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194
MATHEMATICS
Chapter
6
APPLICATION OF
DERIVATIVES
With the Calculus as a key, Mathematics can be successfully applied
to the explanation of the course of Nature
...
1 Introduction
In Chapter 5, we have learnt how to find derivative of composite functions, inverse
trigonometric functions, implicit functions, exponential functions and logarithmic functions
...
g
...
For instance, 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 turning points on
the graph of a function which in turn will help us to locate points at which largest or
smallest value (locally) of a function occurs
...
Finally, we use the derivative to find
approximate value of certain quantities
...
2 Rate of Change of Quantities
ds
, we mean the rate of change of distance s with
dt
respect to the time t
...
Further, if two variables x and y are varying with respect to another variable t, i
...
,
if x = f ( t ) and y = g ( t ) , then by Chain Rule
dy
dy
=
dt
dx
dx
dx
≠0
, if
dt
dt
APPLICATION OF DERIVATIVES
195
Thus, the rate of change of y with respect to x can be calculated using the rate of
change of y and that of x both with respect to t
...
Example 1 Find the rate of change of the area of a circle per second with respect to
its radius r when r = 5 cm
...
Therefore, the rate
of change of the area A with respect to its radius r is given by
When r = 5 cm,
dA d
= ( π r 2 ) = 2π r
...
Thus, the area of the circle is changing at the rate of
dr
10π cm2/s
...
How fast is the surface area increasing when the length of an edge is 10
centimetres ?
Solution Let x be the length of a side, V be the volume and S be the surface area of
the cube
...
dV
= 9cm3/s (Given)
dt
Now
Therefore
9=
dV d 3
d
dx
= ( x ) = ( x3 ) ⋅
(By Chain Rule)
dt dt
dx
dt
2
= 3x ⋅
dx
dt
or
3
dx
= 2
x
dt
Now
dS
d
d
dx
(6x2 ) = (6x 2) ⋅
=
dt
dt
dx
dt
3
= 12x ⋅ 2
x
Hence, when
x = 10 cm,
dS
= 3
...
(1)
36
=
x
(By Chain Rule)
(Using (1))
196
MATHEMATICS
Example 3 A stone is dropped into a quiet lake and waves move in circles at a speed
of 4cm per second
...
Therefore, the rate
of change of area A with respect to time t is
dA
d
d
dr
dr
(π r 2 ) = (π r 2) ⋅
=
= 2π r
dt
dt
dr
dt
dt
It is given that
(By Chain Rule)
dr
= 4cm/s
dt
dA
= 2π (10) (4) = 80π
dt
Thus, the enclosed area is increasing at the rate of 80π cm2/s, when r = 10 cm
...
Note
Example 4 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 2cm/minute
...
Solution Since the length x is decreasing and the width y is increasing with respect to
time, we have
dx
dy
= −3 cm/min
= 2 cm/min
and
dt
dt
(a) The perimeter P of a rectangle is given by
P = 2 (x + y)
dP
dx dy
= 2 + = 2 ( −3 + 2) = −2 cm/min
dt dt
dt
(b) The area A of the rectangle is given by
A=x
...
005 x3 – 0
...
Solution Since marginal cost is the rate of change of total cost with respect to the
output, we have
dC
= 0
...
02(2 x) + 30
dx
Marginal
cost (MC) =
When
2
x = 3, MC = 0
...
04(3) + 30
= 0
...
12 + 30 = 30
...
02 (nearly)
...
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
...
Marginal Revenue
(MR) =
EXERCISE 6
...
Find the rate of change of the area of a circle with respect to its radius r when
(a) r = 3 cm
(b) r = 4 cm
2
...
How fast is the
surface area increasing when the length of an edge is 12 cm?
3
...
Find the rate
at which the area of the circle is increasing when the radius is 10 cm
...
An edge of a variable cube is increasing at the rate of 3 cm/s
...
A stone is dropped into a quiet lake and waves move in circles at the speed of
5 cm/s
...
The radius of a circle is increasing at the rate of 0
...
What is the rate of
increase of its circumference?
7
...
When x = 8cm and y = 6cm, find
the rates of change of (a) the perimeter, and (b) the area of the rectangle
...
A balloon, which always remains spherical on inflation, is being inflated by pumping
in 900 cubic centimetres of gas per second
...
9
...
Find the rate at
which its volume is increasing with the radius when the later is 10 cm
...
A ladder 5 m long is leaning against a wall
...
How fast is its height
on the wall decreasing when the foot of the ladder is 4 m away from the wall ?
11
...
Find the points on the curve at
which the y-coordinate is changing 8 times as fast as the x-coordinate
...
At what rate is the
2
volume of the bubble increasing when the radius is 1 cm?
12
...
A balloon, which always remains spherical, has a variable diameter
3
(2 x + 1)
...
14
...
The falling sand forms a cone
on the ground in such a way that the height of the cone is always one-sixth of the
radius of the base
...
The total cost C (x) in Rupees associated with the production of x units of an
item is given by
C (x) = 0
...
003x2 + 15x + 4000
...
16
...
Find the marginal revenue when x = 7
...
17
...
The total revenue in Rupees received from the sale of x units of a product is
given by
R(x) = 3x2 + 36x + 5
...
3 Increasing and Decreasing Functions
In this section, we will use differentiation to find out whether a function is increasing or
decreasing or none
...
The graph of this function is a
parabola as given in Fig 6
...
Values left to origin
x
f (x) = x
–2
3
−
2
0
f (x) = x2
0
1
2
1
3
2
2
1
1
4
0
x
4
9
4
–1
1
−
2
Values right to origin
2
0
1
4
1
9
4
4
as we move from left to right, the
height of the graph increases
as we move from left to right, the
height of the graph decreases
Fig 6
...
1) to the right of the origin
...
For this reason, the function is said to be increasing for the real numbers x > 0
...
Consequently, the function is said to be decreasing for the real numbers x < 0
...
Definition 1 Let I be an interval contained in the domain of a real valued function f
...
(ii) strictly increasing on I if x1 < x2 in I ⇒ f (x1) < f (x2) for all x1, x2 ∈ I
...
(iv) strictly decreasing on I if x1 < x2 in I ⇒ f (x1) > f (x2) for all x1, x2 ∈ I
...
2
...
2
We shall now define when a function is increasing or decreasing at a point
...
Then f is said to be increasing, strictly increasing, decreasing or strictly decreasing at
x0 if there exists an open interval I containing x0 such that f is increasing, strictly
increasing, decreasing or strictly decreasing, respectively, in I
...
A function f is said to be increasing at x0 if there exists an interval I = (x0 – h, x0 + h),
h > 0 such that for x1, x2 ∈ I
x1 < x2 in I ⇒ f (x1 ) ≤ f (x2)
Similarly, the other cases can be clarified
...
Solution Let x1 and x2 be any two numbers in R
...
We shall now give the first derivative test for increasing and decreasing functions
...
Theorem 1 Let f be continuous on [a, b] and differentiable on the open interval
(a,b)
...
Then, by Mean Value Theorem (Theorem 8 in Chapter 5), there exists a point c
between x1 and x2 such that
f (x2) – f(x1) = f ′(c) (x2 – x1)
i
...
f (x2) – f(x1) > 0
i
...
Thus, we have
(as f ′(c) > 0 (given))
f (x2 ) > f (x 1)
x1 < x2 ⇒ f ( x1 ) < f ( x2 ), for all x1 , x2 ∈[ a , b]
Hence, f is an increasing function in [a,b]
...
It is left as an exercise to the reader
...
Similarly, if f ′(x) < 0 for x in an interval excluding the end
points and f is continuous in the interval, then f is strictly decreasing
...
However, the converse need not
be true
...
Solution Note that
f ′(x) = 3x2 – 6x + 4
202
MATHEMATICS
= 3(x2 – 2x + 1) + 1
= 3(x – 1)2 + 1 > 0, in every interval of R
Therefore, the function f is strictly increasing on R
...
Solution Note that f ′(x) = – sin x
(a) Since for each x ∈ (0, π), sin x > 0, we have f ′(x) < 0 and so f is strictly
decreasing in (0, π)
...
(c) Clearly by (a) and (b) above, f is neither increasing nor decreasing in (0, 2π)
...
Now
the point x = 2 divides the real line into two
disjoint intervals namely, (– ∞, 2) and (2,
∞) (Fig 6
...
In the interval (– ∞, 2), f ′(x)
= 2x – 4 < 0
...
3
Therefore, f is strictly decreasing in this interval
...
Example 11 Find the intervals in which the function f given by f (x) = 4x3 – 6x2 – 72x + 30
is (a) strictly increasing (b) strictly decreasing
...
The
points x = – 2 and x = 3 divides the real line into
three disjoint intervals, namely, (– ∞, – 2), (– 2, 3)
Fig 6
...
In the intervals (– ∞, – 2) and (3, ∞), f ′(x) is positive while in the interval (– 2, 3),
f ′(x) is negative
...
However, f is neither increasing nor decreasing in R
...
Solution We have
or
f (x) = sin 3x
f ′(x) = 3cos 3x
π 3π
π
Therefore, f ′(x) = 0 gives cos 3x = 0 which in turn gives 3x = ,
(as x ∈ 0,
2 2
2
π
π
π
3π
implies 3x ∈ 0, )
...
The point x = divides the interval
6
2
6
2
π
π π
into two disjoint intervals 0, and ,
...
5
π
0, 2
π
π
π
Now, f ′( x ) > 0 for all x ∈ 0, as 0 ≤ x < ⇒ 0 ≤ 3 x < and f ′( x ) < 0 for
6
6
2
π π
π
π
π
3π
all x ∈ , as < x < ⇒ < 3 x <
...
6
6 2
Also, the given function is continuous at x = 0 and x =
π
f is increasing on 0, and decreasing on
6
π
...
Example 13 Find the intervals in which the function f given by
f (x) = sin x + cos x, 0 ≤ x ≤ 2π
is strictly increasing or strictly decreasing
...
4 4 4
4
Note that
or
π 5π
,
as 0 ≤ x ≤ 2π
4 4
Fig 6
...
2
1
...
2
...
3
...
Find the intervals in which the function f given by f (x) = 2x2 – 3x is
(a) strictly increasing
(b) strictly decreasing
5
...
Find the intervals in which the following functions are strictly increasing or
decreasing:
(a) x2 + 2x – 5
(c) –2x3 – 9x2 – 12x + 1
(e) (x + 1)3 (x – 3)3
7
...
8
...
9
...
(2 + cos θ)
2
206
MATHEMATICS
10
...
11
...
π
12
...
On which of the following intervals is the function f given by f(x) = x100 + sin x –1
strictly decreasing ?
(A) (0,1)
π
(B) , π
2
π
(C) 0,
2
(D) None of these
14
...
15
...
Prove that the function f given by
f (x) = x +
1
is strictly increasing on I
...
Prove that the function f given by f (x) = log sin x is strictly increasing on 0,
2
π
and strictly decreasing on , π
...
Prove that the function f given by f (x) = log cos x is strictly decreasing on
π
π
0, and strictly increasing on , π
...
Prove that the function given by f (x) = x3 – 3x2 + 3x – 100 is increasing in R
...
The interval in which y = x2 e–x is increasing is
(A) (– ∞, ∞)
(B) (– 2, 0)
(C) (2, ∞)
(D) (0, 2)
6
...
Recall that the equation of a straight line passing through a given point (x0, y0)
having finite slope m is given by
y – y0 = m (x – x0)
APPLICATION OF DERIVATIVES
207
Note that the slope of the tangent to the curve y = f (x)
at the point (x0, y0) is given by
dy
( = f ′( x0 ))
...
Therefore, the equation of the
f ′ ( x0 )
Fig 6
...
e
...
dx
Particular cases
(i) If slope of the tangent line is zero, then tan θ = 0 and so θ = 0 which means the
tangent line is parallel to the x-axis
...
π
(ii) If θ → , then tan θ → ∞, which means the tangent line is perpendicular to the
2
x-axis, i
...
, parallel to the y-axis
...
Example 14 Find the slope of the tangent to the curve y = x3 – x at x = 2
...
dx x =2
208
MATHEMATICS
Example 15 Find the point at which the tangent to the curve y = 4 x − 3 − 1 has its
2
...
3
2
2
So
=
4x − 3
3
or
4x – 3 = 9
or
x=3
Now y = 4 x − 3 − 1
...
Therefore, the required point is (3, 2)
...
x−3
Solution Slope of the tangent to the given curve at any point (x,y) is given by
dy
2
=
dx
( x − 3) 2
But the slope is given to be 2
...
Thus, there are two tangents to the
given curve with slope 2 and passing through the points (2, 2) and (4, – 2)
...
Solution Differentiating
x 2 y2
+
= 1 with respect to x, we get
4 25
x 2 y dy
+
=0
2 25 dx
−25 x
dy
=
4 y
dx
(i) Now, the tangent is parallel to the x-axis if the slope of the tangent is zero which
or
x 2 y2
−25 x
+
= 1 for x = 0 gives
= 0
...
Then
4 25
4 y
y2 = 25, i
...
, y = ± 5
...
gives
(ii) The tangent line is parallel to y-axis if the slope of the normal is 0 which gives
x 2 y2
4y
+
= 1 for y = 0 gives x = ± 2
...
e
...
Therefore,
4 25
25 x
points at which the tangents are parallel to the y-axis are (2, 0) and (–2, 0)
...
Solution Note that on x-axis, y = 0
...
Thus, the curve cuts the x-axis at (7, 0)
...
Hence, the equation of the
20
tangent at (7, 0) is
y−0=
1
( x − 7)
20
20 y − x + 7 = 0
or
Example 19 Find the equations of the tangent and normal to the curve
at (1, 1)
...
dx (1,1)
So the equation of the tangent at (1, 1) is
y – 1 = – 1 (x – 1)
or
y+x–2=0
Also, the slope of the normal at (1, 1) is given by
−1
=1
slope of the tangent at (1,1)
Therefore, the equation of the normal at (1, 1) is
y – 1 = 1 (x – 1)
or
y–x=0
Example 20 Find the equation of tangent to the curve given by
x = a sin3 t ,
y = b cos3 t
at a point where t =
π
...
(1)
APPLICATION OF DERIVATIVES
211
dy
dy dt
−3bcos2 t sin t −b cos t
=
=
=
dx dx
3asin2 t cos t
a sin t
dt
or
Therefore, slope of the tangent at t =
π
is
2
π
−b cos
dy
2=0
dx t= π =
π
a sin
2
2
Also, when t =
curve at t =
π
, x = a and y = 0
...
e
...
e
...
EXERCISE 6
...
Find the slope of the tangent to the curve y = 3x4 – 4x at x = 4
...
x−2
3
...
4
...
2
...
Find the slope of the normal to the curve x = a cos3 θ, y = a sin 3 θ at θ =
...
Find the slope of the normal to the curve x = 1 − a sin θ, y = b cos 2 θ at θ =
...
Find points at which the tangent to the curve y = x3 – 3x2 – 9x + 7 is parallel to
the x-axis
...
Find a point on the curve y = (x – 2)2 at which the tangent is parallel to the chord
joining the points (2, 0) and (4, 4)
...
Find the point on the curve y = x3 – 11x + 5 at which the tangent is y = x – 11
...
Find the equation of all lines having slope – 1 that are tangents to the curve
1
, x ≠ 1
...
Find the equation of all lines having slope 2 which are tangents to the curve
y=
1
, x ≠ 3
...
Find the equations of all lines having slope 0 which are tangent to the curve
y=
y=
1
...
14
...
Find points on the curve
(i)
(ii)
(iii)
(iv)
y = x4 – 6x3 + 13x2 – 10x + 5 at (0, 5)
y = x4 – 6x3 + 13x2 – 10x + 5 at (1, 3)
y = x3 at (1, 1)
y = x2 at (0, 0)
π
4
15
...
17
...
19
...
Show that the tangents to the curve y = 7x3 + 11 at the points where x = 2 and
x = – 2 are parallel
...
For the curve y = 4x3 – 2x5, find all the points at which the tangent passes
through the origin
...
20
...
APPLICATION OF DERIVATIVES
213
21
...
22
...
23
...
24
...
25
...
Choose the correct answer in Exercises 26 and 27
...
The slope of the normal to the curve y = 2x2 + 3 sin x at x = 0 is
1
1
(C) –3
(D) −
3
3
2
27
...
5 Approximations
In this section, we will use differentials to approximate values of certain quantities
...
Let ∆x denote a small
increment in x
...
We
define the following
(i) The differential of x, denoted by dx, is
defined by dx = ∆x
...
dx
Fig 6
...
214
MATHEMATICS
In case dx = ∆x is relatively small when compared with x, dy is a good approximation
of ∆y and we denote it by dy ≈ ∆y
...
8
...
8, we may note that the
differential of the dependent variable is not equal to the increment of the variable
where as the differential of independent variable is equal to the increment of the
variable
...
6
...
Let x = 36 and let ∆x = 0
...
Then
∆y =
x + ∆x − x = 36
...
6 − 6
or
36
...
6) =
(0
...
05
dx
2 x
2 36
Thus, the approximate value of
36
...
05 = 6
...
Example 22 Use differential to approximate
1
(25) 3
...
Let x = 27 and let ∆x = – 2
...
074
27
1
Thus, the approximate value of (25) 3 is given by
3 + (– 0
...
926
(as y = x )
APPLICATION OF DERIVATIVES
215
Example 23 Find the approximate value of f (3
...
Solution Let x = 3 and ∆x = 0
...
Then
f (3
...
Therefore
or
f (x + ∆x) = f (x) + ∆y
≈ f (x) + f ′(x) ∆x
(as dx = ∆x)
2
f (3
...
02) (as x = 3, ∆x = 0
...
02)
= 45 + 0
...
46
Hence, approximate value of f (3
...
46
...
Solution Note that
V = x3
or
dV
dV =
∆x = (3x2) ∆x
dx
= (3x2) (0
...
06x3 m3
Thus, the approximate change in volume is 0
...
(as 2% of x is 0
...
03 cm,
then find the approximate error in calculating its volume
...
Then r = 9 cm and ∆r = 0
...
Now, the volume V of the sphere is given by
V=
or
4 3
πr
3
dV
= 4πr 2
dr
dV
2
dV =
∆r = (4πr ) ∆r
dr
= 4π(9)2 (0
...
72πcm 3
Thus, the approximate error in calculating the volume is 9
...
Therefore
216
MATHEMATICS
EXERCISE 6
...
Using differentials, find the approximate value of each of the following up to 3
places of decimal
...
3
(iii)
49
...
6
1
1
1
(iv) (0
...
999) 10
(vi) (15) 4
1
1
1
(vii) (26)3
(viii) (255) 4
(ix) (82) 4
1
1
1
(x) (401) 2
(xi) (0
...
57) 3
1
3
1
(xiii) (81
...
968) 2
(xv) (32
...
Find the approximate value of f (2
...
3
...
001), where f (x) = x3 – 7x2 + 15
...
Find the approximate change in the volume V of a cube of side x metres caused
by increasing the side by 1%
...
Find the approximate change in the surface area of a cube of side x metres
caused by decreasing the side by 1%
...
If the radius of a sphere is measured as 7 m with an error of 0
...
7
...
03 m, then find the
approximate error in calculating its surface area
...
If f(x) = 3x2 + 15x + 5, then the approximate value of f (3
...
66
(B) 57
...
66
(D) 77
...
The approximate change in the volume of a cube of side x metres caused by
increasing the side by 3% is
(A) 0
...
6 x3 m3 (C) 0
...
9 x3 m3
6
...
In fact, we will find the ‘turning points’ of the
graph of a function and thus find points at which the graph reaches its highest (or
APPLICATION OF DERIVATIVES
217
lowest) locally
...
Further, we will also find the absolute maximum and absolute minimum
of a function that are necessary for the solution of many applied problems
...
(i) The profit from a grove of orange trees is given by P(x) = ax + bx2 , where a,b
are constants and x is the number of orange trees per acre
...
What is the maximum height the ball will
reach?
(iii) An Apache helicopter of enemy is flying along the path given by the curve
f (x) = x2 + 7
...
What is the nearest distance?
In each of the above problem, there is something common, i
...
, we wish to find out
the maximum or minimum values of the given functions
...
Definition 3 Let f be a function defined on an interval I
...
given by h ( x) = 60 + x −
The number f (c) is called the maximum value of f in I and the point c is called a
point of maximum value of f in I
...
The number f (c), in this case, is called the minimum value of f in I and the point
c, in this case, is called a point of minimum value of f in I
...
The number f (c), in this case, is called an extreme value of f in I and the point c
is called an extreme point
...
9(a), (b) and (c), we have exhibited that graphs of certain particular
functions help us to find maximum value and minimum value at a point
...
218
MATHEMATICS
Fig 6
...
Solution From the graph of the given function (Fig 6
...
Also
f (x) ≥ 0, for all x ∈ R
...
Further, it may be observed
from the graph of the function that f has no maximum
value and hence no point of maximum value of f in R
...
10
Note If we restrict the domain of f to [– 2, 1] only,
then f will have maximum value(– 2)2 = 4 at x = – 2
...
Solution From the graph of the given function
(Fig 6
...
Therefore, the function f has a minimum value 0
and the point of minimum value of f is x = 0
...
Fig 6
...
APPLICATION OF DERIVATIVES
219
(ii) One may note that the function f in Example 27 is not differentiable at
x = 0
...
Solution The given function is an increasing (strictly) function in the given interval
(0, 1)
...
12) of the function f , it
seems that, it should have the minimum value at a
point closest to 0 on its right and the maximum value
at a point closest to 1 on its left
...
It is not possible to locate
such points
...
Also, if x1 is
2
x1 + 1
> x1 for all x1 ∈(0,1)
...
12
2
Therefore, the given function has neither the maximum value nor the minimum
value in the interval (0,1)
...
e
...
Infact, we have the following
results (The proof of these results are beyond the scope of the present text)
Every monotonic function assumes its maximum/minimum value at the end
points of the domain of definition of the function
...
Note By a monotonic function f in an interval I, we mean that f is either
increasing in I or decreasing in I
...
Let us now examine the graph of a function as shown in Fig 6
...
Observe that at
points A, B, C and D on the graph, the function changes its nature from decreasing to
increasing or vice-versa
...
Further, observe that at turning points, the graph has either a little hill or a little
valley
...
13
valleys
...
For this reason, the points A and C
may be regarded as points of local minimum value (or relative minimum value) and
points B and D may be regarded as points of local maximum value (or relative maximum
value) for the function
...
We now formally give the following definition
Definition 4 Let f be a real valued function and let c be an interior point in the domain
of f
...
(b) c is called a point of local minima if there is an h > 0 such that
f (c) < f (x), for all x in (c – h, c + h)
The value f (c) is called the local minimum value of f
...
14(a)
...
e
...
e
...
This suggests that f′(c) must be zero
...
14
APPLICATION OF DERIVATIVES
221
Similarly, if c is a point of local minima of f , then the graph of f around c will be as
shown in Fig 6
...
Here f is decreasing (i
...
, f ′(x) < 0) in the interval (c – h, c) and
increasing (i
...
, f ′(x) > 0) in the interval (c, c + h)
...
The above discussion lead us to the following theorem (without proof)
...
Suppose c ∈ I be any
point
...
Remark The converse of above theorem need
not be true, that is, a point at which the derivative
vanishes need not be a point of local maxima or
local minima
...
But 0 is neither a point of
local maxima nor a point of local minima (Fig 6
...
Note A point c in the domain of a function
f at which either f ′(c) = 0 or f is not differentiable
is called a critical point of f
...
Fig 6
...
Theorem 3 (First Derivative Test) Let f be a function defined on an open interval I
...
Then
(i) If f ′(x) changes sign from positive to negative as x increases through c, i
...
, if
f ′(x) > 0 at every point sufficiently close to and to the left of c, and f ′(x) < 0 at
every point sufficiently close to and to the right of c, then c is a point of local
maxima
...
e
...
(iii) If f ′(x) does not change sign as x increases through c, then c is neither a point of
local maxima nor a point of local minima
...
15)
...
Similarly, if c is a point of local minima of f , then f(c) is a local minimum value of f
...
15 and 6
...
Fig 6
...
Solution We have
f (x) = x3 – 3x + 3
or
f ′(x) = 3x2 – 3 = 3 (x – 1) (x + 1)
or
f ′(x) = 0 at x = 1 and x = – 1
Thus, x = ± 1 are the only critical points which could possibly be the points of local
maxima and/or local minima of f
...
Note that for values close to 1 and to the right of 1, f ′(x) > 0 and for values close
to 1 and to the left of 1, f ′(x) < 0
...
In the case of x = –1, note that
f ′(x) > 0, for values close to and to the left of –1 and f ′(x) < 0, for values close to and
to the right of – 1
...
Values of x
Sign of f ′ (x) = 3(x – 1) (x + 1)
Close to 1
to the right (say 1
...
)
to the left (say 0
...
)
>0
<0
Close to –1
to the right (say − 0
...
)
to the left (say − 1
...
)
<0
>0
APPLICATION OF DERIVATIVES
223
Example 30 Find all the points of local maxima and local minima of the function f
given by
f (x) = 2x3 – 6x2 + 6x +5
...
We shall now examine this point for local
maxima and/or local minima of f
...
Therefore, by first
derivative test, the point x = 1 is neither a point of local maxima nor a point of local
minima
...
Remark One may note that since f ′(x), in Example 30, never changes its sign on R,
graph of f has no turning points and hence no point of local maxima or local minima
...
This test is often easier to apply than the first derivative test
...
Let f be twice differentiable at c
...
(ii) x = c is a point of local minima if f ′( c) = 0 and f ″(c) > 0
In this case, f (c) is local minimum value of f
...
In this case, we go back to the first derivative test and find whether c is a point of
local maxima, local minima or a point of inflexion
...
Example 31 Find local minimum value of the function f
given by f (x) = 3 + | x |, x ∈ R
...
So, second derivative test fails
...
Note that 0 is a critical point of f
...
Also
Fig 6
...
Therefore, by first derivative test,
x = 0 is a point of local minima of f and local minimum value of f is f (0) = 3
...
f ″(x) = 36x2 + 24x – 24 = 12 (3x2 + 2x – 1)
f ′′(0) = −12 < 0
f ′′(1) = 48 > 0
f ′′(−2) = 84 > 0
Therefore, by second derivative test, x = 0 is a point of local maxima and local
maximum value of f at x = 0 is f (0) = 12 while x = 1 and x = – 2 are the points of local
minima and local minimum values of f at x = – 1 and – 2 are f (1) = 7 and f (–2) = –20,
respectively
...
Solution We have
f(x) = 2x3 – 6x2 + 6x +5
or
f ′( x) = 6 x2 − 12 x + 6 = 6( x − 1) 2
f ′′( x) = 12( x − 1)
Now f ′(x) = 0 gives x =1
...
Therefore, the second derivative test
fails in this case
...
We have already seen (Example 30) that, using first derivative test, x =1 is neither
a point of local maxima nor a point of local minima and so it is a point of inflexion
...
Solution Let one of the numbers be x
...
Let S(x)
denote the sum of the squares of these numbers
...
Also S′ = 4 > 0
...
Hence the sum of squares of numbers is
2
minimum when the numbers are
15
15 15
and 15 − =
...
2
2
Example 35 Find the shortest distance of the point (0, c) from the parabola y = x2,
where 0 ≤ c ≤ 5
...
Let D be the required distance
between (h, k) and (0, c)
...
(1)
Since (h, k) lies on the parabola y = x2, we have k = h 2
...
e
...
Also when
2
2c − 1
2c − 1
, then D′( k ) > 0
...
2
2
Hence, the required shortest distance is given by
k>
226
MATHEMATICS
2
2 c − 1 2c − 1
4c − 1
2c − 1
D
+
− c =
=
2
2
2
2
Note The reader may note that in Example 35, we have used first derivative
test instead of the second derivative test as the former is easy and short
...
If AP = 16 m, BQ = 22 m
and AB = 20 m, then find the distance of a point R on
AB from the point A such that RP2 + RQ 2 is minimum
...
Then RB = (20 – x) m (as AB = 20 m)
...
18,
we have
RP2 = AR2 + AP2
Fig 6
...
Therefore
S′(x) = 4x – 40
...
Also S″(x) = 4 > 0, for all x and so S″(10) > 0
...
Thus, the
distance of R from A on AB is AR = x =10 m
...
Solution The required trapezium is as given in Fig 6
...
Draw perpendiculars DP and
Fig 6
...
Let AP = x cm
...
Therefore, QB = x cm
...
Let A be the area of the trapezium
...
e
...
Since x represents distance, it can not be negative
...
Now
100 − x2 ( −4 x −10) − ( −2 x 2 − 10 x + 100)
A″(x) =
=
100 − x
2 x3 − 300 x − 1000
(100 −
or
A″(5) =
3
x2 ) 2
(100
2 100 − x2
2
(on simplification)
2(5) 3 − 300(5) − 1000
3
− (5) 2 ) 2
( −2 x )
=
−2250 −30
=
<0
75 75
75
Thus, area of trapezium is maximum at x = 5 and the area is given by
A(5) = (5 + 10) 100 − (5) 2 = 15 75 = 75 3 cm 2
Example 38 Prove that the radius of the right circular cylinder of greatest curved
surface area which can be inscribed in a given cone is half of that of the cone
...
Let a cylinder
with radius OE = x inscribed in the given cone (Fig 6
...
The height QE of the cylinder
is given by
228
MATHEMATICS
QE
EC
=
OA
OC
or
(since ∆QEC ~ ∆AOC)
QE
r −x
=
h
r
h(r − x)
r
Let S be the curved surface area of the given
cylinder
...
20
2πh
′
S ( x) = r ( r − 2 x)
S′ ( x) = − 4 πh
r
r
r
r
...
So x = is a
2
2
2
point of maxima of S
...
Now S′(x) = 0 gives x =
6
...
1 Maximum and Minimum Values of a Function in a Closed Interval
Let us consider a function f given by
f (x) = x + 2, x ∈ (0, 1)
Observe that the function is continuous on (0, 1) and neither has a maximum value
nor has a minimum value
...
However, if we extend the domain of f to the closed interval [0, 1], then f still may
not have a local maximum (minimum) values but it certainly does have maximum value
3 = f (1) and minimum value 2 = f (0)
...
Similarly, the minimum value 2 of f at x = 0 is called the absolute minimum
value (global minimum or least value) of f on [0, 1]
...
21 of a continuous function defined on a closed
interval [a, d]
...
21
minimum value is f (b)
...
Also from the graph, it is evident that f has absolute maximum value f (a) and
absolute minimum value f (d)
...
We will now state two results (without proof) regarding absolute maximum and
absolute minimum values of a function on a closed interval I
...
Then f has the
absolute maximum value and f attains it at least once in I
...
Theorem 6 Let f be a differentiable function on a closed interval I and let c be any
interior point of I
...
(ii) f ′(c) = 0 if f attains its absolute minimum value at c
...
Working Rule
Step 1: Find all critical points of f in the interval, i
...
, find points x where either
f ′( x ) = 0 or f is not differentiable
...
Step 3: At all these points (listed in Step 1 and 2), calculate the values of f
...
This maximum value will be the absolute maximum (greatest) value of
f and the minimum value will be the absolute minimum (least) value of f
...
Solution We have
f (x) = 2x3 – 15x2 + 36x + 1
or
f ′(x) = 6x2 – 30x + 36 = 6 (x – 3) (x – 2)
Note that f ′(x) = 0 gives x = 2 and x = 3
...
e
...
So
f (1) = 2 (13) – 15(12 ) + 36 (1) + 1 = 24
f (2) = 2 (23) – 15(22 ) + 36 (2) + 1 = 29
f (3) = 2 (33) – 15(32 ) + 36 (3) + 1 = 28
f (5) = 2 (53) – 15(52 ) + 36 (5) + 1 = 56
Thus, we conclude that absolute maximum value of f on [1, 5] is 56, occurring at
x =5, and absolute minimum value of f on [1, 5] is 24 which occurs at x = 1
...
Further note that f ′(x) is not defined at x = 0
...
Now evaluating the value of f at critical points
8
x = 0,
1
and at end points of the interval x = –1 and x = 1, we have
8
4
1
f (–1) = 12 ( −1) 3 − 6 ( −1) 3 = 18
f (0) = 12 (0) – 6(0) = 0
APPLICATION OF DERIVATIVES
4
231
1
1
3
3
f = 12 1 − 6 1 = − 9
8
4
8
8
4
1
f (1) = 12 (1) 3 − 6 (1) 3 = 6
Hence, we conclude that absolute maximum value of f is 18 that occurs at x = – 1
and absolute minimum value of f is
−9
1
that occurs at x =
...
A soldier, placed at (3, 7), wants to shoot down the helicopter when it is
nearest to him
...
Solution For each value of x, the helicopter’s position is at point (x, x 2 + 7)
...
e
...
f (x) = (x – 3)2 + x4
f ′(x) = 2(x – 3) + 4x3 = 2(x – 1) (2x2 + 2x + 3)
Thus, f ′(x) = 0 gives x = 1 or 2x2 + 2x + 3 = 0 for which there are no real roots
...
e
...
The value of f at this point is given by
f (1) = (1 – 3)2 + (1)4 = 5
...
Note that
5 is either a maximum value or a minimum value
...
Hence,
5 is the minimum value of
distance between the soldier and the helicopter
...
5
1
...
Find the maximum and minimum values, if any, of the following functions
given by
(i) f (x) = |x + 2 | – 1
(ii) g (x) = – | x + 1| + 3
(iii) h (x) = sin(2x) + 5
(iv) f (x) = | sin 4x + 3|
(v) h (x) = x + 1, x ∈ (– 1, 1)
3
...
Find
also the local maximum and the local minimum values, as the case may be:
(i) f (x) = x2
(ii) g (x) = x3 – 3x
(iii) h (x) = sin x + cos x, 0 < x <
π
2
(iv) f (x) = sin x – cos x, 0 < x < 2π
(v) f (x) = x3 – 6x2 + 9x + 15
(vi) g (x) =
x 2
+ , x >0
2 x
1
(viii) f ( x) = x 1 − x , 0 < x < 1
x +2
4
...
Find the absolute maximum value and the absolute minimum value of the following
functions in the given intervals:
(i) f (x) = x3, x ∈ [– 2, 2]
(ii) f (x) = sin x + cos x , x ∈ [0, π]
(vii) g ( x) =
2
1
9
(iii) f (x) = 4 x − x2 , x ∈ −2, (iv) f ( x) = ( x − 1) 2 + 3, x ∈[ −3,1]
2
2
6
...
Find both the maximum value and the minimum value of
3x4 – 8x3 + 12x2 – 48x + 25 on the interval [0, 3]
...
At what points in the interval [0, 2π], does the function sin 2x attain its maximum
value?
9
...
Find the maximum value of 2x3 – 24x + 107 in the interval [1, 3]
...
APPLICATION OF DERIVATIVES
233
11
...
Find the value of a
...
Find the maximum and minimum values of x + sin 2x on [0, 2π]
...
Find two numbers whose sum is 24 and whose product is as large as possible
...
Find two positive numbers x and y such that x + y = 60 and xy3 is maximum
...
Find two positive numbers x and y such that their sum is 35 and the product x2 y5
is a maximum
...
Find two positive numbers whose sum is 16 and the sum of whose cubes is
minimum
...
A square piece of tin of side 18 cm is to be made into a box without top, by
cutting a square from each corner and folding up the flaps to form the box
...
18
...
What should be
the side of the square to be cut off so that the volume of the box is maximum ?
19
...
20
...
21
...
A wire of length 28 m is to be cut into two pieces
...
What should be the length of the
two pieces so that the combined area of the square and the circle is minimum?
23
...
27
24
...
25
...
26
...
3
234
MATHEMATICS
Choose the correct answer in the Exercises 27 and 29
...
The point on the curve x2 = 2y which is nearest to the point (0, 5) is
(A) (2 2, 4)
(B) (2 2, 0)
(C) (0, 0)
28
...
The maximum value of [ x( x
1
− 1) + 1]3
(D)
1
3
, 0 ≤ x ≤ 1 is
1
1 3
(A)
3
(B)
1
2
(C) 1
(D) 0
Miscellaneous Examples
Example 42 A car starts from a point P at time t = 0 seconds and stops at point Q
...
Solution Let v be the velocity of the car at t seconds
...
Now v = 0 at P as well as at Q and at P, t = 0
...
Thus, the car will
reach the point Q after 4 seconds
...
Its semi-vertical angle is tan–1 (0
...
Water is poured
into it at a constant rate of 5 cubic metre per hour
...
r
Solution Let r, h and α be as in Fig 6
...
Then tan α =
...
h
–1
α = tan (0
...
5
h
or
r=
h
2
Let V be the volume of the cone
...
e
...
22
(by Chain Rule)
π 2 dh
h
4
dt
dV
= 5 m3/h and h = 4 m
...
88
Example 44 A man of height 2 metres walks at a uniform speed of 5 km/h away from
a lamp post which is 6 metres high
...
236
MATHEMATICS
Solution In Fig 6
...
Then,
MS is the shadow of the man
...
∆MSN ~ ∆ASB
Note that
MS
MN
=
AS
AB
or
or
Thus
So
AS = 3s (as MN = 2 and AB = 6 (given))
AM = 3s – s = 2s
...
23
5
dl
= 5 km/h
...
2
dt
Example 45 Find the equation of the normal to the curve x2 = 4y which passes through
the point (1, 2)
...
Now, slope of the tangent at (h, k) is given by
dy
h
dx ( h, k ) = 2
−2
h
Therefore, the equation of normal at (h, k) is
Hence, slope of the normal at (h, k) =
−2
( x − h)
h
Since it passes through the point (1, 2), we have
y–k=
2− k =
−2
2
(1 − h ) or k = 2 + (1 − h )
h
h
...
(2)
APPLICATION OF DERIVATIVES
237
Since (h, k) lies on the curve x2 = 4y, we have
h 2 = 4k
...
Substituting the values of h and k in (1),
we get the required equation of normal as
y −1 =
−2
( x − 2) or x + y = 3
2
Example 46 Find the equation of tangents to the curve
y = cos (x + y), – 2π ≤ x ≤ 2π
that are parallel to the line x + 2y = 0
...
Thus, tangents to the
2
2
−3 π
π
, 0 and , 0
...
f (x) =
Solution We have
f (x) =
Therefore
3 4 4 3
36
x − x − 3x2 + x + 11
10
5
5
f ′(x) =
3
4
36
(4 x3 ) − (3 x2 ) − 3(2 x) +
10
5
5
=
6
( x −1) ( x + 2) ( x − 3)
5
Now f ′(x) = 0 gives x = 1, x = – 2, or x = 3
...
24)
...
24
Consider the interval (– ∞, – 2), i
...
, when – ∞ < x < – 2
...
(In particular, observe that for x = –3, f ′(x) = (x – 1) (x + 2) (x – 3) = (– 4) (– 1)
(– 6) < 0)
Therefore,
f ′(x) < 0 when – ∞ < x < – 2
...
Consider the interval (– 2, 1), i
...
, when – 2 < x < 1
...
Thus,
f is strictly increasing in (– 2, 1)
...
e
...
In this case, we have
x – 1 > 0, x + 2 > 0 and x – 3 < 0
...
Thus,
f is strictly decreasing in (1, 3)
...
e
...
In this case, we have x – 1 > 0,
x + 2 > 0 and x – 3 > 0
...
Thus, f is strictly increasing in the interval (3, ∞)
...
4
Solution We have
f (x) = tan–1 (sin x + cos x), x > 0
Therefore
f ′(x) =
=
1
(cos x − sin x)
1+ (sin x + cos x)2
cos x − sin x
2 + sin 2 x
(on simplification)
π
Note that 2 + sin 2x > 0 for all x in 0,
...
e
...
4
Example 49 A circular disc of radius 3 cm is being heated
...
05 cm/s
...
2 cm
...
Then
A = πr 2
dA
dr
= 2 πr
dt
dt
or
(by Chain Rule)
dr
∆t = 0
...
dt
Therefore, the approximate rate of increase in area is given by
Now approximate rate of increase of radius = dr =
dA =
dA
dr
( ∆t ) = 2 πr ∆t
dt
dt
= 2π (3
...
05) = 0
...
2 cm)
Example 50 An open topped box is to be constructed by removing equal squares from
each corner of a 3 metre by 8 metre rectangular sheet of aluminium and folding up the
sides
...
Solution Let x metre be the length of a side of the removed squares
...
25)
...
25
V (x) = x (3 – 2x) (8 – 2x)
= 4x3 – 22x2 + 24x
V′( x) = 12 x2 − 44 x + 24 = 4( x − 3)(3x − 2)
V′ ( x) = 24 x − 44
Therefore
Now
V′(x) = 0 gives x = 3,
Thus, we have x =
2
...
Now V′ = 24 − 44 = − 28 < 0
...
e
...
The
100
x
cost price of x items is Rs + 500
...
Solution Let S (x) be the selling price of x items and let C (x) be the cost price of x
items
...
e
...
Also P′′( x) =
−1
−1
...
Hence, the manufacturer can earn maximum
profit, if he sells 240 items
...
Using differentials, find the approximate value of each of the following:
1
1
4
(a) 17
81
(b)
( 33) − 5
log x
has maximum at x = e
...
The two equal sides of an isosceles triangle with fixed base b are decreasing at
the rate of 3 cm per second
...
Find the equation of the normal to curve x 2 = 4y which passes through the point
(1, 2)
...
Show that the normal at any point θ to the curve
2
...
6
...
f ( x) =
7
...
8
...
9
...
If building of tank costs
Rs 70 per sq metres for the base and Rs 45 per square metre for sides
...
The sum of the perimeter of a circle and square is k, where k is some constant
...
APPLICATION OF DERIVATIVES
243
11
...
The total perimeter of the window is 10 m
...
12
...
2
2
3
Show that the maximum length of the hypotenuse is (a 3 + b 3 ) 2
...
Find the points at which the function f given by f (x) = (x – 2) 4 (x + 1)3 has
(i) local maxima
(ii) local minima
(iii) point of inflexion
14
...
Show that the altitude of the right circular cone of maximum volume that can be
4r
...
Let f be a function defined on [a, b] such that f ′(x) > 0, for all x ∈ (a, b)
...
17
...
Also find the maximum volume
...
Show that height of the cylinder of greatest volume which can be inscribed in a
right circular cone of height h and semi vertical angle α is one-third that of the
4
πh 3 tan 2 α
...
19
...
Then the depth of the wheat is increasing at the rate of
(A) 1 m3/h
(B) 0
...
1 m3/h
(D) 0
...
The slope of the tangent to the curve x = t2 + 3t – 8, y = 2t2 – 2t – 5 at the point
(2,– 1) is
(A)
22
7
(B)
6
7
(C)
7
6
(D)
−6
7
244
MATHEMATICS
21
...
The normal at the point (1,1) on the curve 2y + x2 = 3 is
(A) x + y = 0
(B) x – y = 0
(C) x + y +1 = 0
(D) x – y = 1
2
23
...
The points on the curve 9y
intercepts with the axes are
8
(A) 4, ±
3
−8
(B) 4,
3
3
(C) 4, ±
8
3
(D) ± 4,
8
Summary
If a quantity y varies with another quantity x, satisfying some rule y = f ( x) ,
then
dy
(or f ′( x ) ) represents the rate of change of y with respect to x and
dx
dy
dx x =x0 (or f ′( x0 ) ) represents the rate of change of y with respect to x at
x = x0
...
e
...
, if
dt
dt
A function f is said to be
(a) increasing on an interval (a, b) if
x1 < x2 in (a, b) ⇒ f (x1) ≤ f(x2) for all x1, x2 ∈ (a, b)
...
Alternatively, if f ′(x) ≤ 0 for each x in (a, b)
The equation of the tangent at (x0 , y0) to the curve y = f (x) is given by
y − y0 =
dy
(x − x0 )
dx ( x0 ,y 0 )
dy
does not exist at the point ( x0 , y0 ) , then the tangent at this point is
dx
parallel to the y-axis and its equation is x = x0
...
dx x =x0
Equation of the normal to the curve y = f (x) at a point ( x0 , y0 ) is given by
y − y0 =
If
−1
( x − x0 )
dy
dx ( x0 , y0 )
dy
at the point ( x0 , y0 ) is zero, then equation of the normal is x = x0
...
Let y = f(x), ∆x be a small increment in x and ∆y be the increment in y
corresponding to the increment in x, i
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, ∆y = f (x + ∆x) – f (x)
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dx
is a good approximation of ∆y when dx = ∆x is relatively small and we denote
it by dy ≈ ∆y
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MATHEMATICS
First Derivative Test Let f be a function defined on an open interval I
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Then
(i) If f ′(x) changes sign from positive to negative as x increases through c,
i
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, if f ′(x) > 0 at every point sufficiently close to and to the left of c,
and f ′(x) < 0 at every point sufficiently close to and to the right of c,
then c is a point of local maxima
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e
...
(iii) If f ′(x) does not change sign as x increases through c, then c is neither
a point of local maxima nor a point of local minima
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Second Derivative Test Let f be a function defined on an interval I and
c ∈ I
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Then
(i) x = c is a point of local maxima if f ′(c) = 0 and f ″(c) < 0
The values f (c) is local maximum value of f
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(iii) The test fails if f ′(c) = 0 and f ″(c) = 0
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Working rule for finding absolute maxima and/or absolute minima
Step 1: Find all critical points of f in the interval, i
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, find points x where
either f ′(x) = 0 or f is not differentiable
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Step 3: At all these points (listed in Step 1 and 2), calculate the values of f
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This maximum value will be the absolute maximum
value of f and the minimum value will be the absolute minimum value of f