lecture slides differentiation

8/12/2019 Lecture Slides Differentiation

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DifferentiationLecture No. 04

IT302: Mathematical Methods

Spring2014

Institute of Information TechnologyQuaid-i-Azam University Islamabad

Monday 24th February, 2014

Differentiability at a point

A functionf is said to be differentiable atx0if thefollowing limit exists

limh0

f(x0 + h) f(x0)h

The above limit is called the derivative offatx0and isdenoted byf(x0), i.e.,

f(x0)= limh0

f(x0+ h) f(x0)h

Example

Iff(t)= 16t2, then findf(1).

Solution

The difference quotient is

f(1 + h) f(1)h

=16(1 + h)2 16(1)2

h =16(h + 2)

When we take the limit as h 0f(1)= lim

h016(h + 2)= 32

IT302: M.M. Lecture No. 04 (Differentiation) 2


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Interpretation as slope of tangent

In reference to the definition

f(x0)= limh0

f(x0+ h) f(x0)h

consider the following illustration.

Showing thatf(x0) is the slope of the curve y = f(x) atx= x0.

Example

Find the slope of the curvey = 1

xat any pointx = a 0.

What is the slope of the curve atx =

1?

Solution

Heref(x)= 1

xand the slope atx = ais

f(a) = limh0

f(a + h) f(a)h

= limh0

1

a + h 1

ah

= limh0

1a(a + h)

= 1a2

Moreoverf(1)= 1.


Instantaneous rate of change

The difference quotient

f(x0+ h) f(x0)h

is the ratio of thechange in dependent variableto thechange in the independent variable.

So the quotient represents the average rate of change off w.r.t. x whenxvaries in the closed interval [x0,x0+ h].

When we take the limit as h

0, the interval [x0, x0 + h]

reduces to the singleton {x0}.

Showing that

f(x0)= limh0

f(x0+ h) f(x0)h

is the instantaneous rate of change offw.r.t. xatx = x0.

Example

An object is dropped from the top of a 100-m-hightower. Its height above ground after t s is 100 4.9t2 m.What is the average velocity of the object in the first twoseconds? How fast is it falling 2 s after it is dropped?

Solution

The heighthof the object as a function of time is

h(t)= 100 4.9t2The average velocity of the object is

vavg =

h

t =

h(2)

h(0)

2 =

80.4

100

2 =9.8 m/sThe instantaneous velocity of the object is

v(t)= dh

dt =9.8t

Hencev(2)=19.6 m/s.



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Example

Find the linearization off(x)=

1 + xatx = 3.

Solution

The linearization offwith centerx = 3 is given by

L(x)= f(3) +f(3)(x 3)= 5

4 +x

4

Atx = 3.2 the linearization gives

1 + x=

1 + 3.2 5

4+

3.2

4 =2.050

which differs from the true value

4.22.049390153 byless than 0.1%

The linearization with center at x = 0

L(x)= 1 +x

2

gives the following approximation atx = 3.2

1 + x =

1 + 3.21 + 3.2

2 =2.6

a result that is offby more than 25%

Example

Find the linearization off(x)= sin(x) with center atx= 0.

Solution

The linearization offwith center atx = 0 is

L(x) = f(0) +f(0)(x 0)= sin(0) + x cos(0)= L(x) = x

Showing that for values ofxclose to 0 we have

sin(x)x


Differentials

Now we investigate the change in the dependentvariable corresponding to some change in theindependent variable. Consider the adjacent figure,when there is changedxin the independent variable atx = a, the corresponding exact change in the dependentvariabley = f(x) is

y= f(a + dx) f(a)

And the corresponding change in the linearizationLoffwith center atx = ais

L= L(a + dx) L(a)

AsL(x)= f(a) + (x a)f(a), thereforeL= f(a) + (a + dx a)f(a)

L(a+dx)

f(a)

L(a)

=f(a)dx

This changef(a)dxin linearization is anapproximationto the change infand is called the differential off.

Thus, the differential of the function, denoted by df is

df=f(a)dx

Showing that the differentialdfdepends not only on thechangedxin the independent variable but also on thepointx = a, where the changedxtakes place.



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Estimating with differentials

Suppose we know the value of a differentiable functionf(x) at a pointaand want to estimate how much thisvalue will change if we move to a nearby point a + dx.Since

f(a + dx)= f(a) + y,

the differential approximation gives

f(a + dx)f(a) + dfThus the approximationydycan be used to estimatef(a + dx) whenf(a) is known anddxis small.

Example

The radiusrof a circle increases froma = 10 m to 10.1 m.UsedAto estimate the increase in the circles area A.Estimate the area of the enlarged circle and compareyour estimate to the true area found by directcalculation.

Solution

AsA(r)= r2, therefore the estimated increase is

dA= A(a)dr= 2adr= 2(10)(0.1)= 2 m2

SinceA(a + dr)A(a) + dA, we haveA(10 + 0.1)A(10) + 2= 102 + 2= 102

The area of a circle of radius 10.1 m is approximately

102 m2.


The true area is

A(10.1)= (10.1)2 =102.01 m2

The error in our estimate is 0.01 m2, which is thedifference A dA.

Remark

As we move fromato a nearby pointa + dx, we candescribe the change infin three ways:

True Estimated

Absolute f=f(a + dx) f(a) df=f(a)dx

Relative ff(a)

dff(a)

Percentagef

f(a) 100 df

f(a) 100

Sensitivity to change

The equationdf=f(a)dxtells howsensitivethe outputoffis to a change in input ata. The larger the value off(a), the greater the effect of a given changedx.

Example

You want to calculate the depth of a well from the

equation s(t)= 16t2 ft by timing how long it takes aheavy stone you drop to splash into the water below.How sensitive will your calculations be to a 0.1 s error inmeasuring the time?

Solution

The size ofds = 32tdtdepends upon how bigtis. Att= 2 s, the error caused bydt = 0.1 is only

ds= 32(2)(0.1)= 6.4 ft

Three seconds later att = 5 s, the error caused by thesamedtis

ds= 32(5)(0.1)= 16ft


lecture slides differentiation

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