6 - 1 © 2012 Person Education, Inc.. All rights reserved. Chapter 6 Applications of the Derivative.

6 - 1 © 2012 Person Education, Inc.. All rights reserved.

Chapter 6

Applications of the Derivative


Section 6.1

Absolute Extrema


Local extrema and Absolute extrema





Your Turn 1

Find the absolute extrema of the function

on the interval [0,8].

Solution: First look for critical numbers in the interval (0,8).

Continued

2/3 5/3( ) 3 3f x x x

1/3 2/32 5( ) 3 3

3 3f x x x

1/3 2/32 5x x

1/3(2 5 ) Factor.x x

1/3

2 5x

x

Set ( ) 0 and solve for . f x x


Your Turn 1 continued

Notice that

at x = 2/5 and 2/5 is in the

interval (0,8) .

The derivative

is undefined at x = 0 but the

function is defined there, so 0

is also a critical number.

Evaluate the function at the

critical numbers and the

endpoints. Compare.

Conclusions?

x - Value Extrema Candidates Value of the Function

0 0

2/5 0.977 Absolute Maximum

8 -84 Absolute Minimum

( ) 0f x


Your Turn 2

Find the locations and values of the absolute extrema, if they

exist, for the function

Solution: In this example, the extreme value theorem does not

apply since the domain is an open interval (−∞,∞), which has

no endpoints. Begin as before by finding any critical numbers.

x = 0 or x = − 4 or x = 1.

Evaluate the function at the critical numbers. Continued

4 3 2( ) 4 8 20.f x x x x

3 2( ) 4 12 16 0f x x x x 24 ( 3 4) 0

4 ( 4)( 1) 0

x x x

x x x


Your Turn 2 continued

For an open interval, rather

than evaluating the function at

the endpoints, we evaluate

the limit of the function when

the endpoints are approached.

Because the negative x4

term dominates the other

terms as x becomes large,

f approaches negative infinity, so it has no absolute minimum.

The absolute maximum 148,

occurs at x = − 4.

x - Value

Extrema Candidates Value of the Function

-4 148 Absolute Maximum

0 20

1 23


Practical example (textbook p.309)

• As the number of workers (hours of labor) varies, the output varies.

• The maximum output occurs at 320 hours of labor (that’s 8 employees).

• When is output per hour best? Ans: when the slope is the “most

positive”, that’s the point where the tangent slope is maximum.This happens around 270 hours of labor (see graph).

Graph of production output versus hours at work in a business.


Section 6.2

Applications of Extrema:

Optimization



Your Turn 1

Find two nonnegative numbers x and y for which x + 3y = 30,

such that x2y is maximized.

Solution: Step 1: reading and understanding the problem Step 2: does not apply in this example; there is nothing to draw. Step 3: we determine what is to be maximized and assign a name to that

quantity. Here x2y, is to be maximized, so let’s call it the function M

M = x2y.

we must express M in terms of just one variable, which can be done using the equation x + 3y = 30 and solving for either x or y.

Solving for y gives:

So now:

3010 .

3 3

x xy

32 2(10 ) 10 .

3 3

x xM x x


Your Turn 1 Continued

Step 4: Find the domain of the function.

Because of the nonnegativity requirement, x must be at least 0. Since y must also be at least 0, this requires:

So the domain of x is: [0,30].

Step 5: find the critical points for M by finding

and then solving

10 03

103

103

30.

x

x

x

x

dM

dx0 for .

dMx

dx



x = 0 or x = 20

Step 6: find M for the critical numbers and endpoint of the domain.

Conclusion: the maximum value of the function occurs when x = 20, y = 10/3 and the maximum value is 1333.3

2 2320 20 0

3

dMx x x x

dx

(20 ) 0 Factor out h . t e x x x

x 0 20 30

M’ + -

M


Your Turn 2

A math professor participating in the sport of

orienteering must get to a specific tree in the woods

as fast as possible. He can get there by traveling east

along the trail for 300 m and then north through the

woods for 800 m. He can run 160 m per minute along

the trail but only 40 m per minute through the woods.

Running directly through the woods toward the tree

minimizes the distance, but he will be going slowly

the whole time. He could instead run 300 m along

the trail before entering the woods, maximizing the

total distance but minimizing the time in the woods.

Perhaps the fastest route is a combination, as shown.

Find the path that will get him to the tree

in the minimum time.



Solution :

Step 1:We are trying to minimize the total amount of time, which is the sum of the time on the trail plus the time through the woods. We must express this time as a function of x.

Since: Time = Distance/ Speed, the total time is

Notice in this equation that 0 ≤ x ≤ 300.

2 2300 800( ) .

160 40

x xT x

2 2 1/21 1 1( ) (800 ) (2 ) 0

160 40 2T x x x



2 2

1

160 40 800

x

x

2 2

2 2

16 4 800

4 800

x x

x x

2 2 2

2 2

22

16 800

15 800

800

15800

20715

x x

x

x

x

x T(x)

0 21.88

207 21.24 Absolute Minimum

300 21.36

Conclusion:We see from the table that the time is minimized when x = 207 m. That is, when the professor goes 93 m along thetrail and then heads into the woods.


Find x for maximum volume of box


Find r and h for minimum surface area – (Volume=1000 cm3 )



Section 6.4

Implicit Differentiation


Implicit Differentiation

So far we worked with functions where one variable is expressed in terms of another variable—for example:

y = or y = x sin x (in general: y = f (x). )

Some functions, however, are defined implicitly by a relation between x and y, example:

x3 + y3 = 6xy

We say that f is a function defined implicitly - Equation above means: x3 + [f (x)]3 = 6x f (x)


Practice:

For x3 + y3 = 6xy find:



Your Turn 1

Solution: Differentiate with respect to x on both sides of the

equation.

Now differentiate each term on the left side of the equation.

Think of xy as the product (x)(y) and use the product rule and

the chain rule.

Continued

2 2Find if .dy

x y xydx

2 2

2 2( ) ( )

x y xy

d dx y xy

dx dx

2 2 1dy dy

x y x ydx dx



Now solve this result for .

dy

dx

2 2

(2 ) 2

2

2

dy dyy x y x

dx dxdy

y x y xdxdy y x

dx y x


Your Turn 2


Answer:


Your Turn 3

The graph of

is called the devil’s curve. Find the equation of the tangent line

at the point (1, 1).

Solution: We can calculate by implicit differentiation.

Continued

4 4 2 2 0y x y x

3 34 4 2 2 0 Chain Ru le dy dy

y x y xdx dx

3 3

3 3

4 2 4 2

(4 2 ) 4 2

dy dyy y x x

dx dxdy

y y x xdx



To find the slope of the tangent line at the point (1, 1), let x = 1

and y = 1 The slope is

The equation of the tangent line is then found by using the

point-slope form of the equation of a line.

3 3 3

3 3 3

4 2 2(2 ) (2 )

4 2 2(2 ) (2 )

dy x x x x x x

dx y y y y y y

3

3

2(1) 11.

2(1) 1m

1 1( )y y m x x 1 1( 1)

1 1

y x

y x

y x


See the practice examples in the book. Such as:Example 3 p.333: Find tangent line equation at (2,4)

Ans: y=(4/5)x+12/5


Section 6.5

Related Rates



Your Turn 1

Solution: We start by taking the derivative of the relationship,

using the product and chain rules. Keep in mind that both x and

y are functions of t. The result is

Continued

3 2Suppose and are both functions of and 2 1.

If 1, 2, and 6, then find .

x y t x xy y

dx dyx y

dt dt

23 2 2 2 0.dx dy dx dy

x x y ydt dt dt dt

Now substitute 1, 2, and = 6 to get dx

x ydt



23(1) 6 2(1) 2( 2)(6) 2( 2) 0

dy dy

dt dt

18 2 24 4 0dy dy

dt dt

6 2 0dy

dt

2 6dy

dt

3.dy

dt


Application problem:

A 25-ft ladder is placed against a

building. The base of the ladder is

slipping away from the building at a

rate of 3 ft per minute. Find the rate

at which the top of the ladder is

sliding down the building when the

bottom of the ladder is 7 ft

from the base of the building.

Continued


Solution:

Solution: Starting with Step 1, let y be the height of the top of

the ladder above the ground, and let x be the distance of the

base of the ladder from the base of the building. We

are trying to find dy/dt when x = 7. To perform Step 2, use the

Pythagorean theorem to write

Both x and y are functions of time t (in minutes) after the

moment that the ladder starts slipping. According to Step 3,

take the derivative of both sides of the equation with respect to

time, getting

Continued

2 2 225 .x y

2 2 2( ) (25 )

2 2 0.

d dx y

dt dtdx dy

x ydt dt


To complete Step 4, we need to find the values of x, y, and dx/dt.

Once we find these, we can substitute them into Equation to find

dy/dt.

Since the base is sliding at the rate of 3 ft per minute, dx/dt = 3.

Also, the base of the ladder is 7 ft from the base of the building,

so x =7 . Use this to find y.

Continued

2 2 2

2 2

2

2

25

7 625

49 625

576

24

x y

y

y

y

y


In summary, x = 7, y = 24, and dx/dt = 3.

2 2 0

2(7)(3) 2(24) 0

dx dyx y

dt dtdy

dt

42 48 0

48 42

42 7 ft/min

48 8

dy

dtdy

dtdy

dt


Problem 1An oil rig leak spreads in a circular pattern with a radius increasing at the rate of 30m/hr. How fast is the area of the oil increasing when it has a radius of 100m?

Application problems:


Problem 2Sand falls from an overhead bin and forms a cone with a radius that remains always 3 times the height. If the sand is falling at the rate of 120 ft3/min, how fast is the height of the sand pile changing when it is 10 ft high?


Problem 3Two planes approach an airport, one flying due West at 120 mi/hr and the other flying due North at 150mi/hr. Assuming they stay at the same constant elevation, how fast is the distance between them changing when the westbound plane is 180 mi from the airport, and the northbound one is at 225 mi from the airport?


See more examples solved in the textbook and on the class website!

6 - 1 © 2012 Person Education, Inc.. All rights reserved. Chapter 6 Applications of the Derivative.

Documents

person education

continued slide

negative x

absolute minimum slide

derivative slide

absolute extrema

optimization slide

absolute maximum