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4.0

APPLICATIONS OF

DIFFERENTIATION

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4.1 TANGENT AND NORMAL

EQUATION

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At the end of this topic, students

should be able :

a) To find the equations of tangent andnormal to a curve including the parametric

equation

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1 . is the gradient to the function ofdxdyxf )('

)(xfy

2. At point ( ) on the curve )(xfy the equation of tangent is

)( 111 xxmyy

11, yx

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where

is the gradient of the tangent to

the curve at the point

dxdym 1

),( 11 yx

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)(xfy

normaltangent

11, yx

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EXAMPLE

Find the tangent and normal equation to

the curve at x= 2322 xy

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Solution,

If

Differentiate

so the gradient for tangent at x = 2 is

4(2) = 8

xdx

dy

4

32 2 xy

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If x = 2

The tangent equation at point (2, 11) is

113)2(22

y

)2(811 xy

58

16811

xy

xy

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tangent is perpendicular to the

normal thus,

121 mm

2m8

1

1)(8 2 m

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normal equation at point (2,11)

9082888

8

2

8

111

)2(8

111

xyxy

xy

xy

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32 2 xy

908 xynormaltangent

58 xy

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EXAMPLE

Find the tangent and normal equation to

the curve at P(1,1)

Solution,

Differentiate by using the implicit derivative

method,

223 xyyx

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)3()2(

02)1(3

223

223

yxyxyxdx

dy

dxdyxyyyx

dxdyx

xyxyxy

dxdy

2)3( 3

22

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the gradient for tangent at P(1,1) is

_

so tangent equation at P(1,1) is

743

3

4

3

41

)1(3

41

xy

xy

xy

3

4

)1)(1(21

))1()1(31(3

22

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gradient for normal at P(1,1) is

_

4

3

13

4

2

2

m

m

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normal equation at P(1,1)

134

4

3

4

31

)1(4

31

xy

xy

xy

E i Fi d th t t d

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Exercise: Find the tangent and

normal equation to the curve xy = 8

at P(4,2)

Answer;

Tangent equation 2y = -x + 8Normal equation y = 2x 6

xy

dxdy

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EXAMPLE

Find the equation of normal to the curve

given parametrically by

and

at point where t= 1

tx 2 13

2 ty

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so

when t=1

the gradient for tangent is -3(1) = -3

32

3)

2

(6 tt

t

dx

dy

3

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the normal equation at (2,2) is

43

3

2

3

12

)2(3

12

xy

xy

xy

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EXAMPLE

Given parametric equation

and wheret 1

Find the tangent equation which is parallelto line 3y = x

1

3

tx2)1(

2

ty

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Solution

2)1(

3

tdtdx

3)1(

4

tdtdy

)1(34

3

1

)1(

4 2

3

t

t

tdx

dt

dt

dy

dx

dy

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from 3y = x the gradient for the line is

So

4 = 1+ t

t = 3

3

1

31

)1(34

tdx

dy

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when t = 3

tangent equation is

43

133

x81

)31(2

2 y

3824

123

318

)4

3(

3

1

8

1

xy

xy

xy

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Exercise:

Given parametric equation x =

and y =

Find the tangent equation when t= 0

23tt

2

42

t

t

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Answer:

3y = 4x - 6