1 §11.6 related rates §11.7 elasticity of demand the student will be able to solve problems...

1

§11.6 Related Rates §11.7 Elasticity of Demand

The student will be able to solve problems involving

■ Implicit Differentiation

■ Related rate problems and applications.

■ Relative rate of change, and

■ Elasticity of demand

Function Review and New Notation

So far, the equation of a curve has been specified in the form y = x2 – 5x or f (x) = x2 – 5x (for example).

This is called the explicit form. y is given as a function of x.

However, graphs can also be specified by equations of the form F(x, y) = 0, such as

F(x, y) = x2 + 4xy - 3y2 +7.

This is called the implicit form. You may or may not be able to solve for y.

Explicit and Implicit Differentiation

Consider the equation y = x2 – 5x.

To compute the equation of a tangent line, we can use the derivative y’ = 2x – 5. This is called explicit differentiation.

We can also rewrite the original equation as

F(x, y) = x2 – 5x – y = 0

and calculate the derivative of y from that. This is called implicit differentiation.

4

Example 1

Consider the equation x2 – y – 5x = 0.

We will now differentiate both sides of the equation with respect to x, and keep in mind that y is supposed to be a function of x.

This is the same answer we got by explicit differentiation on the previous slide.

52'

052

052

xydx

dydx

dyx

dx

dxyx

dx

d

5

Example 2

Consider x2 – 3xy + 4y = 0 and differentiate implicitly.

6

Example 2

0432

dx

dy

dx

dxy

dx

dx

dx

d

Consider x2 – 3xy + 4y = 0 and differentiate implicitly.

Solve for y’: yxyx 32'43

43

32'

x

yxy

Notice we used the product rule for the xy term.

0'43'32 yyyxx

7

Example 3

Consider x2 – 3xy + 4y = 0. Find the equation of the tangent at (1, -1).

Solution:1. Confirm that (1, -1) is a point on the graph.

2. Use the derivative from example 2 to find the slope of the tangent.

3. Use the point slope formula for the tangent.

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Example 3

Consider x2 – 3xy + 4y = 0. Find the equation of the tangent at (1, -1).

Solution:1. Confirm that (1, -1) is a point on the graph.

12 – 31(- 1) + 4(-1) = 1 + 3 – 4 = 02. Use the derivative from example 2 to find the slope of the tangent.

3. Use the point slope formula for the tangent.

51

5

413

1312

m

45

)1(5)1(

xy

xy

9

Example 3 (continued)

This problem can also be done with the graphing calculator by solving the equation for y and using the draw tangent subroutine.The equation solved for y is

43

2

x

xy

10

Example 4

Consider xex + ln y – 3y = 0 and differentiate implicitly.

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Example 4

03lndx

dy

dx

dy

dx

dxe

dx

d x

0'3'1

yyy

exe xx

Consider xex + ln y + 3y = 0 and differentiate implicitly.

Solve for y’:Notice we used both the product rule (for the xex term) and the chain rule (for the ln y term)

y

eexyeexyy

y

xxxx

13

'or'3'1

Notes

Why are we interested in implicit differentiation? Why don’t we just solve for y in terms of x and differentiate directly? The answer is that there are many equations of the form F(x, y) = 0 that are either difficult or impossible to solve for y explicitly in terms of x, so to find y’ under these conditions, we differentiate implicitly. Also, observe that:

1and' xdx

dyy

dx

d

13

Related Rate (11.6) Introduction

Related rate problems involve three variables: an independent variable (often t = time), and two dependent variables.

The goal is to find a formula for the rate of change of one of the independent variables in terms of the rate of change of the other one.

These problems are solved by using a relationship between the variables, differentiating it, and solving for the term you want.

14

Example 1

A weather balloon is rising vertically at the rate of 5 meters per second. An observer is standing on the ground 300 meters from the point where the balloon was released. At what rate is the distance between the observer and the balloon changing when the balloon is 400 meters high?

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Example 1

A weather balloon is rising vertically at the rate of 5 meters per second. An observer is standing on the ground 300 meters from the point where the balloon was released. At what rate is the distance between the observer and the balloon changing when the balloon is 400 meters high?

y

300

x

Solution: Make a drawing.

The independent variable is time t. Which quantities that change with time are mentioned in the problem?

We use x = distance from observer to balloon, and y = height of balloon.

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dy/dt is given as 5 meters per second. dx/dt is the unknown.

We need a relationship between x and y:

3002 + y2 = x2 (Pythagoras)

Differentiate the equation with respect to t:

We are looking for dx/dt, so we solve for that:

Example 1

dt

dxx

dt

dyy 220

y

300

x

dt

dy

x

y

dt

dx

17

Now we need values for x, y and dy/dt. Go back to the problem statement:

y = 400

dy/dt = 5

dx/dt = (400/500)5 = 4 m/sec

Example 1

dt

dy

x

y

dt

dx

y

300

x500300 22 yx

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Solving Related Rate Problems

■ Step 1. Make a sketch.

■ Step 2. Identify all variables, including those that are given and those to be found.

■ Step 3. Express all rates as derivatives.

■ Step 4. Find an equation connecting variables.

■ Step 5. Differentiate this equation.

■ Step 6. Solve for the derivative that will give the unknown rate.

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Example 2

A rock is thrown into a still pond and causes a circular ripple. If the radius of the ripple is increasing at 2 feet per second, how fast is the area changing when the radius is 10 feet? [Use A = R2 ]

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Example 2

A rock is thrown into a still pond and causes a circular ripple. If the radius of the ripple is increasing at 2 feet per second, how fast is the area changing when the radius is 10 feet? [Use A = R2 ]

Solution: Make a drawing. Let R = radius, A = area.

Note dR/dt = 2 and R = 10 are given. Find dA/dt.

Differentiate A = R2. R

dt

dRR

dt

dA2 402102

dt

dA

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Related Rates in Business

Suppose that for a company manufacturing transistor radios, the cost and revenue equations are given by

C = 5,000 + 2x

and

R = 10x – 0.001x2,

where the production output in 1 week is x radios.

If production is increasing at the rate of 500 radios per week when production is 2,000 radios, find the rate of increase in (a) Cost (b) Revenue

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Related Rates in Business(continued)

Solution:

These are really two related rates problems, one involving C, x and time t, and one involving R, x, and t.

Differentiate the equations for C and R with respect to time.(a)

Cost is increasing at the rate of $1,000 per week.

100050022

25000

dt

dx

dt

dC

xC

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Related Rates in Business(continued)

(b)

Revenue is increasing at the rate of $3,000 per week.

3000

5002000002.050010

002.010

001.010 2

dt

dxx

dt

dx

dt

dR

xxR

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Objectives for Section 4.7 Elasticity of Demand

The student will be able to solve problems involving

■ Relative rate of change, and

■ Elasticity of demand

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Remember that f ’(x) represents the rate of change of f (x).

The relative rate of change is defined as

By the chain rule, this equals the derivative of the logarithm of f (x):

The percentage rate of change of a function f (x) is

Relative and Percentage Rates of Change

)(

)('

xf

xf

)(ln100)(

)('100 xf

dx

d

xf

xf

)(ln)(

)('xf

dx

d

xf

xf

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Example 1

Find the relative rate of change of

f (x) = 50x – 0.01x2

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Example 1(continued)

Find the relative rate of change of

f (x) = 50x – 0.01x2

Solution: The derivative of

ln (50x – 0.01x2)

is xxx

02.05001.050

12

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Example 2A model for the real GDP (gross domestic product expressed in billions of 1996 dollars) from 1995 to 2002 is given by

f (t) = 300t + 6,000,

where t is years since 1990.

Find the percentage rate of change of f (t) for 5 < t < 12.

Solution: If p(t) is the percentage rate of change of f (t), then

The percentage rate of change in 1995 (t = 5) is 4%

( ) 100 ln(300 6,000)

30,000 100

300 6,000 20

dp t t

dx

t t

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Elasticity of Demand

Elasticity of demand describes how a change in the price of a product affects the demand. Assume that

f (p) describes the demand at price p. Then we define

relative rate of change in demandElasticity of Demand = relative rate of change in price

Notice the minus sign. f and p are always positive, but f ’ is negative (higher cost means less demand). The minus sign makes the quantity come out positive.

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Elasticity of Demand Formula

Elasticity of Demand = relative rate of change of demand

relative rate of change of price

'( )ln ( )

'( )( )1 ( )ln

d f pf p

pf pdp f pd f ppdp p

Given a price-demand equation x = f (p) (that is, we can sell amount x of product at price p), the elasticity of demand is given by the formula

)(

)(')(

pf

pfppE

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Elasticity of Demand, Interpretation

E(p) Demand Interpretation

E(p) < 1 Inelastic Demand is not sensitive to changes in price. A change in price produces a small change in demand.

E(p) > 1 Elastic Demand is sensitive to changes in price. A change in price produces a large change in demand.

E(p) = 1 Unit A change in price produces the same change in demand.

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Example

For the price-demand equation

x = f (p) = 1875 - p2,

determine whether demand is elastic, inelastic, or unit for p = 15, 25, and 40.

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Example (continued)

For the price-demand equation

x = f (p) = 1875 - p2,

determine whether demand is elastic, inelastic, or unit for p = 15, 25, and 40.

2

2

2 1875

2

1875

2

)(

)(')(

p

p

p

pp

pf

pfppE

If p = 15, then E(15) = 0.27 < 1; demand is inelastic

If p = 25, then E(25) = 1; demand has unit elasticity

If p = 40, then E(40) = 11.64; demand is elastic

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Revenue and Elasticity of Demand

■ If demand is inelastic, then consumers will tend to continue to buy even if there is a price increase, so a price increase will increase revenue and a price decrease will decrease revenue.

■ If demand is elastic, then consumers will be more likely to cut back on purchases if there is a price increase. This means a price increase will decrease revenue and a price decrease will increase revenue.

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Elasticity of Demand for Different Products

Different products have different elasticities. If there are close substitutes for a product, or if the product is a luxury rather than a necessity, the demand tends to be elastic. Examples of products with high elasticities are jewelry, furs, or furniture.

On the other hand, if there are no close substitutes or the product is a necessity, the demand tends to be inelastic. Examples of products with low elasticities are milk, sugar, and lightbulbs.

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Summary

The relative rate of change of a function f (x) is

)(ln)(

)('xf

dx

d

xf

xf

The percentage rate of change of a function f (x) is

)(ln100)(

)('100 xf

dx

d

xf

xf

Elasticity of demand is

)(

)(')(

pf

pfppE

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Chapter Review

4.1. The Constant e and Continuous Compound Interest

– The number e is defined as either one of the limits

– If the number of compounding periods in one year is increased without limit, we obtain the compound interest formula A = Pert, where P = principal, r = annual interest rate compounded continuously, t = time in years, and A = amount at time t.

n

n ne

11lim s

sse

1

01lim

4.2. Derivatives of Exponential and Logarithmic Functions

– For b > 0, b 1

– The change of base formulas allow conversion from base e to any base b > 0, b 1: bx = ex ln b, logb x = ln x/ln b.

xx eedx

d

xx

dx

d 1ln

ln

1 1log ( )

ln

x x

b

db b b

dx

dx

dx b x

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• 4.3. Derivatives of Products and Quotients

– Product Rule: If f (x) = F(x) S(x), then

– Quotient Rule: If f (x) = T (x) / B(x), then

• 4.4. Chain Rule

– If m(x) = f [g(x)], then m’(x) = f ’[g(x)] g’(x)

Sdx

dF

dx

dSFxf )('

2)]([

)(')()(')()('

xB

xBxTxTxBxf

• 4.4. Chain Rule (continued)

– A special case of the chain rule is the general power rule:

– Other special cases of the chain rule are the following general derivative rules:

)('1 xfxfnxfdx

d nn

)(')(

1)]([ln xf

xfxf

dx

d

)(')()( xfeedx

d xfxf

• 4.5. Implicit Differentiation

– If y = y(x) is a function defined by an equation of the form F(x, y) = 0, we can use implicit differentiation to find y’ in terms of x, y.

4.6. Related Rates

– If x and y represent quantities that are changing with respect to time and are related by an equation of the form F(x, y) = 0, then implicit differentiation produces an equation that relates x, y, dy/dt and dx/dt. Problems of this type are called related rates problems.

• 4.7. Elasticity of Demand

– The relative rate of change, or the logarithmic derivative, of a function f (x) is f ’(x) / f (x), and the percentage rate of change is 100 (f ’(x) / f (x).

– If price and demand are related by x = f (p), then the elasticity of demand is given by

– Demand is inelastic if 0 < E(p) < 1. (Demand is not sensitive to changes in price). Demand is elastic if E(p) > 1. (Demand is sensitive to changes in price).

priceofchangeofraterelative

demandofchangeofraterelative

)(

)(')(

pf

pfppE