317 CHAPTER 11 CHAPTER 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 ah + ! + . This approximation is called a linear approximation or a linear tangent approximation. This is often written: ( ) () () ' f x x f x f x x + ! " + ! .
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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,