Unit 16 Curve Tracing

UNIT 16
CURVE TRACING

Structure Page No.

No.
16.1 Introduction 137
Objectives
16.2 Tracing a Curve: Cartesian Equation 138
16.3 Tracing a Curve: Parametric Equation 156
16.4 Tracing a Curve: Polar Equation 160
16.5 Summary 165
16.6 Solutions/Answers 166

16.1 INTRODUCTION
A picture is worth a thousand words. A curve which is the visual image of a
function gives us a lot of information. Of course, we can also obtain this
information by analysing the equation which defines the functional relation.
But studying the associated curve is often easier and quicker. In addition to
this, a curve which represents a relation between two quantities also helps us
to easily find the value of one quantity corresponding to a specific value of the
other. In Sec. 16.2, we shall try to understand what is meant by the picture or
the graph of a function like y = f ( x ) and the curve with more than one
branches at any point, expressed in the form f ( x, y) = 0 and how to sketch
them. In Sec. 16.3 and Sec. 16.4, we shall discuss the tracing of a curve in
parametric and polar form, respectively. We shall be using many results from
the earlier units here. With this unit we come to the end of Block 4, in which
we have studied various geometrical features of functional relations with the
help of differential calculus.

Now we shall list the objectives of this unit. After going through the unit, please
read this list again and make sure that you have achieved the objectives.

Objectives
After studying this unit, you should be able to:
• list the properties which can be used for tracing a curve;
• trace some curves whose equations are given in cartesian, parametric or
polar form.

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Block 4 Applications of Differential Calculus

16.2 TRACING A CURVE: CARTESIAN EQUATION

You may recall from Unit 2, that by the graph of a function f : D → R , we mean
the set of points {( x, f ( x )) : x ∈ D} . Graphing a function means showing the
points of the corresponding set in a plane. Thus, essentially curve tracing
means plotting the points which satisfy a given relation. However, there are
some difficulties involved in this. Let’s see what these are and how to overcome
them.

It is often not possible to plot all the points on a curve. The standard technique
is to plot some suitable points and to get a general idea of the shape of the
curve by considering tangents, asymptotes, singular points, extreme points,
inflection points, concavity, monotonicity, periodicity etc. Then, we draw a free
hand curve as nearly satisfying the various properties as is possible.

The curve or graph that we draw has a limitation. If the range of values of
either (or both) variables is not finite, then it is not possible to draw the
complete graph. In such cases, the graph is not only approximate, but is also
incomplete. For example, consider the simplest curve, a straight line. Suppose
we want to draw the graph of f : R → R such that f ( x ) = 1 . We know that this
is a line parallel to the x -axis. But it is not possible to draw a complete graph
as this line extends infinitely on both sides. We indicate this by arrows at both
ends as in Fig 1.

Fig. 1

Now, we shall take up the problem of graphing a function by hand, when the
equation is given in the cartesian form.

Let us list some properties which, when taken, will simplify our job of tracing
this curve. We have discussed all these properties. Now, we shall summarize
these one by one.
i) Simplify: If possible, simplify the function y = f ( x ), you wish to sketch.
x2 + x − 2
For example, if f is defined by f ( x ) = , x ≠ 1 , you must write it
( x − 1)
as f ( x ) = x + 2, x ≠ 1 before beginning the procedure listed here.
ii) Domain and Range: In case of y = f ( x ), we find the domain and range
and mark the regions accordingly.
iii) Periodicity: Recall Unit 6, wherein we discussed periodic function.
Periodicity is the tendency of a function to repeat itself in a regular
pattern at a fixed interval. For example, all trigonometric functions have
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Unit 16 Curve Tracing
periodicity. If f ( x + p) = f ( x ) for all x in D, where p is a positive constant,
then, f is called a periodic function and smallest p is called the period of
the function. While tracing a curve, if we know that the function is
periodic and the period is p, we can keep on translating to sketch the
entire curve [Recall from Unit 3 for translation].
iv) Symmetry: The next step is to find out if the curve is symmetrical about
any line, or about the origin. A curve is symmetrical about a line if, when
we fold the curve on the line, the two positions of the curve exactly
coincide. A curve is symmetrical about the origin if we get the same
curve after rotating it through 180o. We have already discussed
symmetry of curves in Unit 6. Fig. 2, shows you some examples of
symmetric curves.

(a) Symmetric about the x -axis. (b) Symmetric about the origin.

(c) Symmetric about the line y = x .

Fig. 2

Here, we give you some hints which will help you to determine the symmetry
of a curve.
a) Symmetry about y -axis: The graph of a function y = f ( x ) is said
to be symmetric about y -axis, if f is an even function, that is, the
equation of the curve is unchanged when x is replaced by − x . For
example, y = cos x , y = x 2 , y = x , etc. This means that our work is
cut in half. If we know what the curve looks like for x ≥ 0 , then we
only need to reflect about the y − axis to obtain the complete
curve.
b) Symmetry about origin: Recall Unit 6, wherein we learnt that odd
functions are symmetric about origin. If f ( x ) = −f (− x ) , then the
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Block 4 Applications of Differential Calculus
curve is symmetrical about the origin. In such cases, it is enough
to draw the part of the graph above the x -axis and rotate it
through 180o to get the complete graph. Some such functions are:
y = x 3 , y = sin x , y = x, etc.
c) Symmetry about the line y = x : If the equation of the curve does
not change when we interchange x and y , then the curve is
symmetric about the line y = x .
v) Points of intersection with axes: The next step is to determine the
points where the curve intersects the axes. If we put y = 0 in y = f ( x ) ,
and solve the resulting equation for x , we get the points of intersection
with the x -axis. Similarly, putting x = 0 and solving the resulting
equation for y , we can find the points of intersection with the y -axis.
For example, in the curve y = 3x 2 − x 3 , if y = 0, we get x = 0, 3 and if
x = 0, we get y = 0 . Thus, the curve intersects axes at (0, 0) and (3, 0) .
You can omit this step, if the equation is difficult to solve.
vi) Points of discontinuity: Try to locate the points where the function is
discontinuous.
vii) Intervals of increasing and decreasing functions: For this, calculate
dy
. This will help you in locating the portions where the curve is rising
dx
 dy   dy 
 > 0  or falling  < 0  . You may recall Unit 13.
 dx   dx 
viii) Concavity and point(s) of inflection: Recall from Unit 14, and
d2 y
calculate second derivative of w.r.t. x . From , you can find
dx 2
d2 y
concavity. The curve is concave upward where > 0 and concave
dx 2
d2 y
downward where 2 < 0 . Inflection point occurs where the direction of
dx
concavity changes. These will give you a good idea about the shape of
the curve.
ix) Relative extrema: Recall from Unit 13, we use the second-derivative
test to find the relative maxima or minima. We substitute the first-order
critical numbers x 0 in the following test:

 d2y 
• If  
2 
> 0, then relative minimum at x 0 .
 dx  x= x
0

 d2y 
• If  
2 
< 0, then relative maximum at x 0 .
 dx  x= x
0

 d2y 
• If  2  = 0, then the test fails.
 dx  x= x0

We can also use first derivative test.

x) Tangents and normals: Compute the equations of the vertical tangents
and corresponding normals. You may recall Unit 14.
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Unit 16 Curve Tracing
xi) Asymptotes: The next step is to find the asymptote(s), if there are any.
We can find asymptotes parallel to axes and oblique as discussed in
Unit 15. They indicate the trend of the branches of the curve extending
to infinity.
xii) Singular point: Another important step is to determine the singular
points. The shape of the curve at these points is, generally, more
complex, as more than one branch of the curve passes through them.
[Recall from Unit 14].
xiii) Plot points: Plot points where f has a relative maxima, minima or
point(s) of inflection, x -intercepts, y -intercepts, etc.
xiv) Sketch the curve: Now, try to draw tangents to the curve at some of
these plotted points. Now join the plotted points by a smooth curve
(except at points of discontinuity). The tangents will guide you in this, as
they give you the direction of the curve. Sketch the asymptotes by dash
lines. Finally, draw the curve using the information in items i) to xiii).

We shall now illustrate this procedure through a number of examples. You will
notice that it may not be necessary to take all the steps mentioned above, in
each case. We begin by tracing some functions which were introduced in
Unit 2 and Unit 6.

Example 1: Sketch the graph of the function y = | x | .

Solution: Let us begin using steps of curve tracing. We can rewrite y as
 x, x≥0
y= .
 − x , x < 0
i) Domain and Range: The domain of this function is R and the range is
non-negative reals. Therefore, y can take only positive values. Thus, the
graph lies above the x -axis.
ii) Symmetry: Since, x = − x , therefore, the function y = | x | is
symmetric about the y -axis.

iii) Points of Intersection with axes: If x = 0, then, y = 0, therefore, the

curve meets the axes only at the origin. On the right of the y -axis,
x > 0 , and, so, | x | = x . Thus, the graph reduces to that of y = x and
you know that this a straight line equally inclined to the axes (Fig. 3(a)
below).Taking its reflection in the y -axis, we get the complete graph as
shown in Fig 3(b).

(a) Graph on the right of the y -axis. (b) Complete graph.

Fig 3
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Block 4 Applications of Differential Calculus
Example 2: Sketch the greatest integer function y = [ x ] .
Solution: Let us see which properties of curve tracing will be used to trace
greatest integer function.
i) Domain and Range: The domain of the function is R and the range is
set of all integers. The curve lies in the first and third quadrant, because
either x ≥ 0 and y ≥ 0 or x ≤ 0 and y ≤ 0 .
ii) Symmetry: If we replace x by − x , we get different value of y .
Therefore, [x ] is not symmetrical about y -axis. Also, y = [ x ] is not an
odd function, thus, not symmetrical about origin.
iii) Points of intersection with axes: When x = 0, y = 0, thus, the curve
passes through the origin. Also, when y = 0, 0 ≤ x < 1 therefore, the
graph lies on x -axis.
iv) Points of Discontinuity: y = [ x ] is discontinuous at every integer point.
Hence, there is a break in the graph at every integer point n . In every
interval [n, n + 1[ its value is constant, and is equal to n .
v) Relative extrema: No maximum or minimum.
vi) Asymptotes: There are no asymptotes.
vii) Concavity: The graph is neither concave upward nor concave
downward.
Hence, the graph is as shown in Fig 4. Note that a hollow circle around
a point indicates that the point is not included in the graph.

Fig 4: Graph of y = [ x ] .

***

Example 3: Sketch the graph of y = x 3 .

Solution: Let us check for the properties for the curve y = x 3 .
i) Domain and Range: The domain and range of the function is R . When
x > 0, y > 0 and when x < 0, y < 0 . Thus, there is no portion of the graph
in the second and fourth quadrants.
ii) Symmetry: The function is an odd function. This means that the curve is
symmetric about the origin. Thus, it is sufficient to draw the graph above
the x -axis and join it to the portion obtained by rotating it through 180o .
iii) Points of intersection with axes: If x = 0, then, y = 0 . Therefore, the
curve meets the axes only at the origin.
dy
iv) Tangents at origin: We have = 3x 2 , which is 0 at the origin. Thus,
dx
the tangent at origin is the x -axis.
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Unit 16 Curve Tracing
dy
v) Monotonicity: We find, = 3x 2 , which is always non-negative. This
dx
means that as x increases, so does y . Thus, the graph keeps on
rising.

dy d2 y
vi) Relative extrema: Here, = 0 at (0, 0) and = 6x is 0 at (0, 0) .
dx dx 2
Since the second derivative test is not helpful to find extrema, let us look
dy
at the sign of on each side.
dx
dy > 0, for x > 0
= 3x 2 = 
dx > 0, for x < 0
dy
Since, the sign of does not change, therefore, there are no extreme
dx
points.
d2 y
vii) Concavity and Point of Inflection: Here, = 0 at the origin. Also,
dx 2
d2 y d2y
< 0 when x < 0 and > 0 when x > 0 . Therefore, the curve is
dx 2 dx 2
concave upward when x > 0 and concave downward when x < 0 . Since
the concavity is changing at origin, therefore, the point of inflection is
(0, 0) .

viii) Asymptotes: The graph has no asymptotes parallel to the axes.

y
Further lim = lim x 2 and this does not exist. This means that the
x →∞ x x →∞

curve does not have any oblique asymptote.

ix) Singular Points: The curve has no singular points.
The graph is shown in Fig 5.

3
Fig 5: Graph of y = x .

***

1
Example 4: Sketch the graph of y = .
x2
Solution: Let us list the properties to trace the curve.
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Block 4 Applications of Differential Calculus
i) Domain and Range: The domain of the function is R − {0} and the
range of the function is non-negative reals. The y -coordinates of any
point on the curve cannot be negative. So, the curve must be above the
x -axis.
ii) Symmetric: Here, f ( x ) = f (− x ), thus the curve is symmetric about the
y -axis. Hence, we shall draw the graph to the right of the y -axis first.
iii) Points of intersection on axes: The curve does not intersect at the
axes at all.

dy 2 d2 y 6 dy
iv) Monotonicity: We have = − 3 and 2
= 4 . Since < 0 for all
dx x dx x dx
x > 0 , therefore, the function is decreasing in ]0, ∞[ , that is, the graph
dy
keeps on falling as x increases. Also, since > 0 for all x < 0 ,
dx
therefore, the function is increasing in ] − ∞, 0[ .
v) Discontinuity: The graph of y is continuous in the domain of the
function.
dy
vi) Relative extrema: Further, since, is non-zero for all x in the
dx
domain, thus, there is no extreme point.

d2y
vii) Concavity and point of inflection: Since, is positive in the
dx 2
domain, therefore, the function is concave upward everywhere in the
domain. Since, the concavity does not change, therefore, there is no
point of inflection.
1
viii) Asymptotes: Since lim = ∞, therefore, x = 0 is the vertical
x →0 x 2

asymptote. Also, lim f (x ) = 0, thus, y = 0 is the horizontal asymptote.

x →∞

The curve is shown in Fig 6.

2
Fig 6: Graph of y = 1 /x .

***

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Unit 16 Curve Tracing
1
Example 5: Sketch the graph of y = .
x
Solution:
i) Doman and Range: The domain of the function is R − {0} and the range
is R . Here, we can see that either x and y both will be positive or both
will be negative. This means that the curve lies in the first and the third
quadrants.
1
ii) Symmetric: Here, f ( x ) = , and f is not an even function, therefore, it
x
is not symmetric about y -axis. Further, it is symmetric about the origin
and hence, it is sufficient to trace it in the first quadrant and rotate this
through 180o to get the portion of the curve in the third quadrant.
dy − 1
iii) Interval of increasing or decreasing: Here, = , which means
dx x 2
that y < 0 for all values of x in the domain. Hence, as x increases, y
decreases.
iv) Asymptotes: Since, lim f ( x ) = 0 , therefore, y = 0 is the horizontal
x →∞

asymptote. Also, lim f ( x ) = ∞, therefore, x = 0 is the vertical asymptote.

x →0

dy − 1
v) Relative extrema: We have = ≠ 0 for any x in the domain. That
dx x 2
is, there are no extrema.

Considering all these points we can trace the curve in the first quadrant (see
Fig 7(a)). Fig 7(b) gives the complete curve.

(a) Graph of xy = 1 in the first Quadrant. (b) Complete graph.

Fig. 7
***

The curve traced in Example 5 is a hyperbola. If we cut a double cone by a

plane as in Fig 8(a), we get a parabola. It is a section of a cone. For this
reason, it is also called a conic section. Fig. 8(b) and Fig. 8(c) show a circle
and an ellipse respectively. The curve in Fig 8(d) is called a hyperbola and
that in Fig 8(e) is the pair of straight lines.

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Block 4 Applications of Differential Calculus

(a) (b) (c) (d) (e)

Fig 8: Conic Sections [(a) Parabola, (b) Circle, (c) Ellipse (d) Hyperbola (e) Pair of
straight lines.]

The earliest mention of these curves is found in the works of a Greek

Mathematician Menaechmas (fourth century B.C.). Later Apollonius (third
century B.C.) studied them extensively and gave them their current names.
In the seventeenth century, Rene Descartes discovered that the conic sections
can be characterised as curves which are governed by a second degree
equation in two variables. Blaise Pascal (1623-1662) presented them as
projections of a circle. (Why don’t you try this experiment? Throw the light of
a torch on a wall at different angles and watch the different conic sections on
the wall). Galileo (1564-1642) showed that the path of a projectile thrown
obliquely (Fig 9) is a parabola.

Fig 9: Projectile path

Paraboloid curves are also used in arches and suspension bridges (Fig 10).
Paraboloid surfaces are used in telescopes, search lights, solar heaters and
radar receivers.

Fig 10

In the seventeenth century, Johannes Kepler discovered that planets move in

elliptical orbits around the sun. Halley’s comet is also known to move along a
146 very elongated ellipse. A comet or meteorite coming into the solar system
Unit 16 Curve Tracing
from a great distance moves in a hyperbolic path. Hyperbolas are also used in
sound ranging and navigation systems.
Let’s look at the next example now.

Example 6: Sketch the graph of y = x 3 + x 2 .

Solution:
i) Domain and Range: The domain and range of the function are R .
ii) Symmetry: The function is neither even nor odd, thus, not symmetric
about y -axis and origin.
iii) Points of intersection: If x = 0, then y = 0, and if y = 0, x = 0,−1 . Thus,
the curve meets the axes at (0, 0) and (−1, 0) .

dy
iv) Tangents: We have = 3x 2 + 2x . The x -axis is the tangent at the
dx
dy dy
origin as = 0 , at x = 0 . Since, = 1 when x = −1 , therefore, the
dx dx
tangent at (−1, 0) makes an angle of 45o with the x -axis (Fig 11(a)).

dy −2
v) Relative extrema: Further, = 0 gives x = 0 and x = . Now,
dx 3
d2 y d2 y
= 6 x + 2 . Since, > 0 at (0, 0) , therefore, the point (0, 0) has a
dx 2 dx 2
−2 4 
relative minimum. The point  ,  has a relative maximum as
 3 27 
d2 y −2
2
< 0 at x = . Thus, in Fig 11(b), O is a valley and P is a peak.
dx 3
d2 y 1
vi) Point of inflection: Here, 2
= 0 at x = − and changes sign from
dx 3
 −1 2 
negative to positive as x passes through − 1 / 3 . Hence,  ,  is a
 3 27 
 1
2
d y
point of inflection. Since, 2
< 0 on  − ∞, −  , therefore, the curve is
dx  3
 1 
2
d y
concave downward. Also, since 2
> 0 on  − , ∞  , therefore, the
dx  3 
curve is concave upward in this interval.
2
vii) Interval of increasing or decreasing function: If − < x < 0 , then
3
dy
< 0 . Thus, the graph rises in ] − ∞, − 2 / 3[ and ] 0, ∞ [ , but falls in
dx
] − 2 / 3, 0 [ .
viii) Asymptotes: As x tends to infinity, so does y . As x → −∞ , so does
y . There is neither horizontal nor vertical asymptote.For oblique
asymptote, lim [(x 3 + x 2 ) − (mx + c)] does not exist, therefore, no oblique
x →∞
asymptote.

Hence, the graph is as shown in Fig 11(c).

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Block 4 Applications of Differential Calculus

(a) (b) (c)

Fig. 11
***

3x 2
Example 7: Sketch the curve y = .
x2 −1
Solution:
i) Domain and Range: The domain is R − {−1,1} .
ii) Symmetry: Since the powers of x are even, therefore, the curve is
symmetric about the y -axis.
iii) Point of intersection with the axes: The curve passes through origin.

3x 2 3
iv) Asymptotes: Since, lim = lim =3
x → ±∞ x 2 − 1 x → ±∞ 1
1− 2
x
Therefore, the line y = 3 is the horizontal asymptote.
3x 2 3x 2 3x 2
Also, lim+ 2 = ∞ , lim− 2 = −∞ , lim+ 2 = ∞ and
x →1 x − 1 x →1 x − 1 x →−1 x − 1

3x 2
lim − 2 = −∞
x → −1 x − 1

Therefore, the lines x = 1 and x = −1 are vertical asymptotes. We can

draw these asymptotes as shown in Fig.12 (a).

5x ( x 2 − 1) − 3x 2 ⋅ 2x − 6x
v) Monotonicity: Here, y′ = 2 2
= 2 . Since,
( x − 1) ( x − 1) 2
y′ > 0 when x < 0 and y′ < 0 when x > 0 , therefore, f is increasing on
] − ∞,−1[ and ] − 1, 0 [ and decreasing on ] 0,1[ and ]1, ∞ [ .
vi) Relative Extrema: When y′ = 0, x = 0 . Since, y′ changes from positive
to negative at 0, therefore, there is a local maximum by the first
derivative test.
− 6( x 2 − 1) 2 + 6x ⋅ 2( x 2 − 1)2 x 6(1 + 3x 2 )
vii) Concavity: We have f ′′( x ) = =
( x 2 − 1) 4 (x 2 − 1)3
Since, 6(1 + 3x 2 ) > 0 for all x , we have y′′ > 0 ⇔ x 2 − 1 > 0 ⇔ x > 1 and
y′′ < 0 ⇔ x < 1 . Thus, the curve is concave upward on the intervals
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Unit 16 Curve Tracing
] − ∞, − 1[ and ]1, ∞[ and concave downward on ] − 1,1[ . Since, 1 and
− 1 are not in the domain of f , therefore, there is no point of inflection.
Using the information in i) to vii), we sketch the curve in Fig. 12(b).

(a) (b)
Fig. 12
***

In the next example, we will trace a curve with exponential function.

Example 8: Trace the curve y = xe x .

Solution:
i) Domain and Range: The domain is R .
ii) Points of intersection with the axes: The curve passes through origin.
iii) Symmetry: There is no symmetry.
iv) Asymptotes: Because both x and e x become large as x → ∞, we have
lim xe x = ∞ . As x → −∞, however, e x → 0 and so, we have an
x →∞
indeterminate product that requires the use of L’Hôpital’s Rule:
x 1
lim xe x = lim x
= lim = lim (−e x ) = 0
x → −∞ x →−∞ e − x →−∞ − e − x x →−∞

Thus, the x − axis is a horizontal asymptote.

v) Monotonicity: We have y′ = xe x + e x = (x + 1)e x . Since, e x is always
positive, therefore, y′ > 0 when x + 1 > 0, and y′ < 0 when x + 1 < 0 . So,
y is increasing on ] − 1, ∞ [ and decreasing on ] − ∞,−1[ .
 dy  dy
vi) Relative Extrema: Since   = 0 and changes from negative to
 dx  x =−1 dx
positive at x = −1 , therefore, (−1, −e −1 ) a local minimum.

vii) Concavity: We have y′′ = ( x + 1)e x + e x = (x + 2)e x . Since, y′′ > 0 if

x > −2 and y′′ < 0 if x < −2, therefore, the curve is concave upward on
] − 2, ∞ [ and concave downward on ] − ∞,−2 [ . The inflection point is
] − 2,−2e −2 [ .

We use this information to trace the curve in Fig. 13.

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Block 4 Applications of Differential Calculus

Fig. 13
***

In the following example, we will trace a curve with trigonometric functions.

cos x
Example 9: Trace the curve y = .
2 + sin x
Solution:
i) Domain and Range: The domain is R .
ii) Points of intersection with the axes: The curve passes through
 1  (2n + 1) π 
 0,  and  , 0  where n is an interger.
 2  2 
iii) Symmetry and Periodicity: The curve is not symmetric about any of
the axes. We have, f ( x + 2π) = f ( x ) for all x and so f is periodic and
has period 2π . Thus, we need to consider only 0 ≤ x ≤ 2π and then
extend the curve by translation.
iv) Asymptotes: There is no asymptote.
dy (2 + sin x ) (− sin x ) − cos x (cos x )
v) Monotonicity: We have = .
dx (2 + sin x ) 2

2 sin x + 1
=−
(2 + sin x ) 2

dy 1 7π 11π
Thus, > 0 when 2 sin x + 1 < 0 ⇔ sin x < − ⇔ <x< . So,
dx 2 6 6
 7 π 11π   7π 
f is increasing on  ,  and decreasing on  0,  and
 6 6   6
 11π 
 6 , 2π  .

vi) Relative Extrema: From the first derivative test, we see that the local
−1 1
minimum value is and the local maximum value is .
3 3
vii) Concavity: If we differentiate f ( x ) again and simplify, we get

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Unit 16 Curve Tracing
2
d y 2 cos x (1 − sin x )
2
=−
dx (2 + sin x ) 3

Since, (2 + sin x )3 > 0 and 1 − sin x ≥ 0 for all x , also we know that
d2 y π 3π
2
> 0 when cos x < 0, that is, < x < , therefore, the curve is concave
dx 2 2
 π 3π   π  3π 
upward on  ,  and concave downward on  0,  and  , 2π  . The
2 2  2  2 
π   3π 
inflection points are  , 0  and  , 0  .
2  2 
We draw the graph of the function only to 0 ≤ x ≤ 2π is shown in Fig. 14 (a).
Then, we extend it, using periodicity, to the complete graph in Fig. 14 (b).

(a) (b)
Fig. 14
***

So far, all our curves were graphs of functions. We shall now trace some
curves which are not the graphs of functions, but have more than one branch.
These curves are of the form f ( x, y) = 0 .

Example 10: Trace the semi cubical parabola y 2 = x 3 .

Solution:
i) Regions where the curve lies: We note that x 3 is always non-negative
for points on the curve. This means that x is always non-negative and
no portion of the curve lies on the left of the y -axis.
ii) Symmetry: There is symmetry about the x -axis (even powers of y ).
iii) Point of intersection with axes: The curve meets the axes only at the
origin.
dy 3
iv) Double point: Here, y = ± x 3 / 2 . The derivative = ± x 1/ 2 . Here,
dx 2
y′ → 0 as x → 0 + and does not exist as x → 0 − . There are two real and
equal tangents at origin, therefore, origin is a cusp.
v) Increasing and decreasing behaviour: In the first quadrant y
increases with x and y → ∞ as x → ∞ .
vi) Asymptotes: There are no asymptotes.
151
Block 4 Applications of Differential Calculus
We first draw the curve in the first quadrant as shown in Fig. 15 (a), and then
take its reflection in the x -axis and we get the complete graph as shown in
Fig. 15 (b).

(a) (b)
Fig 15
***

Example 11: Trace the curve y 2 = ( x − 2) ( x − 3) (x − 4) .

Solution:
i) Region where curve lies: We can see that ( x − 2) (x − 3) (x − 4) is non-
negative. If x < 2 , we get a negative value for y 2 which is impossible.
So, no portion of the curve lies to the left of the line x = 2 . For the same
reason, no portion of the curve lies between the lines x = 3 and x = 4 .
Therefore, the curve lies between the lines x = 2 and x = 3 and right to
the line x = 4 .
ii) Symmetry: Since, y occurs with even powers alone, therefore, the
curve is symmetrical about the x -axis. Thus, we draw the curve above
x − axis, and then get a reflection below the x -axis to complete the
graph.
iii) Point of intersection with axes: The curve meets the axes at points
(2, 0), (3, 0) and (4, 0) .
dy 1
iv) Tangents and normals: Here, = [(x − 2) ( x − 3) + ( x − 2) ( x − 4)
dx 2 y
+ (x − 3) (x − 4)] . Thus, the curve has vertical tangent at (2, 0), (3,0) and
(4, 0) . Combining these facts, the shape of the curve near
A(2, 0), B(3, 0), C(4, 0) must be as shown in Fig 16 (a).
v) Interval of increasing or decreasing: Let us take y > 0 (i.e., consider
point of the curve above the x -axis). Then,
dy 3x 2 − 18x + 26
= . This is zero at x = 3 ± 1 / 3 . If
dx 2 ( x − 2) (x − 3) ( x − 4)
α = 3 + 1 / 3 and β = 3 − 1 / 3 then α lies between 3 and 4 , and can
therefore be ignored. Also, 3x 2 − 18x + 26 = 3( x − β) ( x − α ) and
2 < β < 3 < α < 4 . For x ∈ ] 2, 3[, x − α remains negative. Hence, for

152
Unit 16 Curve Tracing
dy
2 < x < β, > 0 since, (x − α) and ( x − β) are both negative.
dx
dy
Similarly, for β < x < 3, < 0 . Hence, the graph rises in ] 2, β [ and
dx
falls in ]β, 3[ . Thus, the shape of the curve is oval above the x -axis,
and by symmetry about the x -axis, we can complete the graph between
x = 2 and x = 3 as in Fig 16. (b).
vi) Concavity: Now let us consider the portion of the graph to the right of
x = 4 . Shifting the origin to (4, 0) , the equation of the curve becomes
y 2 = x ( x + 1) ( x + 2) = x 2 + 3x 2 + 2x .
As x increases, so does y . As x → ∞ , so does y (considering points
above the x -axis). When x is very small, x 3 and 3x 2 are negligible as
compared to 2x , so that near the (new) origin, the curve is
approximately of the shape of y 2 = 2 x . The large values of x , 3x 2 and
2x are negligible as compared to x 3 , so that the curve shapes like
y 2 = x 3 for large x . Thus, at some point the curve changes its
convexity.
This conclusion could also be drawn by showing the existence of a point
of inflection.
vii) Asymptotes: There are no asymptotes.
viii) Multiple point: There is no multiple point.

Considering the reflection along the x -axis, we have the complete graph as
shown in Fig 16(c).

(a) (b) (c)

Fig. 16
***

Example 12: Trace the curve ( x 2 − 1) ( y 2 − 4) = 4 .

Solution:
4
i) Region, where the curve lies: Here, y 2 = 2
+ 4, therefore,
x −1
4
x ∉] − 1,1[ . Similarly, x 2 = 2
+ 1, therefore, y ∉] − 2, 2[ .
y −4
ii) Symmetry: There is symmetry about both axes. We can therefore,
sketch the graph in the first quadrant only and then take its reflection in 153
Block 4 Applications of Differential Calculus
the y -axis to get the graph above the x -axis. The reflection of this
graph in the x -axis will give the complete graph.
iii) Point of intersection with axes: Notice that the origin is a point on the
curve. The curve does not meet the axes at any other points.
iv) Tangent at origin: The curve has tangents at origin and these are given
by 4x 2 + y 2 = 0 . These being imaginary, the origin is an isolated point
on the graph.

v) Asymptotes: Equating to zero the coefficients of the highest powers of x

and y , we get y = ± 2 and x = ± 1 as asymptotes of the curve. Thus,
the portion of the curve in the first quadrant approaches the lines x = 1
and y = 2 in the region far away from the origin. As x → ∞, y → 2 and
as y → ∞, x → 1 .
vi) Increasing and decreasing: In the first quadrant, as x increases, so
4
does x − 1 , and since x 2 − 1 =
2
2
, therefore, y decreases as x
( y − 4)
increases.
vii) Relative extrema: There are no extreme points.

There are no singular points or points of inflection. Hence, the graph is as

shown in Fig 17.

Fig 17
***

Example 13: Trace the curve y 2 = ( x − 1) (x − 2) 2 .

Solution:
i) Symmetry: Since, the power of y is even, therefore, there is symmetry
about the x -axis.
ii) Region: No portion of the curve lies to the left of x = 1 , as y 2 cannot be
negative.
iii) Points of intersection with axes: Points of intersection with the axes
are (1, 0) and (2, 0) .
154
Unit 16 Curve Tracing
iv) Tangent: The tangent at (1, 0) is vertical. Shifting the origin to (2, 0) ,
the curve transforms into y 2 = x 2 (x + 1) . The tangents at the new origin
are given by y 2 = x 2 . This means that the point (2, 0) is a node, and the
tangents at (2,0) are equally inclined to the axes.
Let us try to build up the graph above the x -axis between x = 1 and
x = 2 . Differentiating the equation of the curve with respect to x , we get
2 yy′ = ( x − 2) 2 + 2(x − 1) ( x − 2) = ( x − 2) (3x − 4)
( x − 2) (3x − 4)
or, y′ =
2y
when 1 < x < 2, (x − 2) < 0 . If y is positive, then y′ > 0 provided
 4
3x − 4 < 0 . Thus, y′ > 0 when x ∈  1,  and y′ < 0 when
 3
4 
x ∈  , 2  . The tangent is parallel to the x -axis when 3x − 4 = 0 , that
3 
is, when x = 4 / 3 (see Fig 18 (a)). Hence, for 1 < x < 2 , the curve
shapes as in Fig 18 (b).
v) Intervals for increasing and decreasing: As x → ∞, y → ∞, in the
first quadrant. Note that when (2, 0) is taken as the origin, the equation
of the curve reduces to
y 2 = x 2 ( x + 1) = x 3 + x 2 .
This shows that when x > 0 and y > 0 , the curve lies above the line
y = x (on which y 2 = x 2 ). Hence, the final sketch (Fig 18 (c)) shows
the complete graph.

(a) (b) (c)

Fig. 18
***

If you have gone through Examples 1-13 carefully, you should be able to do
the following exercise.

E1) Trace the following curves by stating all the properties you use to trace:
i) y = x2 ii) y 2 = ( x − 2)3
iii) y(1 + x 2 ) = x iv) y 2 = x 2 (1 − x 2 )
v) y = x e −1 / x vi) y = sin 3 x
vii) y = x (ln x ) viii) y = x − 5x1 / 3
155
Block 4 Applications of Differential Calculus
ix) y = ln(sin x ) x) y = x 5− x

x2 x3 + 4
xi) y= xii) y=
x −1 x2

E2) Find the oblique asymptotes of the curve y = x − tan −1 x and hence,
trace the curve using this fact.
m0
E3) In the theory of relativity, the mass of a particle is m = ,
1 − v2 / c 2
where m 0 is the mass of the particle, m is the mass when the particle
moves, with speed v relative to the observer, and c is the speed of light.
Trace the curve for m as a function of v .

In the following section, we will trace the curves which are in parametric form.

16.3 TRACING A CURVE: PARAMETRIC

EQUATION
Sometimes a functional relationship may be defined with the help of a
parameter. In such cases, we are given a pair of equations which relate x
and y with the parameter. For example, imagine a particle that moves along a
curve and the x and y coordinates are defined in terms of time t as shown in
Fig. 19.

Fig. 19

In this case, we write x = f ( t ) and y = g (t ) , where t is the third variable called

parameter. The equations x = f ( t ) and y = g (t ) are known as parametric
equations. The parameter t does not necessarily represent time. You may
recall what you learnt in Appendix I of Block 3.

Now, we shall see how we can trace a curve whose equation is in the
parametric form.

We shall illustrate the process with an example.

Example 14: Trace the cycloid x = a (t + sin t ), y = a (1 − cos t ) as t varies

from − π to π .
156
Unit 16 Curve Tracing
dx dy dy
Solution: Here = a (1 + cos t ), = a sin t , so that = tan( t / 2) . Since,
dt dt dx
dy
> 0 for all t ∈ ] − π, π [, x increases with t from − aπ (at t = − π) to 0 (at
dx
t = 0 ) to aπ (at t = π ).
dy
Also, is negative when t ∈] − π, 0 [ and positive when t ∈] 0, π [ . Hence,
dx
y decreases from 2a to 0 in [−π, 0] and increases from 0 to 2a in [0, π] .
Let us tabulate this data in Table 2.

Table 2

t ∈ [− π, 0] t ∈ [0, π]
i) x increases from − a to 0 i) x increases from 0 to a
ii) y decreases from 2a to 0 ii) y increases from 0 to 2a
iii) Hence, the curve falls iii) Hence, the curve rises

Also, at the terminal points − π, 0 and π of the intervals [−π, 0] and [0, π] ,
we summarize this in Table 3.

Table 3
t (x , y) dy dx Tangent
dx dy
−π (−aπ, 2a ) not defined 0 vertical
0 (0, 0) 0 not defined horizontal
π (aπ, 2a ) not defined 0 vertical

On the basis of the data tabulated in Table 3, the graph is drawn in Fig 20.

Fig. 20

If t is increased by 2π, x is increased by 2πa and y does not change.

Thus, the complete graph can be obtained in intervals
K[−5π, − 3π], [−3π, − π], [π, 3π], [3π, 5π]K by translation through a proper
distance.
***
The cycloid is known as the Helen of geometry because it was the cause of
many disputes among mathematicians. It has many interesting properties.
157
Block 4 Applications of Differential Calculus
We shall describe just one of them here. Consider this question: What shape
should be given to a trough connecting two points A and B , so that a ball
rolls from A to B in the shortest possible time?

Now, we know that the shortest distance between A and B would be along
the line AB (Fig. 21). But since we are interested in the shortest time rather
than distance, we must also consider the fact that the ball will roll quicker, if
the trough is steeper at A . The Swiss mathematician Jakob and Johann
Bernoulli proved by exact calculations that the trough should be made in the
form of an arc of a cycloid. Because of this, a cycloid is also called the curve
of the quickest descent.

Fig. 21

The cycloid is used in clocks and in teeth for gear wheels. It can be obtained
as the locus of a fixed point on a circle as the circle rolls along a straight line.

Now, let us trace another curve in parametric form.

Example 15: The position of a particle at time t is given by the parametric

equations x = t 2 − 2t and y = t + 1 . Sketch and identify the path along which
the particle moves.
dx dy dy 1
Solution: We have = 2t − 2 and = 1, so that = . Here,
dt dt dx 2t − 2
dy dy
> 0, when t > 1 and < 0 , when t < 1 . That means y is increasing when
dx dx
t > 1 and y is decreasing when t < 1 . At t = 1, the tangent of the curve is
vertical. In Fig. 22 we plot the curve.

Fig. 22

We can also mark the points given in the Table 4.

158
Unit 16 Curve Tracing
Table 4
t x y
−2 8 −1
−1 3 0
0 0 1
1 −1 2
2 0 3
3 3 4
4 8 5

A particle whose position is given by the parametric equations moves along

the curve in the direction of the arrows as t increases. Notice that the
consecutive points marked on the curve appear at equal time intervals but not
at equal distances. That is because the particle slows down and then speeds
up as t increases.

It appears from Fig. 22 that the curve traced out by the particle may be a
parabola. We can confirm this by eliminating the parameter t as follows:
We obtain t = y − 1 from the second equation and substitute into the first
equation. This gives x = t 2 − 2 t = ( y − 1) = y 2 − 4 y + 3 and so the curve
represented by the given parametric equations is the parabola x = y 2 − 4 y + 3 .
***
Example 16: What curve is represented by the following parametric
equations?
x = cos t , y = sin t , where 0 ≤ t ≤ 2π
Solution: If we plot points, it appears that the curve is a circle. We can confirm
this impression by eliminating t . Observe that
x 2 + y 2 = cos 2 t + sin 2 t = 1
Thus, the point ( x, y) moves on the unit circle x 2 + y 2 = 1 . Notice that in this
example the parameter t can be interpreted as the angle (in radians) shown in
Fig. 23. As t increases from 0 to 2π , the point P (cos t, sin t ) moves once
around the circle in the counterclockwise direction starting from the point
(1, 0) .

Fig. 23
***
159
Block 4 Applications of Differential Calculus

See if you can do this exercise now.

E4) Trace the following curves:

i) x = a ( t + sin t ), y = a (1 + cos t ), − π ≤ t ≤ π, a > 0 .
ii) x = a sin 2t (1 + cos 2 t ), y = a cos 2t (1 − cos 2 t ), 0 ≤ t ≤ π, a > 0 .
iii) x = at 2 , y = 2at , 0 ≤ t ≤ 1 .
iv) x = sin 2t , y = cos 2 t, 0 ≤ t ≤ 2π .

v) x = sin t, y = sin 2 t .

So far, we have discussed tracing of curves in cartesian and parametric forms.

In the following section, we will discuss tracing of a curve in polar form.

16.4 TRACING A CURVE: POLAR EQUATION

In this section, we shall consider the problem of tracing those curves, whose
equations are given in the polar form. You may recall Unit 3 for polar
coordinates. In such a coordinate system, we can associate each point P in
the plane with a pair of polar coordinates (r, θ) , where r is the number of units
between P and the pole and θ is an angle from the polar axis to the ray OP as
shown in Fig. 24. If r is negative, then the point is located on the opposite side
of the origin. Thus, r is a position on a rotated axis.

Fig. 24

The following considerations can be useful in this connection.

i) Symmetry: If the equation remains unchanged when θ is replaced by
− θ , then the curve is symmetric with respect to the initial line.
If the equation does not change when r is replaced by − r , then the
curve is symmetric about the pole (or the origin).
Finally, if the equation does not change when θ is replaced by π − θ ,
then the curve is symmetric with respect to the line θ = π / 2 .
ii) Region: Find the limits within which r must lie for the permissible values
of θ . If r < a (r > a ) for some a > 0 , then the curve lies entirely within
(outside) the circle r = a . If r 2 is negative for some values of θ , then
the curve has no portion in the corresponding region.
iii) Angle between the line joining a point of the curve to the origin and
the tangent: At suitable points, this angle can be determined easily. It

160
Unit 16 Curve Tracing
helps in knowing the shape of the curve at these points. You may recall
dθ
from Unit 14 that the angle φ is given by the relation tan φ = r .
dr
We shall illustrate the procedure through some examples of graphing
equations of the form r = f (θ) in polar coordinates, where θ is assumed to be
measured in radians. Study them carefully, so that you can trace some curves
on your own later.

Example 17: Trace the cardioid r = a (1 + cos θ) .

Solution: We can make the following observations.
i) Symmetry: Since, cos θ = cos(−θ) , therefore, the curve is symmetric
with respect to the initial line. That means we need to trace the curve
only above the initial line, rest half curve would be the reflection along
the initial line.
ii) Region: Since, − 1 ≤ cos θ ≤ 1 , therefore, the curve lies inside the circle
r = 2a .
dr dr
iii) Tangents: = −a sin θ . Hence, < 0 , when 0 < θ < π . Thus, r
dθ dθ
decreases as θ increases in the interval ] 0, π / 2 [ . Similarly, r
increases with θ in ] π / 2, θ [ . Some corresponding values of r and θ
are given in Table 5.
Table 5

θ 0 π/ 2 π
r 2a a 0

Combining the above facts, we can easily draw the graph above the
initial line. By reflecting this portion in the initial line we can completely
draw the curve as shown in Fig 25 (a). Notice the decreasing radii
2a , r1 , r2 , r3 etc. If we allow a to vary and keep is positive, then the size
of cardioid varies. If a is negative, then the cardioid changes its
direction. These cardioids are shown in Fig. 25 (b).

(a) (b)
Fig. 25: (a) Curve r = a (1 + cos θ) ; (b) Curve r = a (1 + cos θ) for a = 1, 2, 5, − 1 .

This curve is called a cardioid since it resembles a heart.

***

You may note that the equations with any of the four forms r = a ± b sin θ and
r = a ± b cos θ in which a > 0 and b > 0 represent polar curves called limacons
161
Block 4 Applications of Differential Calculus
(from the Latin word “limax” for a snail-like creature that is commonly called a
slug) as shown in Fig. 26 (a) to Fig. 26 (d). There are four possible shapes for
a limacon in each of the four cases that are determined by the ratio a / b (Fig
26(e) to Fig. 26 (h)). If a = b (the case a / b = 1) , then the limacon is called a
cardioid because of its heart-shaped appearance, as noted in Example 17.

(a) r = a + b cos θ (b) r = a − b cos θ (c) r = a − b sin θ

a a
(d) r = a + b sin θ (e) < 1 (with inner loop) (f) = 1 (Cardioid)
b b

a a
(g) 1 < < 2 (Dimpled) (h) > 2 (Convex)
b b
Fig 26

Example 18: Trace the equiangular spiral r = ae θ cot α .

Solution: We proceed as follows.

i) Region: When θ = 0, r = a .
ii) Symmetry: There is no symmetry.
dr
iii) Tangents: = r cot α , which is positive, assuming cot α > 0 . Hence
dθ
dθ
as θ increases so does r . r = tan α . Thus, at every point, the angle
dr
between the line joining a point on the curve to the origin and the
tangent is the same, namely α . Hence the name.
Combining these facts, we get the shape of the curve as shown in
162 Fig 27.
Unit 16 Curve Tracing

Fig. 27: Curve r = a e θ cot α .

The equiangular (or logarithmic) spiral r = ae θ cot α is known as the curve of

pursuit. Suppose four dogs start from the four corners of a square, each
pursues the dog in front with the same uniform velocity (always following the
dog in front), then each will describe an equiangular spiral. Several shells and
fossils have forms which are quite close to equiangular spirals (Fig 28). Seeds
in the sunflower or blades of pine cones are also arranged in this form.

Fig. 28: Spiral.

The first discussion of this spiral occur in letters written by Descartes to

Mersenne in 1638. The name logarithmic spiral is due to Jacques Bernoulli.
He was so fascinated by it that he willed that an equiangular spiral be carved
on his tomb with the words ‘Though changed, I rise unchanged’ inscribed
below it.

The spiral r = aθ is known as the Archimedean spiral. Its study was,

however, initiated by Conon. Archimedes used this spiral to square the circle,
that is, to find a square of area equal to that of a given circle. This spiral is
widely used as a cam to produce uniform linear motion. It is also used as
casings of centrifugal pumps to allow air which increases uniformly in volume
with each degree of rotation of the fan blades to be conducted to the outlet
without creating back-pressure.

The spiral rθ = a , due to Varignon, is known as the reciprocal or hyperbolic

(recall that xy = a is a hyperbola) spiral. It is the path of a particle under a
central force which varies as the cube of the distance.
***
163
Block 4 Applications of Differential Calculus
Now let’s consider the next example.

Example 19: Trace the curve r = a sin 3θ, a > 0 .

Solution:

i) Symmetry: You may note that there is symmetry about the line
θ = π / 2 as the equation is unchanged if θ is replaced by π − θ .

ii) Region: The curve lies inside the circle r = a , because sin 3θ ≤ 1 . The
origin lies on the curve and this is the only point where the initial line
meets the curve.

iii) Tangents: r = 0 ⇒ θ = nπ / 3 , where n is any integer. Hence the origin

π 2π 4π 5π
is a multiple point, the lines θ = 0, , , π, , , 2π etc. being
3 3 3 3
tangents at the pole.

dr
iv) Monotonicity: = 3 cos 3θ . Hence r increases in the intervals
dθ
 ππ π  7 π 3π 
 0, 6 ,  2 , 6  , and  6 , 2  , and decreases in the intervals
 π π   5π 7 π   3π 5π 
 6 , 2 ,  6 , 6  and  2 , 3  . Notice that r is negative when
 π 2π   4π   5π 
θ ∈ ,  or θ ∈  π,  or θ ∈  , 2π . Hence, the curve
3 3   3  3 
consists of three loops as shown in Fig 29. The function is periodic and
the curve retraces itself as θ increases from 2π on.

Fig. 29: Curve r = a sin 3θ .

***
In polar coordinates, equations of the form r = a sin nθ and r = a cos nθ, in
which a > 0 and n is a positive interger represent families of flower-shaped
curves called roses (Fig. 30). The rose consists of n equally spaced petals of
radius a if n is odd and 2n equally spaced petals of radius a if n is even. It can
be shown that a rose with an even number of petals is traced out exactly once
as θ varies over the interval 0 ≤ θ < 2π and a rose with an odd number of
petals is traced out exactly once as θ varies over the interval 0 ≤ θ < π . A
three-petal rose of radius a was graphed in Example 19.
164
Unit 16 Curve Tracing

(a) r = a sin n θ .

(b) r = a cos n θ .
Fig. 30: Rose Curves

Now try to trace a few curves on your own.

E5) Trace the following curves in polar coordinates.

i) r =1
π
ii) θ=
4
iii) r=θ (θ ≥ 0)

E6) Trace the following curves by stating all the properties you used:
i) r = a (1 − cos θ), a > 0 .
ii) r = 2 + 4 cos θ .
iii) r = a cos 3θ, a > 0 .
iv) r = a sin 2θ, a > 0

Now, let us summarize what we have studied in this unit.

16.5 SUMMARY
In this unit, we have covered the following points.

1. Tracing a curve y = f ( x ) or f ( x, y) = 0 means plotting the points which

satisfy this relation.

2. Criteria for symmetry and monotonicity, equations of tangents,

asymptotes and points of inflection are used in curve tracing.
165
Block 4 Applications of Differential Calculus
3. Curve tracing is illustrated by some examples when the equation of the
curve is given in
i) Cartesian form
ii) Parametric form
iii) Polar form.

16.7 SOLUTIONS/ANSWERS
Dotted lines represents tangents or asymptotes throughout.

E1) i) Domain: R and the curve lies in first and second quadrant.
Point of intersection with axes: (0,0)
Symmetry: About y − axis
Asymptotes: None
Monotonicity: Increasing on ] 0, ∞[ and decreasing on ] − ∞,0[
Relative extrema: Minimum at 0 and f (0) = 0 .
The corresponding sketch of the curve is given in Fig. 31.

Fig. 31
ii) Region of existence: [2, ∞ [ and the curve lies in first and fourth
quadrant.
Point of intersection with axes: (2, 0)
Symmetry: About x − axis
Asympotes: None
Double Point: If we shift origin at (2, 0) , then, (2, 0) is double
point and is cusp.
The corresponding curve is traced in Fig. 32.

Fig. 32
166
Unit 16 Curve Tracing
iii) Domain: R and the curve is in first and third quadrant as either
x , y are both positive or both negative.
Symmetry: About origin.
Asymptote: x -axis is an asymptote.
Monotonicity: Function rises in ] − 1, 1[ and falls elsewhere.
Tangents: y = x is the tangent at the origin
 3  3
Concavity: (0, 0),  3 , ,  − 3 , −  are points of inflection.
 4   4 
The graph is shown in Fig. 33.

Fig. 33
y2
iv) Region of existence: The curve = 1 − x 2 shows that the entire
x2
curve lies within the lines x = ±1 .
Point of intersection with axes: (0, 0), (1,0) and (−1,0)
Symmetry: About x − axis, y − axis and origin.
Tangents: Tangents at the origin are y = ± x . Tangents at
x = ±1 are vertical.
( )
Relative extrema: Maxima at ± 1 / 2 , 1 / 4 , and minima at
 1 1
± ,− 
 2 4
Multiple point: y = ± x 1 − x 2 , y is defined if 1 − x 2 ≥ 0, or
− 1 ≤ x ≤ 1 . If we equate lowest degree term to 0, we get
y 2 = x 2 , which gives y = ± x . Therefore, the curve has two
tangents at origin, namely, y = x and y = − x, and the origin is a
node. The curve is sketched in Fig. 34.

Fig. 34
167
Block 4 Applications of Differential Calculus
v) Domain and Range: ] − ∞, 0 [ ∪ ] 0, ∞ [
Symmetry: None
Point of intersection with axes: None
Concavity: Concave up on ] 0, ∞ [ and concave down on ] − ∞, 0 [
Relative extrema: Maxima on − 1 and the relative maximum value
is f (−1) = − e .
Monotonicity: Increasing on ] − ∞,−1[ and on ] 0, ∞ [ .
decreasing on ]− 1, 0 [ .
Point of continuity: Origin is the point of discontinuity
The corresponding curve is given in Fig. 35.

Fig. 35
vi) Domain: R
Symmetry: About the origin
Periodicity: Period 2π
Point of intersection: Origin (0, 0), (n π, 0) (n is integer)
 3π 
Monotonicity: Increasing on ] 0, π / 2 [ and  , 2π  ,decreasing
2 
 π 3π 
on  , 
2 2 
Relative extrema: Maximum at π / 2, f (π / 2 ) = 1 and minima at
3π  3π 
, f   = −1 .
2  2 
Concavity: Concave upward on ] 0, a [, ]π − a , π[ and concave
downward on ] a , π − a [ where a = sin −1 2 / 3 .
Point of inflection: x = 0, π, π − a .
Curve is traced in Fig. 36.

Fig. 36
168
Unit 16 Curve Tracing
vii) Domain: ] 0, ∞ [
Point of intersection with axes: (1, 0)
Symmetry: None
Asymptote: None
1   1
Monotonicity: Increasing on  , ∞  and decreasing on  0,  .
e   e
1 1 1
Relative extrema: Minima at x = ,f  = −
e e e
Concavity: Concave upward on ] 0, ∞ [
The curve is drawn in Fig. 37.

Fig. 37
viii) Domain: R
Point of intersection with axes: (0, 0), ( ±3 3 , 0)
Symmetry: About the origin
Asymptote: None
Monotonicity: Increasing on ] − ∞, − 1[, ]1, ∞ [ and decreasing on
] − 1,1[
Relative extrema: Maximum at − 1 and f (−1) = 2, Minimum at
1 and f (1) = −2
Concavity: Concave upward on ] 0, ∞ [ and concave downward on
] − ∞, 0 [
Point of inflection: (0, 0) is point of inflection.
The corresponding curve is traced in Fig. 38.

Fig. 38
169
Block 4 Applications of Differential Calculus
ix) Domain: x ∈] 2nπ, (2n + 1) π [ ,where n is an interger. The value of
y is always negative. Therefore, the curve lies in third and fourth
quadrant.
π 
Point of intersection with axes:  + 2n π, 0  .
2 
Symmetry: None

Periodicity: Period 2π

Asymptotes: Vertical asymptotes at x = n π

 π 
Monotonicity: Increasing on  2nπ, + 2n π  and decreasing on
2  
π 
 2 + 2n π, (2n + 1) π .
π π 
Relative extrema: Maximum at + 2n π and f  + 2n π  = 0 .
2 2 
The corresponding curve is traced in Fig. 39.

Fig. 39
x) Domain: ] − ∞, 5 ]
Point of intersection with axes: (0, 0), (5, 0)

Symmetry: None

Asymptote: None
 10  10 
Monotonicity: Increasing on  − ∞,  and decreasing on  ,5  .
 3 3 
10  10  10
Relative extrema: Maximum at and f   = 5.
3 3 9

Concavity: Concave downward on ] − ∞, 5[

The corresponding curve is traced in Fig. 40.

170
Unit 16 Curve Tracing

Fig. 40
xi) Domain: ] − ∞,1[ ∪ ] 1, ∞ [
Point of intersection with axes: (0, 0)
Symmetry: None
Asymptotes: x = 1, y = x + 1
Monotonicity: Increasing on ] − ∞, 0 [ and ] 2, ∞ [
Decresing on ] 0,1[ and ]1, 2 [
Relative extrema: Maximum at x = 0 and f (0) = 0 and minimum
at x = 2 and f (2) = 4 .
Concavity: Concave upward on ]1, ∞ [ and concave downward on
] − ∞,1[ .
The corresponding curve is traced in Fig. 41.

Fig. 41
xii) Domain: ] − ∞, 0 [ ∪ ] 0, ∞ [
Point of intersection with axes: (−(4)1 / 3 ,0)
Symmetry: None
Asymptotes: x = 0, y = x
Monotonicity: Increasing on ] − ∞, 0 [ and ] 2, ∞ [ and decreasing
on ] 0, 2 [ .
Relative extrema: Minimum at x = 2 and f (2) = 3
Concavity: Concave upward on ] − ∞,0[ and ] 0, ∞ [ .
The corresponding curve is traced in Fig. 42.
171
Block 4 Applications of Differential Calculus

Fig. 42
π
E2) y = x − tan −1 x oblique asymptotes are y = x ± , which are shown in
2
Fig. 43.

Fig. 43

E3) Domain: [0, c[

Point of intersection with axes: (0, m 0 ) .
Symmetry: None
Asymptotes: v = c
Monotonicity: Increasing on [0, c [
The corresponding curve is traced in Fig. 44.

Fig. 44
172
Unit 16 Curve Tracing
E4) i)

Fig. 45
ii)

Fig. 46
iii)

Fig. 47

iv) Again we have x 2 + y 2 = sin 2 2 t + cos2 2t = 1

So the parametric equations again represent the unit circle
x 2 + y 2 = 1 . But as t increases from 0 to 2π , the point

173
Block 4 Applications of Differential Calculus
( x, y) = (sin 2t , cos 2 t ) starts at (0,1) and moves twice around the
circle in the anti clockwise direction as indicated in Fig. 48

Fig. 48

v) You may observe that y = (sin t ) 2 = x 2 and so the point

( x, y) moves on the parabola y = x 2 . But note also that, since
− 1 ≤ sin t ≤ 1, we have − 1 ≤ x ≤ 1, so the parametric equations
represent only the part of the parabola for which − 1 ≤ x ≤ 1 . Since
sin t is periodic, the point ( x, y) = (sin t , sin 2 t ) moves back and
forth infinitely often along the parabola from (−1,1) to (1,1) . (See
Fig 49).

Fig. 49

E5) i) For all values of θ , the point (1, θ) is 1 unit away from the pole.
Thus, the graph is the circle of radius 1 centered at the pole (Fig.
50 (a))
ii) For all values of r , the point (r, π / 4) lies on a line that makes an
angle of π / 4 with the polar axis (Fig. 50(b)). Positive values of
r correspond to points on the line in the first quadrant and
negative values of r to points on the line in the third quadrant.
Thus, in the absence of any restriction on r , the graph is the
entire line. Observe, however, that had we imposed the restriction
r ≥ 0, the graph would have been just the ray in the first quadrant.

174
Unit 16 Curve Tracing
iii) Observe that as θ increases, so does r ; thus, the graph is a curve
that spirals out from the pole as θ increases. A reasonably
accurate sketch of the spiral can be obtained by plotting the
intersections with the x − and y − axes for values of θ that are
multiples of π / 2 , keeping in mind that the value of r is always
equal to the value of θ (Fig. 50(c)).

(a) r =1 (b) θ = π/4 (c) r=θ

Fig. 50

E6) i)

Fig. 51
ii)

Fig. 52 175
Block 4 Applications of Differential Calculus
iii)

Fig. 53
iv)

Fig. 54

176
Block 4 Miscellaneous Examples and Exercises

MISCELLANEOUS EXAMPLES AND EXERCISES

The examples and exercises given below cover the concepts and processes
you have studied in this block. Doing them will give you a better understanding
of the concepts concerned, as well as practice in solving such problems.
Example 1: Find the intervals on which the following functions are increasing
and the intervals on which they are decreasing.
i) f (x) = x 2 − 6x + 5 ii) f (x) = x 3
Solution: i) The graph of f in Fig. 1 suggests that f is decreasing for
x ≤ 3 and increasing for x ≥ 3 . To confirm this, we differentiate f to obtain
f ′(x ) = 2x − 6 = 2 ( x − 3) .
It follows that f ′( x ) < 0 if x < 3 , and f ′( x ) > 0 if x > 3 .
Since, f is continuous at x = 3 , using the first derivative test, we can say that
f is decreasing on ] − ∞, 3[ and f is increasing on ]3, + ∞[ .
We can also conclude these from the graph of f in Fig. 1.

2
Fig. 1: Graph of x − 6x + 5 .

ii) The graph of f in Fig. 2 suggests that f is increasing above the x -axis.
To confirm this, we differentiate f and obtain f ′(x ) = 3x 2 . Thus,
f ′( x ) > 0 if x < 0 and f ′( x ) > 0 if x > 0 .
Since, f is continuous at x = 0 , therefore, using first derivative test f is
increasing on ] − ∞, 0 [ and ] 0,+∞ [ .
Hence, f is increasing over the entire interval ] − ∞,+∞ [ , we also
conclude the same from the graph in Fig. 2.

3
Fig. 2: Graph of x .
***
177
Block 4 Applications of Differential Calculus
Example 2: Use the graph of the function of defined as
1 4 1 3
f (x) = x − x − x 2 + 5 given in Fig. 3 to mark the intervals on which f is
4 3
increasing or decreasing. Also, verify it using derivatives.

1 4 1 3 2
Fig. 3: Graph of x − x −x +5.
4 3
Solution: The graph suggests that f is decreasing if x ≤ −1 , increasing if
− 1 ≤ x ≤ 0 , decreasing if 0 ≤ x ≤ 2 , and increasing if x ≥ 2 .
On differentiating f we obtain
f ′( x ) = x 3 − x 2 − 2x = x ( x 2 − x − 2) = x ( x + 1) ( x − 2)
The sign of f ′ is given in Table 1, which confirms the conclusion derives from
the graph.

Table 1

Interval Sign of (x) (x + 1) (x − 2) Sign of Conclusion

f ′(x)
x < −1 ( −) ( −) ( −) − f is decreasing on
] − ∞, − 1[
−1 < x < 0 ( −) ( + ) ( −) + f is increasing on
] − 1, 0 [
0<x<2 ( + ) ( + ) ( −) − f is decreasing on
] 0, 2 [
x>2 (+) (+) (+) + f is increasing on
] 2, + ∞[
***
Example 3: Find the intervals on which the following functions are concave
upward and concave downward.
1
i) f (x) = x 2 − 6x + 5 ii) f (x) = x 3 iii) f (x) = x 3 − x 2 + 2
3
Solution: i) Calculating the first two derivatives we obtain f ′( x ) = 2 x and
f ′′( x ) = 2 . Since, f ′′( x ) > 0 for all x , the function f is concave upward on
] − ∞, + ∞[ . You can verify this from the graph given in Fig. 1.
ii) Calculating the first two derivatives, we obtain f ′(x ) = 3x 2 and f ′′( x ) = 6 x .
Since, f ′′( x ) < 0 if x < 0 and f ′′( x ) > 0 if x > 0 , therefore, the function f is
178
Block 4 Miscellaneous Examples and Exercises
concave downward on ] − ∞, 0 [ and concave upward on ] 0, + ∞[ as shown
in Fig. 2.
iii) Calculating the first two derivatives, we obtain f ′( x ) = x 2 − 2 x and
f ′′( x ) = 2 x − 2 = 2 ( x − 1) . Since, f ′′( x ) > 0 if x > 1 and f ′′( x ) < 0 if x < 1 ,
we conclude that f is concave upward on ]1, + ∞ [ and f is concave
downward on ] − ∞,1[ . Fig. 4 shows the graph and verifies this.

1 3 2
Fig. 4: Graph of x −x +2.
3
***

Example 4: Consider the graph of the function f defined by

1 4 1 3
f (x) = x − x − x 2 + 5 as shown in Fig. 3. Find the inflection points from
4 3
the graph, and check your answer by finding the inflection points using
derivatives.
Solution: From the graph shown in Fig. 3, it is clear that the graph changes
from concave upward to concave downward between − 1 and 0 , say roughly at
x = −0.50 , and the graph changes from concave downward to concave upward
somewhere between 1 and 2 , say roughly at x = 1.25 . To check this result with
the exact inflection points, we obtain the second derivative of f . We get
f ′( x ) = x 3 − x 2 − 2x and f ′′( x ) = 3x 2 − 2x − 2 = [3x − (1 + 7 )][3x − (1 − 7 )] .
We check sign of the second derivative at different intervals as given in Table 2.
Thus, from the sign of f ′′ in Table 2, we can say that f has inflection points at
1− 7 1+ 7
the values x = ≈ −0.55 and x = ≈ 1.22 . The graph of f ′′ is shown
3 3
in Fig. 5 verifies this.
Table 2

Interval Sign of f ′′ Result

1− 7 + f is concave upward
x<
3
1− 7 1+ 7 − f is concave downward
<x<
3 3
1+ 7 + f is concave upward
x>
3

179
Block 4 Applications of Differential Calculus

Fig. 5: Graph of f ′′ .
***

Example 5: Find the inflection points of f ( x ) = sin x on the interval [0, 2π] ,
and verify your results with the graph of the function.

Solution: Calculating the first two derivatives of f , we obtain f ′( x ) = cos x and

f ′′( x ) = − sin x .
Thus, f ′′( x ) < 0 if 0 < x < π, and f ′′( x ) > 0 if π < x < 2π , which implies that the
graph is concave downward on 0 < x < π and concave upward on π < x < 2π .
Thus, there is an inflection point at x = π ≈ 3.14 as shown in Fig. 6.

Fig. 6

***

Example 6: Use the graph of y = f ′( x ) given in the Fig. 7 to fill the boxes with
<, =, or > . Give reasons for your answer.

i) f (0) f (1) ii) f (1) f (2) iii) f ′(0) 0

iv) f ′(1) 0 v) f ′′(0) 0 vi) f ′′(2) 0
180
Block 4 Miscellaneous Examples and Exercises

Fig. 7
Solution: i) Since, f ′ > 0 on [0,1], therefore, f is increasing on [0,1] and
f (0) < f (1) .
ii) Since f ′ < 0 on [1, 2], therefore, f is decreasing on [1, 2] and f (1) > f (2) .

iii) f ′(0) > 0

iv) f ′(1) = 0, the graph of f ′ intersects the x -axis at x = 1 .
v) f ′′(0) = 0
vi) f ′′(2) = 0 .
***

3 1
Example 7: Show that 1 + x < 1 + x if x > 0 .
3
x 3
Solution: Let f ( x ) = 1 + − 1+ x .
3
1 
−2
1 1 1 1 1
Then, f ′(x ) = − (1 + x ) 3 = − 2/ 3
= 1 − 2/3 
.
3 3 3 3(1 + x ) 3  (1 + x ) 
Here, f ′( x ) > 0 , when x > 0 , therefore, f is an increasing function on ] 0, ∞[ .
Hence, f (0) < f ( x ) ∀ x ∈] 0, ∞ [ .
x
Which gives 0 < 1 + − 3 1 + x [Qf (0) = 0]
3
x
Thus, 3 1 + x < 1 +
3
***

Example 8: Find the relative maxima and minima of f ( x ) = x 4 − 2x 2 . Mark

these on the graph of f .
Solution: We have f ′(x ) = 4x 3 − 4x = 4 x ( x − 1) ( x + 1) , and f ′′( x ) = 12x 2 − 4 .
On solving f ′(x ) = 0, we get critical points, which are x = 0, x = 1 and x = −1 .
Using second derivative test, we get
f ′′(0) = −4 < 0 , thus maxima at x = 0 .
f ′′(1) = 8 > 0 , thus minima at x = 1 .
f ′′(−1) = 8 > 0 , thus minima at x = −1 .
181
Block 4 Applications of Differential Calculus
So, there is a relative maximum at x = 0 and relative minima at x = 1 and at
x = −1 as shown in Fig. 8.

Fig. 8
***

x3 − x 2 − 8
Example 9: Trace the curve y = by showing all the properties you
x −1
use to trace.
Solution:
i) Symmetry: There is no symmetry about x -axis, y -axis or about origin.
ii) Point of intersection with axes: Setting y = 0 , gives the equation
x 3 − x 2 − 8 = 0 . The LHS of this equation changes its sign in the interval
[2, 3] , therefore, the graph of y intersects x -axis between 2 and 3. Also,
the curve passes through the point (0, 8) .

x3 − x2 − 8
iii) Asymptotes: Here lim tends to ∞, therefore x = 1 is a vertical
x →1 x −1
asymptote. There are no horizontal asymptotes.
dy d  2 8 
iv) Relative maxima or minima: We get  x −
x − 1
=
dx dx 
8
= 2x +
( x − 1) 2
d2y 16
and 2
= 2−
dx (x − 1) 3
8
Here, y′ = 0 when 2x = − or when 2(x 3 − 2 x 2 + x + 4)
( x − 1) 2
= 2(x + 1) ( x 2 − 3x + 4) = 0 . The only real solution to this equation is
x = −1 . Therefore, there is a relative minimum at x = −1 , and the
minimum value of y is 5.
v) Increasing or decreasing function: Here, y′ < 0 , when x < −1 ,
therefore, f is decreasing, and y′ > 0 , when − 1 < x < ∞ , thus, f is
increasing on ] − 1, ∞ [ except at x = 1 .

182
Block 4 Miscellaneous Examples and Exercises
16
vi) Concavity: Here, y′′ = 0 , when 2 = or when ( x − 1)3 = 8 . Then,
( x − 1)3
x − 1 = 2 , so, x = 3 .Thus, there is an inflection point at x = 3 . The
coordinates of the inflection point are (3, 5) .
Combining all the properties, we discussed from (i) to (vi), we can trace
the curve y = f (x ) . Fig. 9 shows the curve.

Fig. 9
***

Example 10: The position function of a moving particle is given by

s( t ) = 2t 3 − 21t 2 + 60 t + 3 . Find the velocity and acceleration of the particle and
also, determine the interval in which velocity and acceleration are increasing
or decreasing.
Solution: The velocity and acceleration at time t are
ds
v( t ) = s′( t ) = = 6 t 2 − 42t + 60 = 6( t − 2) ( t − 5) and
dt
d 2s  7
a ( t ) = v′( t ) = 2 = 12t − 42 = 12 t −  .
dt  2
At each instant we can determine the direction of the motion from the sign of
v( t ) and whether the particle is speeding up or slowing down from the signs of
v( t ) and a ( t ) together (Fig. 10 (a) and (b)).

Table 3

Time Velocity v( t ) Acceleration a ( t ) Interpretation

0<t≤2 v(0) = 60 m / s a (0) = −42 m / s 2 Since, the acceleration

is negative, the speed
of the particle is
decreasing.

v(2) = 0 m / s a (2) = −18 m / s 2 The particle continuous

moving with decreasing
speed.

183
Block 4 Applications of Differential Calculus

Time Velocity v( t ) Acceleration a ( t ) Interpretation

7 v(2) = 0m / s a (2 ) = −18 m / s 2 The particle begins to

2≤ t≤ slow down.
2
7 27 7
v  = − m /s a   = 0 m / s2
2 2 2

7  7  − 27 7 The particle continues

≤t≤5 v  = m/s a   = 0 m / s2 moving until time t = 5 ,
2 2 2 2
when it stops at t = 5 ,
v(5) = 0 m / s s(5) = 28m, it reverses
a (5) = 18 m / s 2
direction again, and
begins to speed up with
acceleration a (5) = 18 .
The particle then
continues moving right
thereafter with
increasing speed.

(a) Graph of v(t) (b) Graph of a(t)

Fig. 10

The motion of the particle is described schematically by the curved line in

Fig. 11.

Fig. 11
***

Example 11: Find all absolute extrema of the function f ( x ) = x 3 − 3x 2 + 4 on the

interval ] − 1, 2 ] .
184
Block 4 Miscellaneous Examples and Exercises
Solution: Here, f ( x ) = x 3 − 3x 2 + 4 . On differentiating, we get
f ′( x ) = 3x 2 − 6 x and f ′′( x ) = 6 x − 6 .
f ′( x ) = 0 gives 3x 2 − 6x = 0 , which implies x = 0, 2 .
Since, f ′′(0) = −6 , therefore, f has a relative maxima at x = 0 . The maximum
value is f (0) = 4 .
Since, f ′′(2) = 6 > 0 , therefore, f has a relative minima at x = 2 . The
minimum value is f (2) = 0 .
Since, f has only one relative maximum value and one relative minimum value
on ] − 1, 2 [ , therefore, the relative extrema would be absolute extrema. Thus,
f has an absolute maximum at x = 0, and the absolute minimum at x = 2 . You
can see the graph of f in Fig. 12.

Fig. 12
***

Example 12: Find the radius and height of the right circular cylinder of the
largest volume that can be inscribed in a right circular cone with radius 12 cm
and height 20 cm.

Solution: Let r be the radius (in cm) of the cylinder, h be the height (in cm) of
the cylinder and V be the volume (in cubic cm) of the cylinder as shown in
Fig. 13 (a).
The volume of the inscribed cylinder is V = πr 2 h .
Since, the volume has two variables, we can eliminate one of the variables
using relationship between r and h . For this, we use similar triangles (Fig. 13
(b)),
BC BO
We obtain =
CD OA
20 − h 20 5
= or h = 20 − r
r 12 3

Putting h in terms of r in the formula of V , we get

 5  5
V = πr 2  20 − r  = 20πr 2 − πr 3
 3  3
which expresses V in terms of r alone. Because r represents a radius, it
cannot be negative, and because the radius of the inscribed cylinder cannot
exceed the radius of the cone, the variable r must satisfy 0 ≤ r ≤ 12 .
185
Block 4 Applications of Differential Calculus
dV
On differentiating V with respect to r, we get = 40πr − 5πr 2 = 5πr (8 − r ) and
dr
d 2V
= 40π − 10πr .
dr 2
Setting, dV / dr = 0 gives 5πr(8 − r) = 0 , so, r = 0 and r = 8 are critical
numbers. Since, these lie on the interval [0, 12] , the maximum must occur at
one of the values r = 0, r = 8, r = 12 .
(V) at r =0 = 0
1280π
(V) at r=8 =
3
(V) at r=12 =0
1280π 3
Here, the maximum volume V = cm occurs when the inscribed cylinder
3
20
has radius 8 cm. When r = 8cm, h = cm . Thus, the inscribed cylinder of the
3
20
largest volume has radius 8 cm and height cm .
3

(a) (b)
Fig. 13
***

Example 13: A pharmaceutical firm sells liquid form of penincillin in units at a

price of `100 per unit. The total production cost (in `) for x units is
C( x ) = 100,000 + 20 x + 0.004x 2 and the production capacity of the firm is at
most 20,000 units in a specified time. Find the number of sells a liquid form of
penicillin at units of penicillin, manufactured and sold in that time to maximise
the profit?

Solution: The total revenue for selling x units is R ( x ) = 100 x, the profit
P( x ) on x units will be P(x ) = R (x ) − C( x )
= 100x − (100,000 + 20 x + 0.004x 2 ) = 80 x − 100,000 − 0.004x 2
dP
On differentiating, P( x ) with respect to x , we get, = 80 − 0.008x .
dx
dP
Setting, = 0 gives 80 − 0.008x = 0, which gives x = 10,000 .
dx
186
Block 4 Miscellaneous Examples and Exercises
Since, the capacity is at most 20,000 units, the critical number lies in the
interval [0, 20,000] . Hence, the maximum profit must occur at one of the
values x = 0, x = 10,000, or x = 20,000
Now, the value of P( x ) at each critical number is
P(0) = −100,000
P(10,000) = 300000
P(20,000) = −100,000
Thus, the firm must manufacture 10,000 units to maximise the profit.
***
2
Example 14: Trace the curve y = e − x /2
by stating all the properties you use to
trace.

Solution: i) Symmetry: Since, the power of x is even, therefore, the curve is

symmetrical about the y -axis.

2
ii) Points of intersection with axes: Setting y = 0 , we get e − x / 2 = 0 ,
which has no solution, because all powers of e have positive values.
Thus, there are no x -intercepts. Now, setting x = 0 gives y = 1 .
Therefore, the curve passes through the point (0, 1) .
2
iii) Asymptotes: There are no vertical asymptotes, since e − x /2
is defined
and continuous on ] − ∞, +∞ [ .
2 2
Also, lim e − x /2
= lim e − x /2
=0
x →−∞ x →+∞
2
Thus, the curve y = e − x /2
has horizontal asymptote, which is y = 0 .

iv) Increasing and decreasing function: On differentiating, we get

dy −x2 / 2 d  x2  2
=e  −  = − xe − x / 2
dx dx  2 
Here, y′ > 0, when x < 0, thus, y is increasing on ] − ∞, 0[ , and
y′ < 0, when x > 0, thus, y is decreasing on ] 0, ∞ [ .

2
v) Relative extrema: Since, e − x /2
> 0 for all x , the sign of
2
dy / dx = − xe − x /2
is the same as the sign of − x .
Therefore, y has a relative minimum e 0 = 1 at x = 0 .

d2y d −x 2 / 2 2/2 d
vi) Concavity: Here,
dx 2
= −x
dx
e [
+ e−x
dx
]
[−x ]
2 2 2
= x 2e − x / 2 − e − x / 2 = ( x 2 − 1)e − x / 2
2 2
Since, e − x /2
> 0 for all x , the sign of d 2 y / dx 2 = ( x 2 − 1)e − x / 2 is the same
as the sign of ( x 2 − 1) , and the sign of ( x 2 − 1) would change at
x = 1 and at x = −1 . Thus, the inflection points occur at x = −1 and at
x = 1 . These inflection points are (−1, e −1/ 2 ) ≈ (−1, 0.607) and
(1, e −1 / 2 ) ≈ (1,0.607) .
We combine all these points, and trace the curve as shown in Fig. 14.
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Block 4 Applications of Differential Calculus

− x 2 /2
Fig. 14: Curve y = e .

***

Example 15: Find the type of the indeterminate forms in the following limits.
Also, find the limit.
x2 − 9 1 − sin x
i) lim ii) lim
x →3 x −3 x →π / 2 cos x

ex −1 x −4 / 3
iii) lim iv) lim
x →∞ x3 x → +∞ sin(1 / x )

Solution: i) The numerator and denominator are 0 as x → 3 . Therefore, the

limit is an indeterminate form of type 0 / 0 . Applying L’Hôpital’s rule, we get
d 2
2 [x − 9]
x −9 dx 2x
lim = lim = lim =6
x →3 x − 3 x →3 d x →3 1
[x − 3]
dx
Alternatively, you may find this limit by factoring
x2 − 9 ( x − 3) ( x + 3)
lim = lim = lim ( x + 3) = 6 .
x →3 x − 3 x →3 x −3 x →3

π
ii) The numerator and denominator are ∞ as x → , therefore, the limit is an
2
indeterminate form of type 0 / 0 . Applying L’Hôpital’s rule, we get
d
[1 − sin x ]
1 − sin x dx − cos x 0
limπ = limπ = limπ = =0
x→ cos x x→ d 3 x → − sin x −1
2 2 [ x ] 2
dx
iii) The numerator and denominator are 0 as x → ∞, therefore, the limit is an
∞
indeterminate form of type . Applying L’Hôpital’s rule repeatedly, we get
∞
d x
[e − 1]
ex − 1 dx ex ex ex
lim = lim = lim = lim = lim =∞.
x →∞ x3 x →∞ d 3
[x ]
x →∞ 3x 2 x →∞ 6 x x →∞ 6

dx
iv) The numerator and denominator are 0 as x → ∞ , so the limit is an
indeterminate form of type 0 / 0 . Applying L’Hôpital’s rule, we get
188
Block 4 Miscellaneous Examples and Exercises
4 4 −1/ 3
− x −7 / 3 x
x −4 / 3 3 3 0
lim = lim 2
= lim = = 0.
x → +∞ sin(1 / x ) x →+∞ ( −1 / x ) cos(1 / x ) x →+∞ cos(1 / x ) 1
***

Example 16: Show that lim (1 + x )1 / x = e .

x →0

Solution: Let y = (1 + x )1 / x and taking the natural logarithm of both the sides.
1 ln(1 + x )
ln y = ln(1 + x )1 / x = ln(1 + x ) =
x x
ln(1 + x )
Thus, lim ln y = lim , which is an indeterminate form of type 0 / 0 .
x →0 x →0 x
ln(1 + x ) 1 /(1 + x )
Using L’Hôpital’s rule, we get lim ln y = lim = lim =1
x →0 x →0 x x →0 1
Since, we have ln y → 1 as x → 0 , the continuity of the exponential function
implies that e ln y → e1 as x → 0 , and this implies that y → e as x → 0 . Thus,
lim (1 + x )1/ x = e .
x →0

***

Now you may try the following exercises.

E1) Find the inflection points, if any, for the curve y = x 4 .

E2) Use the graph of y = f ( x ) given in Fig. 15 to find the following:

i) The intervals on which f is increasing.
ii) The intervals on which f is decreasing.
iii) The intervals on which f is concave upward.
iv) The intervals on which f is concave downward.
v) The values of x at which f has an inflection point.

Fig. 15

E3) Show that x < tan x if 0 < x < π / 2 .

E4) Find the intervals on which f is increasing, the intervals on which f is

decreasing, the open intervals on which f is concave upward, the open
189
Block 4 Applications of Differential Calculus
intervals on which f is concave downward, and the x -coordinates of all
inflection points for the functions defined as follows:

i) f ( x ) = ( x + 2)3 iii) f (x) = 3 x + 2

x2
ii) f (x) = 2 iv) f ( x ) = x1 / 3 ( x + 4)
x +2

E5) Prove that a general cubic polynomial f ( x ) = ax 3 + bx 2 + cx + d

(a ≠ 0) has exactly one inflection point.

E6) Find the relative extrema of the functions defined as follows using both
the first and second derivative tests.
i) f ( x ) = 1 − 4x − x 2
ii) f ( x ) = 2 x 3 − 9x 2 + 12 x
iii) f ( x ) = sin 2 x , 0 < x < 2π

iv) f (x) = x 2 − 4

E7) Trace the following curves by stating all the properties you use to trace.
i) y = ( x − 4) 2 / 3
ii) y = 6x1/ 3 + 3x 4 / 3 .

E8) Let f ( x ) = x 2 + px + q . Find the values of p and q such that f (1) = 3 is

an extreme value of f on [0, 2] . Is this value a maximum or minimum?

E9) Find the absolute maximum and minimum values of f , if any, on the
stated interval.
i) f ( x ) = ( x 2 − 1) on ] − ∞,+∞ [
ii) f ( x ) = x 2 / 3 (20 − x ) on [−1, 20]
iii) f ( x ) = 2 sec x − tan x on [0, π / 4]
iv) f ( x ) = sin(cos x ) on [0, 2π]

E10) Suppose that the equations of motion of a paper aeroplane during the
first 12 seconds of flight are x = t − 2 sin t , y = 2 − 2 cos t , 0 ≤ t ≤ 12 .
What are the highest and lowest points in the trajectory, and when is the
aeroplane at those points?

E11) A closed cylindrical can is to be made to hold 1 litre (1000cm 3 ) of liquid.

What should be the height and radius of the can to minimize the amount
of material needed to manufacture the can?

E12) Find a point on the curve y = x 2 that is closest to the point (18, 0) .

E13) A firm determines that x units of its product can be sold daily at p rupees
per unit, where x = 1000 − p . The cost of producing x units per day is
C( x ) = 3000 + 20x .
190
Block 4 Miscellaneous Examples and Exercises
i) Find the revenue function R ( x ) .

ii) Find the profit function P( x ) .

iii) Assuming that the production capacity is at most 500 units per day,
determine how many units the company must produce and sell each
day to maximise the profit.
iv) Find the maximum profit.
v) What price per unit must be charged to obtain the maximum profit?

E14) Find the point on the curve y = (1 + x 2 ) −1 , at which the tangent line has
the greatest slope?

(ln x )
E15) Trace the curve y = by stating all the properties you use to trace
x
it.

L
E16) Trace the curve y = , where y is the population at time
1 + Ae −kt
t ( t ≥ 0) and A, k and L are positive constants.

E17) Suppose that a hollow tube rotates with a constant angular velocity of ω
rad/s about a horizontal axis at one end of the tube, as shown in the Fig.
16. Assume that an object is free to slide without friction in the tube while
the tube is rotating. Let r be the distance from the object to the pivot
point at time t ≥ 0 , and assume that the object is at rest and r = 0 when
t = 0 . If the tube is horizontal at time t = 0 and rotating, then
g
r= [sinh( ωt ) − sin( ωt )] during the period that the object is in the
2ω 2
tube. Assume that t is in seconds and r is in meters, and use
g = 9.8 m / s 2 and ω = 2 rad / s .
i) Trace the curve r = f (t ) for 0 ≤ t ≤ 1 .

ii) If the length of the tube is 1 m , then find the limit taken by the object
to reach the end of the tube?

Fig. 16

E18) Trace the following curves given in polar coorindates:

3π
i) θ=− iii) r 2 = sin 2θ
4 191
Block 4 Applications of Differential Calculus
ii) r − 2 = 2 cos θ iv) r = 4θ

E19) Find the slope of the tangent line to the following polar curves for the
given value of θ .
i) r = 2 cos θ; θ = π / 3
1
ii) r= ; θ=2
θ

E20) Show that the curve with parametric equations x = t 3 − 4 t ,

y = t 2 intersects itself at the point (0, 4) , and find equations for the two
tangent lines to the curve at the point of intersection.

SOLUTIONS/ANSWERS
E1) Calculating the first two derivatives of f we obtain f ′( x ) = 4 x 3 , f ′′( x ) = 12 x 2 .
Here, f ′′( x ) > 0 for x < 0 and for x > 0, which implies that f is concave
upward for x < 0 and for x > 0 . In fact, f is concave upward on ] − ∞,+∞ [ .
Thus, there are no inflection points, and in particular, there is no inflection
point at x = 0 , even though f ′′(0) = 0 . (See Fig. 17)

Fig. 17

E2) i) The function f is increasing on [d, f ] .

ii) The function f is decreasing on [a , d ] and [f , g ] .

iii) The function f is concave upward on the intervals ] a , b [ and ] c, e[ .

iv) The function f is concave downward on ] b, c [ and ] e, g [ .
v) The points of inflection are at x = b, x = c and x = d .

E3) Let f ( x ) = tan x − x

f ′( x ) = sec 2 x − 1
π
f ′( x ) ≥ 0, as sec 2 x ≥ 1 on 0 < x <
.
2
π
Therefore, f is increasing on 0 < x < .
2
Hence, f (0) < f ( x )
192
Block 4 Miscellaneous Examples and Exercises
Here, f (0) = 0 . Thus, 0 < tan x − x which gives, x < tan x .

E4) i) Increasing on ] − ∞,+∞ [ .

Not decreasing anywhere on R .
Concave upward on ] − 2,+∞ [ .
Concave downward on ] − ∞, − 2 [ .
Point of inflection is at x = −2 .
ii) Increasing on [0, + ∞[ .
Decreasing on ] − ∞, 0 ]
 2 2
Concave upward on  − , 
 3 3
 2  2 
Concave downward on  − ∞,−  and on  ,+∞  .
 3  3 
 2 1   2 1 
Points of inflection are  + ,  and  − ,
 3 3  3 3 

iii) Increasing on ] − ∞, + ∞[ .
Not Decreasing on R .
Concave upward on ] − ∞, − 2[ .
Concave downward on ] − 2, +∞ [ .
Point of inflection at x = −2 .
iv) Increasing on [−1,+∞ [
Decreasing on ] − ∞, −1]
Concave upward on ] − ∞,0[ and ] 2,+∞ [ .
Concave downward on ] 0, 2[ .
Points of inflection are (0, 0) and (2, 6 (2)1/ 3 ) .

E5) f ( x ) = ax 3 + bx 2 + cx + d
f ′( x ) = 3ax 2 + 2bx + c
f ′′(x ) = 6ax + 2b
2b
f ′′(x ) = 0 ⇒ x = − .
6a
Thus, f has exactly one inflection point.

E6) i) Relative maximum at x = −2 and f (−2) = 5 .

ii) Relative maximum of 5 at x = 1 and relative minimum of 4 at x = 2 .
iii) Relative minimum of 0 at x = π and relative minima of 1 at
π 3π
x= , .
2 2
iv) Relative maximum of 4 at x = 0 and relative minima of 0 at
x = 2 and − 2 .

E7) i) The properties to trace the curve are as follows:

i) Symmetry: There are no symmetries about the coordinate
axes or the origin. However, the graph of y = ( x − 4) 2 / 3 is
symmetric about the line x = 4 , since, it is a translation (four
193
Block 4 Applications of Differential Calculus
units to the right) of the graph of y = x 2 / 3 , which is symmetric
about the y -axis.

ii) Point of intersection with axes: (4, 0) and (0, 2.52)

iii) Asymptotes: None, since f ( x ) = (x − 4) 2 / 3 is continuous

everywhere and also, lim (x − 4) 2 / 3 = +∞ and
x →+∞
2/ 3
lim (x − 4) = +∞ .
x →−∞

iv) Relative extrema: The derivatives are

dy 2 2
= f ′( x ) = ( x − 4) −1 / 3 =
dx 3 3( x − 4)1 / 3
d2y 2 2
2
= f ′′( x ) = − ( x − 4) −4 / 3 = −
dx 9 9( x − 4)4 / 3
There is a critical number at x = 4, since f is not differentiable
there; and by the first derivative test there is a relative minimum
at that critical number, since f ′( x ) < 0 if x < 4 and f ′( x ) > 0 if
x > 4.
v) Concavity: Since f ′′( x ) < 0 if x ≠ 4, the graph is concave down
for x < 4 and for x > 4 .
vi) Vertical tangent lines: Since f ( x ) = (x − 4) 2 / 3 is continuous at
x = 4 and
2
lim f ′( x ) = lim+ = +∞
x →4+ x →4 3( x − 4)1 / 3
2
lim− f ′(x ) = lim− 1/ 3
= −∞ , therefore is a vertical tangent
x →4 x →4 3( x − 4)

line and cusp at x = 4 .

Combining all the properties, we can trace the curve as shown

in Fig. 18.

Fig. 18
ii) The properties used to trace the given curve are as follows:
i) Symmetry: There are no symmetry about the coordinate axes
or the origin.
194
Block 4 Miscellaneous Examples and Exercises
ii) Points of intersection with axes: (0, 0) and (−2, 0) .

iii) Asymptotes: None, since f ( x ) = 6 x1 / 3 + 3x 4 / 3 is continuous

everywhere, also, since
lim (6x1/ 3 + 3x 4 / 3 ) = lim 3x1/ 3 (2 + x ) = +∞
x →+∞ x →+∞
1/ 3 4/3
lim (6x + 3x ) = lim 3x1/ 3 (2 + x ) = +∞
x →−∞ x →−∞

iv) Relative extrema: The derivatives are

dy 2(2 x + 1)
= f ′(x ) = 2x −2 / 3 + 4 x1 / 3 = 2x −2 / 3 (1 + 2 x ) =
dx x2/ 3
and
d2 y 4 4 4 4( x − 1)
2
= f ′′( x ) = − x −5 / 3 + x −2 / 3 = x −5 / 3 (−1 + x ) = .
dx 3 3 3 3x 5 / 3
1
The critical number is x = − . The sign of dy / dx changes at
2
−1
x= from negative to positive, therefore,from the first
2
1
derivative test, there is a relative minimum at x = − .
2
v) Increasing and decreasing: The function f is decreasing
when x < −0.5 and increasing when x > −0.5 .
vi) Tangents: There is a point of vertical tangency at x = 0 , since
2(2 x + 1)
lim f ′( x ) = lim+ = +∞
x →0+ x →0 x2/ 3
2(2x + 1)
lim− f ′( x ) = lim− = +∞
x →0 x →0 x2/ 3
d2y
vii) Concavity: Here > 0, when x < 0, therefore, the curve is
dx 2
d2y
concave upward and < 0, when 0 < x < 1, therefore, the
dx 2
curve is concave downward, and the curve is again concave
upward for x > 1 . There are inflection points at (0, 0) and (1, 9) .
Combining all the properties, we trace the curve as shown in
Fig. 19.

Fig. 19 195
Block 4 Applications of Differential Calculus
E8) f (1) = 3 = 1 + p + q ⇒ p + q = 2
f ′( x ) = 2x + p = 0 ⇒ x = −p / 2
Since, x = 1 is an extreme value, therefore, p = −2 , which gives q = 4 .
Now, f ′′(x ) = 2 and f ′′(1) = 2 > 0 , therefore, this extreme value is
minimum value.

E9) i) Minimum value − 1 at x = 0

No maximum.
ii) Maximum value 48 at x = 8
Minimum value 0 at x = 0, 20 .
iii) Maximum value 2 at x = 0
π
Minimum value 3 at x =
.
6
iv) Maximum value 0.841 at x = 0, 2π .
Minimum value − 0.841 at x = π .

E10) Maximum y = 4 at t = π, 3π .
Minimum y = 0 at t = 0, 2π .

E11) The surface area of the can S = 2πr 2 + 2πrh , where r and h are the
radius and height of the can, respectively. Also, πr 2 h = 1000 cm 3 .
2000 dS 2000
Thus, S = 2πr 2 + , which gives = 4πr − 2
r dr r
dS 10
=0⇒r = cm.
dr (2π)1/ 3
d 2S 4000  d 2S 
Now, = 4π + 3 and  2  > 0, therefore minimum.
dr 2 r  dr at r = 10
( 2 π )1 / 3

20
Hence, the height h = cm.
(2π)1/ 3

E12) Let the point be ( x, y) . The distance between ( x, y) and (18, 0) is

D = (x − 18) 2 + ( y − 0) 2 . Since, the point ( x, y) lies on y = x 2 ,
therefore, D = ( x − 18) 2 + x 4 .
2
Suppose, L = D
L = ( x − 18) 2 + x 4
dL
= 2( x − 18) + 4x 3
dx
The critical number is x = 2 .
d 2L
2
= 2 + 12 x 2
dx
 d 2L 
 2  = 50 > 0 .
 dx  at x =2
Thus, the distance is minimum, when x = 2 . Thus, the point is (2, 4) .

E13) i) The revenue function R ( x ) = xp = x (1000 − x )

196
Block 4 Miscellaneous Examples and Exercises
ii) The profit function P( x ) = R ( x ) − C(x ) = 980 x − x 2 − 3000

iii) It is given that x ≤ 500 . Here, P′(x ) = 980 − 2 x

P′(x ) = 0 gives x = 490 .
P′′( x ) = −2 < 0 , thus, maxima.
Therefore, the company must produce and sell 490 units to
maximise the profit.
iv) P(490) = `237100
v) p = 1000 − 490 = `510

 1 3
E14) The required point is  − , .
 3 4

E15) i) Symmetry: None

ii) Points of intersection with axes: (1, 0)

1
iii) Asymptotes: Since, lim+ = +∞ and lim+ ln x = −∞ it follows that
x →0 x x →0

ln x 1
values of y = = (ln x ) will decrease without bound as x → 0+ ,
x x
ln x
so, lim+ = −∞ and the graph has a vertical asymptote x = 0 .
x →0 x
(ln x ) ln x
You may note that > 0 for x > 1 . The limit lim = 0 . Thus,
x x →+∞ x

(ln x) / x is asymptotic to y = 0 as x → +∞ .
iv) Increasing and decreasing function: The derivatives are
dy x (1 / x ) − (ln x ) (1) 1 − ln x
= = and
dx x2 x2
d 2 y x 2 (−1 / x ) − (1 − ln x ) (2 x ) 2 x ln x − 3x 2 ln x − 3
= = = .
dx 2 x4 x4 x3
dy 1 − ln x
Since, x 2 > 0 for all x > 0 , the sign of = is the same as
dx x2
the sign of 1 − ln x . But ln x is an increasing function with ln e = 1 ,
so, 1 − ln x is positive for x < e and negative for x > e .
v) Relative extrema: There is a relative maximum
(ln e)
= 1 / e ≈ 0.37 at x = e .
e
d 2 y 2 ln x − 3
vi) Concavity: Since, x 3 > 0 for all x > 0 , the sign of = is
dx 2 x3
the same as the sign of 2 ln x − 3 . Now, 2 ln x − 3 = 0 when
3
ln x = , or x = e3 / 2 . Again, since, ln x is an increasing function,
2
2 ln x − 3 is negative for x < e 3 / 2 and positive for x > e3 / 2 .
 3 
Thus, an inflection point occurs at  e 3 / 2 , e −3 / 2  ≈ (4.48, 0.33) .
 2 
Fig. 20 shows the curve.

197
Block 4 Applications of Differential Calculus

Fig. 20

E16) You may like to trace the curve by assuming values of L, A and k .

E17) You may like to trace the curve yourself.

E18) Fig. 21(a), 21 (b), 21 (c) and 21 (d) shows the graphs of the polar curves
of (i), (ii), (iii) and (iv) respectively.

Fig. 21
198
Block 4 Miscellaneous Examples and Exercises

1
E19) i)
3
tan 2 − 2
ii)
2 tan 2 + 1

x
E20) y = ± + 4.
2

199