317
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 ' (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 .
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Example
Question:
Find, without the use of a calculator, an approximate value for
Answer:
Let f (x ) = x , let a = 4 and h = 0.01 .
Note that f ' (x ) =
Then
1
2 x
4.01 = f (4.01)
= f (4 + 0.1)
! 4+
1
2 4
(0.01)
= 2 + 0.0025
= 2.0025
!
4.01 ! 2.0025 (This is an excellent approximation)
Example
Evaluate, approximately, the value of x10 + 5 x5 + x when x = 1.01 .
Let
f (x ) = x10 + 5 x5 + x .
Then
f ' (x ) = 10 x9 + 25 x 4 + 1
f ' (1.01) ! f (1) + f ' (1)(0.01)
= 7 + 36 (0.01)
= 7.36
In fact, f (1.01) = 7.36967 (approx.)
4.01 .
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Example
The relation x 2 y + 2 xy 3 = 8 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 f (1.92 ).
f (1.92 ) ! f (2 ) + f ' (2 )("0.08 )
To find f ' (2 ) we need to find
*
dy
when x = 2 and y = 1 .
dx
Differentiate with respect to x .
y2x + x2
dy
dy "
!
+ 2 # y 3 + x3 y 2 $ = 0
dx
dx &
%
At ( 2,1 )
4+4
i.e.
dy
dy "
!
+ 2 #1 + 6 $ = 0
dx
dx &
%
dy
3
=!
dx
8
Substituting in
*
yields:
# 3&
f 1.92 ! 1 + % " ( "0.08
$ 8'
(
)
(
= 1.03
)
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Worksheet 1
1. Without the use of a calculator find the approximate value of
1
a) (8.02 )3
b) sin 31°
c) (4.1)
d)
1.5
3
(note 1° = 0.01745 radians)
0.126
2. Find an approximate value for x3 ! 3x 2 + 2 x ! 1 when x = 1.998 without the aid
of a calculator.
3. The surface area of a sphere is 4! r 2 . 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 x 2 y + y 2 x = 2 . 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 x 4 + y + xy 4 = 1 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 x = !0.9 .
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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) f ' < 0
(B) f ' > 0
(C) f " > 0
(D) f " < 0
c) 8.3
d) 0.5013
Answers to Worksheet 1
1. a) 2.0017
b) 0.5151
2. -1.004
3. 8!
4. 4.04%
5. 0.72! cubic cm
6. 3.03%
7. 0.99
8. 0.9
9. D
(E) f ' = f " = 0
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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 f ' (x ) =
1
2 x
, a = 4 and h = 0.02 ,
4.02 = f (4.02 ) ! f (4 ) + f ' (4 )(0.02 )
we have
= 2+
1
2 4
(0.02 ) = 2.005
If we use Euler’s Method where we do the approximation twice we have
!x = 0.01 repeated.
i.e.
4.01 = f (4.01) ! f (4 ) + f ' (4 )(0.01)
= 2 + 0.0025
Then
4.02 = f (4.02 ) ! f (4.01) + f ' (4.01)(0.01)
= 2.0025 +
1
(0.01)
2 4.01
= 2.0025 +
1
(0.01)
4.005
= 2.004997
In fact 4.02 = 2.0049938 (approx.)
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Example
Given f ' (x ) =
1
and f (1) = 0 find f (1.5 ) using Euler’s Method with two
x
iterations of !x = 0.25 .
f (1.25 ) ! f (1) + f ' (1)(0.25 )
= 0+
1
(0.25)
1
= 0.25
f (1.5 ) = f (1.25 ) +
= 0.25 +
1
(0.25)
1.25
4
(0.25) = 0.45
5
Graphically Euler’s Method can be viewed and follows:
To evaluate f (1.5 ) : 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
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Worksheet 2
1.
dy
= 0.5 xy . Use Euler’s method to find y when x = 2 given that y = 2 when
dx
x = 0 . Use steps each of size 1.
2. The solution of the differential equation
dy
x2
=!
contains the point ( 3,-2 ).
dx
y
Find using approximation methods the value of y when x = 2.7 with
!x = "0.3 .
3.
dy x ! y
=
and y = !2 when x = 3 .
dx
2y
An estimate for the value of y when x = 3.2 using a linear tangent
approximation is:
(A) -2
(B) -2.15
(C) -2.2
(D) -2.25
(E) -2.30
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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