Transcript 1-1 Functions - North Carolina Central University

§11.6 Related Rates
§11.7 Elasticity of Demand
The student will be able to solve problems involving
■
■
■
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Implicit Differentiation
Related rate problems and applications.
Relative rate of change, and
Elasticity of demand
1
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.
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.


d 2
d
x  y  5x 
0
dx
dx
dy
This is the same answer we
2x 
5  0
got by explicit
dx
differentiation on the
dy
 y'  2 x  5
previous slide.
dx
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Example 2
Consider x2 – 3xy + 4y = 0 and differentiate
implicitly.
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Example 2
Consider x2 – 3xy + 4y = 0 and differentiate
implicitly.
d 2
d
d
d
x 
3 xy 
4y 
0
dx
dx
dx
dx
2 x  3x y '  3 y  4 y '  0
Solve for y’:
3x  4 y'  2x  3 y
Notice we used the
product rule for the
xy term.
2x  3y
y' 
3x  4
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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.
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 – 31(- 1) + 4(-1) = 1 + 3 – 4 = 0
2. Use the derivative from example 2 to find the slope of
the tangent.
2 1  3   1
5
m
3 1  4

1
 5
3. Use the point slope formula for the tangent.
y  (1)   5 ( x  1)
y  5 x  4
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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
2
x
y
3x  4
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Example 4
Consider xex + ln y – 3y = 0 and differentiate
implicitly.
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Example 4
Consider xex + ln y + 3y = 0 and differentiate
implicitly.
d
d
d
d
x
xe 
ln y 
3y 
0
dx
dx
dx
dx
1
x
x
xe  e  y '3 y '  0
y
Solve for y’:
Notice we used both the product rule (for the xex
term) and the chain rule (for the ln y term)
1
y'  3 y '   x e x  e x
y
or
 x ex  ex
y' 
1
3
y
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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:
d
y  y'
dx
and
d
x 1
dx
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.
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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?
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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?
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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y
x
300
Example 1
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:
dy
dx
0  2 y  2x
dt
dt
x
dx y dy

dt x dt
300
y
We are looking for dx/dt, so we solve for that:
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Example 1
dx y dy

dt
x dt
Now we need values for x, y and dy/dt.
Go back to the problem statement:
y = 400
dy/dt = 5
x  300  y  500
2
2
y
x
dx/dt = (400/500)5 = 4 m/sec
300
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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.
dA
dR
  2R
dt
dt
R
dA
  2 10  2  40 
dt
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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.
C  5000 2 x
(a)
dC
dx
2
 2  500  1000
dt
dt
Cost is increasing at the rate of $1,000 per week.
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Related Rates in Business
(continued)
(b)
R  10x  0.001x 2
dR
dx
dx
 10
 0.002x
dt
dt
dt
 10  500  0.002 200 0 500
 300 0
Revenue is increasing at the rate of $3,000 per week.
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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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Relative and Percentage Rates of Change
Remember that f ’(x) represents the rate of change of
f (x).
f ' ( x)
f ( x)
The relative rate of change is defined as
f ' ( x) d
 ln f ( x)
f ( x) dx
By the chain rule, this equals the derivative of the
logarithm of f (x):
f ' ( x)
d
100 
f ( x)
 100 
dx
ln f ( x)
The percentage rate of change of a function f (x) is
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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
is
ln (50x – 0.01x2)
1
 50  0.02x 
2
50 x  0.01x 
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Example 2
A 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
d
ln(300t  6, 000)
dx
30, 000
100


300t  6, 000 t  20
p (t )  100 
The percentage rate of change in 1995 (t = 5) is 4%
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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
Elasticity of Demand =
relative rate of change in 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
d
f '( p )
ln f ( p )
pf '( p )
dp
f ( p)



d
1
f ( p)
ln p
dp
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
p  f ' ( p)
E ( p)  
f ( p)
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Elasticity of Demand, Interpretation
E(p)
Demand
E(p) < 1
Inelastic
E(p) > 1
Elastic
E(p) = 1
Unit
Interpretation
Demand is not sensitive to changes in
price. A change in price produces a
small change in demand.
Demand is sensitive to changes in price.
A change in price produces a large
change in demand.
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.
p  f ' ( p)
p   2 p 
2 p2
E ( p)  


2
f ( p)
1875  p
1875  p 2
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
f ' ( x)
d

ln f ( x)
f ( x) dx
The percentage rate of change of a function f (x) is
f ' ( x)
d
100 
 100 
ln f ( x)
f ( x)
dx
Elasticity of demand is
p  f ' ( p)
E ( p)  
f ( p)
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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
1

e  lim 1  
n
n  
n
e  lim 1  s 
1
s
s 0
– 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
37
time t.
4.2. Derivatives of Exponential and Logarithmic
Functions
– For b > 0, b  1
d x
e  ex
dx
d
1
ln x 
dx
x
d x
b  b x ln b
dx
d
1 1
log b x 
( )
dx
ln b x
– 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.
• 4.3. Derivatives of Products and Quotients
– Product Rule: If f (x) = F(x)  S(x), then
f ' ( x)  F
dS dF

S
dx dx
– Quotient Rule: If f (x) = T (x) / B(x), then
B ( x)  T ' ( x)  T ( x)  B ' ( x)
f ' ( x) 
[ B( x)] 2
• 4.4. Chain Rule
– If m(x) = f [g(x)], then m’(x) = f ’[g(x)] g’(x)
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• 4.4. Chain Rule (continued)
– A special case of the chain rule is the general power
rule:
d
 f x  n  n  f x  n 1  f ' ( x)
dx
– Other special cases of the chain rule are the
following general derivative rules:
d
1
ln [ f ( x)] 
 f ' ( x)
dx
f ( x)
d f ( x)
e
 e f ( x ) f ' ( x)
dx
• 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
p  f ' ( p)
relativerate of change of demand
E ( p)  

f ( p)
relativerate of change of price
– 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).