317

CHAPTER 11CHAPTER 11

Linear Tangent Approximations and Euler’s Method

Before the arrival of calculators, a method for estimating values by extrapolation was

sometimes effected by the use of the fact that for small changes in x , dy y

dx x

!"!

.

Graphically, this meant that on the graph below

provided h was small, then points Q and R were virtually the same point. This

meant that their y co-ordinates were approximately equal.

i.e. ( )MQ MR MT TR! = +

This means that ( ) ( ) ( )'f a h f a f a h+ ! + .

This approximation is called a linear approximation or a linear tangent

approximation. This is often written: ( ) ( ) ( )'f x x f x f x x+ ! " + ! .

318

Example

Question: Find, without the use of a calculator, an approximate value for 4.01 .

Answer: Let ( )f x x= , let 4a = and 0.01h = .

Note that ( )1

'2

f xx

=

Then ( )4.01 4.01f=

( )4 0.1f= +

( )1

4 0.012 4

! +

2 0.0025= +

2.0025=

! 4.01 2.0025! (This is an excellent approximation)

Example

Evaluate, approximately, the value of 10 55x x x+ + when 1.01x = .

Let ( ) 10 55f x x x x= + + .

Then ( ) 9 4' 10 25 1f x x x= + +

( ) ( ) ( )( )' 1.01 1 ' 1 0.01f f f! +

( )7 36 0.01= +

7.36=

In fact, ( )1.01 7.36967f = (approx.)

319

Example

The relation 2 32 8x y xy+ = defines y as a function of x near to ( 2,1 ).

Call this function ( )y f x= . Use the linear tangent approximation to find an

approximate value for ( )1.92f .

( ) ( ) ( )( )1.92 2 ' 2 0.08f f f! + "

To find ( )' 2f we need to find dydx

when 2x = and 1y = .

Differentiate with respect to x .

2 3 22 2 3 0

dy dyy x x y x y

dx dx

! "+ + + =# $

% &

At ( 2,1 )

4 4 2 1 6 0dy dy

dx dx

! "+ + + =# $

% &

i.e. 3

8

dy

dx= !

Substituting in yields:

f 1.92( ) ! 1+ "3

8

#$%

&'("0.08( )

1.03=

*

*

320

Worksheet 1

1. Without the use of a calculator find the approximate value of

a) ( )1

38.02 b) sin31° (note 1 0.01745° = radians)

c) ( )1.5

4.1 d) 3 0.126

2. Find an approximate value for 3 23 2 1x x x! + ! when 1.998x = without the aid

of a calculator.

3. The surface area of a sphere is 24 r! . If the radius of the sphere is increased

from 10 cm to 10.1 cm, what is the approximate increase in area?

4. One side of a rectangle is three times another side. If the perimeter increases

by 2% what is the approximate percentage increase in area?

5. A new spherical ball bearing has a 3 cm radius. What is the approximate value

of the metal lost when the radius wears down to 2.98 cm?

6. Find the percentage error in the volume of a cube if an error of 1% is made in

measuring the edge of the cube.

7. ( 1,1 ) is a point on the graph of 2 22x y y x+ = . Find a reasonable

approximation for the y co-ordinate of a point near ( 1,1 ) whose

x co-ordinate is 1.01.

8. The equation 4 41x y xy+ + = defines y implicitly in terms of x near the point

( -1,1 ). Use the tangent line approximation at the point ( -1,1 ) to estimate the

value of y when 0.9x = ! .

321

9. The local linear approximation of a function f will always be greater than the

function’s value if, for all x in the interval containing the point of tangency,

(A) ' 0f < (B) ' 0f > (C) " 0f > (D) " 0f < (E) ' " 0f f= =

Answers to Worksheet 1

1. a) 2.0017 b) 0.5151 c) 8.3 d) 0.5013

2. -1.004

3. 8!

4. 4.04%

5. 0.72! cubic cm

6. 3.03%

7. 0.99

8. 0.9

9. D

322

Euler’s Method

Euler’s Method involves the use of the linear tangent approximation more than once

and is essentially the same. It is somewhat more accurate.

For example to find 4.02 where ( )f x x= and ( )1

'

2

f xx

= , 4a = and 0.02h = ,

we have ( ) ( ) ( )( )4.02 4.02 4 ' 4 0.02f f f= ! +

( )1

2 0.02 2.0052 4

= + =

If we use Euler’s Method where we do the approximation twice we have

0.01x! = repeated.

i.e. ( ) ( ) ( )( )4.01 4.01 4 ' 4 0.01f f f= ! +

2 0.0025= +

Then ( ) ( ) ( )( )4.02 4.02 4.01 ' 4.01 0.01f f f= ! +

( )1

2.0025 0.012 4.01

= +

( )1

2.0025 0.014.005

= +

2.004997=

In fact 4.02 2.0049938= (approx.)

323

Example

Given ( )1

'f xx

= and ( )1 0f = find ( )1.5f using Euler’s Method with two

iterations of 0.25x! = .

( ) ( ) ( )( )1.25 1 ' 1 0.25f f f! +

( )1

0 0.251

= +

0.25=

( ) ( ) ( )1

1.5 1.25 0.251.25

f f= +

( )4

0.25 0.255

= + = 0.45

Graphically Euler’s Method can be viewed and follows:

To evaluate ( )1.5f : we want the y co-ordinate of P

Euler’s method yields the y co-ordinate of Q

One linear approximation yields the y co-ordinate of R

324

Worksheet 2

1. 0.5dy

xydx

= . Use Euler’s method to find y when 2x = given that 2y = when

0x = . Use steps each of size 1.

2. The solution of the differential equation 2dy x

dx y= ! contains the point ( 3,-2 ).

Find using approximation methods the value of y when 2.7x = with

0.3x! = " .

3. 2

dy x y

dx y

!= and 2y = ! when 3x = .

An estimate for the value of y when 3.2x = using a linear tangent

approximation is:

(A) -2 (B) -2.15 (C) -2.2 (D) -2.25 (E) -2.30

325

4.

In the figure above PC is tangent to the graph of ( )y f x= at point P. Points

A and B are on PC and points D and E are on the graph of ( )y f x= .

Which statement is true?

I. Euler’s method uses the y co-ordinate of point A to

approximate the y co-ordinate of point E.

II. Euler’s method uses the x co-ordinate of point B to

approximate the root of the function ( )f x at point D.

III. Euler’s method uses the x co-ordinate of C to approximate a

root of the function ( )f x at point D.

(A) I only (B) II only (C) III only (D) I and II (E) I and III

Answers to Worksheet 2

1. 3 2. -3.35 3. D 4. A