In the name of Allah most gracious most merciful

• بسم لرحمن١ال الرحيم

Tangent and Normal:

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,

)( xXdx

dyyY −=−

( , ) ( )dy

at P Y Xdx

α β β α⇒ − = −Or,

The equation of the normal at P(x,y) on the curve, y=f(x) is

)(1

xX

dxdy

yY −−=−Or,

1( , ) ( )at P y x

dydx

α β β α−⇒ − = −

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

ydy dx

y

y

ψ

ψ

=

= ×

= ×

=

(ii) Length of the sub-normal,

1

tanMN PM

dyydx

yy

ψ=

= ×

=

(iii) Length of the tangent, )1( 2

11

yy

yPT +=

(iv) Length of the normal, )1( 2

1yyPN +=

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

dyx

dxdy

At xdx =

= + +

∴ = = × + =

The length of the tangent of (1, -4) is,

073

073

334

)1(3)4(

)(

=−−∴=−−⇒−=+⇒

−=−−⇒

−=−

YX

YX

XY

XY

xXdx

dyyY

And, the equation of normal is as follows:1( )

1( 4) ( 1)

33( 4) 1

3 11 0

3 11 0

Y y X xdydx

Y X

Y X

X Y

X Y

− = − −

⇒ − − = − −

⇒ + = − +⇒ + + =∴ + + =

Length of the sub tangent is:1y

y=

3

4−=3

4=

∴

Length of the sub normal 1y y=

34 ×−=

12−=

.12=

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=

21

2

1

rrr

r +=

21

2 rr +=

Question # 03:Compute the length of the polar sub tangent, sub normal, tangent and also normal, of the curve at .θcos42 =r

6

πθ =

)(cos42 ir →= θ

6

πθ =

Solution: Given,

At,

32

32

2

34

6cos4

2

2

2

=∴

=⇒

×=⇒

=

r

r

r

rπ

Differentiating (i) w. r. to

, we getθ 2 4( sin )

2

drrddrr Sind

θθ

θθ

= −

⇒ = −

At, 6

πθ =

1

2 sin61

22

1

1

1

2 3

drrddrrddrrddr

d r

r

πθ

θ

θ

θ

= − ×

⇒ = − ×

⇒ = −

⇒ = −

⇒ = −

Therefore, the length of the sub tangent is:

( )

2

1

1

2

4

2 31

2 3

2 3 2 3

2 6 3

r

r=

=

= ×

=Length of the subnormal , 1

1

2 3r =

Length of the tangent , 2 2

11

26 3rr r

r+ =

Length of the normal, 2 21

13

2 3r r+ =

Curvature:

Sδλ

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

The average curvature or average bending of the arc

Thus, Curvature at is:

SPQ

δδψ=

P 0sLimS

d

ds

δδψλδ

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

ψλρ

d

ds==

Question#01: Find the radius of curvature for y = f (x)Solution: We know, ψtan=

dx

dy

22

2

2

2

3

sec

sec

1sec sec

1sec

d y d

dxdxd ds

ds dx

ψψ

ψψ

ψ ψρ

ψρ

⇒ =

=

=

=

ψ

ψ

ψ

sec

cos

1

cos

=

=

=

dx

ds

dx

dsds

dx

( )

( )

( )( )

( )

( ) 23

23

23

23

23

23

21

2

2

21

2

21

212

22

22

2

1

1

1

1

11

tan11

sec1

y

y

y

y

y

y

yy

y

dx

yd

+==∴

+=∴

+=⇒

+=⇒

+=⇒

=⇒

ρλ

ρ

ρ

ρ

ψρ

ψρ

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 +−=

21 3 4 7

dyy x x

dx= = − +

2

226 4

d yy x

dx== = −

( ) ( )21 3 0 4 0 7 7y = − + =

( )2 6 0 4 4y = − = −

Thus, radius of curvature at (0, 0) is:

( )2

21

23

1

y

y+=ρ

( )

( ) ( )

32 2

333 32 22

2

1 7

4

50 25 2 5 2 125

4 4 22

+=

−

× ×= = − = − = −−

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+ =

2a

b−

3 3,

2 2

a a ÷ 3 3 3x y axy+ = 8 2

3a

−

Question#4: Find the radius of curvature at of the curve

Solution: Given,

Differentiating w. r. to , we get

( ),r θθ2cos22 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

10

1.2

1

1

→−=⇒

−=⇒

×−×+=

θ

θ

θθθ

rr

rr

d

dr

r

θ2tan2221 rr =∴

Again differentiating w. r. to , we have

θ

( )

( ) ( )

2

2

21

2

tan 2

tan 2 sec 2 .2.

tan 2 2 sec 2

tan 2 tan 2 2 sec 2

dr r

ddr

rd

r r

r r

θθ

θ θθ

θ θ

θ θ θ

= −

= − −

= − −

= − − −

θθ 2sec22tan 222 rrr −=⇒

Therefore, radius of curvature at is, ( ),r θ

( )3

2 2 21

2 21 22

r r

r r r rρ

+=

+ −

( )( )

( ){ }

( )

32 2 2 2

2 2 2 2 2

322 2

2 2 2 2 2 2 2

33 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

θθ θ

θ θθ θ θ θ

θθ

+=

+ − −

+=

+ − +

=+ +

= =+ + +

= = ×2

2

2 2 22 2

2 2

sec23

1cos2 sec2

3 cos2

r a

r

a a ar a

r r r

θ

ρ θ θθ

= ×

∴ = = ⇒ = ⇒ =

Q

1

2 22

tan 2

tan 2 2 sec 2

r r

r r r

θ

θ θ

= −

= −

Q

Question#5: Find the radius of curvature at of the curve

Ans:

( ),r θcosm mr a mθ=

1( 1)

m

m

a

m rρ −=

+

Centre of Curvature: Let be the centre of curvature at P(x, y) of curve y = f (x).

Then,

where

),( βαC

21 1

2

21

2

(1 ),

1

y yx

y

yy

y

α

β

+= −

+= +1

2

2 2

dyy

dx

d yy

dx

=

=

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

ix

y

xy

→=⇒

=

)(1621 iix

y →−=

1 2

161.

4y = − = −

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

16y

x

= − Q

32

32

xy =

2

1

4

3232 ==y

),( βαC

8

21

)11)(1(4

)1(

2

211 =+−−=+−=

y

yyxα

( )

( )

1 4, 4

2 4, 4

1

1

2

y

y

= −

=

Q

8

2111

41 2

2

21 =++=++=

y

yyβ

M. M. Billah, Assistant Professor of Mathematics

AUST