Mathematics Introduction And Rate of Change of Quantities For CBSE-NCERT
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Class 12 Chapter 6 - APPLICATION OF DERIVATIVES

Introduction And Rate of Change of Quantities For CBSE-NCERT

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`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.
 
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