Ch-6 (Applications of Derivatives) PDF
Ch-6 (Applications of Derivatives) PDF
Ch-6 (Applications of Derivatives) PDF
APPLICATION OF DERIVATIVES
6.1 Overview
6.1.1 Rate of change of quantities
For the function y = f (x),
d
(f (x)) represents the rate of change of y with respect to x.
dx
ds
represents the rate of
dt
dy
that point and its equation is given y y1 =
( x , y ) ( x x1 ) .
dx 1 1
The normal to the curve is the line perpendicular to the tangent at the point of contact,
and its equation is given as:
y y1 =
1
( x x1 )
dy
(x , y )
dx 1 1
The angle of intersection between two curves is the angle between the tangents to the
curves at the point of intersection.
6.1.3 Approximation
Since f (x) = lim
x 0
f ( x + x) f ( x)
, we can say that f (x) is approximately equal
x
f ( x + x) f ( x)
x
approximate value of f (x + x) = f (x) + x .f (x).
to
118
MATHEMATICS
local maxima, if there exists an h > 0 , such that f (c) > f (x), for all x in
(c h, c + h).
local minima if there exists an h > 0 such that f (c) < f (x), for all x in
(c h, c + h).
APPLICATION OF DERIVATIVES
(b)
6.1.8
119
(ii)
(iii)
(ii)
(iii)
dy
= 5 6x2
dx
d dy
dx
= 12x.
dt dx
dt
120
MATHEMATICS
= 12 . (3) . (2)
= 72 units/sec.
Thus, slope of curve is decreasing at the rate of 72 units/sec when x is increasing at the
rate of 2 units/sec.
at the
4
uniform rate of 2 cm2 /sec in the surface area, through a tiny hole at the vertex of the
bottom. When the slant height of cone is 4 cm, find the rate of decrease of the slant
height of water.
Solution If s represents the surface area, then
r
ds
2
d t = 2cm /sec
Example 2 Water is dripping out from a conical funnel of semi-vertical angle
2
.
l
l=
4
2
s = r.l = l . sin
Therefore,
ds 2 dl
l. =
=
2 dt
dt
2l .
l
p/4
dl
dt
dl
1
1
2
when l = 4 cm, dt = 2.4 .2 = 2 2 = 4 cm/s .
Fig. 6.1
y = 0, y = 1
and
x2 = y
2y
dy
= 2x.
dx
dy
=1
dx
1
dy
=
2y
dx
APPLICATION OF DERIVATIVES
121
At (0, 0), the slope of the tangent to the curve y2 = x is parallel to y-axis and the
tangent to the curve x2 = y is parallel to x-axis.
angle of intersection =
2
tan =
1
and that of x2 = y is 2.
2
3
= tan1 4
, .
Example 4 Prove that the function f (x) = tanx 4x is strictly decreasing on
3 3
Solution f (x) = tan x 4x f (x) = sec2x 4
When
, .
Hence f is strictly decreasing on
3 3
Example 5 Determine for which values of x, the function y = x4
and for which values, it is decreasing.
Solution y = x4
4 x3
3
dy
= 4x3 4x2 = 4x2 (x 1)
dx
4 x3
is increasing
3
122
MATHEMATICS
Now,
dy
= 0 x = 0, x = 1.
dx
3
f (x) = 0 x = 2 (critical point)
Since f (x) > 0 for all x <
Hence x =
x=
3
3
and for all x >
2
2
3
is a point of inflexion i.e., neither a point of maxima nor a point of minima.
2
3
is the only critical point, and f has neither maxima nor minima.
2
0.082
0.082 =
1
0.008
0.09 0.008 . 2 0.09 = 0.3
0.6
APPLICATION OF DERIVATIVES
123
x2 y 2
= 1; xy = c2 to intersect
a 2 b2
orthogonally.
Solution Let the curves intersect at (x1, y1). Therefore,
2 x 2 y dy
x2 y 2
dy b 2 x
=
=
1
=
0
a 2 b 2 dx
dx a 2 y
a 2 b2
b 2 x1
slope of tangent at the point of intersection (m1) = 2
a y1
Again xy = c2 x
y1
dy
dy y
+ y= 0
=
m2 = x .
dx
dx
x
1
b2
For orthoganality, m1 m2 = 1 2 = 1 or a2 b2 = 0.
a
Example 9 Find all the points of local maxima and local minima of the function
3 4
3 45 2
f (x) = x 8 x x +105 .
4
2
Solution f (x) = 3x3 24x2 45x
= 3x (x2 + 8x + 15) = 3x (x + 5) (x + 3)
f (x) = 0 x = 5, x = 3, x = 0
f (x) = 9x2 48x 45
= 3 (3x2 + 16x + 15)
f (0) = 45 < 0. Therefore, x = 0 is point of local maxima
f (3) = 18 > 0. Therefore, x = 3 is point of local minima
f (5) = 30 < 0. Therefore x = 5 is point of local maxima.
124
MATHEMATICS
1
is less than local minimum
x
value.
Solution Let y = x +
1
dy
1
=1 2,
dx
x
x
dy
2
dx = 0 x = 1 x = 1.
d2y
d2y
2
d2y
= + 3 , therefore
(at x = 1) > 0 and
(at x = 1) < 0.
dx 2
dx 2
x
dx 2
Hence local maximum value of y is at x = 1 and the local maximum value = 2.
Local minimum value of y is at x = 1 and local minimum value = 2.
Therefore, local maximum value (2) is less than local minimum value 2.
Long Answer Type (L.A.)
Example 11 Water is dripping out at a steady rate of 1 cu cm/sec through a tiny hole
at the vertex of the conical vessel, whose axis is vertical. When the slant height of
water in the vessel is 4 cm, find the rate of decrease of slant height, where the vertical
angle of the conical vessel is
.
6
dv
= 1 cm 3/s, where v is the volume of water in the
dt
conical vessel.
From the Fig.6.2, l = 4cm, h = l cos
Therefore, v =
1 2
l2 3
l
r h =
3
3 4 2
l
3
l and r = l sin = .
=
6
6 2
2
3 3
l .
24
APPLICATION OF DERIVATIVES
dv
3 2 dl
=
l
dt
8
dt
Therefore, 1 =
125
3
dl
16.
8
dt
l
p/6
dl
1
=
cm/s.
dt 2 3
1
2 3
cm/s.
Fig. 6.2
Example 12 Find the equation of all the tangents to the curve y = cos (x + y),
2 x 2, that are parallel to the line x + 2y = 0.
Solution Given that y = cos (x + y)
dy
= sin (x + y)
dx
dy
1+ dx
sin ( x + y )
dy
=
dx
1+ sin ( x + y )
or
sin ( x + y )
1
Therefore, 1+ sin x + y = sin (x + y) = 1
(
)
2
Since
...(i)
1
2
.... (ii)
1 = y2 + 1 or y = 0.
Therefore, cosx = 0.
Therefore, x = (2n + 1)
, n = 0, 1, 2...
2
126
MATHEMATICS
Thus, x =
3
, , but x = , x =
satisfy equation (ii)
2
2
2
2
3
,0 .
Hence, the points are , 0 ,
2 2
1
Therefore, equation of tangent at , 0 is y = x or 2x + 4y = 0, and
2
2
2
3
3
1
,0 is y = x +
equation of tangent at
2
2
2
or 2x + 4y + 3 = 0.
Example 13 Find the angle of intersection of the curves y2 = 4ax and x2 = 4by.
Solution Given that y2 = 4ax...(i) and x2 = 4by... (ii). Solving (i) and (ii), we get
2
x2
= 4ax x4 = 64 ab2 x
4b
1
x (x 64 ab ) = 0 x = 0, x = 4a 3 b 3
or
2 1
1 2
Again, y2 = 4ax
dy 4a 2a
= =
and x2 = 4by
dx 2 y y
dy 2 x x
= =
dx 4b 2b
Therefore, at (0, 0) the tangent to the curve y2 = 4ax is parallel to y-axis and tangent
to the curve x2 = 4by is parallel to x-axis.
Angle between curves =
2
1
2 1
1 2
a 3
3 b 3 , 4a 3 b 3
4
a
, m1 (slope of the tangent to the curve (i)) = 2
At
2a
2
4a 3 b 3
1 a 3
4a 3 b 3
a 3
= , m2 (slope of the tangent to the curve (ii)) =
=2
2 b
2b
b
APPLICATION OF DERIVATIVES
m2 m1
Therefore, tan = 1+ m m =
1 2
127
a 3 1 a 3
2
b 2 b
1
3
1
3
a 1a
1+ 2
b 2 b
1
1
3
3a . b 3
2
2
2 a 3 + b3
1
1
3a 3 . b 3
Hence, = tan1
2
2
3
3
+
2
a
b
Example 14 Show that the equation of normal at any point on the curve
x = 3cos cos3, y = 3sin sin3 is 4 (y cos3 x sin3) = 3 sin 4.
Solution We have x = 3cos cos3
Therefore,
dx
= 3sin + 3cos2 sin = 3sin (1 cos2) = 3sin3 .
d
dy
= 3cos 3sin2 cos = 3cos (1 sin2) = 3cos3
d
dy
cos3
sin 3
=
+
.
Therefore,
slope
of
normal
=
dx
sin 3
cos3
Hence the equation of normal is
y (3sin sin3) =
sin 3
[x (3cos cos3)]
cos3
y cos3 3sin cos3 + sin3 cos3 = xsin3 3sin3 cos + sin3 cos3
y cos3 xsin3 = 3sin cos (cos2 sin2)
128
MATHEMATICS
or
3
sin2 . cos2
2
3
sin4
4
1
2
or
x=
or
x = and
5
3
is a point of minima.
f = 2 (4 + 3 4) = 6 > 0. Therefore, x =
3
3
5
5
is a point of minima.
f = 2 (4 + 3 4) = 6 > 0. Therefore, x =
3
3
APPLICATION OF DERIVATIVES
Maximum Value of y at x = 0 is
1+0=1
Maximum Value of y at x = is
1 + 0 = 1
Minimum Value of y at x =
is
3
2 + 2 log
1
= 2 (1 log2)
2
Minimum Value of y at x =
5
is
3
2 + 2 log
1
= 2 (1 log2)
2
129
Example 16 Find the area of greatest rectangle that can be inscribed in an ellipse
x2 y 2
+
= 1.
a 2 b2
Solution Let ABCD be the rectangle of maximum area with sides AB = 2x and
BC = 2y, where C (x, y) is a point on the ellipse
x2 y 2
+
= 1 as shown in the Fig.6.3.
a 2 b2
The area A of the rectangle is 4xy i.e. A = 4xy which gives A2 = 16x2y2 = s (say)
x2 2
16b 2
Therefore, s = 16x 1 2 . b =
(a2x2 x4)
a2
a
2
(0, b)
ds 16b 2
= 2 . [2a2x 4x3].
dx
a
y
(0, 0)
(a, 0)
A
Again,
a
b
ds
and y =
=0 x=
2
2
dx
Now,
d 2 s 16b 2
= 2 [2a2 12x2]
dx 2
a
At
x=
a
d 2 s 16b 2
16b 2
2
2
,
=
[2
a
6
a
]
=
( 4a 2 ) < 0
2
2
2
a
a
2 dx
(0, b)
Fig. 6.3
(a, 0)
B
130
MATHEMATICS
Thus at x =
a
2
,y=
b
2
a b
.
= 2ab sq units.
2 2
Example 17 Find the difference between the greatest and least values of the
function f (x) = sin2x x, on , .
2 2
Solution f (x) = sin2x x
f (x) = 2 cos2 x 1
1
2x is
2
or
f = sin ( ) + =
2 2
2
2
f = sin + = 3 +
6 6
6
2
6
2
f = sin = 3
6 6
6
2
6
f = sin ( ) =
2
2
2
Clearly,
Therefore, difference =
+ =
2
2
3x=
or
6 6
APPLICATION OF DERIVATIVES
131
.
6
Solution Let ABC be an isosceles triangle inscribed in the circle with radius a such
that AB = AC.
AD = AO + OD = a + a cos2 and BC = 2BD = 2a sin2 (see fig. 16.4)
Therefore, area of the triangle ABC i.e. =
1
BC . AD
2
1
2a sin2 . (a + a cos2)
2
= a2sin2 (1 + cos2)
= a2sin2 +
Therefore,
1 2
a sin4
2
d
= 2a2cos2 + 2a2cos4
d
= 2a2(cos2 + cos4)
d
= 0 cos2 = cos4 = cos ( 4)
d
Therefore, 2 = 4 =
d 2
2
).
2 = 2a (2sin2 4sin4) < 0 (at =
6
d
Therefore, Area of triangle is maximum when =
.
6
132
MATHEMATICS
(B)
1
3
(C) 2
(D)
1
2
Solution Let (x1, y1) be the point on the given curve 3y = 6x 5x3 at which the normal
dy
= 2 5 x12 . Again the equation of
passes through the origin. Then we have
dx
( x1 , y1 )
the normal at (x 1, y1) passing through the origin gives 2 5 x12 =
x1
3
=
.
y1 6 5 x12
Solution From first equation of the curve, we have 3x2 3y2 6xy
x2 y 2
dy
=
= (m1) say and second equation of the curve gives
2 xy
dx
6xy + 3x2
dy
dy
3y2
=0
dx
dx
2 xy
dy
= 2
= (m2) say
x y2
dx
dy
=0
dx
APPLICATION OF DERIVATIVES
133
makes
4
Solution
(B)
(C)
(D)
dx
dy
= et . sint + etcost,
= etcost + etsint
dt
dt
dy
cos t + sin t
2
=
and hence the correct answer is (D).
Therefore, dx t = =
cos t sint
4
0
Example 22 The equation of the normal to the curve y = sinx at (0, 0) is:
(A) x = 0
Solution
(B) y = 0
(C) x + y = 0
(D) x y = 0
1
dy
= cosx. Therefore, slope of normal = cos x = 1. Hence the equation
x =0
dx
of normal is y 0 = 1(x 0) or x + y = 0
Therefore, correct answer is (C).
Example 23 The point on the curve y2 = x, where the tangent makes an angle of
with x-axis is
4
1 1
(A) ,
2 4
Solution
1 1
(B) ,
4 2
(C) (4, 2)
dy 1
1
1
=
= tan = 1 y =
x=
dx 2 y
2
4
4
(D) (1, 1)
134
MATHEMATICS
dy
= 0 2x + a = 0
dx
Solution
a2
a
+ a + 25 = 0
4
2
Therefore,
i.e.
a
x= ,
2
a = 10
1
, then its maximum value is _______.
4 x + 2 x +1
2
1 2
) +
4
1
3
1 giving the minimum value of 4x2 + 2x + 1 = .
4
4
4
.
3
Example 27 Minimum value of f if f (x) = sinx in , is _____.
2 2
Solution 1
Example 28 The maximum value of sinx + cosx is _____.
Solution
2.
APPLICATION OF DERIVATIVES
135
Example 29 The rate of change of volume of a sphere with respect to its surface
area, when the radius is 2 cm, is______.
Solution 1 cm3/cm2
v=
ds
4 3 dv
dv r
= = 1 at r = 2.
r = 4r 2 , s = 4r 2
= 8r
dr
3
ds 2
dr
6.3 EXERCISE
Short Answer (S.A.)
1.
A spherical ball of salt is dissolving in water in such a manner that the rate of
decrease of the volume at any instant is propotional to the surface. Prove that
the radius is decreasing at a constant rate.
2.
If the area of a circle increases at a uniform rate, then prove that perimeter
varies inversely as the radius.
3.
4.
Two men A and B start with velocities v at the same time from the junction of
two roads inclined at 45 to each other. If they travel by different roads, find
the rate at which they are being seperated..
5.
6.
7.
Find the approximate volume of metal in a hollow spherical shell whose internal
and external radii are 3 cm and 3.0005 cm, respectively.
8.
2
A man, 2m tall, walks at the rate of 1 m/s towards a street light which is
3
1
5 m above the ground. At what rate is the tip of his shadow moving? At what
3
136
MATHEMATICS
1
rate is the length of the shadow changing when he is 3 m from the base of
3
the light?
9.
10.
The volume of a cube increases at a constant rate. Prove that the increase in
its surface area varies inversely as the length of the side.
11.
x and y are the sides of two squares such that y = x x2 . Find the rate of
change of the area of second square with respect to the area of first square.
12.
Find the condition that the curves 2x = y2 and 2xy = k intersect orthogonally.
13.
14.
y = 4 at which tangent
16.
17.
Find the equation of the normal lines to the curve 3x2 y2 = 8 which are
parallel to the line x + 3y = 4.
18.
19.
20.
x
x y
= 1, touches the curve y = b . e a at the point where
a b
the curve intersects the axis of y.
1+ x 2 x is increasing in R.
APPLICATION OF DERIVATIVES
1, f (x) =
137
21.
22.
23.
24.
If the sum of the lengths of the hypotenuse and a side of a right angled triangle
is given, show that the area of the triangle is maximum when the angle between
them is
26.
Find the points of local maxima, local minima and the points of inflection of the
function f (x) = x5 5x4 + 5x3 1. Also find the corresponding local maximum
and local minimum values.
27.
A telephone company in a town has 500 subscribers on its list and collects
fixed charges of Rs 300/- per subscriber per year. The company proposes to
increase the annual subscription and it is believed that for every increase of
Re 1/- one subscriber will discontinue the service. Find what increase will
bring maximum profit?
28.
29.
x2
If the straight line x cos + y sin = p touches the curve 2
a
prove that a2 cos2 + b2 sin2 = p2.
An open box with square base is to be made of a given quantity of card board
of area c2. Show that the maximum volume of the box is
30.
y2
= 1, then
b2
c3
6 3
cubic units.
Find the dimensions of the rectangle of perimeter 36 cm which will sweep out
a volume as large as possible, when revolved about one of its sides. Also find
the maximum volume.
138
MATHEMATICS
31.
If the sum of the surface areas of cube and a sphere is constant, what is the
ratio of an edge of the cube to the diameter of the sphere, when the sum of
their volumes is minimum?
32.
AB is a diameter of a circle and C is any point on the circle. Show that the
area of ABC is maximum, when it is isosceles.
33.
A metal box with a square base and vertical sides is to contain 1024 cm3. The
material for the top and bottom costs Rs 5/cm2 and the material for the sides
costs Rs 2.50/cm2 . Find the least cost of the box.
34.
x
and a sphere is given to be constant. Prove that the sum of their volumes
3
is minimum, if x is equal to three times the radius of the sphere. Also find the
minimum value of the sum of their volumes.
and
The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. The
rate at which the area increases, when side is 10 cm is:
(A) 10 cm2/s
36.
(B)
3 cm2/s
(D)
10 2
cm /s
3
1
radian/sec
10
(D) 10 radian/sec
(A)
(B)
1
radian/sec
20
37.
(C) 10 3 cm2/s
(C) 20 radian/sec
APPLICATION OF DERIVATIVES
139
The equation of normal to the curve 3x2 y2 = 8 which is parallel to the line
x + 3y = 8 is
(A) 3x y = 8
(C) x + 3y
39.
8=0
42.
43.
(C) 6
(D) .6
(B) .032
(C) 5.68
(D) 5.968
(B) x 5y = 2
(C) 5x y = 2
(D) 5x + y = 2
The points at which the tangents to the curve y = x3 12x + 18 are parallel to
x-axis are:
(A) (2, 2), (2, 34)
The tangent to the curve y = e2x at the point (0, 1) meets x-axis at:
(A) (0, 1)
44.
(B) 0
41.
(D) x + 3y = 0
If the curve ay + x2 = 7 and x3 = y, cut orthogonally at (1, 1), then the value of
a is:
(A) 1
40.
(B) 3x + y + 8 = 0
(B)
1
,0
2
(C) (2, 0)
(D) (0, 2)
140
MATHEMATICS
(A)
45.
48.
(B)
(C)
(B) [2, 1]
(D)
(C)
, 2]
(D) [1, 1]
(C) x > 0
3
2
(C) decreasing in
,
2 2
(B) decreasing in ,
2
(D) decreasing in 0,
51.
(D) 6
3
(A) increasing in ,
50.
6
7
(C)
The interval on which the function f (x) = 2x3 + 9x2 + 12x 1 is decreasing is:
(A) [1,
47.
6
7
(B)
46.
22
7
(B) tanx
(C) cosx
(D) cos 3x
APPLICATION OF DERIVATIVES
52.
53.
(B) 0
55.
57.
(C)
135
(D) 160
1
4
At x =
(B)
1
2
(C)
(D) 2 2
5
, f (x) = 2 sin3x + 3 cos3x is:
6
(A) maximum
(B) minimum
(C) zero
(C) 16
(D) 32
59.
(B) 0
(A) 0
58.
(D) 2
(A)
56.
(C) 1
54.
141
(B) x =
1
e
1
The maximum value of
x
(A) e
(B) e
(C) x = 1
(D) x =
is:
(C) e
1
e
(D)
1
e
1
e
142
MATHEMATICS
61.
62.
The values of a for which the function f (x) = sinx ax + b increases on R are
______.
63.
64.
2 x 2 1
, x > 0, decreases in the interval _______.
x4
b
(a > 0, b > 0, x > 0) is ______.
x