Differentials

Intro

• The device on the first slide is called a micrometer….it is used for making precision measurements of the size of various objects…..a small metal cube, the diameter of a ball bearing, etc….

• However, even a precision instrument like a micrometer has an error in measuring things….

• Errors in measurement, like a small error in the diameter of a ball bearing, can lead to major problems in an engine, if the ball bearing’s size is too far off from it’s allowed variation.

• To measure the precise change in a mathematical function, such as a change in an equation, we use Δy, the change in y.

• But this can be messy to calculate ……. it would be nice to use a simpler way to represent the change in y, and using Calculus, we can !!!

After this lesson, you should be able to:

• Understand the concept of a tangent line approximation.

• Compare the value of the differential, dy, with the actual change in y, Δy.

• Estimate a propagated error using a differential.

• Find the differential of a function using differentiation formulas.

A Simple Example

Differentials

Let S(x) be the area of a square of side length x. Or, Now we give the side a change Δx , then the corresponding change of area will be

S(x) = x2

ΔS = S(x + Δx) – S(x) = (x + Δx)2 – x2

= 2xΔx + (Δx)2

Δx

Δxx

x

xΔx

xΔx (Δx)2

Compare with Δx , (Δx)2 is the infinite smaller change. Or,

0)(

lim2

0

x

xx

A Simple Example

DifferentialsΔx

Δxx

x

xΔx

xΔx (Δx)2

For any function of g(Δx), if 0

)(lim

0

x

xgx

We call g(Δx) is higher order of infinitesimal of Δx, denoted as ε(Δx). In other words, the change of area can be approximated to the first part of ΔS = S(x + Δx) – S(x)

= 2xΔx + ε(Δx)

= 2xΔx + (Δx)2The first part above is linear to Δx. We

call the first part as “Linear Principal Part ”.

Find the equation of the line tangent to

at (1,1).

2( )f x x y

Try This

2 1y x

(1,1)

( )f x

Analysis

2 1y x

(1,1)

2( )f x x x 0.9 1.0 1.1

f(x)

y

Complete the table

Analysis

2 1y x

(1,1)

2( )f x x x 0.9 1.0 1.1

f(x) .81 1 1.21

y .8 1 1.2

Complete the table

Linear Approximation to a Function

Linear Approximation or TANGENT LINE!!!!!

y f x

0 0 0'y f x f x x x

Important IdeaThe equation of the line tangent to f(x) at c can be used to approximate values of f(x) near f(c).

Definition of Differentials

x

y

y = f(x) =x2 y = 2x + 1

Δx

dyΔy

ε(Δx)

In many types of application, the differential of y can be used as an approximation of the change of y.

dyyΔ

dxxfyΔ )('

xΔxfyΔ )('

or

or

(1,1)

f(x+Δx)

f(x)

Animated Graphical View

• Note how the "del y" and the dy in the figure get closer and closer

'

In general, ( ) ( )

which is the ACTUAL CHANGE IN y!!!!

Can be approximated by

( )

which is the APPROXIMATED CHANGE IN y!!!!

y f c x f c

f c dx

2Let . Find when 1 and 0.01.

Compare this value with for 1 and 0.01.

y x dy x dx

y x x

2Let . Find when 1 and 0.01.

Compare this value with for 1 and 0.01.

y x dy x dx

y x x

2Let . Find when 1 and 0.01.

Compare this value with for 1 and 0.01.

y x dy x dx

y x x

2Let . Find when 1 and 0.01.

Compare this value with for 1 and 0.01.

y x dy x dx

y x x

2Let . Find when 1 and 0.01.

Compare this value with for 1 and 0.01.

y x dy x dx

y x x

2Let . Find when 1 and 0.01.

Compare this value with for 1 and 0.01.

y x dy x dx

y x x

2Let . Find when 1 and 0.01.

Compare this value with for 1 and 0.01.

y x dy x dx

y x x

2Let . Find when 1 and 0.01.

Compare this value with for 1 and 0.01.

y x dy x dx

y x x

2Let . Find when 1 and 0.01.

Compare this value with for 1 and 0.01.

y x dy x dx

y x x

Example 1

Find dy for f (x) = x2 + 3x and evaluate dy for x = 2 and dx = 0.1.

Example 1

Find dy for f (x) = x2 + 3x and evaluate dy for x = 2 and dx = 0.1.

Solution:

dy = f ’(x) dx = (2x + 3) dx

When x = 2 and dx = 0.1, dy = [2(2) + 3] 0.1 = 0.7.

Derivatives in differential form

Function derivative differential

y = x n y ’ = n x n-1 dy = n x n-1 dx

y = UV y ′ = UV ′ + VU ′ dy = U dV + V dU

y = U y ’ = V U ′ – U V ′ dy = V dU – U dV

V V 2 V

2

Finding Differentials

Function DerivativeDifferential

y = x2

y = 2 sin x

y = x cosx

Finding Differentials

Function DerivativeDifferential

y = x2

y = 2 sin x

y = x cosx

Finding Differentials

Function DerivativeDifferential

y = x2

y = 2 sin x

y = x cosx

Finding Differentials

Function DerivativeDifferential

y = x2

y = 2 sin x

y = x cosx

Finding Differentials

Function DerivativeDifferential

y = x2

y = 2 sin x

y = x cosx

Finding Differentials

Function DerivativeDifferential

y = x2

y = 2 sin x

y = x cosx

Finding Differentials

Function DerivativeDifferential

y = x2

y = 2 sin x

y = x cosx

Try This

Find the differential dy of the function:

2sin 2y x

4cos 2dy xdx

Linear Approximation to a Function

Linear Approximation or TANGENT LINE!!!!!

y f x

0 0 0'y f x f x x x

Find the tangent line approximation of

( ) 1 sin( )f x x at the point (0,1).

Find the tangent line approximation of

0 0 0'y f x f x x x

'(0) (0)( 0)

1 (1)( 0)

1

y f f x

x

x

at the point (0,1).

( ) 1 sin( )f x x

Example

SolutionLet’s consider the function and set x = 1, Δx = 0.02, then

Example 1 Find the approximate value of the cubic root 3 02.1

3)( xxf

xΔxfxfxΔxf )(')()(02.13

006.102.03

11

xΔxx -2/33

3

1

Example

Solution

Practice Find the approximate value of square root 98.48

Let’s consider the function and set x = 49, Δx = –0.02, then

xxf )(

xΔxfxfxΔxf )(')()(98.48

9986.6)02.0(14

17

xΔx

x2

1

Example 1

• Estimate ( 64.3 ) 1/3

Let y = ( x ) 1/3

. 64.3 = 64 + 0.3

Let x = 64 and Δx = 0.3 = 3/10

So y + Δy ≈ y + dy and dy = 1/3 x – 2/3

dx

y + dy = ( 64 ) 1/3

+ 1/3 ( 64 ) -2/3

( 3/10 )

= 4 + 1/3 ( 1/16 ) ( 3/ 10 )

• = 4 + 1 / 160 = 4 = 4.006251

160

By calculator, ( 64.3 ) ( 64.3 ) 1/31/3 = =

4.006240264.00624026

Not bad….considering they had no calculator back then !!!

Error PropagationPhysicists and engineers tend to make liberal use of the approximation of Δy by dy. One way this occrus in practice is in the estimation of errors propagated by physical measuring devices.

x – measured value of a variablex + Δx – the exact valueΔx – the error in measurement

f(x + Δx) – f(x) = Δy

Measurement Error

Measured Value

Exact Value

Propagated Error

If the measured value x is used to compute another value f(x), then

Propagated Error

• Consider a rectangular box with a square base– Height is 2 times length

of sides of base– Given that x = 3.5– You are able to measure with 3% accuracy

• What is the error propagated for the volume?

xx

2x

Propagated Error

• We know that

• Then dy = 6x2 dx = 6 * 3.52 * 0.105 = 7.7175This is the approximate propagated error for the volume

32 2

3% 3.5 0.105

V x x x x

dx

Propagated Error

• The propagated error is the dy– sometimes called the df

• The relative error is

• The percentage of error – relative error * 100%

7.71750.09

( ) 85.75

dy

f x

Example

Solution

It is obvious that from the given informationr = 6 inches and Δr = dr = ±0.02 inches

Example 2 The radius of a sphere is measured to be 6 inches, with a possible error of 0.02 inch. Use differentials to approximate the maximum possible error in calculating

(a)The volume of the sphere(b)The surface area of the sphere(c)The relative errors in parts (a) and (b)

3

3

4rV

(a) drrdV 24 )02.0(64 2

88.2 in3

Example

Solution

It is obvious that from the given informationr = 6 inches and Δr = dr = ±0.02 inches

Example 2 The radius of a sphere is measured to be 6 inches, with a possible error of 0.02 inch. Use differentials to approximate the maximum possible error in calculating

(a)The volume of the sphere(b)The surface area of the sphere(c)The relative errors in parts (a) and (b)

24 rS (b) rdrdS 8 )02.0)(6(8

96.0 in2

Example

Solution

Example 2 The radius of a sphere is measured to be 6 inches, with a possible error of 0.02 inch. Use differentials to approximate the maximum possible error in calculating

(a)The volume of the sphere(b)The surface area of the sphere(c)The relative errors in parts (a) and (b)(c)

r

dr

r

drr

V

dV 3

)3/4(

43

2

02.06

3

01.0

r

dr

r

rdr

S

dS 2

4

82

02.06

2

.....00666.0

Calculating DifferentialsThe differentiation rules we learned in Chapter 2 can be migrate to here

Constant Multiple

Sum or Difference

Product

Quotient

dxcucducud '][

dxvdxudvduvud ''][

udxvvdxuudvvduuvd ''][

22

''

v

dxuvdxvu

v

udvvdu

v

ud

Calculating DifferentialsDon’t forget the chain rule in calculating the derivatives. The same rule is true in calculating the differentials

dx

du

du

dy

dx

dy

)( and ),( xguufy

dxxgxgfdxuufdy )('))(('')('

Example

Example 3 Find the differentials

xxxf sin)(

Solution

( ) sin sindf x dy xdx xd x xdxxxdx cossin

dxxxx )cos(sin

Example

Example 4 Find the differentials2

2 )1(2)(

xxxf

Solution2

2 2(1 )( )

xdf x dy dx d

2)1(2

2

xdxdx

)1(2)1(222

xd

xxdx

xdxxdx 12

)1(42

dxxxdx )1(2

)1(42

dxx

xdx

)1(82

dx

xx

)1(42

Example

Example 5 Find the differentials

2

2sin

x

xy

Solution

2

2sin

x

xddy

4

22 2sin2sin

x

xdxxdx

4

2 2sin22cos2

x

xdxxxdxx

4

)2sin2cos(2

x

dxxxxx

3

)2sin2cos(2

x

dxxxx

Example

Example 6 Find the differential dy in an implicit function if y is a function of x and 122 xyyx

Solution

0)1()( 22 dxyyxd

xyx

yxy

dx

dy

2

22

2

0)()( 22 xydyxd

0][][ 2222 xdydxydyxydx

0]2[]2[ 22 xydydxydyxxydx

0]2[]2[ 22 dxyxydyxyx

dxyxydyxyx ]2[]2[ 22