UNIT
1
Differential Calculus—I
1.1 INTRODUCTION
In many practical situations engineers and scientists come across problems which involve quantities
of varying nature. Calculus in general, and differential calculus in particular, provide the analyst with
several mathematical tools and techniques in studying how the functions involved in the problem
behave. The student may recall at this stage that the derivative, obtained through the basic operation
of calculus, called differentiation, measures the rate of change of the functions (dependent variable)
with respect to the independent variable. In this chapter we examine how the concept of the derivative
can be adopted in the study of curvedness or bending of curves.
1.2 RADIUS OF CURVATURE
Let P be any point on the curve C. Draw the tangent at P to the
circle. The circle having the same curvature as the curve at P
touching the curve at P, is called the circle of curvature. It is also
called the osculating circle. The centre of the circle of the curvature is called the centre of curvature. The radius of the circle
of curvature is called the radius of curvature and is denoted
by ‘ρ’.
Y
C
O
P
Note : 1. If k (> 0) is the curvature of a curve at P, then the radius
1
. This follows from the definition
k
of radius of curvature and the result that the curvature of a circle is the
reciprocal of its radius.
O
of curvature of the curve of ρ is
Note : 2. If for an arc of a curve, ψ decreases as s increases, then
But the radius of a circle is non-negative. So to take ρ =
i.e., k =
dψ
.
ds
1
X
Fig. 1.1
dψ
is negative, i.e., k is negative.
ds
1
ds
=
some authors regard k also as non-negative
dψ
k
2
ENGINEERING MATHEMATICS—II
dψ
indicates the convexity and concavity of the curve in the neighbourhood of
ds
ds
the point. Many authors take ρ =
and discard negative sign if computed value is negative.
dψ
The sign of
∴ Radius of curvature ρ =
1
·
k
1.2.1 Radius of Curvature in Cartesian Form
Suppose the Cartesian equation of the curve C is given by y = f (x) and A be a fixed point on it. Let
P(x, y) be a given point on C such that arc AP = s.
Then we know that
dy
= tan ψ
dx
where ψ is the angle made by the tangent to the curve C at P with the x-axis and
ds
=
dx
R|1 + FG dy IJ
S| H dx K
T
2
Differentiating (1) w.r.t x, we get
U|
V|
W
...(1)
1
2
...(2)
dψ
d2y
sec 2 ψ ⋅
2 =
dx
dx
d
2
= 1 + tan ψ
i ddsψ ⋅ dxds
LM1 + FG dy IJ OP 1 LM1 + FG dy IJ OP
MN H dx K PQ ρ MN H dx K PQ
1 R| F dy I U|
S1 + G J V
ρ |T H dx K |W
R|1 + FG dy IJ U|
S| H dx K V|
W
T
1
2 2
2
=
2
=
2
Therefore,
where y1 =
ρ=
dy
d2y
and y2 =
.
dx
dx 2
d2y
dx 2
[By using the (1) and (2)]
3
2
3
2
...(3)
3
DIFFERENTIAL CALCULUS—I
Equation (3) becomes,
ρ=
3
2 2
1
o1 + y t
y2
This is the Cartesian form of the radius of curvature of the curve y = f (x) at P (x, y) on it.
1.2.2 Radius of Curvature in Parametric Form
Let x = f (t) and y = g (t) be the Parametric equations of a curve C and P (x, y) be a given point
on it.
Then
dy dt
dy
= d x dt
dx
...(4)
RS
T
UV
W
d2y
d dy / dt dt
⋅
2 =
dt dx / dt dx
dx
and
dx d 2 y dy d 2 x
⋅
−
⋅
dt dt 2
dt dt 2 ⋅ 1
=
2
dx
dx
dt
dt
FG IJ
H K
dx d 2 y dy d 2 x
–
⋅
⋅
d2y
dt dt 2 dt dt 2
=
3
dx 2
dx
dt
FG IJ
H K
Substituting the values of
...(5)
dy
d2y
and
in the Cartesian form of the radius of curvature of the
dx
dx 2
curve y = f (x) [Eqn. (3)]
∴
ρ=
=
R|1 + FG dy IJ
o1 + y t = S|T H dx K
3
2 2
1
d2y
dx 2
y2
|RS1 + FG dy / dt IJ
|T H dx / dt K
2
2
2
3
2
3
2
3
2
R|FG dx IJ + FG dy IJ U|
S|H dt K H dt K V|
W
T
2
ρ=
U|
V|
W
|UV
|W
RS dx ⋅ d y – dy ⋅ d x UV / FG dx IJ
T dt dt dt dt W H dt K
2
∴
2
2
3
2
dx d 2 y dy d 2 x
⋅
−
⋅
dt dt 2
dt dt 2
...(6)
4
ENGINEERING MATHEMATICS—II
2
2
dx
dy
, y′ =
, x″ = d x , y″ = d y
2
dt
dt
dt
dt 2
where x′ =
ρ=
o
x′ 2 + y′ 2
t
3
2
x ′ y ″ – y′ x ″
This is the cartesian form of the radius of curvature in parametric form.
WORKED OUT EXAMPLES
1. Find the radius of curvature at any point on the curve y = a log sec
Solution
Radius of curvature ρ =
Here,
o
t
1 + y12
3
2
y2
y = a log sec
FG x IJ
H aK
1
y1 = a ×
sec
FG x IJ
H aK
⋅ sec
FG x IJ tan FG x IJ ⋅ 1
H aK H aK a
FG x IJ
H aK
F xI 1
= sec G J ⋅
H aK a
y 1 = tan
y2
2
RS1 + tan FG x IJ UV
H aKW
T
ρ =
1
F xI
sec G J
H aK
a
RSsec FG x IJ UV
T H aKW =
=
1
F xI
sec G J
H aK
a
3
2
2
Hence
2
32
2
2
∴
Radius of curvature = a sec
FG x IJ
H aK
FG x IJ
H aK
F xI
sec G J
H aK
F xI
= a sec G J
H aK
a sec 3
2
FG x IJ .
H aK
5
DIFFERENTIAL CALCULUS—I
FG x IJ , show that ρ =
H cK
d1 + y i
ρ =
2. For the curve y = c cos h
y2
·
c
3
2 2
1
Solution
y2
FG x IJ
H cK
F xI 1
F xI
y = c sin h G J × = sin h G J
H cK c
H cK
F xI 1
y = cos h G J ×
H cK c
RS1 + sin h FG x IJ UV c FG cos h x IJ
H cKW = H cK
T
ρ =
x
1
F xI
cos h
cos h G J
c
H cK
c
F xII
F xI 1 F
= c cos h G J = G c cos h G J J
H cKK
H cK c H
y = c cos h
Here,
1
and
2
3
2
2
2
3
2
2
2
=
1 2
⋅y
c
y2
ρ =
·
c
∴
Hence proved.
3. Find the radius of curvature at (1, –1) on the curve y = x2 – 3x + 1.
d1 + y i
ρ =
3
2 2
1
Solution. Where
y2
Here,
Now,
( y1)(1,
( y2)(1,
∴
ρ(1,
at (1, – 1)
y = x2 – 3x + 1
y 1 = 2x – 3, y2 = 2
–1) = – 1
–1) = 2
–1)
b1 + 1g
=
3
2
2
=
=
2 2
2
2
4. Find the radius of curvature at (a, 0) on y = x3 (x – a).
Solution. We have
ρ =
d
1 + y12
y2
i
3
2
at (a, 0)
6
ENGINEERING MATHEMATICS—II
y = x3(x – a) = x4 – x3a
y 1 = 4x3 – 3ax2
y 2 = 12x2 – 6ax
( y1)(a, 0) = 4a3 – 3a3 = a3
( y2)(a, 0) = 12a2 – 6a2 = 6a2
Here,
and
Now
RS1 + da i UV
T
W
=
2
3
∴
ρ(a,
0)
6a 2
o
1 + a6
=
6a
2
t
3
2
d
1 + y12
Solution. We have
ρ =
Here
y = a sec
y1
y1
and
y2
i
·
πa
on y = a sec
4
5. Find the radius of curvature at x =
∴
3
2
3
2
at x =
y2
πa
4
FG x IJ
H aK
F xI F xI 1
= a sec G J ⋅ tan G J ×
H aK H aK a
F xI F xI
= sec G J tan G J
H aK H aK
x 1
F xI F xI 1
× + sec G J ⋅ tan G J ⋅
= sec
H aK H aK a
a a
1L
F xI F xI F xIO
= Msec G J + sec G J tan G J P
H aK H aK H aKQ
aN
3
2
3
and y2 =
2
π
π
πa
, y = sec ⋅ tan = 2
1
4
4
4
At x =
e
j
1
3 2
2 2+ 2 =
a
a
RS1 + e 2 j UV
W
T
2
∴
ρ
x=
πa
4
FG x IJ .
H aK
=
=
3 2
a
3
a.
2
3
2
=
3 3
3 2
⋅a
7
DIFFERENTIAL CALCULUS—I
FG x IJ .
H 2K
π
on y = 2 log sin
3
6. Find ρ at x =
Solution. We have ρ =
d1 + y i
3
2 2
1
y2
y = 2 log sin
The curve is
at x =
FG x IJ
H 2K
1
× cos
sin
y2
At
x=
π
,y
3 1
and
2
y2 =
–1
π
cosec 2 = – 2
2
6
=
RS1 + e 3j UV
T
W
=
b1 + 3g
2
∴
ρ
x=
π
3
Solution. We have ρ =
3
2
–2
3
2
–2
=
o
1 + y12
t
y2
4×2
= – 4.
–2
FG 3a , 3a IJ
H 2 2K
7. Find the radius of curvature at
x3
FG x IJ × 1
H 2K 2
FG x IJ
H 2K
F xI
= cot G J
H 2K
F xI 1
= – cosec G J ×
H 2K 2
F πI
= cot G J = 3
H 6K
y1 = 2 ⋅
and
π
3
3
2
at
on x3 + y3 = 3axy.
FG 3a , 3a IJ .
H2 2K
y3
+
= 3axy
Here,
Differentiating with respect to x
3x2 + 3y2 y1 = 3a (xy1 + y)
3 ( y2 – ax) y1 = 3 (ay – x2)
⇒
y1 =
ay – x 2
y 2 − ax
...(1)
8
ENGINEERING MATHEMATICS—II
Again differentiating w.r.t x.
⇒
y2 =
Now, from (1), at
dy
2
ib
g d
− ax ⋅ ay1 − 2 x − ay − x 2
FG 3a , 3a IJ
H 2 2K
F 3a I F 3a I
aG J −G J
H 2K H 2K
y =
FG 3a IJ − a FG 3a IJ
H 2K H 2K
dy
2
− ax
i
i b2 yy
1
−a
g
2
2
2
1
=
6a 2 − 9a 2
9a 2 − 6a 2
d
– 9a 2 − 6a 2
=
From (2), at
FG 3a , 3a IJ
H 2 2K
y2 =
d9 a
F 9a
GH 2
2
− 6a
−
2
3a 2
2
i
i
= –1
I b– a − 3ag − F 3a
JK
GH 2
F 9a − 3a I
GH 4 2 JK
2
2
–
3 2
3a 2
a × 4a –
× 4a
4
4
F 3a I
GH 4 JK
=
=
2
2
2
– 6a 3 – 32
=
3a
9a 4
16
Using these
ρ F 3a
GH 2 , 32a IJK
{1 + b– 1g }
FG − 32 IJ
H 3a K
3
2 2
=
= –
∴ Radius of curvature at
2 2 × 3a − 3a
=
32
8 2
FG 3a , 3a IJ
H 2 2K
is
3a
·
8 2
2
2
−
9a 2
4
I b– 3a − ag
JK
...(2)
9
DIFFERENTIAL CALCULUS—I
8. Find the radius of curvature of b2x2 + a2y2 = a2b2 at its point of intersection with the y-axis.
Solution. We have ρ =
Here,
o1 + t
3
2 2
y1
at x = 0
y2
b2x2 + a2y2 = a2b2
When x = 0,
a2y2 = a2b2
y2 = b 2
⇒
y =±b
i.e., the point is (0, b) or (0, – b)
The curve is b2x2 + a2y2 = a2b2.
Differentiating w.r. to x
2b2x + 2a2yy1 = 0
b2 x
a2 y
Differentiating again w.r. to x
y1 = –
y2 =
F y − xy I
GH y JK
– b b0g
=0
a bbg
– b F b − 0I
G J
a H b K
– b2
a2
1
2
2
Now at (0, b),
y1 =
2
2
and
y2 =
2
2
–b
a2
i.e., Radius of curvature at (0, b) is
=
∴
ρ(0,
b1 + 0g
FG – b IJ
Ha K
3
2
b)
=
=
– a2
b
2
∴ Radius of curvature is
a2
b
Next consider (0, – b),
y1 =
– b2
0
×
=0
2
–b
a
y2 =
– b2
a2
FG – b – 0IJ = a
H b K b
2
2
10
ENGINEERING MATHEMATICS—II
ρ(0,
– b)
=
b1 + 0g
FG b IJ
Ha K
3
2
=
a2
b
2
∴ Radius of curvature of (0, – b) is
a2
.
b
9. Show that at any point P on the rectangular hyperbola xy = c2, ρ =
r3
where r is the
2c 2
distance of the point from the origin.
Solution. The curve is xy = c2
Differentiating w.r. to x
xy1 + y = 0
y
x
y1 = –
Again differentiating w.r.t. x
=
RS xy − y UV
T x W
R| – xy − y U|
2y
–S x
=
V
x
|W x
|T
R|1 + FG y IJ U|
S| H x K V|
W
T
=
d
=
dx
1
y2 = –
=
ρ=
where
x2
∴
+
y2
=
r2
2
2
2
d1 + i
and
2
3
2 2
y1
y2
xy =
3
2
2y
x2
x2 + y2
i
3
2
i
3
2
2y
x3 × 2
x
2
+ y2
2 xy
c2.
ρ =
10. Show that, for the ellipse
r3
.
2c2
x2 y2
a 2b 2
+ 2 = 1, ρ =
where p is the length of the perpen2
p3
a
b
dicular from the centre upon the tangent at (x, y) to the ellipse.
11
DIFFERENTIAL CALCULUS—I
Solution. The ellipse is
x2 y2
+
= 1
a 2 b2
Differentiating w.r.t. x
2 x 2 yy1
+ 2 = 0
a2
b
⇒
y1 =
Again Differentiating w.r. to x
y2 =
b2 x
a2 y
–
LM y − xy OP
a N y
Q
LM y + b ⋅ x OP
–b M
y P
a
a M
MN y PPQ
b Ly
x O
–
+ P
M
a y Nb
a Q
LM3 x
–b
a y
N a
– b2
1
2
2
2
2
=
2
2
2
4
=
y2 =
ρ =
Now,
2
2
2
3
2
2
2
2
4
2
3
2
o1 + t
3
y12 2
y2
+
F1 + b x I
GH a x JK
F– b I
GH a y JK
4 2
=
ρ =
da y
–
4
4
4
da y
4
Taking magnitude
ρ =
The tangent at (x0, y0) to the ellipse
x x0 y y0
+ 2 = 1
a2
b
2
+ b4 x 2
2
a6 y3
2
3
2
4 2
=
2
da y
−
OP
Q
y2
=1
b2
i
3
2
i
3
2
+ b4 x 2
×
3
a2 y3
b4
a 4b 4
+ b4 x 2
a 4b 4
i
3
2
x2 y2
+
= 1 is
a 2 b2
...(1)
12
ENGINEERING MATHEMATICS—II
Length of perpendicular from (0, 0) upon this tangent
1
=
FG x IJ + FG y IJ
H a K Hb K
2
0
2
2
0
2
a 2b2
=
a 4 y02 + b 4 x02
So, the length of perpendicular from the origin upon the tangent at (x, y) is
a 2b 2
p =
a 4 y02 + b 4 x02
By replacing x0 by x and y0 by y
a 2b 2
p =
a4 y2 + b4 x2
Reciprocal and cube on both sides, we get,
⇒
1
=
p3
=
da y
4
+ b4 x 2
2
i
3
2
i
3
2
a 6b 6
d
a 4 y 2 + b4 x2
4 4
a b
×
1
a b
2 2
By using eq. (1), we get
ρ
1
2 2
3 =
p
a b
⇒
ρ =
a 2b 2
·
p3
ax
,
11. Show that, for the curve y =
a+x
Solution. Here,
y =
FG 2ρ IJ
HaK
2
3
F xI F yI
= G J + GH JK
H yK x
2
ax
a+x
Differentiating w.r.t. x
LM ba + xgb1g − x OP a
MN ba + xg PQ = ba + xg
2
y1 = a
Again Differentiating w.r.t. x
2
y2 = a
2
–2
=
– 2a 2
b a + x g ba + x g
3
3
2
2
·
13
DIFFERENTIAL CALCULUS—I
ρ =
Now,
d1 + y i
3
2 2
1
y2
Substituting y1 and y2, we get
R|1 + a U|
S| ba + xg V|
W
T
R| − 2a U|
S| ba + xg V|
W
T
{ba + xg + a }
–
2a b a + x g
FG x IJ + FG y IJ
H yK H xK
R|
|S 2 {ba + xg + a }
|| 2a ba + xg
T
ba + x g + a
a ba + x g
ba + x g + a
a
ba + x g
3
2
4
4
=
2
3
3
4 2
4
ρ =
To show that
L.H.S.
FG 2ρIJ
HaK
23
FG 2ρ IJ
HaK
2
=
2
3
2
4
=
4
=
=
using (1)
2
2
FG x IJ + FG y IJ
H yK H xK
R| x U|
S| FG ax IJ V|
TH a + xK W
ba + x g +
a2
∴ L.H.S. = R.H.S. using (2) and (3).
...(2)
2
2
=
2
3
2
2
2
R.H.S.
U|
|V
||
W
4
2
=
3
2
3
3
4
=
...(1)
3
2
2
2
R| F ax I U|
S GH a + x JK V|
+ |
T x W
a2
ba + x g
2
2
...(3)
14
ENGINEERING MATHEMATICS—II
12. Find ρ at any point on x = a (θ + sinθ) and y = a (1 – cosθ).
Solution. Here
x = a (θ + sinθ), y = a (1 – cosθ)
Differentiating w.r.t. θ
dx
dy
= a (1 + cos θ),
= a sin θ
dθ
dθ
dy
dy
a sin θ
= dθ =
y1 =
dx
dx
a 1 + cos θ
dθ
b
g
θ
θ
cos
2
2
2 θ
2 cos
2
2 sin
=
y 1 = tan
Again differentiating w.r.t. θ
y2 =
=
=
θ
2
FG IJ
H K
d F
G tan θ2 IJK × ddxθ
dθ H
1
F θI 1
sec G J × ×
H 2 K 2 a b1 + cos θg
d
θ
tan
dx
2
2
θ
2
sec 2
=
2a × 2 cos2
y2 =
ρ =
1
4a cos4
θ
2
o1 + y t
3
2 2
1
y2
RS1 + tan θ UV
2W
T
R|
U|
1
S|
θV
4a cos |
2W
T
2
=
θ
2
4
3
2
15
DIFFERENTIAL CALCULUS—I
=
=
RSsec FG θ IJ UV
T H 2K W
2
3
2
1
× 4a cos
θI
F
cos G J
H 2K
F θI
4 a cos G J ·
H 2K
3
ρ =
FG θ IJ
H 2K
FG θ IJ
H 2K
× 4a cos4
4
13. Find the radius of curvature at the point ′θ′ on the curve x = a log sec θ,
y = a (tan θ – θ).
Solution
x = a log sec θ, y = a (tan θ – θ)
Differentiating w.r.t. θ
dy
dx
1
⋅ sec θ ⋅ tan θ ,
= a
= a (sec2 θ – 1)
dθ
dθ
sec θ
= a tan θ
= a tan2 θ
∴
dy
dy
= dθ
y1 =
dx
dx
dθ
a tan 2 θ
=
a tan θ
y 1 = tan θ
b g
d2y d
tanθ
=
dx 2 dx
d
dθ
=
tan θ ⋅
dθ
dx
y2 =
b g
2
= sec θ ×
=
Now,
ρ =
sec 2 θ
a tan θ
o1 + y t
3
2 2
1
y2
d1 + tan θi
F sec θ I
GH tan θ JK
2
=
1
a tan θ
2
3
2
16
ENGINEERING MATHEMATICS—II
sec 3 θ
× a tan θ
sec 2 θ
ρ = a sec θ tan θ.
=
14. For the curve x = a (cos θ + θ sin θ), y = a (sin θ – θ cos θ), show that the radius of
curvature at ′θ′ varies as θ.
Solution
x = a (cos θ + θ sin θ)
dx
= a (– sin θ + θ cos θ + sin θ) = a θ cos θ
dθ
y = a (sin θ – θ cos θ)
⇒
dy
= a (cos θ + θ sin θ – cos θ) = a θ sin θ
dθ
⇒
y1 =
y2 =
=
dy a θ sin θ
=
= tan θ
dx a θ cos θ
FG IJ b
H K
d
dθ
tan θg ⋅
b
dθ
dx
d dy
d
=
tan θ
dx dx
dx
2
= sec θ ×
=
ρ =
Now,
=
g
1
a θ cos θ
1
a θ cos3 θ
d
1 + y12
i
3
2
y2
d
i
F 1 I
GH a θ cos θ JK
1 + tan 2 θ
3
2
3
= sec3 θ × a θ cos3 θ
= aθ
ρ ∝ θ.
i.e.,
15. If ρ1 and ρ2 are the radii of curvatures at the extremities of a focal chord of the parabola
b g .
Solution. If dat , 2at i and dat , 2at i are the extremities of a focal chord of the parabola
y2 = 4ax, then show that ρ 1
2
1
y2 = 4ax.
Then
1
–2
3
+ ρ2
–2
3
= 2a
2
2
t1 · t2 = – 1
2
−2
3
17
DIFFERENTIAL CALCULUS—I
The parametric equations to
x
x′
x″
∴
the parabola are
y = 2at
= at2,
= 2at
y′= 2a
= 2a
y″ = 0
ρ =
ox ′
2
+ y′2
t
3
2
x′ y ″ – x″ y′
{b2at g + b2ag }
d – 4a i
3
2 2
2
=
2
= –
d
8a 3 1 + t 2
4a
2
d
i
2
ρ = − 2a 1 + t
ρ
–2
3
–2
ρ3
=
=
i
3
2
3
2
b2ag d1 + t i
–2
3
1
2
−1
1
×
b2ag d1 + t i
2
3
2
Let t1 and t2 be extremities of a focal chord. Then t2 = –
–2
–2
ρ1 3
Now,
3
= ρt = t1 =
–2
ρ 23
= ρ
=
–2
Adding
–2
ρ1 3 + ρ23
=
=
i.e.,
–2
3
1
ρ
–2
+ ρ 23
=
–2
3
t=
–1
t1
1
1
×
1
b2ag d1 + t i
2
3
=
×
1
2
1
1
×
b2ag d1 + t i
2
3
2
1
t12
b2ag d1 + t i
1 R 1
ST1 + t + 1 +t t UVW
b2 a g
b2ag × d11 ++ tt i
d i
2
3
2
1
2
1
2
3
–2
3
b2 a g
–2
3
2
1
2
1
2
1
2
1
· Hence proved.
1
·
t1
18
ENGINEERING MATHEMATICS—II
EXERCISE 1.1
1. Find ρ at any point on y = log sin x.
2. Find ρ at x = 1 on y =
Ans. cosec x
log x
.
x
FG x IJ ·
H 2K
π
3. Find the radius of curvature at x =
on y = log tan
4
x2 y2
+
4. Find the radius of curvature of (3, 4) on
= 2.
9
16
5. Find the radius of curvature at (x1, y1) on b2x2 – a2y2 = a2b2.
LMAns.
N
OP
2 2Q
LM
O
3I P
F
MNAns. 2 × GH 2 JK PQ
LMAns. 125OP
N 12 Q
LM b x + a y OP
MNAns. e a b j PQ
3
2
2 2
1
Ans. 4 2
d4 x
7. Show that ρ at any point on 2xy = a2 is
4
+ a4
i
3
2
8a 2 x 3
8. Show that ρ at (0, 0) on y2 = 12x is 6.
9. Find radius of curvature at x = 2 on y2 =
b
·
g·
LMAns. 1 OP
N 3Q
LMAns. a OP
2Q
N
LMAns. 5 5 a OP
6 PQ
MN
x x−2
x −5
10. Find ρ at (a, a) on x3 + y3 = 2a3.
11. Find the radius of curvature at (a, 2a) on x2y = a(x2 + a2).
12. Find the radius of curvature at (–2a, 2a) on x2y = a (x2 + y2).
13. Show that ρ at (a
a
sin3θ)
on
3
4 2 2
1
4 4
6. Find ρ at (4, 2) on y2 = 4 (x – 3).
cos3θ,
3
2
x3
+
2
y3
=
2
a3
Ans. 2a
is 3a sin θ cos θ.
π
14. Find radius of curvature at θ =
on x = a sin θ, y = b cos 2θ.
3
LM a
MNAns. e
15. Find ρ for x = t – sin ht cos ht, y = 2cos ht.
Ans. 2 cos h 2 t sin ht
2
+ 12b 2
4ab
O
j PP
Q
3
2
19
DIFFERENTIAL CALCULUS—I
1.2.3 Radius of Curvature in Pedal Form
Let polar form of the equation of a curve be r = f (θ) and
P(r, θ) be a given point on it. Let the tangent to the curve
at P subtend an angle ψ with the initial side. If the angle
between the radius vector OP and the tangent at P is φ then
we have ψ = θ + φ (see figure).
Let p denote the length of the perpendicular from the
pole O to the tangent at P. Then from the figure,
r = f (G)
OM p
=
OP r
p = r sin φ
B
r
O
P (r, G)
O
G
X
sin φ =
Hence,
∴
dψ
dθ dφ dθ dφ dr
1
+
=
+
⋅
=
=
ds
ds ds ds dr ds
ρ
We know that tan φ = r ⋅
Hence,
sin φ = r ⋅
From (2),
dθ
ds
dr
ds
LM
N
1
dφ
sin φ + r cos φ ⋅
r
dr
b
1 d
⋅
r sin φ
r dr
r sin φ = p
=
Since,
Fig. 1.2
dθ
dr
sin φ
dφ
1
+ cos φ ⋅
=
r
dr
ρ
=
g
OP
Q
dr
dp
This is the Pedal form of the radius of curvature.
Therefore,
M
dθ
ds
dr
ds
i.e.,
cos φ =
...(2)
p
r⋅
sin φ
=
cos φ
and
...(1)
ρ = r⋅
...(3)
1.2.4 Radius of Curvature in Polar Form
Let r = f (θ) be the equation of a curve in the polar form and p(r, θ) be a point on it. Then we know
that
1
1
1
Differentiating w.r.t. r, we2 get
= 2 + 4
p
r
r
FG dr IJ
H dθ K
2
...(4)
20
ENGINEERING MATHEMATICS—II
FG IJ d– 4r i + 1 ⋅ 2 ⋅ dr ⋅ d FG dr IJ
H K
dθ dr H dθ K
r
– 2 4 F dr I
2 dr d r dθ
⋅
⋅
⋅
− .G J +
r
r H dθ K
r dθ dθ dr
– 2 4 F dr I
2 d r
+
⋅
−
G
J
H
K
r
r dθ
r dθ
L 1 2 FG dr IJ – 1 d r OP
p M +
MN r r H dθ K r dθ PQ
2
– 2 dp
–2
dr
⋅
+
=
3
3
p dr
dθ
r
–5
4
2
=
3
2
5
4
2
2
=
Hence,
Now,
dp
=
dr
3
2
5
4
2
2
2
3
3
ρ = r⋅
5
dr
=
dp
=
r2 + 2
4
2
r
p3
R| 1 + 2 FG dr IJ
S| r r H dθ K
T
r6 ⋅
1
p3
3
FG dr IJ
H dθ K
5
2
−r⋅
2
−
1 d 2r
r 4 dθ 2
U|
V|
W
d 2r
dθ 2
By using equation (4),
R| 1 1 FG dr IJ U|
r ⋅S +
V
T| r r H dθ K W|
F dr I d r
r +2G J –r⋅
H dθ K dθ
3
2
2
6
2
=
4
2
2
2
2
R|r + FG dr IJ U|
S| H dθ K V|
T
W
F dr I d r
+2G J −r⋅
H dθ K dθ
2
2
ρ=
2
r
where
∴
r1 =
ρ=
2
3
2
2
2
dr
d 2r
, r =
·
dθ 2
dθ 2
o
r 2 + r12
r +
2
2r12
t
3
2
− r r2
This is the formula for the radius of curvature in the polar form.
...(5)
21
DIFFERENTIAL CALCULUS—I
WORKED OUT EXAMPLES
1. Find the radius of the curvature of each of the following curves:
(i) r3 = 2ap2 (Cardiod)
(ii) p2 = ar
1
1
1
r2
=
+
−
(Ellipse).
p 2 a 2 b 2 a 2b 2
(iii)
Solution. (i) Here
r3 = 2ap2
Differentiating w.r.t. p, we get
3r 2 ⋅
dr
= 4ap
dp
4ap
dr
=
dp
3r 2
⇒
ρ = r⋅
Hence,
Fr I
where p = G J
H 2a K
3
dr
4ap 4ap
=r⋅ 2 =
dp
3r
3r
1
2
Fr I
4a ⋅ G J
H 2a K
3
ρ =
1
2
3r
3
4a r 2
2 2ar
=
=
3
3r 2a
(ii) Here
p2 = ar
Differentiating w.r.t. p, we get
dr
dp
Then
2p = a ⋅
⇒
2p
dr
=
a
dp
where p =
ar ·
3
dr
2 ⋅ ar 2r 2
=r⋅
=
ρ = r
dp
a
a
(iii) Given
1
p2
=
1
1
r2
+
–
a2
b 2 a 2b 2
Differentiating w.r.t. p, we get
–2
–1
dr
= 2 2 2r ⋅
p3
dp
a b
22
ENGINEERING MATHEMATICS—II
Hence
Therefore,
dr
a 2b 2
=
dp
p 3r
ρ = r.
dr
a 2b 2 a 2b 2
=r. 3 = 3
dp
p r
p
2. Find the radius of curvature of the cardiod r = a (1 + cos θ) at any point (r, θ) on it. Also
ρ2
is a constant.
r
Solution. Given
r = a (1 + cos θ)
Differentiating w.r.t. θ
dr
= – a sin θ
r1 =
dθ
prove that
d 2r
= – a cos θ
and
r2 =
dθ 2
∴ The radius of curvature in the polar form
ρ =
or
2
r +
2
+
t
3
2 2
r1
2r12
− r r2
g + a sin θ}
{ b
a b1 + cos θg + 2a sin θ − ab1 + cos θgb – a cos θg
a o1 + 2 cos θ + cos θ + sin θt
a o1 + 2 cos θ + cos θ + 2 sin θ + cos θ + cos θt
a m2 b1 + cos θgr
3 b1 + cos θg
2 2 a b1 + cos θg
a 2 1 + cos θ
=
2
2
2
2
2
2
3
2
2
2
3
=
2
2
2
3
2
2
2
3
2
=
1
2
=
3
2
=
F θI
2 a G 2 cos J
H 2K
2
3
4
θ
a cos
ρ =
3
2
Squaring on both sides, we get
ρ2 =
16 2
θ
a cos2
9
2
b
8 a2
1 + cos θ
=
9
1
2
LM where cos θ = 1 b1 + cos θgOP
2
2
Q
N
LM where 1 + cos θ = r OP
aQ
N
2
g
23
DIFFERENTIAL CALCULUS—I
ρ2
8a 2 r 8ar
⋅ =
=
9 a
9
8a
ρ2
=
which is constant.
9
r
Hence,
3. Show that for the curve rn = an cos nθ the radius of curvature is
rn
an
Solution. Here
=
cos nθ
Taking logarithms on both sides, we get
n log r = n log a + log cos nθ
Differentiating w.r.t. θ, we have
an
·
n + 1 rn –1
b g
n dr
n sin nθ
= 0–
r dθ
cos nθ
dr
= – r tan nθ
dθ
Differentiating w.r.t. θ again, we obtain
r1 =
r2 =
RS
T
– onr sec
dr
d 2r
− rn sec 2 nθ + tan nθ ⋅
2 =
dθ
dθ
=
t
nθ − r tan 2 nθ
2
= r tan2 nθ – nr sec2 nθ
Using the polar form of ρ, we get
ρ =
=
=
=
=
=
o
r 2 + r12
r +
2
2r12
t
3
2
− rr2
or
b
2
r 2 + 2 – r tan nθ
t
+ r 2 tan 2 nθ
g
2
d
3
2
− r r tan 2 nθ − nr sec 2 nθ
i
r 3 sec 3 nθ
r 2 1 + 2 tan 2 nθ − tan 2 nθ + n sec 2 nθ
r sec 3 nθ
n + 1 sec 2 nθ
b g
r
n + 1 cos nθ
b g
r
bn + 1g FGH ar IJK
n
n
=
UV
W
an
·
n + 1 rn–1
b g
LMwhere cos nθ = r OP
a Q
N
n
n
24
ENGINEERING MATHEMATICS—II
4. Find the radii of curvature of the following curves:
(ii) r (1 + cos θ) = a
(i) r = aeθ cot α
(iii) θ =
r2 − a2
a
– cos –1
FG a IJ ·
H rK
Solution. (i) Here
r = aeθ
Differentiating w.r.t θ
cot α
dr
= aeθ cot α · cot α
dθ
= r · cot α
So that,
tan φ =
=
r
dr
dθ
r
= tan α
r cot α
Hence,
φ = α, since p = r sin φ
We get,
p = r sin α.
This is the Pedal equation of the given curve. From which, we get
dr
1
=
dp
sin α
Hence,
p = r⋅
dr
= r cosec α.
dp
(ii) Given equation of the curve is
r (1 + cos θ) = a
Differentiating w.r.t. θ, we get
r (– sin θ) + (1 + cos θ) ·
dr
=0
dθ
dr
r sin θ
=
dθ
1 + cos θ
or
We have,
1
p2
1
1
= 2 + 4
r
r
=
FG dr IJ
H dθ K
2
r 2 sin 2 θ
1
1
+
⋅
r 2 r 4 1 + cos θ 2
b
g
1 L
sin θ O
M
P
1+
r MN b1 + cos θg P
Q
1 L b1 + cos θg + sin
M
r MN
b1 + cos θg
2
=
2
2
2
=
2
2
2
θ
OP
PQ
25
DIFFERENTIAL CALCULUS—I
=
=
where 1 + cos θ =
LM 2 b1 + cos θg OP
MN b1 + cos θg PQ
1
r2
r
2
2
1 + cos θ
b
2
g
a
r
1
p2
=
2
a
r2 ⋅
r
=
2
ar
ar
which is the pedal equation of the curve.
2
Differentiating w.r.t. p, we get
Hence,
p2 =
2p =
a dr
⋅
2 dp
⇒
4p
dr
=
a
dp
∴
ρ = r⋅
dr
dp
= r⋅
4p
where p =
a
= r.
4
a
ar
2
ar
2
3
a r2 .
= 2 2
(iii) Here,
Then,
θ =
r2 − a2
− cos–1
a
dθ
2r
+
=
dr
a ⋅ 2 ⋅ r2 − a2
=
=
r
a r −a
2
2
–
r2 − a2
ar r 2 − a 2
FG a IJ
H rK
1
F1 − a
GH r
a
r r − a2
2
2
2
FG – a IJ
I HrK
JK
1
2
26
ENGINEERING MATHEMATICS—II
so that
dθ
=
dr
r2 − a2
ar
dr
=
dθ
ar
r2 − a2
We have the Pedal equation, we get
1
p2
1
p2
FG dr IJ
H dθ K
2
=
1
1
+ 4
2
r
r
=
a 2r 2
1
1
+
⋅
r2 r4 r2 − a2
=
a2
1
1
+
r2
r2 − a2
=
1
r − a2
RS
T
d
UV
W
i
2
p2 = r2 – a2
Hence
From this we get
p
dr
=
r
dp
ρ = r⋅
∴
p
= p = r2 − a2 ⋅
r
EXERCISE 1.2
1. Find the radius curvature at the point ( p, r) on each of the following curves:
LMAns. r OP
N aQ
LMAns. a OP
N 3r Q
LM Ans. a OP
MN bn + 1g r PQ
LM a + r OP
MMAns. d r + 2ai PP
MN
PQ
3
(i) pr =
a2
(ii)
r3
a2p
(iii)
pan
(Hyperbola)
2
2
=
(Lemniscate)
n
(iv) p =
=
r n+1
(Sine spiral)
r4
(Archimedian spiral)
r2 + a2
n −1
2
3
2 2
2
2
27
DIFFERENTIAL CALCULUS—I
2. Find the radius of curvature at (r, θ) on each of the following curves:
LM r a + r OP
MM Ans. d a i PP
MN
PQ
LMAns. a OP
N 3rQ
LMAns. r OP
N aQ
LMAns. r OP
N 3p Q
3
2 2
2
(i) r =
a
θ
2
(iii)
r2
=
a2
cos 2θ
3
(v) r2 cos 2θ = a2
2
4
(vii) r = a sec 2θ
2
LMAns. a OP
N 2Q
(ii) r = a cos θ
3
LM Ans. a OP
(iv)
=
sin nθ
MN bn + 1g r PQ
LMAns. 2 ar OP
a
(vi) r = b1 − cos θg
3 PQ
2
MN
LM Ans. na OP
(viii) r = a sin nθ
N 2Q
n
rn
an
n −1
3. If ρ1 and ρ2 are the radii of curvature at the extremities of any chord of the cardiode
2
2
r = a (1 + cos θ) which passes through the pole. Prove that ρ1 + ρ2
=
16a 2
·
9
1.3 SOME FUNDAMENTAL THEOREM
1.3.1 Rolle’s Theorem
If a function f (x) is
1. continuous in a closed interval [a, b],
2. differentiable in the open interval (a, b) and
3. f (a) = f (b).
Then there exists at least one value c of x in (a, b) such that f ′ (c) = 0
(No proof).
1.3.2 Lagrange’s Mean Value Theorem
Suppose a function f (x) satisfies the following two conditions.
1. f (x) is continuous in the closed interval [a, b].
2. f (x) is differentiable in the open interval (a, b).
Then there exists at least one value c of x in the open interval (a, b), such that
bg
f (b) – f a
b–a
= f ′ (c)
28
ENGINEERING MATHEMATICS—II
Proof. Let us define a new function
φ(x) = f (x) – k·x
...(1)
where k is a constant. Since f (x), kx and φ (x) is continuous in [a, b], differentiable in (a, b).
From (1) we have,
φ (a) = f (a) – k·a
φ (b) = f (b) – k·b
∴
φ (a) = φ (b) holds good if
f (a) – k·a = f (b) = k·b
i.e.,
k (b – a) = f (b) – f (a)
or
k =
bg bg
f b − f a
b−a
...(2)
Hence, if k is chosen as given by (2), then φ (x) satisfy all the conditions of Rolle’s theorem.
Therefore, by Rolle’s theorem there exists at least one point c in (a, b) such that φ′(c) = 0.
Differentiating (1) w.r.t. x we have,
φ′(x) = f ′(x) – k
φ′(c) = 0 gives f ′(c) – k = 0
and
k = f ′(c)
i.e.,
...(3)
Equating the R.H.S. of (2) and (3) we have
bg bg
f b − f a
b−a
= f ′(c)
...(4)
This proves Lagrange’s mean value theorem.
1.3.3 Cauchy’s Mean Value Theorem
If two functions f (x) and g (x) are such that
1. f (x) and g (x) are continuous in the closed interval [a, b].
2. f (x) and g (x) are differentiable in the open interval (a, b).
3. g′ (x) ≠ 0 for all x ∈ (a, b).
Then there exists at least one value c ∈ (a, b) such that
bg
g bbg – g b a g
f (b) – f a
=
b g·
g ′ b cg
f′ c
Proof: Let us define a new function
φ (x) = f (x) – kg (x)
...(1)
where k is a constant. From the given conditions it is evident that φ (x) is also continuous in [a, b],
differentiable in (a, b).
Further (1), we have
φ (a) = f (a) – k g (a); φ (b) = f (b) – k g (b)