mathrm({v})=\frac{4}{3} \pi r^{3}$
Given: $r=2cm$ and the volume of sphere is increasing at the rate of $\text { 3 cubic cm/second }$
Solution: Let r be the radius and v be the volume of then
$\mathrm({v})=\frac{4}{3} \pi r^{3}$
$\frac{d v}{d t}=\frac{4}{3} \pi \times 3 r^{2} \times \frac{d r}{d t}$
$\Rightarrow \frac{d r}{d t}=\frac{1}{4 \pi r^{2}} \times \frac{d V}{d t}$
$\begin{aligned} &\frac{d r}{d t}=\frac{3}{4 \pi(2)^{2}}[r=2 \mathrm{~cm}] \\\\ &\frac{d r}{d t}=\frac{3}{16 \pi} \mathrm{cm} / \mathrm{sec} \end{aligned}$
Now,let S be the surface area of sphere
So, $S=4 \pi r^{2}$
$\Rightarrow \frac{d s}{d t}=8 \pi r \frac{d r}{d t}$
$\Rightarrow \frac{d S}{d t}=8 \pi \times 2 \times \frac{3}{16 \pi}$ (because r=2)
$\Rightarrow \frac{d S}{d t}=3 \mathrm{~cm}^{2} / \mathrm{sec}$

Derivative as a rate measure exercise Very short answer type question 3

Answer: $10 \sqrt{3} \mathrm{~cm}^{2} / \mathrm{s}$
Hint: Here, we all know the formula about equilateral triangle are increasing at the rate
Of 2cm/sec, x=10cm. Here, x stands for edge.
Solution: Let the side of an equilateral triangle be x cm
Area of equilateral triangle $A=\frac{\sqrt{3}}{4} x^{2}$ .................(1)

Differentiating equation (1) along with time, we get
$\begin{aligned} &\frac{d A}{d t}=\frac{\sqrt{3}}{4} \times 2 x \times \frac{d x}{d t} \\\\ &=\frac{\sqrt{3} x}{2} \frac{d x}{d t} \end{aligned}$
Let's put the value of x where (x=10 cm)
$\frac{d A}{d t}=\frac{\sqrt{3} \times 10}{2} \times 2=10 \sqrt{3} \mathrm{~cm}^{2} / \mathrm{sec}$

Derivative as a rate measure exercise Very short answer type question 4

Answer: $0.4 \mathrm{~cm} / \mathrm{sec}$
Hint: Here we use the basic concept of rate of change of bodies and quantities
Given: The side of square is increasing at the rate of 0.1cm/sec
Solution: Let a be side of square
So, $\frac{d a}{d t}=0.1 \mathrm{~cm} / \mathrm{s}$
We know about the formula of square perimeter (p)
So
$\begin{gathered} \mathrm{p}=4 \mathrm{a} \\ \frac{d p}{d t}=4 \frac{d a}{d t} \end{gathered}$ (taking differentiate of a with respect to time)
$\begin{aligned} &=4 \times 0.1 \quad\left(\frac{d a}{d t}=0.1\right) \\ &=0.4 \end{aligned}$
So, the rate of increase of square’s perimeter is 0.4cm/sec

Derivative as a rate measure exercise Very short answer type question 5

Answer: $3.14 \mathrm{~cm} / \mathrm{sec}$
Hint: Here, we use basic concept of rate change of bodies and quantities
Given: The radius of a circle is increasing at the rate of 0.5cm/sec
Solution: Let r be the radius of a circle
So,
$\frac{d r}{d t}=0.5 \mathrm{~cm} / \mathrm{sec}$
We know the formula about circumference of circle
$\begin{aligned} &\mathrm{C}=2 \pi \mathrm{r} \\ &\frac{d c}{d t}=2 \pi \frac{d r}{d t} \end{aligned}$ (differentiate with respect to t)
$\begin{aligned} &=2 \pi \times 0.5\left(\frac{d r}{d t}=0.5\right) \\ &=\pi \\ &=3.14 \mathrm{~cm} / \mathrm{sec} \end{aligned}$
So, the rate of increase about circle’s circumference is 3.14 cm/sec.

Derivative as a rate measure exercise Very short answer type question 6

Answer: $1 \mathrm{~cm} / \mathrm{sec}$
Hint: Here, we use basic concept of rate of change of bodies and quantities.
Given: The side of an equilateral triangle is increasing $\frac{1}{3} \mathrm{~cm} / \mathrm{sec}$
Solution: Let a be the side of an equilateral triangle is
So, $\frac{d a}{d t}=\frac{1}{3} \mathrm{~cm} / \mathrm{sec}$
We know the formula about perimeter of equilateral triangle
$\begin{aligned} &p=3 a \\\\ &\frac{d p}{d t}=3 \frac{d a}{d t} \end{aligned}$ (differentiate with respect to time (t))
$\begin{aligned} &\frac{d p}{d t}=3 \times \frac{1}{3}\left(\frac{d a}{d t}=\frac{1}{3}\right) \\\\ &\frac{d p}{d t}=1 \mathrm{~cm} / \mathrm{sec} \end{aligned}$
So, the rate change of the equilateral triangle’s perimeter is 1cm/sec

Derivative as a rate measure exercise Very short answer type question 7

Answer: $1 \text { square/unit }$
Hint: Here, we use basic formula of sphere’s area and volume
Given: Volume is changing at the same rate of it’s radius of sphere
$\text { So, } \frac{d v}{d t}=\frac{d r}{d t}(\text { Given) }$
Solution: We know the volume of sphere,
$v=\frac{4}{3} \pi r^{3}$
$\begin{aligned} &\Rightarrow \frac{d V}{d t}=4 \pi r^{2} \times \frac{d r}{d t} \\\\ &\Rightarrow \frac{d V}{d t}=4 \pi r^{2} \times \frac{d V}{d t} \quad \text { because } \frac{d V}{d t}=\frac{d r}{d t} \quad(\text { given }) \end{aligned}$
$\begin{gathered} \mathrm{So}, 1=4 \pi r^{2} \\\\ r^{2}=\frac{1}{4 \pi} \end{gathered}$ .......................(1)
Surface of sphere is
$\begin{aligned} &S=4 \pi r^{2} \\ &=4 \pi \times \frac{1}{4 \pi}\left(r^{2}=\frac{1}{4 \pi} \text { Frome\; quation(1) }\right) \\ &S=1 \text { square /unit } \end{aligned}$
So, surface of sphere is 1 square/unit

Derivative as a rate measure exercise Very short answer type question 10

Answer: $\frac{1}{20} \mathrm{rad} / \mathrm{s}$
Hint: Here we use the basic concept of rate of change about bodies and quantities.
Given:

Solution: let us consider the above ?ABC,right angled B
In above diagram,the right angled is at B and left angle floor and ladder be $\theta$
Let at any time ‘t’ AB=x cm and BC=y cm and we know that AC = 500 cm.
$\sin \theta=\frac{x}{500} \text { and } \cos \theta=\frac{y}{500}$
So,
$\begin{aligned} &\Rightarrow x=500 \sin \theta \\ &\Rightarrow y=500 \cos \theta \end{aligned}$
Also, it is given that
$\begin{aligned} &\frac{d x}{d t}=10 \mathrm{~cm} / \mathrm{sec} \\\\ &\frac{d(500 \sin \theta)}{d t}=10 \mathrm{~cm} / \sec \end{aligned}$
$\begin{aligned} &\text { for } Y=2 m=200 \mathrm{~cm} \text { (given) } \\\\ &\Rightarrow \frac{d \theta}{d t}=\frac{1}{50 \times \frac{y}{500}}=\frac{10}{y}=\frac{10}{200}=\frac{1}{20} \mathrm{rad} / \sec \end{aligned}$

Also see,

RD Sharma Solutions Class 12 Mathematics Chapter 12 exercise 12.1
RD Sharma Solutions Class 12 Mathematics Chapter 12 exercise 12.2
RD Sharma Solutions Class 12 Mathematics Chapter 12 exercise MCQ
RD Sharma Solutions Class 12 Mathematics Chapter 12 exercise FBQ

The RD Sharma class 12th exercise VSA contain concise answers of utmost importance. In RD Sharma class 12 solution chapter 12 exercise VSA, the questions will cover many important concepts that must be learned by everyone. Using RD Sharma Class 12th exercise VSA solutions, students will be prepared for their exams and even impress their peers by solving homework questions faster and with ease.

RD Sharma Class 12 solution of Derivative as a rate measure exercise VSA will follow the question pattern of NCERT Books and provide accurate answers every time. Since the RD Sharma class 12th exercise VSA has been prepared by experts, students will find some alternative methods of solving problems and add to their existing knowledge. RD Sharma Class 12 Chapter 12 exercise VSA has 10 questions and It includes the following concepts:

Meaning of derivative as a rate measurer
Theorems and remarks on derivative as a rate measurer.
Rate of change of TSA of cylinder, volume of cylinder and many more.
Application based question on derivative as a rate measurer

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RD Sharma Chapter-wise Solutions

Chapter 1: Relations

Chapter 2: Functions

Chapter 3: Inverse Trigonometric Functions

Chapter 4: Algebra of Matrices

Chapter 5: Determinants

Chapter 6: Adjoint and Inverse of a Matrix

Chapter 7: Solution of Simultaneous Linear Equations

Chapter 8: Continuity

Chapter 9: Differentiability

Chapter 10: Differentiation

Chapter 11: Higher Order Derivatives

Chapter 12: Derivative as a Rate Measurer

Chapter 13: Differentials, Errors, and Approximations

Chapter 14: Mean Value Theorems

Chapter 15: Tangents and Normals

Chapter 16: Increasing and Decreasing Functions

Chapter 17: Maxima and Minima

Chapter 18: Indefinite Integrals

Chapter 19: Definite Integrals

Chapter 20: Areas of Bounded Regions

Chapter 21: Differential Equations

Chapter 22 Algebra of Vectors

Chapter 23: Scalar Or Dot Product

Chapter 24: Vector or Cross Product

Chapter 25: Scalar Triple Product

Chapter 26: Direction Cosines and Direction Ratios

Chapter 27 Straight line in space

Chapter 28: The plane

Chapter 29: Linear programming

Chapter 30: Probability

Chapter 31: Mean and variance of a random variable

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