Mathematics Introduction And Rate of Change of Quantities For CBSE-NCERT
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### Topic covered

star Introduction
star Rate of Change of Quantities

### Introduction

\color{green} ✍️ we will learn how the derivative can be used

(i) to determine rate of change of quantities,
(ii) to find the equations of tangent and normal to a curve at a point,
(iii) to find intervals on which a function is increasing or decreasing.

### Rate of Change of Quantities

\color{green} ✍️ color{red}{(dy)/(dx)} or f'(x) represents the rate of color{red}{"change of" \ \ y\ \ "with respect" \ \ t o \ \ x}

=>color{blue}{(dy)/(dx)]]_(x=0) =f′(x_0) represents the rate of change of y with respect to x at x = x_0.

\color{green} ✍️ Further, if two variables x and y are varying with respect to another variable t, i.e., if x = f (t) and y = g(t ), then by Chain Rule

color{blue}{(dy) / (dx) =((dy)/(dt))/ ((dx)/(dt))  If (dx)/(dt) ne 0}

color{red}{"Concept :"} : color{green}{(dy)/(dx) "is positive if"\ \ y\ "increases as" \ x \ "increases and is negative if" \ y \ "decreases as" \ x \ "increases"}.
Q 3185734667

Find the rate of change of the area of a circle per second with respect to
its radius r when r = 5 cm.
Class 12 Chapter 6 Example 1
Solution:

The area A of a circle with radius r is given by A = πr^2. Therefore, the rate

of change of the area A with respect to its radius r is given by  (dA)/(dr) = d/(dr) (pi r^2) = 2 pi r

When r = 5 cm,  (dA)/(dr) = 10 pi . Thus, the area of the circle is changing at the rate of
10π (cm^2)/s .
Q 3115734669

The volume of a cube is increasing at a rate of 9 cubic centimetres per
second. How fast is the surface area increasing when the length of an edge is 10
centimetres ?
Class 12 Chapter 6 Example 2
Solution:

Let x be the length of a side, V be the volume and S be the surface area of
the cube. Then, V = x^3 and S = 6x^2 , where x is a function of time t.

Now  (dV)/(dt) = 9 cm^3/s (Given)

Therefore  9 = (dV)/(dt) = d/(dt) (x^3) = d/(dx) (x^3 ) * (dx)/(dt) (By Chain Rule)

= 3 x^2 * (dx)/(dt)

or  (dx)/(dt) = 3/x^2 ........(1)

Now  (dS)/(dt) = d/(dt) (6 x^2) = d/(dx) (6 x^2) * (dx)/(dt) (By Chain Rule)

= 12 x * (3/x^2) = 36/x (Using (1))

Hence, when  x = 10 cn , (dS)/(dt) = 3.6 cm^2/s
Q 3105834768

A stone is dropped into a quiet lake and waves move in circles at a speed
of 4cm per second. At the instant, when the radius of the circular wave is 10 cm, how
fast is the enclosed area increasing?
Class 12 Chapter 6 Example 3
Solution:

The area A of a circle with radius r is given by A = πr^2. Therefore, the rate
of change of area A with respect to time t is

 (dA)/(dt) = d/(dt) (pi r^2) =d/(dr) (pi r^2) * (dr)/(dt) = 2 pi r (dr)/(dt) (By Chain Rule)

It is given that  (dr)/( dt) = 4 cm/s

Therefore, when r = 10 cm, (dA)/(dt) = 2pi (10) (4) = 80 pi

Thus, the enclosed area is increasing at the rate of 80π cm^2/s, when r = 10 cm.
Q 3145034863

The length x of a rectangle is decreasing at the rate of 3 cm/minute and
the width y is increasing at the rate of 2 cm/minute. When x =10cm and y = 6cm, find
the rates of change of (a) the perimeter and (b) the area of the rectangle.
Class 12 Chapter 6 Example 4
Solution:

Since the length x is decreasing and the width y is increasing with respect to
time, we have

 (dx)/(dt) = -3 cm/min and  (dy)/(dt) = 2 cm/min

(a) The perimeter P of a rectangle is given by
P = 2 (x + y)

Therefore  (dP)/(dt) = 2 ( (dx)/(dt) + (dy)/(dt) ) = 2 (-3 +2) = -2 cm/min

(b) The area A of the rectangle is given by

A = x . y

Therefore  (dA)/(dt) = (dx)/(dt) * y + x * (dy)/(dt)

= – 3(6) + 10(2) (as x = 10 cm and y = 6 cm)

= 2 cm^2/min
Q 3175034866

The total cost C(x) in Rupees, associated with the production of x units of
an item is given by
C(x) = 0.005 x^3 – 0.02 x2 + 30x + 5000
Find the marginal cost when 3 units are produced, where by marginal cost we
mean the instantaneous rate of change of total cost at any level of output.
Class 12 Chapter 6 Example 5
Solution:

Since marginal cost is the rate of change of total cost with respect to the
output, we have

Marginal cost (MC) = (dC)/(dx) = 0.005 (3x^2) - 0.02 (2x) +30

when  x = 3, MC = 0.015(3^2 ) − 0.04(3) + 30

= 0.135 – 0.12 + 30 = 30.015
Hence, the required marginal cost is Rs 30.02 (nearly).
Q 3105134968

The total revenue in Rupees received from the sale of x units of a product
is given by R(x) = 3x^2 + 36x + 5. Find the marginal revenue, when x = 5, where by
marginal revenue we mean the rate of change of total revenue with respect to the
number of items sold at an instant.
Class 12 Chapter 6 Example 6
Solution:

Since marginal revenue is the rate of change of total revenue with respect to
the number of units sold, we have

Marginal Revenue  (MR ) = (dR)/(dx) = 6x + 36

when x = 5, MR = 6(5) + 36 = 66
Hence, the required marginal revenue is Rs 66.