2. Economic Applications of Single-Variable Calculus

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2. Economic Applications of Single-Variable Calculus. Derivative Origins Single Variable Derivatives Economic uses of Derivatives. 2. Economic Applications of Single-Variable Calculus. 2.1 Derivatives of Single-Variable Functions 2.2 Applications using Derivatives.

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2. Economic Applications of Single-Variable Calculus
• Derivative Origins
• Single Variable Derivatives
• Economic uses of Derivatives
2. Economic Applications of Single-Variable Calculus

2.1 Derivatives of Single-Variable Functions

2.2 Applications using Derivatives

2. Economic Applications of Single-Variable Calculus

In economics, derivatives are used in various ways:

• Marginal amounts (slope)
• Maximization
• Minimization
• Sketching Graphs
• Estimation
2.1 Derivatives of Single-Variable Functions

Slope:

-consider the following graph

-the curved movement between A and B is approximated by the red straight line

2.1 – Quadratic example

Slope Approximation:

Slope = a decrease of 65 units over 4 time periods, or an average decrease of 16.25

A

B

2.1 – Which is the slope at point B?

Slope = rise/run

=Δq/ Δp = (q1-q0)/(p1-p0)

Slope AB

=(q1-q0)/(p1-p0)

=(35-100)/(3-1)

=-65/2

=-32.5

Slope BC

=(q1-q0)/(p1-p0)

=(20-35)/(5-3)

=-15/2

=-7.5

A

B

C

2.1 Derivatives of Single-Variable Functions

-the slopes of these secants (AB and BC) reveal the rate of change of q in response to a change in p

-these slopes change as you move along the curve

-in order to find the slope AT B, one must use an INSTANTANEOUS SLOPE

-slope of a tangent line

2.1 – Tangents

A

B

C

The green tangent line represents the instantaneous slope

2.1 Instantaneous Slope

To calculate an instantaneous slope (using calculus), you need:

• A function
• A continuous function
• A smooth continuous function
2.1.1 – A Function

Definitions:

-A function is any rule that assigns a maximum and minimum of one value to a range of another value

-ie y=f(x) assigns one value (y) to each x

-note that the same y can apply to many x’s, but each x has only one y -ie: y=x1/2 is not a function

x = argument of the function (domain of function)

f(x) or y = range of function

Function:

y=0+2sin(2pi*x/14)+2cos(2pi*x/14)

4

3

2

1

0

x

y

-1

1

3

5

7

9

11

13

15

17

19

-2

-3

-4

x

Each X

Has 1 Y

Not a Function:

Here each x Corresponds to 2 y values

Often called the straight line test

2.1.1 – Continuous
• -if a function f(x) draws close to one finite number L for all values of x as x draws closer to but does not equal a, we say:
• lim f(x) = Lx-> a
• A function is continuous iff (if and only iff)
• f(x) exists at x=aii) Lim f(x) exists x->aiii) Lim f(x) = f(a) x->a
2.1.1 – Limits and Continuity
• In other words:
• The point must exist
• Points before and after must exist
• These points must all be joined
• Or simply:
• The graph can be drawn without lifting one’s pencil.
2.1.2 Smooth

-in order for a derivative to exist, a function must be continuous and “smooth” (have only one tangent)

2.1.2 Derivatives
• -if a derivative exists, it can be expressed in many different forms:
• dy/dx
• df(x)/dx
• f ’(x)
• Fx(x)
• y’
2.1.2 Derivatives and Limits

-a derivative (instantaneous slope) is derived using limits:

This method is known as differentiation by first principles, and determines the slope between A and B as AB collapses to a point (A)

B

f(x+h)

A

f(x)

h

x+h

x

2.1.2 Rules of Derivatives

-although first principles always work, the following rules are more economical:

1) Constant Rule

If f(x)=k (k is a constant),f ‘(x) = 0

2) General Rule

If f(x) = ax+b (a and b are constants)f ‘ (x) = a

2.1.2 Examples of Derivatives

1) Constant Rule

If f(x)=27 f ‘(x) = 0

2) General Rule

If f(x) = 3x+12 f ‘(x) = 3

2.1.2 Rules of Derivatives

3) Power Rule

If f(x) = kxn,f ‘(x) = nkxn-1

If f(x) = g(x) + h(x),f ‘(x) = g’(x) + h’(x)

2.1.2 Examples of Derivatives

3) Power Rule

If f(x) = -9x7,f ‘(x) = 7(-9)x7-1

=-63x6

If f(x) = 32x -9x2f ‘(x) = 32-18x

2.1.2 Rules of Derivatives

5) Product Rule

If f(x) =g(x)h(x),f ‘(x) = g’(x)h(x) + h’(x)g(x)-order doesn’t matter

6) Quotient Rule

If f(x) =g(x)/h(x),f ‘(x) = {g’(x)h(x)-h’(x)g(x)}/{h(x)2}-order matters-derived from product rule (implicit derivative)

2.1.2 Rules of Derivatives

5) Product Rule

If f(x) =(12x+6)x3f ‘(x) = 12x3 + (12x+6)3x2 = 48x3 + 18x26) Quotient Rule

If f(x) =(12x+1)/x2f ‘(x) = {12x2 – (12x+1)2x}/x4= [-12x2-2x]/x4 = [-12x-2]/x3

2.1.2 Rules of Derivatives

7) Power Function Rule

If f(x) = [g(x)]n,f ‘(x) = n[g(x)]n-1g’(x)-work from the outside in-special case of the chain rule

8) Chain Rule

If f(x) = f(g(x)), let y=f(u) and u=g(x), thendy/dx = dy/du X du/dx

2.1.2 Rules of Derivatives

7) Power Function Rule

If f(x) = [3x+12]4,f ‘(x) = 4[3x+12]33 = 12[3x+12]3

8) Chain Rule

If f(x) = (6x2+2x)3 , let y=u3 and u=6x2+2x, dy/dx = dy/du X du/dx = 3u2(12x+2) = 3(6x2+2x)2(12x+2)

2.1.2 More Exciting Derivatives

1) Inverses

If f(x) = 1/x= x-1,f ‘(x) = -x-2=-1/x2

1b) Inverses and the Chain Rule

If f(x) = 1/g(x)= g(x)-1,f ‘(x) = -g(x)-2g’(x)=-1/g(x)2g’(x)

2.1.2 – More Exciting Derivatives

2) Natural Logs

If y=ln(x),

y’ = 1/x

-chain rule may apply

If y=ln(x2)

y’ = (1/x2)2x = 2/x

2.1.2 – More Exciting Derivatives

3) Trig. Functions

If y = sin (x),

y’ = cos(x)

If y = cos(x)

y’ = -sin(x)

-Use graphs as reminders

2.1.2 – More Derivatives

Reminder: derivatives reflect slope:

2.1.2 – More Derivatives

3b) Trig. Functions – Chain Rule

If y = sin2 (3x+2),

y’= 2sin(3x+2)cos(3x+2)3

Exercises:

y=ln(2sin(x) -2cos2(x-1/x))

y=sin3(3x+2)ln(4x-7/x3)5

y=ln([3x+4]sin(x)) / cos(12xln(x))

2.1.2 – More Derivatives

4) Exponents

If y = bx

y’ = bxln(b)

Therefore

If y = ex

y’ = ex

2.1.2 – More Derivatives

4b) Exponents and chain rule

If y = bkx

y’ = bkxln(b)k

Or more generally:

If y = bg(x)

y’ = bg(x)ln(b) X dg(x)/dx

2.1.2 – More Derivatives

4b) Exponents and chain rule

If y = 52x

y’ = 52xln(5)2

Or more complicated:

If y = 5sin(x)

y’ = 5sin(x)ln(5) * cos(x)

2.1.2.1 – Higher Order Derivatives

-Until this point, we have concentrated on first order derivates (y’), which show us the slope of a graph

-Higher-order derivates are also useful

-Second order derivatives measure the instantaneous change in y’, or the slope of the slope

-or the change in the slope:

2.1.2.1 Second Derivatives

Here the slopeincreases as tincreases, transitioningfrom a negativeslope to a positiveslope.

A second derivative would be positive, and confirm a minimum point on the graph.

2.1.2.1 Second Derivatives

Here, the slope moves from positive to negative, decreasingover time.

A second derivative would be negative and indicate a maximum point on the graph.

2.1.2.1 – Second Order Derivatives

To take a second order derivative:

• Apply derivative rules to a function
• Apply derivative rules to the answer to (1)

Second order differentiation can be shown a variety of ways:

• d2y/dx2 b) d2f(x)/dx2
• f ’’(x) d) fxx(x)

e) y’’

2.1.2.1 – Second Derivative Examples

y=12x3+2x+11

y’=36x2+2

y’’=72x

y=sin(x2)

y’=cos(x2)2x

y’’=-sin(x2)2x(2x)+cos(x2)2

y’’’=-cos(x2)2x(4x2)-sin(x2)8x-sin(x2)2x(2)

=-cos(x2)8x3-sin(x2)12x

2.1.2.2 – Implicit Differentiation

So far we’ve examined cases where our function is expressed:

y=f(x) ie: y=7x+9x2-14

Yet often equations are expressed:

14=7x+9x2-y

Which requires implicit differentiation.

-In this case, y can be isolated. Often, this is not the case

2.1.2.2 – Implicit Differentiation Rules
• Take the derivative of EACH term on both sides.
• Differentiate y as you would x, except that every time you differentiate y, you obtain dy/dx (or y’)

Ie: 14=7x+9x2-y

d(14)/dx=d(7x)/dx+d(9x2)/dx-dy/dx

0 = 7 + 18x – y’

y’=7+18x

2.1.2.2 – Implicit Differentiation Examples

Sometimes isolating y’ requires algebra:

xy=15+x

y+xy’=0+1

xy’=1-y

y’=(1-y)/x (this can be simplified further)

= [1-(15+x)/x]/x

= (x-15-x)/x2

=(-15)/x2

2.1.2.2 – Implicit Differentiation Examples

x2-2xy+y2=1

d(x2)/dx+d(2xy)/dx+d(y2)/dx=d1/dx

2x-2y-2xy’+2yy’=0

y’(2y-2x)=2y-2x

y’=(2y-2x) / (2y-2x)

y’=1

2.1.2.2 – Implicit Differentiation Examples

Using the implicit form has advantages:

3x+7y8=18

3+56y7y’=0

56y7y’=-3

y’=-3/56y7

vrs.

y=[(18-3x)/7)1/8

y’=1/8 * [(18-3x)/7)-7/8 * 1/7 * (-3)

Which simplifies to the above.

2.2.1 Derivative Applications - Graphs

Derivatives can be used to sketch functions:

First Derivative:

-First derivative indicates slope

-if y’>0, function slopes upwards

-if y’<0, function slopes downwards

-if y’=0, function is horizontal

-slope may change over time

-doesn’t give shape of graph

2.2.1 Positive Slope Graphs

Linear, Quadratic, and Lin-Log Graphs

2.2.1 Derivative Applications - Graphs

Next, shape/concavity must be determined

Second Derivative:

-Second derivative indicates concavity

-if y’’>0, slope is increasing (convex)

-if y’’<0, slope is decreasing (concave, like a hill or a cave)

-if y’’=0, slope is constant (or an inflection point occurs, see later)

2.2.1 Sample Graphs

x’’= 2, slope is increasing; graph is convex

2.2.1 – Sample Graphs

x’’=-2, slope is decreasing; graph is concave

2.2.1 Derivative Applications - Graphs

Maxima/minima can aid in drawing graphs

Maximum Point:

If 1) f(a)’=0, and

2) f(a)’’<0,

-graph has a maximum point (peak) at x=a

Minimum Point:

If 1) f(a)’=0, and

2) f(a)’’>0,

-graph has a minimum point (valley) at x=a

2.2.1 Sample Graphs

x’’= 2, slope is increasing; graph is convex

x’=-10+2t=0t=5

Minimum

2.2.1 – Sample Graphs

x’’=-2, slope is decreasing; graph is concave

x’=10-2t=0t=5

Maximum

2.2.1 Derivative Applications - Graphs

Inflection Points:

If 1) f(a)’’=0, and

2) the graph is not a straight line

-then an inflection point occurs

-(where the graph switches between convex and concave)

2.2.1 Derivatives and Graphing

Cyclical case:

Maximum (concave)

Inflection

Point

Minimum (convex)

2.2.1 Derivative Applications - Graphs

7 Graphing Steps:

• Evaluate f(x) at extreme points (x=0, ∞, - ∞, or a variety of values)
• Determine where f(x)=0
• Calculate slope: f ’(x) - and determine where it is positive and negative
• Identify possible maximum and minimum co-ordinates where f ‘(x)=0. (Don’t just find the x values)
2.2.1 Derivative Applications - Graphs

7 Graphing Steps:

v) Calculate the second derivative – f ‘’(x) and use it to determine max/min in iv

vi) Using the second derivative, determine the curvature (concave or convex) at other points

vii) Check for inflection points where f ‘’(x)=0

2.2.1 Graphing Example 1

y=(x-5)2-3

• f(0)=22, f(∞)= ∞, f(-∞)=∞
• y=0 when

(x-5)2=3

(x-5) = ± 31/2

x = ± 31/2+5

x = 6.7, 3.3 (x-intercepts)

iii) y’=2(x-5)

y’>0 when x>5

y’<0 when x<5

2.2.1 Graphing Example 1

y=(x-5)2-3

iv) y’=0 when x=5

f(5)=(5-5)2-3=-3

(5,-3) is a potential max/min

v) y’’=2, (5,-3) is a minimum

vi) Function is always positive, it is always convex

vii) y’’ never equals zero

2.2.1 Graphing Example 1

(0,22)

(3.3,0)

(6.7,0)

(5,-3)

2.2.1 Graphing Example 2

y=(x+1)(x-3)=x2-2x-3

• f(0)=-3, f(∞)= ∞, f(-∞)=∞
• y=0 when

(x+1)(x-3)=0

x = 3,-1 (x-intercepts)

iii) y’=2x-2

y’>0 when x>1

y’<0 when x<1

2.2.1 Graphing Example 2

y=(x+1)(x-3)=x2-2x-3

iv) y’=0 when x=1

f(1)=12-2(1)-3=-4

(1,-4) is a potential max/min

v) y’’=2, x=1 is a minimum

vi) Function is always positive, it is always convex

vii) y’’ never equals zero

2.2.2 Optimization

Some claim economists have 3 jobs:

• Analyze what has happened (past)
• Describe the current economy (present)
• Advise on future decisions (future)

For #3, an economist must first calculate the best possible result.

2.2.2 Optimization

Optimization falls into two categories:

• Maximization (ie: production, profits, utility, happiness, grades, health, employment, etc.)
• Minimization (ie: costs, pollution, disutility, unemployment, sickness, homework, etc.)
2.2.2 Optimizing in 3 Steps

There are three steps for optimization:

• FIRST ORDER CONDITION (FOC) Find where f’(x)=0. These are potential maxima/minima.
• SECOND ORDER CONDITION (SOC) Evaluate f’’(x) at your potential maxima/minima. This determines if (1)’s solutions are maxima/minima/inflection points
• Co-Ordinates Obtain the co-ordinates of your maxima/minima
2.2.2 Optimizing Example 1

Cooking is tricky – too long spent cooking, and it burns, too little time spent cooking – and some of it is raw and inedible.

Production of oatmeal is expressed as:

Let x=15+10t-t2

x = bowls of oatmeal

t = 5 minute intervals of time

Maximize Oatmeal Production

2.2.2 Optimous Oatmeal

Let x=15+10t-t2

FOC:

x’ = 10-2t = 0

10 = 2t

• = t

SOC:

x’’ = -2

x’’ < 0, concave, maximum

2.2.2 Optimous Oatmeal

Let x=15+10t-t2

Co-ordinates:

x(5) = 15+10(5)-52

x(5) = 15+50-25

= 40

Gourmet oatmeal production is maximized at 40 bowls when 25 minutes (5X5) are spent cooking (All else held equal).

2.2.2 – Oatmeal for everyone

Production is maximized at (5,40).

Maximum

2.2.2 Marriage and Motorcycles

Steve wants to buy a new motorbike. Being a married man however, he knows that his utility is directly tied to his wife’s opinion of the idea. Furthermore, he knows it’s best to bring it up to Denise (his wife) when she’s at her weakest.

Denise’s daily opposition to a motorcycle is expressed as x=-cos(tπ/12)

Where t = hour of the day (0-24)

When should Steve ask Denise?

2.2.2 M & M

Let x=-cos(tπ/12)

FOC:

x’ = sin(tπ/12)π/12 = 0

0 = sin(tπ/12)

This occurs when

t π/12 = 0, π, 2π

t = 0, 12, 24

2 possible mimima (0=24 on the clock)

2.2.2 Early Bird Gets the Motorcycle

Let x=-cos(tπ/12)

SOC:

x’’ = cos(tπ/12)(π/12)2

x’’ > 0 when 0≤t<6, 18<t≤24

-convex, min (0, 24 are acceptable)

x’’ < 0 when 6<t<18

-concave, max (12 is out)

2.2.2 Early Bird Gets the Motorcycle

Let x=-cos(tπ/12)

Results:

x(0)=-1

x(24)=-1

Denise has her least resistance (of -1) at first thing in the morning or last thing at night. Steve’s best move is to bring up the motorcycle when Denise is tired or asleep.

2.2.2 Necessary and Sufficient

The FOC provides a NECESSARY condition for a maximum or minimum.

The FOC is not a SUFFICIENT condition for a maximum or minimum.

The FOC and SOC together are NECESSARY AND SUFFICIENT conditions for a max. or min.

2.2.2 Marginal Concepts

Marginal Profit = Change in profit from the sale/production of one extra unit.

MP=dπ/dq

If Marginal Profit >0, quantity should increase; as the next unit will increase profit.

If Marginal Profit <0, quantity should decrease, as the last unit decreased profit

2.2.2 Marginal Concepts

Quantity is therefore optimized when Marginal Profit=0;

Ie: when dπ/dq=0.

SOC still confirms that this is a maximum (ie: that the previous unit increased profit and the next unit will decrease profit.)

2.2.2 Marginal Concepts

Alternately, remember that

Profit = Total Revenue – Total Costs

Or

π = TR-TC

therefore

Mπ=MR-MC (dπ/dq=dTR/dq-dTC/dq)

So Mπ = 0 is equivalent to saying that

MR-MC =0

MR=MC

2.2.2 Marrigal Example

Your significant other shows you a new, horrible outfit they just bought and asks how they look. You think of 6 possible lies:

• You look amazing
• You have such great taste
• That colour really brings out your eyes
• Neon is in this year
• I should get a matching outfit
• You should wear that to my office party
2.2.2 Marrigal Example

The benefit lying is

TR=2L3+60L

The cost of lying is

TC=21L2+100

Where L = number of lies

How many lies should you tell?

2.2.2 Marrigal Example

Profit=TR-TC

Profit=2L3+60L-[21L2+100]

Profit=2L3-21L2+60L-100

FOC:

Mπ=6L2-42L+60

Mπ=6(L2-7L+10)

2.2.2 Marrigal Example

Mπ=6(L2-7L+10)

Solving using the quadratic formula:

L=-b±(b2-4ac)1/2 / 2a

L=-7±(49-40)1/2/(2)

L=(-7±3)/2 = 2, 5

2.2.2 Marrigal Example

Profit=2L3-21L2+60L-100

Mπ=6L2-42L+60

SOC: π’’=12L-42

π’’(2)=12(2)-42=-18, concave MAX

π’’(5)=12(5)-42=18, convex MIN

You should tell 2 lies

2.2.2 Marrigal Example

Profit=2L3-21L2+60L-100

Profit(2)=2(2)3-21(2)2+60(2)-100

Profit(2)=16-84+120-100

Profit(2)=-48

You are minimizing the damage at -48 by telling two lies.

2.2.2 Constrained Optimization

Thus far we have considered UNCONSTRAINED optimization. (No budget constraint, time constraint, savings constraint, etc.)

We can also deal with CONSTRAINED optimization.

Ie: Maximize utility with respect to a set income.

Maximize income with respect to a 24 hour day.

Maximize lie effectiveness while keeping a straight face.

To use this, we use a Lagrangean.(chapter 4)

2.2.3 Elasticities

We have already seen how the derivative, or the slope, can change as x and y change

-even if a slope is constant, changes can have different impacts at different points

-For example, given a linear demand for Xbox 720’s, a \$100 price increases affects profits differently at different starting prices:

2.2.3 Xbox 720 Example

Price increase from \$0 to \$100

New Income

2.2.3 Xbox 720- Example

Price increase from \$500 to \$600

Old Income

New Income

2.2.3 Elasticities

\$0 to \$100

Old Revenue: \$0

New Revenue: 4.5 million sold X \$100 each

\$450 million (INCREASE)

\$500 to \$600

Old Revenue: 2.5 million sold X \$500 each

\$1.25 Billion

New Revenue: 1.5 million sold X 600 each

\$0.9 Billion (DECREASE)

2.2.3 Elasticities

-to avoid this problem, economists often utilize ELASTICITIES

-elasticities deal with PERCENTAGES and are therefore more useful across a variety of points on a curve

ELASTICITY = a PROPORTIONAL change in y from a PROPORTIONAL change in x

Example: elasticity of demand:

E = Δy/y / Δx/x

= (Δy/Δx) (x/y)

= (dy/dx) (x/y)

2.2.3 Elasticity Example 1

Let y=12x+7

Find elasticities at x=0, 5, and 10

• dy/dx = 12
• f(0) = 12(0) +7 = 7
• f(5) = 12(5) + 7 = 67
• f(10) = 12(10) +7 = 127

Next we apply the formula:

2.2.3 Elasticity Example 1

Let y=12x+7

Find elasticities at (x,y)=(0,7) (5,67) and (10,127)

E = dy/dx * x/y

• E (0)= 12* 0/7 = 0
• E (5)= 12 * 5/67 = 0.90
• E (10)= 12 * 10/127 = 0.94
2.2.3 Elasticity Interpretation

What does an elasticity of 3 mean?

=> for a 1% increase in x (or the independent variable), there will be a 3% increase in y (or the dependent variable)

In our example, a 1% increase in x caused a:

• 0% increase in y
• 0.90% increase in y
• 0.94% increase in y

Question: Is an increase in y good or bad?

2.2.3 Inelastic vrs. Elastic: The Boxer’s or Briefs debate

An elasticity of less than 1 (in absolute terms) is inelastic.

That is, y responds less than x in percentage terms.

An elasticity of greater than 1 (in absolute terms) is elastic.

That is, y responds more than x in percentage terms.

-This has policy implications…

2.2.3 Elastic Xboxes

Let q=5,000,000-5,000p Or

q=5,000-5p

Where q is Xbox’s demanded in 1000’s

p is price of an Xbox

Find elasticities at p=0, 100, and 500

• dq/dp = -5
• Q(0) =5,000
• Q(100)=4,500
• Q(500)=2,500
2.2.3 Elastic Xboxes

Find elasticities at p,q =(0,5000), (100,4500) and (500,2500)

E = dq/dp * p/q

• E = -5* 0/5000 = 0
• E = -5 * 100/4500 = -0.11 (inelastic)
• E = -5 * 500/2500 = -1 (unit elastic)
2.2.3 Elastic Xboxes

From these values, we know that demand for Xboxes 720’s is INELASTIC below \$500 and ELASTIC above \$500

How does this impact profits?

Total Revenue = p*q(p)

dTR/dp = q(p)+p*dq/dp

= q( 1+p/q*dq/dp)

= q (1+ E)

2.2.3 Making Microsoft Money

If E = -1, dTR/dp = 0; a small change in price won’t affect profits

If |E| < 1 (inelastic), dTR/dp>0, small increases in prices increase profits

If |E| > 1 (elastic), dTR/dp<0, small increases in prices decrease profits

Therefore, price increases are revenue enhancing up to a price of \$500.

2.2.3 Elasticity Production Rule

If demand is inelastic

Raise Price

If demand is elastic

Decrease Price

If demand is unit elastic

Price is perfect (usually)

2.2.3 More Elasticity Exercises

Let q = 100-2p

• Find Elasticities at p=5, 20 and 40
• Formulate Economic Advice at these points

Let q = 200+2p-4p2

• Find Elasticities at p=0, 5 and 10
• Formulate Economic Advice at these points
2.2.3 Elastic Logs

The MAIN reason to use logs in economic formulae is to more easily calculate elasticities:

E = dy/dx * x/y

= (1/y) dy/dx (x)

= (dlny/dy) dy/dx (dx/dlnx)

= (dlny/dlnx) dx/dx (dy/dy)

= (dlny/dlnx)

2.2.3 Examples are a log’s best friend

Let ln(q) = -1ln(p)

E = dln(q)/dln(p)

= -1

Hence demand is unit elastic and change in price would not affect total revenue.

2.2.3 Log Elasticity Exercises

Let ln(q) = 100+ln(30/p)

• Find Elasticities at p=5, 10 and 20
• Formulate Economic Advice at these points

Let ln(q) = 1/2 ln(p2)

• Find Elasticities at p=0.5, 1, and 2
2.2.4 – Linear Approximation

Derivatives can be used to approximate a function in a linear fashion:

The point f(x) can be estimated as f(a) plus a distance along the red secant:

f(x)

f(a)

a

x

2.2.4 – Linear Approximation Example

Note that true values are f(3.1)=19.22, f(4)=32

2.2.4 – Quadratic Approximation

Note that the approximated values differ from the true values as we move away from our starting point.

A further estimation would be a quadratic TAYLOR SERIES EXPANSION:

2.2.4 – Quadratic Approximation

Which is obviously more accurate (since it`s exactly correct).

2.2.4 – Quadratic Approximation 2

Approximate instead f(x)=x0.5 at a=4

Since 50.5=2.236, this is a good approximation