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DASAR-DASAR METODE KUANTITATIF
Oleh : EDY SUHARTONO
Subject
• Differentiation• Integral• Differential Equation
WHAT IS A FUNCTION?
LINEAR FUNCTIONS
Slope and Rate of Change
Linear Functions in General
Which of the following tables of values could represent a linear function?
RATES OF CHANGE
The Derivative at a Point
THE DERIVATIVE FUNCTION
Alternative Notations for the Derivative
Basic Differentiation Rules
1.
Ex.
2.
0 a constantdc c
dx
( ) 5
( ) 0
f x
f x
Ex.
1 a real numbern ndx nx n
dx
7
6
( )
( ) 7
f x x
f x x
Basic Differentiation Rules
3.
Ex.
4.
( ) ( ) a constantd dcf x c f x c
dx dx
8( ) 3f x x
Ex.
( ) ( ) d d d
f x g x f x g xdx dx dx
12( ) 7f x x
7 7( ) 3 8 24f x x x
11 11( ) 0 12 12f x x x
More Differentiation Rules5.
Ex.
3 7 2( ) 2 5 3 8 1f x x x x x
( ) ( ) ( ) ( ) d d d
f x g x f x g x g x f xdx dx dx
Product Rule
2 7 2 3 6( ) 3 2 3 8 1 2 5 21 16f x x x x x x x x
9 7 6 4 230 48 105 40 45 80 2x x x x x x
Derivative of first
Derivative of Second
More Differentiation Rules6.
2( ) ( ) ( ) ( )
( ) ( )
d dg x f x f x g xf xd dx dx
dx g x g x
Quotient Rule
lo hi hi lohi
lo lo lo
d dd
dx
Sometimes remembered as:
More Differentiation Rules6.
Ex.
2
3 5( )
2
xf x
x
Quotient Rule (cont.)
2
22
3 2 2 3 5( )
2
x x xf x
x
2
22
3 10 6
2
x x
x
Derivative of numerator
Derivative of denominator
More Differentiation Rules
7. The Chain Rule
If ( ) ( ) thenh x g f x
( ) ( ) ( )h x g f x f x
Note: h(x) is a composite function.
If ( ) with ( ) theny h x g u u f x
dy dy du
dx du dx
Another Version:
Copyright (c) 2003
Brooks/Cole, a division of Thomson Learning, Inc.
More Differentiation Rules
The General Power Rule:
If ( ) ( ) , real thenn
h x f x n
1( ) ( ) ( )
nh x n f x f x
Ex.
1 22 2( ) 3 4 3 4f x x x x x
1 221
( ) 3 4 6 42
f x x x x
2
3 2
3 4
x
x x
The Chain Rule leads to
Copyright (c) 2003
Brooks/Cole, a division of Thomson Learning, Inc.
Chain Rule Example
72 1
( )3 5
xG x
x
6
2
3 5 2 2 1 32 1( ) 7
3 5 3 5
x xxG x
x x
66
2 8
91 2 12 1 13( ) 7
3 5 3 5 3 5
xxG x
x x x
Ex.
Copyright (c) 2003
Brooks/Cole, a division of Thomson Learning, Inc.
Chain Rule Example
5 2 8 2, 7 3y u u x x Ex. dy dy du
dx du dx
3 2 7556 6
2u x x
3 28 2 757 3 56 6
2x x x x
3 27 8 2140 15 7 3x x x x
Sub in for u
Higher DerivativesThe second derivative of a function f is the derivative of the derivative of f at a point x in the domain of the first derivative.
Derivative Notations
nf
f
f
(4)f
Second
Third
Fourth
nth
2
2
d y
dx3
3
d y
dx4
4
d y
dxn
n
d y
dx
Example of Higher Derivatives5 3( ) 3 2 14f x x x Given find ( ).f x
4 2( ) 15 6f x x x
3( ) 60 12f x x x
2( ) 180 12f x x
Example of Higher Derivatives
Given2 1
( )3 2
xf x
x
find (2).f
2
2 2
2 3 2 3 2 1 7( ) 7 3 2
3 2 3 2
x xf x x
x x
3
3
42( ) 14 3 2 3
3 2f x x
x
3 3
42 42 21(2)
3243(2) 2f
Implicit Differentiation33 4 17y x x
y is explicitly a function of x.
3 3 1y xy x
y is implicitly a function of x.
To differentiate the implicit case we write f (x) in place of y to get:
3( ) ( ) 3 1f x x f x x
Implicit Differentiation (cont.)Now differentiate
using the chain rule:
3( ) ( ) 3 1f x x f x x
23 ( ) ( ) ( ) ( ) 3f x f x f x xf x
23 3y y y xy Which is
subbing in y
23 3y y x y
2
3
3
yy
y x
Solve for y’
Practice 1
• A cylinder tank is being filled with water. The tank has height 40 m and radius 3 m. If water is being pumped in at a constant rate of 2 cubic meter per minute, find the rate at which the height of the cylinder changes at 30 minute.
Practice 2
• The length of a rectangle increases by 3 meter per minute while the width decreases by 2 meter per minute. When the length is 40 meter and the width is 15 meter, what is the rate at which the changes area
Practice 30,5 /
dym dtk
dt
?dx
dt
y= 15 m
X = 10 m
Practice 4
y= 15 m
X = 10 m0,5 /
dxm dtk
dt
?dy
dt
Practice 5
y= 15 m
X = 15m
0,5 /dx
m dtkdt
?dy
dt
Practice 6• A wire of length 12 inches can be bent into a
circle, a square, or cut to make both a circle and a square. How much wire should be used for the circle if the total area enclosed by the figure(s) is to be a minimum? A maximum?
Practice 7• A window consisting of a rectangle topped by
a semicircle is to have an outer perimeter P. Find the radius of the semicircle if the area of the window is to be a maximum.
Practice 8• A rectangular field as shown is to be bounded
by a fence. Find the dimensions of the field with maximum area that can be enclosed with 1000 feet of fencing. You can assume that fencing is not needed along the river and building.
Practice 9
• A company manufactures cylindrical barrels to store nuclear waste. The top and bottom of the barrels are to be made with material that costs $10 per square foot and the rest is made with material that costs $8 per square foot. If each barrel is to hold 5 cubic feet, find the dimensions of the barrel that will minimize the total cost.
The Integral
• The Indefinite Integral
• Substitution
• The Definite Integral As a Sum
• The Definite Integral As Area
•The Definite Integral: The Fundamental Theorem of Calculus
AntiderivativeAn antiderivative of a function f is a function F such that
F f
Ex.
2( ) 3 2F x x
An antiderivative of ( ) 6f x x
since ( ) ( ).F x f x
is
( )f x dxmeans to find the set of all antiderivatives of f.
The expression:read “the indefinite integral of f with respect to x,”
( )f x dx
Integral sign Integrand
Indefinite Integral
x is called the variable of integration
Every antiderivative F of f must be of the form F(x) = G(x) + C, where C is a constant.
Notice 26 3xdx x C
Constant of Integration
Represents every possible antiderivative of 6x.
Power Rule for the Indefinite Integral, Part I
1
if 11
nn xx dx C n
n
Ex.
43
4
xx dx C
Power Rule for the Indefinite Integral, Part II
1 1lnx dx dx x C
x
x xe dx e C Indefinite Integral of ex and bx
ln
xx bb dx C
b
Sum and Difference Rules
( ) ( )kf x dx k f x dx
f g dx fdx gdx Ex.
( constant)k
4 43 32 2 2
4 2
x xx dx x dx C C
2 2x x dx x dx xdx 3 2
3 2
x xC
Constant Multiple Rule
Ex.
Integral Example/Different Variable
Ex. Find the indefinite integral:
273 2 6ue u du
u
213 7 2 6ue du du u du du
u
323 7 ln 6
3ue u u u C
Position, Velocity, and Acceleration Derivative Form
If s = s(t) is the position function of an object at time t, then
Velocity = v = Acceleration = a = ds
dtdv
dt
Integral Form
( ) ( )s t v t dt ( ) ( )v t a t dt
Integration by SubstitutionMethod of integration related to chain rule differentiation. If u is a function of x, then we can use the formula
/
ffdx du
du dx
Integration by Substitution
Ex. Consider the integral: 92 33 5x x dx3 2pick +5, then 3 u x du x dx
10
10
uC
9u du 103 5
10
xC
Sub to get Integrate Back Substitute
23
dudx
x
2Let 5 7 then 10
duu x dx
x
Ex. Evaluate
3/ 21
10 3/ 2
uC
3/ 225 7
15
xC
25 7x x dx
2 1/ 215 7
10x x dx u du
Pick u, compute du
Sub in
Sub in
Integrate
3ln
dx
x xLet ln then u x xdu dx
Ex. Evaluate
3
3ln
dxu du
x x
2
2
uC
2ln
2
xC
3
3 2
t
t
e dt
e 3
3Let +2 then
3t
t
duu e dt
e
Ex. Evaluate
3
3
1 1
32
t
t
e dtduue
ln
3
uC
3ln 2
3
teC
Shortcuts: Integrals of Expressions Involving ax + b
Rule
1
1( 1)
nn ax b
ax b dx C na n
1 1lnax b dx ax b Ca
1ax b ax be dx e Ca
1
lnax b ax bc dx c C
a c
Riemann SumIf f is a continuous function, then the left Riemann sum with n equal subdivisions for f over the interval [a, b] is defined to be
0 1 1( ) ( ) ... ( )nf x x f x x f x x
1
0
n
kk
f x x
0 1where ... are the na x x x b
0 1 1( ) ( ) ... ( )nf x f x f x x
subdivisions and ( ) / .x b a n
The Definite IntegralIf f is a continuous function, the definite integral of f from a to b is defined to be
1
0
( ) limb n
kn
ka
f x dx f x x
The function f is called the integrand, the numbers a and b are called the limits of integration, and the variable x is called the variable of integration.
Approximating the Definite Integral
Ex. Calculate the Riemann sum for the
integral using n = 10.22
0
x dx
1 9
2
0 0
1
5
n
k kk k
f x x x
2 2 2(1/ 5) (2 / 5) ... (9 / 5) (1/ 5) 2.28
The Definite Integral
is read “the integral, from a to b of f(x)dx.”
( )b
a
f x dx
Also note that the variable x is a “dummy variable.”
( ) ( )b b
a a
f x dx f t dt
The Definite Integral As a Total
If r(x) is the rate of change of a quantity Q (in units of Q per unit of x), then the total or accumulated change of the quantity as x changes from a to b is given by
Total change in quantity ( )b
a
Q r x dx
The Definite Integral As a Total
Ex. If at time t minutes you are traveling at a rate of v(t) feet per minute, then the total distance traveled in feet from minute 2 to minute 10 is given by
10
2
Total change in distance ( )v t dt
Area Under a Graph
a b
Idea: To find the exact area under the graph of a function.
( )y f x
Method: Use an infinite number of rectangles of equal width and compute their area with a limit.
Width:b a
xn
(n rect.)
Approximating AreaApproximate the area under the graph of
using n = 4.
2( ) 2 on 0,2f x x
0 1 2 3( ) ( ) ( ) ( )A x f x f x f x f x
1 1 30 1
2 2 2A f f f f
1 1 9 70 2
2 2 2 2A
Area Under a Graph
a b
( )y f x
f continuous, nonnegative on [a, b]. The area is
1
0
Area limn
kn
k
f x x
( )b
af x dx
Geometric Interpretation
(All Functions)
( )b
af x dx Area of R1 – Area of R2 + Area of
R3
a b
( )y f xR1
R2
R3
Area Using Geometry
Ex. Use geometry to compute the integral
5
1
1x dx
Area = 2
5
1
1 4 2 2x dx
Area =4
Fundamental Theorem of Calculus
Let f be a continuous function on [a, b].
2. If F is any continuous antiderivative of f and is defined on [a, b], then
( ) ( ) ( )b
af x dx F b F a
1. If ( ) ( ) , then ( ) ( ). x
a
A x f t dt A x f x
The Fundamental Theorem of Calculus
Ex. 3 4 If ( ) 5 , find ( ). x
a
A x t tdt A x
3 4 ( ) 5A x x x
Evaluating the Definite Integral
Ex. Calculate
5
1
12 1x dx
x
55 2
1 1
12 1 lnx dx x x x
x
2 25 ln 5 5 1 ln1 1
28 ln 5 26.39056
Substitution for Definite Integrals
Ex. Calculate 1 1/ 22
02 3x x dx
2let 3u x x
then 2
dudx
x
1 41/ 22 1/ 2
0 02 3x x x dx u du
43/ 2
0
2
3u
16
3
Notice limits change
Computing Area Ex. Find the area enclosed by the x-axis, the vertical lines x = 0, x = 2 and
the graph of
23
02x dx
Gives the area since 2x3 is nonnegative on [0, 2].
22
3 4
00
12
2x dx x 4 41 1
2 02 2
8
Antiderivative Fund. Thm. of Calculus
22 .y x
Applications of Definite Integrals
Area of between a Curve and the x-axis
b
a
n
ii
x
dxxf
xzf
PQRSofarea
)(
)(lim1
0
0
)(lim1
0
n
ii
xxzf
PQRSofarea
dxxf
dxxf
b
a
b
a
)(
|)(|
1
1
b
c
c
a
b
c
c
a
b
a
dxxfdxxf
dxxfdxxf
xf
areashadedtheofareathe
)()(
)()(
)(
22
22
2
Area of between a Curve and the y-axis
d
c
n
ii
y
dyyg
yzg
PQRSofarea
)(
)(lim1
0
0
)(lim1
0
n
ii
yyzg
PQRSofarea
dyyg
dyyg
d
c
d
c
)(
|)(|
1
1
d
e
e
c
d
e
e
c
d
c
dyygdyyg
dyygdyyg
yg
areashadedtheofareathe
)()(
)()(
)(
22
22
2
Area between Two Curves
b
adxyy
Aareaofareashaded
)( 21
d
cdyxx
Aareaofareashaded
)( 21
Volume of Solid of Revolution about the Coordinate Axes
x
y
0
axis of revolutionsphere
semi-circle
Volume of Solid of Revolution about the Coordinate Axes
x
y
0
axis of revolution
rectangle
cylinder
cylinder
made up of many many different sizes of cylinder
Consider a vertical rectangular stripe on the interval [xi-1, xi] with thickness x, a thin cylinder is formed. This cylinder has radius yi and height x. From the formula for the volume of a cylinder, we have
xyV ii 2
xyV ii 2
n
ii
x
n
ii
x
xy
VV
1
2
0
10
lim
lim
b
adxyV 2
d
cdyxV 2
Volume of Hollow Solid of Revolution about the Coordinate Axes
dxyyVb
a)( 2
22
1
hollow
dyxxVd
c)( 2
22
1
hollow
Differential Equations
• Adifferential equation is an equation involving an unknown function and its derivatives
Differential Equations
• A differential equation is an ordinary differential equation if the unknown function depends on only one independent variable.
• If the unknown function depends on two or more independent variables, the differential equation is a partial differential equation
Standard and Differential Forms
• Standard form for a first-order differential equation in the unknown function y(x) is
• where the y ′ derivative appears only on the left side. Many, but not all, first-order differential equations can be written in standard form by algebraically solving for y ′ and then setting
f (x,y) equal to the right side of the resulting equation M(x, y)dx + N(x, y)dy = 0
' ( , )y f x y
First Order Linear Equations• A first order linear differential equation has the following form:
• The general solution is given by
• Where
• called the integrating factor. If an initial condition is given, use it to find the constant C.
Here are some practical steps to follow• If the differential equation is given as
• rewrite it in the form
• where
Here are some practical steps to follow• Find the integrating factor
• Evaluate the integral
• Write down the general solution
• If you are given an IVP, use the initial condition to find the constant C.
.
Separable Equations• The differential equation of the form
• is called separable, if f(x,y) = h(x) g(y); that is
• In order to solve it, perform the following steps:• Solve the equation g(y) = 0, which gives the constant solutions
of (S);
Separable Equations• Rewrite the equation (S) as
• and, then, integrate
• to obtain
Separable Equations• Write down all the solutions; the constant ones
obtained from (1) and the ones given in (2);• If you are given an IVP, use the initial condition to
find the particular solution. Note that it may happen that the particular solution is one of the constant solutions given in (1). This is why Step 3 is important.
Modeling via Differential Equations
• One of the most difficult problems that a scientist deals with in his everyday research is: "How do I translate a physical phenomenon into a set of equations which describes it?'‘
• It is usually impossible to describe a phenomenon totally, so one usually strives for a set of equations which describes the physical system approximately and adequately.
Modeling via Differential Equations• In general, once we have built a set of equations, we
compare the data generated by the equations with real data collected from the system (by measurement).
• If the two sets of data "agree'' (or are close), then we gain confidence that the set of equations will lead to a good description of the real-world system.
• For example, we may use the equations to make predictions about the long-term behavior of the system.
Modeling via Differential Equations
• It is also important to keep in mind that the set of equations stays only "valid" as long as the two sets of data are close.
• If a prediction from the equations leads to some conclusions which are by no means close to the real-world future behavior, then we should modify and "correct" the underlying equations.
• As you can see, the problem of generating "good" equations is not an easy exercise.
• Note that the set of equations is called a Model for the system
How do we build a Model?
• Clearly state the assumptions on which the model will be based. These assumptions should describe the relationships among the quantities to be studied.
• Completely describe the parameters and variables to be used in the model.
• Use the assumptions (from Step 1) to derive mathematical equations relating the parameters and variables (from Step 2).
Exponential Growth – Population
Let P(t) be a quantity that increases with time t and the rate of increase is proportional to the same quantity P as follows
d P / d t = k P
Where d p / d t is the first derivative of P, k > 0 and t is the time. The solution to the above first order differential equation is given by
P(t) = A ek t
Where A is a constant not equal to 0.
Exponential Growth – Population
If P = P0 at t = 0, then P0 = A e0
which gives A = P0
The final form of the solution is given by
P(t) = P0 ek t
Assuming P0 is positive and since k is positive, P(t) is an increasing exponential. d P / d t = k P is also called an exponential growth model.
Exponential Decay - Radioactive Mate Let M(t) be the amount of a product that decreases with time t and the rate of decrease is proportional to the amount M as follows
d M / d t = - k M
where d M / d t is the first derivative of M, k > 0 and t is the time. Solve the above first order differential equation to obtain
M(t) = A e- k t
where A is non zero constant.
Exponential Decay - Radioactive Mate
If we assume that M = M0 at t = 0, then
M0 = A e0
which gives A = M0
The solution may be written as follows
M(t) = M0e- k t
Assuming M0 is positive and since k is positive, M(t) is an decreasing exponential. d M / d t = - k M is also called an exponential decay model.
Falling Object• An object is dropped from a height at time t = 0. If h(t) is the height
of the object at time t, a(t) the acceleration and v(t) the velocity. The relationships between a, v and h are as follows:
a(t) = dv / dt , v(t) = dh / dt.
For a falling object, a(t) is constant and is equal to g = -9.8 m/s.
Combining the above differential equations, we can easily deduce the following equation
d2h / dt2 = g
Falling Object
• Integrate both sides of the above equation to obtain
dh / dt = g t + v0
Integrate one more time to obtain
h(t) = (1/2) g t + v0 t + h0
The above equation describes the height of a falling object, from an initial height h0 at an initial velocity v0, as a function of time.
Newton's Law of Cooling It is a model that describes, mathematically, the change in temperature of an object in a given environment. The law states that the rate of change (in time) of the temperature is proportional to the difference between the temperature T of the object and the temperature Te of the environment surrounding the object.
d T / d t = - k (T - Te)
Let x = T - Te so that dx / dt = dT / dt
Using the above change of variable, the above differential equation becomes
d x / d t = - k x
Newton's Law of CoolingThe solution to the above differential equation is given by
x = A e - k t substitute x by T - Te
T - Te = A e - k t Assume that at t = 0 the temperature T = To
To - Te = A e 0 which gives A = To - Te The final expression for T(t) i given by
T(t) = Te + (To - Te) e - k t
This last expression shows how the temperature T
of the object changes with time.
Dilution Problems
Consider a tank which initially holds Vo gal of brine that contains a lb of salt. Another solution, containing b lb of salt per gallon, is poured into the tank at the rate of e gal/min while simultaneously, the well-stirred solution leaves the tank at the rate of f gal/min (see Figure).
The problem is to find the amount of salt in the tank at any time t.
Dilution Problems
• Let Q denote the amount (in pounds) of salt in the tank at any time.
• The time rate of change of Q, dQ/ dt, equals the rate at which salt enters the tank minus the rate at which salt leaves the tank.
• Salt enters the tank at the rate of be lb/min. To determine the rate at which salt leaves the tank, we first calculate the volume of brine in the tank at any time t, which is the initial volume Vo plus the volume of brine added et minus the volume of brine removed ft.
• Thus, the volume of brine at any time is
Vo + et − ft
• The concentration of salt in the tank at any time is Q / (V0 + et − ft)• from which it follows that salt leaves the tank at the rate of
Practice 1
• A bacteria culture is known to grow at a rate proportional to the amount present. After one hour, 1000 strands of the bacteria originally in the culture. and after four hours, 3000 strands. Find
(a) an expression for the approximate number of strands of the bacteria present in the culture at any time t and
(b) the approximate number of strands of the bacteria originally in the culture.
Practice 2
• A tank initially holds 100 gal of a brine solution containing 20 lb of salt. At t = 0, fresh water is poured into the tank at the rate of 5 gal/min, while the well-stirred mixture leaves the tank at the same rate.
• Find the amount of salt in the tank at any time t.
