Calculus 12 - Weebly

Calculus 12Chapter 4 – Applications of Differentiation (Part I)(Section A)

Dr. John LoRoyal Canadian College

2020-2021

1. Differential and Linear Approximation

› Consider a generic function 𝑦 = 𝑓(𝑥) and we draw a tangent line at 𝑥 = 𝑐 whose slope is given by 𝑓′(𝑐).

› Choosing a point (𝑥, 𝑦) on the tangent line, we can easily show that

› Since ∆𝑥 = 𝑥 − 𝑐,

CHAPTER 4 - APPLICATIONS OF DIFFERENTIATION (PART I)

2

𝑦 − 𝑓 𝑐 = 𝑓′(𝑐)(𝑥 − 𝑐)

𝑦 − 𝑓 𝑐 = 𝑓′(𝑐)∆𝑥

RCC @ 2020/2021

› We denote 𝑦 − 𝑓(𝑐) by 𝑑𝑦 and ∆𝑥 by 𝑑𝑥. The terms 𝑑𝑥and 𝑑𝑦 are called the differentials of 𝑥 and 𝑦respectively.

› Hence, we can write

› However, the actual change in 𝑦, called increment, when 𝑥 changes from 𝑐 to 𝑐 + ∆𝑥 is

› The differential and increment of 𝑦 are not necessarily equal.


3

∆𝑦 = 𝑓 𝑐 + ∆𝑥 − 𝑓(𝑐)

𝑑𝑦 = 𝑓′ 𝑐 𝑑𝑥

RCC @ 2020/2021

› Example: For the function 𝑦 = 𝑥2:

› (a) Find the differential 𝑑𝑦.

› (b) Find the increment ∆𝑦.

› Solution:

› (a) The differential 𝑑𝑦 is given by

› (b) The increment ∆𝑦 is given by


4

𝑑𝑦 = 𝑓′ 𝑥 𝑑𝑥 = 2𝑥𝑑𝑥

∆𝑦 = 𝑓 𝑥 + ∆𝑥 − 𝑓 𝑥 = 𝑥 + ∆𝑥 2 − 𝑥2 = 2𝑥∆𝑥 + ∆𝑥 2

RCC @ 2020/2021

› How are they different?


5

The differential 𝑑𝑦corresponds to the bluearea in the diagram

The increment ∆𝑦 corresponds to both the blue and red areas in the diagram

RCC @ 2020/2021

› Since 𝑑𝑥 = ∆𝑥, these two expressions differ by the extra term ∆𝑥 2.

› Therefore, they are approximately equal to one another when ∆𝑥 ≈ 0.


6RCC @ 2020/2021

› Example: For the function 𝑦 = 𝑥3 + 4𝑥2 − 5𝑥 − 7:

› (a) Find the differential and increment of 𝑦.

› (b) Calculate 𝑑𝑦 and ∆𝑦 when 𝑥 changes from 1 to 1.01.

› Solution:

› (a) By definition:


7

𝑑𝑦 = 𝑓′ 𝑥 𝑑𝑥 = 3𝑥2 + 8𝑥 − 5 𝑑𝑥

∆𝑦 = 𝑓 𝑥 + ∆𝑥 − 𝑓(𝑥)

= (𝑥 + ∆𝑥)3+4 𝑥 + ∆𝑥 2 − 5 𝑥 + ∆𝑥 − 7 − 𝑓(𝑥)

= 3𝑥2∆𝑥 + 3𝑥 ∆𝑥 2 + ∆𝑥 3 + 8𝑥∆𝑥 + 4 ∆𝑥 2 − 5∆𝑥RCC @ 2020/2021

› (b) Note that 𝑥 = 1 and ∆𝑥 = 0.01. By substitution,


8

𝑑𝑦 = 3 1 2 + 8 1 − 5 0.01 = 0.06

∆𝑦 = 3 1 2 0.01 + 3 1 0.01 2 + (0.01)3

+ 8 1 0.01 + 4(0.01)2 −5(0.01)

= 0.060701

RCC @ 2020/2021

› Example: A metal circle of radius of 50 cm is to be cut from a sheet of metal. The radius can be cut with a maximum error of 0.1 cm.

› (a) Use differentials to estimate the maximum possible error in the area.

› (b) Write the relative error as percent of the circle’s area.


9RCC @ 2020/2021

› An application of differentials is to find an approximate value of a function.

› When ∆𝑥 is small, the following relationship is valid:

› That means

› This leads to the local linear approximation:


10

∆𝑦 ≈ 𝑑𝑦

𝑓 𝑥 + ∆𝑥 − 𝑓(𝑥) ≈ 𝑓′ 𝑥 ∆𝑥

𝑓 𝑥 + ∆𝑥 ≈ 𝑓 𝑥 + 𝑓′(𝑥) ∙ ∆𝑥

RCC @ 2020/2021

› Example: Find the approximate value of 5.02 2 using linear approximation. Compare your answer with the actual value found using a calculator.

› Solution: Assume 𝑓 = 𝑥2. Hence, 𝑓′ 𝑥 = 2𝑥.

› By linear approximation with 𝑥 = 5 and ∆𝑥 = 0.02,

› The actual value of 5.02 2 is 25.2004. The error is about 0.002%.


11

5.02 2 ≈ 5 2 + 2 5 0.02 = 25.2

RCC @ 2020/2021

› Example: Find the approximate value of sin 2°.

› Solution: Let 𝑓 𝑥 = sin 𝑥. We have 𝑓′ 𝑥 = cos 𝑥.

› Therefore, when we choose 𝑥 = 0,

› Note that 𝑥 is given in terms of radian. So

› Hence,


12

sin 0 + ∆𝑥 ≈ sin 0 + cos(0) ∙ ∆𝑥

2° = 2°𝜋

180°=

𝜋

90

sin 2° ≈ 0 + 1𝜋

90= 0.03491

RCC @ 2020/2021

2. Iterations and Newton’s Method

› In mathematics, one of the most central tasks is to obtain the exact solution(s) of an equation algebraically. While it is quite easy and straightforward in some cases (e.g. linear equations), it is indeed very challenging or even impossible in many other cases (e.g. high-order polynomials).

› In those situations, one could only find the approximatesolution(s) using numerical methods.

› Two methods will be introduced in this class: (1) Babylonian square root method; (2) Newton’s method.


13RCC @ 2020/2021

(1) Babylonian algorithm

› This method is useful in solving equations of the form:

› Let’s say 𝑥0 is a first guess for 𝐴. If 𝑥0 is too large, then

› Therefore,


14

𝑥2 − 𝐴 = 0

𝐴

𝑥0<

𝐴

𝐴= 𝐴

𝐴

𝑥0< 𝐴 < 𝑥0

RCC @ 2020/2021

› Based on this condition, we see that the average of these boundaries would be a better guess; that means,

› The next guess, 𝑥2, can be generated using the same relationship.


15

𝑥1 =1

2𝑥0 +

𝐴

𝑥0

𝑥2 =1

2𝑥1 +

𝐴

𝑥1

RCC @ 2020/2021

› This cycle can be repeated successively (called iteration). In general, the iterative formula that connects the two successive guesses looks like:

› When 𝑛 is large (that means, many rounds of iterations), this formula is reduced to

› (You can try to prove it for your own interest!)


16

𝑥𝑛+1 =1

2𝑥𝑛 +

𝐴

𝑥𝑛

𝑥𝑛 ≈ 𝐴

RCC @ 2020/2021

› Example: Find 2 using the Babylonian square root method with an initial guess of 1.

› Solution: Using the formula, we have


17

𝑥1 =1

21 +

2

1= 1.5

𝑥2 =1

21.5 +

2

1.5=17

12

𝑥3 =1

2

17

12+ 2 ∙

12

17=577

408

𝑥4 =1

2

577

408+ 2 ∙

408

577=665857

470832RCC @ 2020/2021

› Note that

› Using a calculator (or computer program) we can obtain

› Amazingly, a simple algorithm offers an answer with an accuracy up to 11 decimal places!


18

𝑥4 =665857

470832≈ 1.4142135623746

2 ≈ 1.4142135623730950488

RCC @ 2020/2021

› Example: Find the solution of 𝑥2 − 5 = 0.


19RCC @ 2020/2021

(2) Newton’s method

› The Babylonian approach works very nicely for quadratic equations 𝑥2 − 𝐴 = 0; however it is not applicable to other cases.

› Instead, we can use the Newton’s method which uses tangent lines to approximate the zeros near an 𝑥-intercept of a function 𝑓(𝑥).


20RCC @ 2020/2021

› Assume 𝑓(𝑥) is a differentiable function on (𝑎, 𝑏) in which one zero exists.

› For example, we make the first guess 𝑥 = 𝑥1, and draw the tangent line at (𝑥1, 𝑓 𝑥1 ).


21RCC @ 2020/2021

› This tangent line is described by the following equation:

› Since it passes through the 𝑥-axis at 𝑥2,

› Rearranging this expression yields


22

𝑦 − 𝑓 𝑥1 = 𝑓′ 𝑥1 𝑥 − 𝑥1

0 − 𝑓 𝑥1 = 𝑓′ 𝑥1 𝑥2 − 𝑥1

𝑥2 = 𝑥1 −𝑓 𝑥1𝑓′ 𝑥1

RCC @ 2020/2021

› The point 𝑥2 is then used as a next guess and construct a new tangent line.

› It can be shown that


23

𝑥3 = 𝑥2 −𝑓 𝑥2𝑓′ 𝑥2

RCC @ 2020/2021

› The relationship between two successive approximation can be generalized as

› Newton’s method assumes that 𝑥𝑛 gets closer to the true zero of 𝑓(𝑥) when 𝑛 gets larger.

› The true solution for 𝑓 𝑥 = 0 is not known a priori; therefore, the computation is usually iterated until a desired level of accuracy is met:


24

𝑥𝑛+1 = 𝑥𝑛 −𝑓 𝑥𝑛𝑓′ 𝑥𝑛

𝑥𝑛+1 − 𝑥𝑛 < 𝐸

RCC @ 2020/2021

› To summarize:


25RCC @ 2020/2021

› Example: Find an approximate, smallest positive root of 𝑥3 + 4𝑥2 − 5𝑥 − 7 = 0. Round to the nearest thousandth.

› Solution: Let 𝑓 𝑥 = 𝑥3 + 4𝑥2 − 5𝑥 − 7.

› Note that

Therefore, there must be a root on [1, 2].

› Let’s choose an initial guess: 𝑥0 = 1.5.

› The derivative of 𝑓(𝑥) is


26

𝑓 1 = (1)3+4(1)2−5 1 − 7 = −7

𝑓 2 = (2)3+4(2)2−5 2 − 7 = 7

𝑓′ 𝑥 = 3𝑥2 + 8𝑥 − 5

RCC @ 2020/2021

› By Newton’s method,

› Find the next approximation:

› Again:


27

𝑥1 = 𝑥0 −𝑓 𝑥0𝑓′ 𝑥0

= 1.5 −𝑓 1.5

𝑓′ 1.5≈ 1.654545

𝑥2 = 𝑥1 −𝑓 𝑥1𝑓′ 𝑥1

= 1.654545 −𝑓 1.654545

𝑓′ 1.654545≈ 1.641979

𝑥3 = 1.641979 −𝑓 1.641979

𝑓′ 1.641979≈ 1.641892

RCC @ 2020/2021

› Summary:

› Since the desired level of accuracy is 0.001, we can stop at 𝑥3.

› Hence, the approximate smallest positive solution of 𝑓(𝑥)is 1.642.


28

𝑛 𝑥𝑛

0 1.5

1 1.654545

2 1.641979

3 1.641892

RCC @ 2020/2021

› Example: Find the approximate smallest positive root of 𝑥3 − 2𝑥 − 1 = 0. Round to 4 decimal places.


29RCC @ 2020/2021

› Example: Find the value of 32 using the Newton’s

method. Round your answer to 6 decimal places.

› Solution: Let 𝑥 =32. This equation is equivalent to

› So, we can set 𝑓 𝑥 = 𝑥3 − 2 and apply the Newton’s method to obtain the root.

› Using IVT, we see that there must be a root on [1, 2]. Hence, we choose the initial guess


30

𝑥3 − 2 = 0

𝑥0 = 1.5

RCC @ 2020/2021

› Standard procedures of Newton’s method yield the following:

› Hence, 32 ≈ 1.259921.


31

𝑛 𝑥𝑛

0 1.5

1 1.2962963

2 1.2609322

3 1.25992186

4 1.25992105

5 1.25992105

RCC @ 2020/2021

› Example: Use Newton’s method to find the approximate value of

470. Round to 4 decimal places.


32RCC @ 2020/2021

› In the previous examples, the values 𝑥0, 𝑥1, 𝑥2, … approach a limit and they form a convergent sequence. This is usually observed in Newton’s method if 𝑓(𝑥) is well-behaved and if 𝑥0 is close to the true value.

› However, sometimes Newton’s method does fail to locate an approximate root of an equation. There are a number of such scenarios.


33RCC @ 2020/2021

Case 1: Zero slope

› Recall that

› If it happens that 𝑓′ 𝑥𝑛 = 0, then the fraction will blow up and the iteration will terminate.


34

𝑥𝑛+1 = 𝑥𝑛 −𝑓(𝑥𝑛)

𝑓′(𝑥𝑛)

RCC @ 2020/2021

Case 2: Non-differentiable at the root

› If a function is not differentiable at the root, Newton’s method will be impossible to identify this root.

› Consider 𝑓 𝑥 = 𝑥1/3. Using 𝑥0 = 0.1 yields


35

𝒏 𝒙𝒏

0.1 -0.2

-0.2 0.4

0.4 -0.8

-0.8 1.6

1.6 -3.2

RCC @ 2020/2021

Case 3: The values are oscillatory

› Sometimes we may encounter a situation in which Newton’s method yields the values bouncing back and forth.

› For instance, consider 𝑥3 − 2𝑥 − 2 = 0. Using an initial guess of 𝑥0 = 0, we obtain the following data:


36

𝑥1 = 0 −𝑓 0

𝑓′ 0= 0 −

(−2)

(−2)= −1

𝑥2 = −1 −𝑓 −1

𝑓′ −1= −1 −

(−1)

(1)= 0

𝑥3 = 0 −𝑓 0

𝑓′ 0= 0 −

−2

−2= −1

RCC @ 2020/2021

› Obviously, the values produced by Newton’s method oscillate between −1 and 0. In this case, the method fails.

› Graphically, it is what’s happening during the iterations.


37RCC @ 2020/2021

3. Related Rate Problems

› The chain rule is a very powerful technique of finding the derivative of a composite function. The idea behind is to develop the connection between the variables 𝑥 and 𝑦through a “chain” 𝑦 → 𝑢 → ⋯ → 𝑥.

› This method has been applied to finding the derivatives of implicit functions.

› Another application of the chain rule is to solve the problems of related rates, in which the rates of change of two or more related variables in terms of time are determined.


38RCC @ 2020/2021

› Consider a conical tank of radius 𝑟, volume 𝑉, and height ℎ. When water is drained out of a conical tank from the bottom, what is the rate of change of volume related to the rates of change of radius and the height?


39RCC @ 2020/2021

› First, we have to find out how 𝑉, 𝑟 and ℎ are related. According to geometry, the volume of a cone is given by

› Then, we differentiate both sides with respect to 𝑡:

› Since both 𝑟 and ℎ vary with time,


40

𝑉 =1

3𝜋𝑟2ℎ

𝑑

𝑑𝑡𝑉 =

𝑑

𝑑𝑡

1

3𝜋𝑟2ℎ

𝑑𝑉

𝑑𝑡=1

3𝜋 ℎ

𝑑

𝑑𝑡𝑟2 + 𝑟2

𝑑

𝑑𝑡ℎ

RCC @ 2020/2021

› This gives

› Hence, to find the rate of change of volume 𝑑𝑉/𝑑𝑡 at any given time 𝑡, we only need to substitute the values of 𝑟, ℎ, 𝑑𝑟/𝑑𝑡, and 𝑑ℎ/𝑑𝑡 at 𝑡 into the expression above.


41

𝑑𝑉

𝑑𝑡=1

3𝜋 2𝑟ℎ

𝑑𝑟

𝑑𝑡+ 𝑟2

𝑑ℎ

𝑑𝑡

RCC @ 2020/2021

› Example: Assume that there is a car whose position, 𝑠, from the starting point at a given time 𝑡, is given by 𝑠 =90𝑡. The gas used by the car, 𝐺 liters for a given distance 𝑠, is given by 𝐺 = 0.07𝑠. How fast is the gas consumed by the car?

› Solution: Recall that

› Hence,

and


42

𝑠 = 90𝑡

𝐺 = 0.07𝑠

𝑑𝐺

𝑑𝑡=𝑑𝐺

𝑑𝑠∙𝑑𝑠

𝑑𝑡

𝑑𝐺

𝑑𝑡= 0.07 90 = 6.3 L/hr

RCC @ 2020/2021

› The general procedures of solving a related rate problem

– Step 1: Identify all the quantities involved in the question, and which quantity is to be determined.

– Step 2: Develop the relationship, i.e., the equation, for the quantities involved.

– Step 3: Differentiate the equation appropriately to find out how the rates of change are connected.

– Step 4: Perform substitution to find the required answer.

– Step 5: Check the final answer (i.e., Is the unit correct? Does the answer make sense?)


43RCC @ 2020/2021

› Example: A raindrop falls in a puddle and the ripples spread in circles. The radii of the circles increase at the rate of 2 inches/second. Find the rate at which the area of a circle will be growing when its radius is 6 inches.

› Solution: There are three quantities in this example; namely, the area 𝐴, the radius 𝑟, and the time 𝑡. These quantities are related by

› Differentiating with respect to 𝑡 gives


44

𝐴 = 𝜋𝑟2

𝑑𝐴

𝑑𝑡= 2𝜋𝑟

𝑑𝑟

𝑑𝑡

RCC @ 2020/2021

› Note that

› When 𝑟 = 6,

› The unit of the answer is in2/sec.


45

𝑑𝑟

𝑑𝑡= 2

𝑑𝐴

𝑑𝑡= 2𝜋 6 2 = 24𝜋

RCC @ 2020/2021

› Example: A spherical balloon is inflated so that its radius is increasing at one inch/minute. Find the rate at which the volume increases when:

(a) the diameter is 2 feet.

(b) the surface area is 324𝜋 in2.


46RCC @ 2020/2021

› Example: A conical flower vase is 30 inches high with a radius of 6 inches at the top. If it is being filled with water at a rate of 10 cubic inches per second, find the rate at which the water level is rising when the depth is 20 inches.

› Solution: The quantities involved are

𝑉: the volume of the vase

ℎ: the depth of the water

𝑟: the radius of water surface

𝑡: time in second


47RCC @ 2020/2021

› From geometry, we know

› The volume depends on both 𝑟 and ℎ, but we want to find only 𝑑ℎ/𝑑𝑡 yet 𝑑𝑟/𝑑𝑡 is unknown. Hence, we need to perform a transformation so that 𝑉 is given solely by ℎ.

› Using similar triangles:

› Thus 𝑟 = ℎ/5.


48

𝑉 =1

3𝜋𝑟2ℎ

6 in

30 inℎ

𝑟𝑟

ℎ=

6

30

RCC @ 2020/2021

› So the volume formula becomes

› Differentiating on both sides gives

› Rearrange it to yield


49

𝑉 =1

3𝜋

ℎ

5

2

ℎ =1

75𝜋ℎ3

𝑑𝑉

𝑑𝑡=

𝜋

753ℎ2

𝑑ℎ

𝑑𝑡=𝜋ℎ2

25∙𝑑ℎ

𝑑𝑡

𝑑ℎ

𝑑𝑡=

25

𝜋ℎ2∙𝑑𝑉

𝑑𝑡

RCC @ 2020/2021

› Given ℎ = 20 and

› Substituting these to the equation yields

› So, the water level is rising at a rate of about 0.20 in/s.


50

𝑑𝑉

𝑑𝑡= 10

𝑑ℎ

𝑑𝑡=

25

𝜋(20)2∙ 10 =

5

8𝜋

RCC @ 2020/2021

› Example: A water tank is built in the shape of a cone with height 5 m and diameter of 6 m. Water is pumped into the tank at a rate of 1.6 m3/min. Find the rate at which the water level is rising when water is 2 m deep.


51RCC @ 2020/2021

› Example: A 26 foot ladder is resting against a vertical wall. The foot of the ladder starts to slip horizontally away from the wall at a rate of 6 in/sec. Find the rate at which the top of the ladder descends when it is 24 feet above the ground.

› Solution: Consider the situation


52RCC @ 2020/2021

› From the diagram we can deduce the relationship between the quantities 𝑥 and 𝑦:

› Differentiating this expression implicitly yields

› Rearranging this we obtain the following


53

𝑥2 + 𝑦2 = 262

2𝑥𝑑𝑥

𝑑𝑡+ 2𝑦

𝑑𝑦

𝑑𝑡= 0

𝑑𝑦

𝑑𝑡= −

𝑥

𝑦∙𝑑𝑥

𝑑𝑡

RCC @ 2020/2021

› To find 𝑑𝑦/𝑑𝑡, we need 𝑥, 𝑦, and 𝑑𝑥/𝑑𝑡. The second and third quantities are already given in the question; namely

› The value of 𝑥 is not provided, but it can be calculated using the Pythagorean theorem:

› Hence,


54

𝑦 = 24𝑑𝑥

𝑑𝑡= 6

𝑥 = 262 − 𝑦2 = 262 − 242 = 10

𝑑𝑦

𝑑𝑡= −

10

24∙ 6 = −2.5 in/s

RCC @ 2020/2021

› Example: A man starts walking north at a speed of 1.5 m/s. At the same time and starting from the same point, a woman begins walking west at a speed of 2 m/s. At what rate is the distance between the man and the woman increasing one minute after they start walking?


55RCC @ 2020/2021

› Example: A spotlight on the ground shines on a wall 10 m away. A 2-m tall man walks from the spotlight toward the wall at a speed of 1.2 m/s. How fast is height of the man’s shadow on the wall decreasing when he is 3 m from the wall?

› Solution: The story can be represented by the following diagram.


56RCC @ 2020/2021

› Note that the quantities 𝑥 and 𝑦 are not related by the Pythagorean theorem! Instead, they are related by similar triangles:

› This gives

› Differentiating both sides with respect to 𝑡 yields


57

2

𝑥=

𝑦

10

𝑦 =20

𝑥

𝑑𝑦

𝑑𝑡= −

20

𝑥2∙𝑑𝑥

𝑑𝑡

RCC @ 2020/2021

› It is given

› Note that when the man is 3 m from the wall, 𝑥 is not 3 but 𝑥 = 10 − 3 = 7. Hence,

› The shadow is decreasing at a rate of about 0.49 m/s.

› Think carefully: Can we solve this question using Pythagorean theorem?


58

𝑑𝑥

𝑑𝑡= 1.2

𝑑𝑦

𝑑𝑡= −

20

721.2 = −

24

49

RCC @ 2020/2021

› Example: A 6-ft figure skater is directly beneath a spotlight 30 feet above the ice. She skates from the light at a rate of 16 ft/s and the spotlight follows her.

(a) How fast is the skater’s shadow lengthening when she is 25 ft from her starting position?

(b) How fast is the top of her shadow’s head moving when she is 25 ft from her starting position?


59RCC @ 2020/2021

› Example: A camera is mounted at a point 3000 feet from the base of a rocket. The rocket blasts off and is rising vertically at 880 feet per second. When the rocket is 4000 feet above the launching pad, how fast must the camera’s angle of elevation change so that it is still aimed at the rocket?

› Solution: We first sketch a diagram describing the motion of the rocket.


60RCC @ 2020/2021

› Note that this question requires us to find 𝑑𝜃/𝑑𝑡. The angle 𝜃 and the height of the rocket, 𝑦, are related by

› Differentiating this equation with respect to 𝑡 gives

› Thus

› To find 𝑑𝜃/𝑑𝑡, we need 𝜃 and 𝑑𝑦/𝑑𝑡 when 𝑦 = 4000.


61

𝑦 = 3000 tan 𝜃

𝑑𝑦

𝑑𝑡= 3000 sec2 𝜃 ∙

𝑑𝜃

𝑑𝑡

𝑑𝜃

𝑑𝑡=

1

3000 sec2 𝜃∙𝑑𝑦

𝑑𝑡=cos2 𝜃

3000∙𝑑𝑦

𝑑𝑡

RCC @ 2020/2021

› When 𝑦 = 4000,

› This gives

› Since 𝑑𝑦/𝑑𝑡 is constantly 880 ft/s, we have

› Question: What is the unit of the answer?


62

tan 𝜃 =4000

3000

cos 𝜃 =3000

5000= 0.6

𝑑𝜃

𝑑𝑡=

0.62

3000880 =

66

625

RCC @ 2020/2021

› Example: A helicopter is holding its position 50 m from a hot air balloon and at the same altitude. An observer in the helicopter watches a parachutist jump from the balloon. The parachutist immediately opens his chute and falls at 5 m/s. How quickly do the observer’s eyes open down to keep the parachutist in sight at the following moments?

(a) When the parachutist jumps.

(b) When the parachutist has fallen 50 m.

(c) When the parachutist has fallen 100 m.


63RCC @ 2020/2021

4. Optimization Problems

› Another common type of applications of differentiation is the determination of maximum or minimum values.

› We have already seen this in PREC 11 indeed. Consider the following case.

› Example: A rancher has 100 m of fencing available to build a rectangular corral. What are the dimensions such that the enclosed area is maximum?

› This question can be solved without using calculus.


64RCC @ 2020/2021

› The equation calculating the area is:

› Rewriting it into vertex form:

› Hence, the largest area can be obtained when 𝑤 = 25 m. The resulting area will be 625 m2.


65

𝐴 = 𝑙𝑤

= 50 − 𝑤 𝑤

= −𝑤2 + 50𝑤

𝐴 = − 𝑤 − 25 2 + 625

RCC @ 2020/2021

› This approach works only for quadratic equations. For other types of relationship, we will need to apply calculus to locate the maximum or minimum (called extremum) quantity.


66RCC @ 2020/2021

› For a function that is continuous in a closed interval 𝑎, 𝑏 , it would have both minimum and maximum on the interval.


67RCC @ 2020/2021

› Depending on where the extrema appear on a closed or open interval, they are called absolute extrema or relative extrema.


68RCC @ 2020/2021

› In general, a relative extremum occur on a “hill” (maximum) or a “valley” (minimum) for a smooth and differentiable function. At this point, the slope of the tangent line is zero.


69RCC @ 2020/2021

› Therefore, to find a relative maximum or minimum of a function 𝑓(𝑥), we differentiate it and set it equal to zero:

› The solution(s) of this equation, called critical value(s), will correspond to the extrema of 𝑓(𝑥).

› Note that the converse of this statement is not guaranteed!

› The identities of these values can be verified by back substitution to 𝑓(𝑥) or by the first derivative test (which will be discussed later in this chapter).


70

𝑓′ 𝑥 = 0

RCC @ 2020/2021

› Example: Find the relative extremum for each of the following functions.

› (a) 𝑓 𝑥 =9 𝑥2−3

𝑥3

› (b) 𝑓 𝑥 = 𝑥

› (c) 𝑓 𝑥 = sin 𝑥


71RCC @ 2020/2021

› We can make a more accurate description of the relationship between relative extrema and critical values:


72RCC @ 2020/2021

› To find an extremum on a closed interval, we use the following guidelines.


73RCC @ 2020/2021

› Example: Find the extrema of 𝑓 𝑥 = 3𝑥4 − 4𝑥3 on the interval −1, 2 .

› Key: critical point is not necessarily an extremum!CHAPTER 4 - APPLICATIONS OF DIFFERENTIATION

(PART I)74

RCC @ 2020/2021

› Example: Find the extrema of 𝑓 𝑥 = 2𝑥 − 3𝑥2/3 on the interval −1, 3 .


75RCC @ 2020/2021

› Example: Find the extrema of 𝑓 𝑥 = 2 sin 𝑥 − cos 2𝑥 on the interval 0, 2𝜋 .


76RCC @ 2020/2021

› The extremum theorem states that both maximum and minimum exist for 𝑓(𝑥) if it is continuous on a closed interval 𝑎, 𝑏 , but these points may occur at the endpoints 𝑥 = 𝑎 or 𝑥 = 𝑏.

› There exists a theorem, called Rolle’s theorem, that ensures the existence of such points in the interior of a closed interval if the following condition is fulfilled:


77RCC @ 2020/2021

› The Rolle’s theorem implies the following:

› Note that such argument is not warranted if 𝑓(𝑥) is not differentiable on 𝑎, 𝑏 .


78RCC @ 2020/2021

› We now have all the tools necessary for us to solve problems about optimization.

› Consider the following example:

› Example: A manufacturer wants to design an open box having a square base and a surface area of 108 in2. What dimensions will produce a box with maximum volume?


79RCC @ 2020/2021

› The volume of the box is:

› This is the primary equation that will be optimized.

› In order to perform the optimization, one of the variables on the right hand side has to be eliminated. Using the condition about area:

› We can write:


80

𝑉 = 𝑥2ℎ

𝑆 = 𝑥2 + 4𝑥ℎ = 108

ℎ =108 − 𝑥2

4𝑥

RCC @ 2020/2021

› Back substitution to the primary equation yields:

› Now we can find the maximum value of 𝑉 by differentiation:

› Solving this gives:


81

𝑉 = 𝑥2108 − 𝑥2

4𝑥= 27𝑥 −

𝑥3

4

𝑑𝑉

𝑑𝑥= 27 −

3

4𝑥2 = 0

𝑥 = ±6

RCC @ 2020/2021

› From the question, we know that the domain of 𝑥 is between 0 and 108. Therefore, we only take the positive answer 6.

› To confirm the answer, we check the endpoints:

› Hence, when 𝑥 = 6, the volume is largest. The dimensions are 6 × 6 × 3 inches.


82

𝑉 0 = 02 × 108 = 0

𝑉 6 = 62 × 3 = 108

𝑉 108 = 1082× 0 = 0

RCC @ 2020/2021


83RCC @ 2020/2021

› Example: Which points on the graph of 𝑦 = 4 − 𝑥2 are closest to the point 0, 2 ?


84RCC @ 2020/2021

› Example: A rectangular page is to contain 24 square inches of print. The margins at the top and bottom of the page are to be 1.5 inches, and the margins on the left and right are to be 1 inch. What should the dimensions of the page be so that the least amount of paper is used?


85RCC @ 2020/2021

› Example: Two posts, one 12 feet high and the other 28 feet high, stand 30feet apart. They are to be stayed by two wires, attached to a single stake, running from ground level to the top of each post. Where should the stake be placed to use the least amount of wire?


86RCC @ 2020/2021

› Example: Find the volume of the largest right circular cone that can be inscribed in a sphere of radius 𝑟.


87RCC @ 2020/2021

5. Economic applications

› To illustrate the applicability of abstract ideas of calculus to solving real-world problems, we will spend a little bit of time discussing the concept of marginality in economics.

› The ultimate goal of operating a firm or an industry is simply to maximize the profit that it can earn within the constraint of budget. To make a good management decision, one has to consider two factors:

1. What is the production cost of an item?

2. What is the revenue made out of it?


88RCC @ 2020/2021

› The cost that a company incurs in producing items of a commodity is called a cost function, denoted by 𝐶(𝑥), in which 𝑥 is the number of units of production.

› Cost functions may take different forms. The following is the most general one:


89RCC @ 2020/2021

› There are two quantities associated with 𝐶(𝑥):

(1) Average cost

› This is the cost per unit when 𝑥 units are produced.

(2) Marginal cost

› This is the rate of change of cost with respect to 𝑥.

› These two values are equal if the average cost is minimum.


90

𝑐 𝑥 =𝐶(𝑥)

𝑥

𝐶′ 𝑥 =𝑑𝐶(𝑥)

𝑑𝑥

RCC @ 2020/2021

› Example: A company estimates that the cost of producing 𝑥 items is 𝐶 𝑥 = 2600 + 2𝑥 + 0.001𝑥2.

a. Find the cost, average cost, and marginal cost of producing 1000 items, 2000 items and 3000 items.

b. At which production level will the average cost be lowest, and what is this minimum average cost?


91RCC @ 2020/2021

› Now, we change our focus to marketing. Let 𝑝(𝑥) be the price per unit that the company can charge if it sells 𝑥units. This function is called the price function.

› If 𝑥 items are sold and the price per unit is 𝑝(𝑥), then the total revenue will be

› The function 𝑅(𝑥) is called the revenue function.

› The rate of change of revenue with respect to 𝑥 is called the marginal revenue:


92

𝑅 𝑥 = 𝑥𝑝(𝑥)

𝑅′ 𝑥 =𝑑𝑅(𝑥)

𝑑𝑥= 𝑝 𝑥 + 𝑥𝑝′(𝑥)

RCC @ 2020/2021

› Hence, the total profit earned by selling 𝑥 units is given by:

› The function 𝑃(𝑥) is called the profit function.

› The marginal profit is defined as

› When the profit is maximum, 𝑃′ 𝑥 = 0, and


93

𝑃′ 𝑥 = 𝑅′ 𝑥 − 𝐶′(𝑥)

𝑃 𝑥 = 𝑅 𝑥 − 𝐶 𝑥 = 𝑥𝑝 𝑥 − 𝐶(𝑥)

𝑅′ 𝑥 = 𝐶′(𝑥)

RCC @ 2020/2021

› Example: Determine the production level that will maximize the profit for a company with cost and price functions:


94

𝐶 𝑥 = 84 + 1.26𝑥 − 0.01𝑥2 + 0.00007𝑥3

𝑝 𝑥 = 3.5 − 0.01𝑥

RCC @ 2020/2021


95RCC @ 2020/2021


96RCC @ 2020/2021


97RCC @ 2020/2021


98RCC @ 2020/2021


99RCC @ 2020/2021


100RCC @ 2020/2021

Calculus 12 - Weebly

Documents

Transcript of Calculus 12 - Weebly