Math By Masumbillah sir

1. In the name of Allah most gracious
most merciful
• ‫الرحيم‬‫ال‬١‫لرحمن‬‫بسم‬

2. Tangent and Normal:

3. Let P be a given point on a curve y = f(x) and Q be another point on it
and let the point Q moves along the curve nearer and nearer to the point
P then the limiting position of the secant PQ provided limit exists, when
Q moves up to and ultimately coincide with P, is called the tangent to
the curve at the point P. The line through the point P perpendicular to
the tangent is called the normal to the curve at the point P.
The equation of the tangent at P(x,y) on the curve, y=f(x) is,
)( xX
dx
dy
yY −=−
( , ) ( )
dy
at P Y X
dx
α β β α⇒ − = −
Or,
The equation of the normal at P(x,y) on the curve, y=f(x) is
)(
1
xX
dx
dy
yY −
−
=− Or,
1
( , ) ( )at P y x
dy
dx
α β β α
−
⇒ − = −

4. Now draw PM perpendicular on x-axis. The projection TM of the
tangent PT on the x-axis, called the sub tangent.
While the projection, MN of the normal PN on the x-axis is called
sub normal.
Formula:
(i) Length of the sub tangent,
1
cot
1
tan
1
/
TM MP
y
y
dy dx
y
y
ψ
ψ
=
= ×
= ×
=

5. (ii) Length of the sub-normal,
1
tanMN PM
dy
y
dx
yy
ψ=
= ×
=
(iii) Length of the tangent,
)1(
2
1
1
y
y
y
PT +=
(iv) Length of the normal,
)1(
2
1yyPN +=

6. Find the equation of the tangent and normal to the curve of
at the point Hence calculate the length of the sub-
tangent and sub-normal.
6)( 2
−+= xxxf
.1=x
Solution:
Given,
At
We have to find the length of the tangent and also normal at the point (1, -4).
Differentiating (1) w.r. to x, we get
)(6)( 2
ixxxfy →−+==
1,x = .4612
−=−+= xy
1
2 1 0
1, 2 1 1 3x
dy
x
dx
dy
At x
dx
=
= + +
∴ = = × + =

7. The length of the tangent of (1, -4) is,
073
073
334
)1(3)4(
)(
=−−∴
=−−⇒
−=+⇒
−=−−⇒
−=−
YX
YX
XY
XY
xX
dx
dy
yY
And, the equation of normal is as follows:1
( )
1
( 4) ( 1)
3
3( 4) 1
3 11 0
3 11 0
Y y X x
dy
dx
Y X
Y X
X Y
X Y
− = − −
⇒ − − = − −
⇒ + = − +
⇒ + + =
∴ + + =

8. Length of the sub tangent is:
1y
y
=
3
4−
=
3
4
=
∴
Length of the sub normal 1y y=
34×−=
12−=
.12=

9. Formulae: (Polar System)
Length of the sub tangent :
Length of the sub normal :
Length of the Tangent :
Length of the Normal :
1
2
r
r
=
1r=
2
1
2
1
rr
r
r
+=
2
1
2
rr +=

10. Question # 03:Compute the length of the polar
sub tangent, sub normal, tangent and also
normal, of the curve at .θcos42
=r 6
π
θ =
)(cos42
ir →= θ
6
π
θ =
Solution: Given,
At,
32
32
2
3
4
6
cos4
2
2
2
=∴
=⇒
×=⇒
=
r
r
r
r
π

11. Differentiating (i) w. r. to
, we getθ 2 4( sin )
2
dr
r
d
dr
r Sin
d
θ
θ
θ
θ
= −
⇒ = −
At,
6
π
θ =
1
2 sin
6
1
2
2
1
1
1
2 3
dr
r
d
dr
r
d
dr
r
d
dr
d r
r
π
θ
θ
θ
θ
= − ×
⇒ = − ×
⇒ = −
⇒ = −
⇒ = −

12. Therefore, the length of the sub tangent is:
( )
2
1
1
2
4
2 3
1
2 3
2 3 2 3
2 6 3
r
r
=
=
= ×
=
Length of the subnormal , 1
1
2 3
r =
Length of the tangent ,
2 2
1
1
26 3
r
r r
r
+ =
Length of the normal, 2 2
1
13
2 3
r r+ =

13. Curvature:
Sδ
λ
The curvature at a given point P is the limit (if it exists) of the
average curvature (bending) of arc PQ when the length of this arc
approaches zero. The curvature at P is denoted by
The angle is called the angle of contingence of P .
.
δψ

14. The average curvature or average bending of the arc
Thus, Curvature at is:
S
PQ
δ
δψ
=
P 0sLim
S
d
ds
δ
δψ
λ
δ
ψ
→=
=
Therefore, the curvature is the rate at which the curve
curve's or how much the curve is curving.
Radius of curvature:
The reciprocal of the curvature is called the radius of
the curvature of the curve at P.
It is usually denoted by,
Thus,
λ
ρ
.
1
ψλ
ρ
d
ds
==

15. Question#01: Find the radius of curvature for y = f (x)
Solution:
We know, ψtan=
dx
dy
2
2
2
2
2
3
sec
sec
1
sec sec
1
sec
d y d
dxdx
d ds
ds dx
ψ
ψ
ψ
ψ
ψ ψ
ρ
ψ
ρ
⇒ =
=
=
=
ψ
ψ
ψ
sec
cos
1
cos
=
=
=
dx
ds
dx
ds
ds
dx

16. ( )
( )
( )
( )
( )
( ) 2
3
2
3
2
3
2
3
2
3
2
3
2
1
2
2
2
1
2
2
1
2
12
2
2
2
2
2
1
1
1
1
1
1
tan1
1
sec
1
y
y
y
y
y
y
yy
y
dx
yd
+
==∴
+
=∴
+
=⇒
+=⇒
+=⇒
=⇒
ρ
λ
ρ
ρ
ρ
ψ
ρ
ψ
ρ

17. Question#02: Find the radius of curvature at (0, 0) of
the curve
Solution:
Given,
Differentiating w. r. to. x, we have
Again, differentiating w. r. to. x , we get
Now at (0, 0),
And,
xxxy 72 23
+−=
xxxy 72 23
+−=
2
1 3 4 7
dy
y x x
dx
= = − +
2
22
6 4
d y
y x
dx
== = −
( ) ( )2
1 3 0 4 0 7 7y = − + =
( )2 6 0 4 4y = − = −

18. Thus, radius of curvature at (0, 0) is:
( )
2
2
1
2
3
1
y
y+
=ρ
( )
( ) ( )
3
2 2
333
32 22
2
1 7
4
50 25 2 5 2 125
4 4 22
+
=
−
× ×
= = − = − = −
−

19. Question#02: Find the curvature and radius of
curvature at (0, b) of the curve
Ans:
Question#03: Show that the curvature at of
the curve is
2 2
1.
x y
a b
+ =
2
a
b
−
3 3
,
2 2
a a 
 ÷
 3 3
3x y axy+ = 8 2
3a
−

20. Question#4: Find the radius of curvature at of
the curve
Solution: Given,
Differentiating w. r. to , we get
( ),r θ
θ2cos22
ar =
2 2
2 2
2
cos2
ln ln( cos2 )
2ln ln ln(cos2 )
r a
r a
r a
θ
θ
θ
=
⇒ =
⇒ = +
θ
)1(2tan
2tan.
1
2)2sin(
2cos
1
0
1
.2
1
1
→−=⇒
−=⇒
×−×+=
θ
θ
θ
θθ
rr
r
r
d
dr
r
θ2tan222
1 rr =∴

21. Again differentiating w. r. to , we haveθ
( )
( ) ( )
2
2
2
1
2
tan 2
tan 2 sec 2 .2.
tan 2 2 sec 2
tan 2 tan 2 2 sec 2
d
r r
d
dr
r
d
r r
r r
θ
θ
θ θ
θ
θ θ
θ θ θ
= −
= − −
= − −
= − − −
θθ 2sec22tan 22
2 rrr −=⇒

22. Therefore, radius of curvature at is,( ),r θ
( )
3
2 2 2
1
2 2
1 22
r r
r r r r
ρ
+
=
+ −
( )
( )
( ){ }
( )
3
2 2 2 2
2 2 2 2 2
3
22 2
2 2 2 2 2 2 2
3
3 2 2
2 2 2 2 2
3 3 3 3
2 2 2 2 2 2 2 2
3 3
2 2
tan 2
2 tan 2 tan 2 2 sec 2
1 tan 2
2 tan 2 tan 2 2 sec 2
sec 2
tan 2 2 sec 2
sec 2 sec 2
(1 tan 2 ) 2 sec 2 sec 2 2 sec 2
sec 2
33 sec 2
r r
r r r r r
r
r r r r
r
r r r
r r
r r r r
r r
r
θ
θ θ θ
θ
θ θ θ
θ
θ θ
θ θ
θ θ θ θ
θ
θ
+
=
+ − −
+
=
+ − +
=
+ +
= =
+ + +
= = ×
2
2
2 2 2
2 2
2 2
sec 2
3
1
cos 2 sec 2
3 cos 2
r a
r
a a a
r a
r r r
θ
ρ θ θ
θ
= ×
 
∴ = = ⇒ = ⇒ = 
 
Q
1
2 2
2
tan 2
tan 2 2 sec 2
r r
r r r
θ
θ θ
= − 
 
= −  
Q

23. Question#5: Find the radius of curvature at of
the curve
Ans:
( ),r θ
cosm m
r a mθ=
1
( 1)
m
m
a
m r
ρ −
=
+

24. Centre of Curvature:
Let be the centre of curvature at P(x, y) of curve
y = f (x).
Then,
where
),( βαC
2
1 1
2
2
1
2
(1 )
,
1
y y
x
y
y
y
y
α
β
+
= −
+
= +
1
2
2 2
dy
y
dx
d y
y
dx
=
=

25. Question#06: Find the centre of curvature of
corresponding to the point (4, 4).
Solution: Given the equation of the curve is,
Differentiating w. r. to. x, we have,
At (4, 4),
16=xy
)(
16
16
i
x
y
xy
→=⇒
=
)(
16
21 ii
x
y →−=
1 2
16
1.
4
y = − = −

26. Again differentiating w.r.to. x we get,
At (4, 4),
If be the centre of curvature at P(x, y) of curve
y= f (x, y)
then,
Therefore, the centre of the curvature is (8, 8).
1 2
16
y
x
 
= − 
 
Q
32
32
x
y =
2
1
4
32
32 ==y
),( βαC
8
2
1
)11)(1(
4
)1(
2
2
11
=
+−
−=
+
−=
y
yy
xα
( )
( )
1 4, 4
2 4, 4
1
1
2
y
y
 = −
 
 
= 
 
Q
8
2
1
11
4
1 2
2
2
1
=
+
+=
+
+=
y
y
yβ

27. M. M. Billah,
Assistant Professor of Mathematics
AUST