Calculus 12Chapter 4 β Applications of Differentiation (Part I)(Section A)
Dr. John LoRoyal Canadian College
2020-2021
1. Differential and Linear Approximation
βΊ Consider a generic function π¦ = π(π₯) and we draw a tangent line at π₯ = π whose slope is given by πβ²(π).
βΊ Choosing a point (π₯, π¦) on the tangent line, we can easily show that
βΊ Since βπ₯ = π₯ β π,
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π¦ β π π = πβ²(π)(π₯ β π)
π¦ β π π = πβ²(π)βπ₯
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βΊ We denote π¦ β π(π) by ππ¦ and βπ₯ by ππ₯. The terms ππ₯and ππ¦ are called the differentials of π₯ and π¦respectively.
βΊ Hence, we can write
βΊ However, the actual change in π¦, called increment, when π₯ changes from π to π + βπ₯ is
βΊ The differential and increment of π¦ are not necessarily equal.
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βπ¦ = π π + βπ₯ β π(π)
ππ¦ = πβ² π ππ₯
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βΊ Example: For the function π¦ = π₯2:
βΊ (a) Find the differential ππ¦.
βΊ (b) Find the increment βπ¦.
βΊ Solution:
βΊ (a) The differential ππ¦ is given by
βΊ (b) The increment βπ¦ is given by
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ππ¦ = πβ² π₯ ππ₯ = 2π₯ππ₯
βπ¦ = π π₯ + βπ₯ β π π₯ = π₯ + βπ₯ 2 β π₯2 = 2π₯βπ₯ + βπ₯ 2
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βΊ How are they different?
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The differential ππ¦corresponds to the bluearea in the diagram
The increment βπ¦ corresponds to both the blue and red areas in the diagram
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βΊ Since ππ₯ = βπ₯, these two expressions differ by the extra term βπ₯ 2.
βΊ Therefore, they are approximately equal to one another when βπ₯ β 0.
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βΊ Example: For the function π¦ = π₯3 + 4π₯2 β 5π₯ β 7:
βΊ (a) Find the differential and increment of π¦.
βΊ (b) Calculate ππ¦ and βπ¦ when π₯ changes from 1 to 1.01.
βΊ Solution:
βΊ (a) By definition:
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ππ¦ = πβ² π₯ ππ₯ = 3π₯2 + 8π₯ β 5 ππ₯
βπ¦ = π π₯ + βπ₯ β π(π₯)
= (π₯ + βπ₯)3+4 π₯ + βπ₯ 2 β 5 π₯ + βπ₯ β 7 β π(π₯)
= 3π₯2βπ₯ + 3π₯ βπ₯ 2 + βπ₯ 3 + 8π₯βπ₯ + 4 βπ₯ 2 β 5βπ₯RCC @ 2020/2021
βΊ (b) Note that π₯ = 1 and βπ₯ = 0.01. By substitution,
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ππ¦ = 3 1 2 + 8 1 β 5 0.01 = 0.06
βπ¦ = 3 1 2 0.01 + 3 1 0.01 2 + (0.01)3
+ 8 1 0.01 + 4(0.01)2 β5(0.01)
= 0.060701
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βΊ 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.
βΊ (a) Use differentials to estimate the maximum possible error in the area.
βΊ (b) Write the relative error as percent of the circleβs area.
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βΊ An application of differentials is to find an approximate value of a function.
βΊ When βπ₯ is small, the following relationship is valid:
βΊ That means
βΊ This leads to the local linear approximation:
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βπ¦ β ππ¦
π π₯ + βπ₯ β π(π₯) β πβ² π₯ βπ₯
π π₯ + βπ₯ β π π₯ + πβ²(π₯) β βπ₯
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βΊ Example: Find the approximate value of 5.02 2 using linear approximation. Compare your answer with the actual value found using a calculator.
βΊ Solution: Assume π = π₯2. Hence, πβ² π₯ = 2π₯.
βΊ By linear approximation with π₯ = 5 and βπ₯ = 0.02,
βΊ The actual value of 5.02 2 is 25.2004. The error is about 0.002%.
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5.02 2 β 5 2 + 2 5 0.02 = 25.2
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βΊ Example: Find the approximate value of sin 2Β°.
βΊ Solution: Let π π₯ = sin π₯. We have πβ² π₯ = cos π₯.
βΊ Therefore, when we choose π₯ = 0,
βΊ Note that π₯ is given in terms of radian. So
βΊ Hence,
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sin 0 + βπ₯ β sin 0 + cos(0) β βπ₯
2Β° = 2Β°π
180Β°=
π
90
sin 2Β° β 0 + 1π
90= 0.03491
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2. Iterations and Newtonβs Method
βΊ 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).
βΊ In those situations, one could only find the approximatesolution(s) using numerical methods.
βΊ Two methods will be introduced in this class: (1) Babylonian square root method; (2) Newtonβs method.
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(1) Babylonian algorithm
βΊ This method is useful in solving equations of the form:
βΊ Letβs say π₯0 is a first guess for π΄. If π₯0 is too large, then
βΊ Therefore,
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π₯2 β π΄ = 0
π΄
π₯0<
π΄
π΄= π΄
π΄
π₯0< π΄ < π₯0
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βΊ Based on this condition, we see that the average of these boundaries would be a better guess; that means,
βΊ The next guess, π₯2, can be generated using the same relationship.
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π₯1 =1
2π₯0 +
π΄
π₯0
π₯2 =1
2π₯1 +
π΄
π₯1
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βΊ This cycle can be repeated successively (called iteration). In general, the iterative formula that connects the two successive guesses looks like:
βΊ When π is large (that means, many rounds of iterations), this formula is reduced to
βΊ (You can try to prove it for your own interest!)
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π₯π+1 =1
2π₯π +
π΄
π₯π
π₯π β π΄
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βΊ Example: Find 2 using the Babylonian square root method with an initial guess of 1.
βΊ Solution: Using the formula, we have
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π₯1 =1
21 +
2
1= 1.5
π₯2 =1
21.5 +
2
1.5=17
12
π₯3 =1
2
17
12+ 2 β
12
17=577
408
π₯4 =1
2
577
408+ 2 β
408
577=665857
470832RCC @ 2020/2021
βΊ Note that
βΊ Using a calculator (or computer program) we can obtain
βΊ Amazingly, a simple algorithm offers an answer with an accuracy up to 11 decimal places!
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π₯4 =665857
470832β 1.4142135623746
2 β 1.4142135623730950488
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βΊ Example: Find the solution of π₯2 β 5 = 0.
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(2) Newtonβs method
βΊ The Babylonian approach works very nicely for quadratic equations π₯2 β π΄ = 0; however it is not applicable to other cases.
βΊ Instead, we can use the Newtonβs method which uses tangent lines to approximate the zeros near an π₯-intercept of a function π(π₯).
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βΊ Assume π(π₯) is a differentiable function on (π, π) in which one zero exists.
βΊ For example, we make the first guess π₯ = π₯1, and draw the tangent line at (π₯1, π π₯1 ).
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βΊ This tangent line is described by the following equation:
βΊ Since it passes through the π₯-axis at π₯2,
βΊ Rearranging this expression yields
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π¦ β π π₯1 = πβ² π₯1 π₯ β π₯1
0 β π π₯1 = πβ² π₯1 π₯2 β π₯1
π₯2 = π₯1 βπ π₯1πβ² π₯1
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βΊ The point π₯2 is then used as a next guess and construct a new tangent line.
βΊ It can be shown that
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π₯3 = π₯2 βπ π₯2πβ² π₯2
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βΊ The relationship between two successive approximation can be generalized as
βΊ Newtonβs method assumes that π₯π gets closer to the true zero of π(π₯) when π gets larger.
βΊ The true solution for π π₯ = 0 is not known a priori; therefore, the computation is usually iterated until a desired level of accuracy is met:
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π₯π+1 = π₯π βπ π₯ππβ² π₯π
π₯π+1 β π₯π < πΈ
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βΊ To summarize:
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βΊ Example: Find an approximate, smallest positive root of π₯3 + 4π₯2 β 5π₯ β 7 = 0. Round to the nearest thousandth.
βΊ Solution: Let π π₯ = π₯3 + 4π₯2 β 5π₯ β 7.
βΊ Note that
Therefore, there must be a root on [1, 2].
βΊ Letβs choose an initial guess: π₯0 = 1.5.
βΊ The derivative of π(π₯) is
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π 1 = (1)3+4(1)2β5 1 β 7 = β7
π 2 = (2)3+4(2)2β5 2 β 7 = 7
πβ² π₯ = 3π₯2 + 8π₯ β 5
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βΊ By Newtonβs method,
βΊ Find the next approximation:
βΊ Again:
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π₯1 = π₯0 βπ π₯0πβ² π₯0
= 1.5 βπ 1.5
πβ² 1.5β 1.654545
π₯2 = π₯1 βπ π₯1πβ² π₯1
= 1.654545 βπ 1.654545
πβ² 1.654545β 1.641979
π₯3 = 1.641979 βπ 1.641979
πβ² 1.641979β 1.641892
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βΊ Summary:
βΊ Since the desired level of accuracy is 0.001, we can stop at π₯3.
βΊ Hence, the approximate smallest positive solution of π(π₯)is 1.642.
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π π₯π
0 1.5
1 1.654545
2 1.641979
3 1.641892
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βΊ Example: Find the approximate smallest positive root of π₯3 β 2π₯ β 1 = 0. Round to 4 decimal places.
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βΊ Example: Find the value of 32 using the Newtonβs
method. Round your answer to 6 decimal places.
βΊ Solution: Let π₯ =32. This equation is equivalent to
βΊ So, we can set π π₯ = π₯3 β 2 and apply the Newtonβs method to obtain the root.
βΊ Using IVT, we see that there must be a root on [1, 2]. Hence, we choose the initial guess
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π₯3 β 2 = 0
π₯0 = 1.5
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βΊ Standard procedures of Newtonβs method yield the following:
βΊ Hence, 32 β 1.259921.
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π π₯π
0 1.5
1 1.2962963
2 1.2609322
3 1.25992186
4 1.25992105
5 1.25992105
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βΊ Example: Use Newtonβs method to find the approximate value of
470. Round to 4 decimal places.
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βΊ In the previous examples, the values π₯0, π₯1, π₯2, β¦ approach a limit and they form a convergent sequence. This is usually observed in Newtonβs method if π(π₯) is well-behaved and if π₯0 is close to the true value.
βΊ However, sometimes Newtonβs method does fail to locate an approximate root of an equation. There are a number of such scenarios.
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Case 1: Zero slope
βΊ Recall that
βΊ If it happens that πβ² π₯π = 0, then the fraction will blow up and the iteration will terminate.
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π₯π+1 = π₯π βπ(π₯π)
πβ²(π₯π)
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Case 2: Non-differentiable at the root
βΊ If a function is not differentiable at the root, Newtonβs method will be impossible to identify this root.
βΊ Consider π π₯ = π₯1/3. Using π₯0 = 0.1 yields
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π ππ
0.1 -0.2
-0.2 0.4
0.4 -0.8
-0.8 1.6
1.6 -3.2
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Case 3: The values are oscillatory
βΊ Sometimes we may encounter a situation in which Newtonβs method yields the values bouncing back and forth.
βΊ For instance, consider π₯3 β 2π₯ β 2 = 0. Using an initial guess of π₯0 = 0, we obtain the following data:
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π₯1 = 0 βπ 0
πβ² 0= 0 β
(β2)
(β2)= β1
π₯2 = β1 βπ β1
πβ² β1= β1 β
(β1)
(1)= 0
π₯3 = 0 βπ 0
πβ² 0= 0 β
β2
β2= β1
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βΊ Obviously, the values produced by Newtonβs method oscillate between β1 and 0. In this case, the method fails.
βΊ Graphically, it is whatβs happening during the iterations.
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3. Related Rate Problems
βΊ 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 π₯ and π¦through a βchainβ π¦ β π’ β β― β π₯.
βΊ This method has been applied to finding the derivatives of implicit functions.
βΊ 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.
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βΊ Consider a conical tank of radius π, volume π, and height β. 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?
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βΊ First, we have to find out how π, π and β are related. According to geometry, the volume of a cone is given by
βΊ Then, we differentiate both sides with respect to π‘:
βΊ Since both π and β vary with time,
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π =1
3ππ2β
π
ππ‘π =
π
ππ‘
1
3ππ2β
ππ
ππ‘=1
3π β
π
ππ‘π2 + π2
π
ππ‘β
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βΊ This gives
βΊ Hence, to find the rate of change of volume ππ/ππ‘ at any given time π‘, we only need to substitute the values of π, β, ππ/ππ‘, and πβ/ππ‘ at π‘ into the expression above.
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ππ
ππ‘=1
3π 2πβ
ππ
ππ‘+ π2
πβ
ππ‘
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βΊ Example: Assume that there is a car whose position, π , from the starting point at a given time π‘, is given by π =90π‘. The gas used by the car, πΊ liters for a given distance π , is given by πΊ = 0.07π . How fast is the gas consumed by the car?
βΊ Solution: Recall that
βΊ Hence,
and
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π = 90π‘
πΊ = 0.07π
ππΊ
ππ‘=ππΊ
ππ βππ
ππ‘
ππΊ
ππ‘= 0.07 90 = 6.3 L/hr
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βΊ The general procedures of solving a related rate problem
β Step 1: Identify all the quantities involved in the question, and which quantity is to be determined.
β Step 2: Develop the relationship, i.e., the equation, for the quantities involved.
β Step 3: Differentiate the equation appropriately to find out how the rates of change are connected.
β Step 4: Perform substitution to find the required answer.
β Step 5: Check the final answer (i.e., Is the unit correct? Does the answer make sense?)
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βΊ 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.
βΊ Solution: There are three quantities in this example; namely, the area π΄, the radius π, and the time π‘. These quantities are related by
βΊ Differentiating with respect to π‘ gives
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π΄ = ππ2
ππ΄
ππ‘= 2ππ
ππ
ππ‘
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βΊ Note that
βΊ When π = 6,
βΊ The unit of the answer is in2/sec.
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ππ
ππ‘= 2
ππ΄
ππ‘= 2π 6 2 = 24π
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βΊ 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.
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βΊ 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.
βΊ Solution: The quantities involved are
π: the volume of the vase
β: the depth of the water
π: the radius of water surface
π‘: time in second
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βΊ From geometry, we know
βΊ The volume depends on both π and β, but we want to find only πβ/ππ‘ yet ππ/ππ‘ is unknown. Hence, we need to perform a transformation so that π is given solely by β.
βΊ Using similar triangles:
βΊ Thus π = β/5.
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π =1
3ππ2β
6 in
30 inβ
ππ
β=
6
30
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βΊ So the volume formula becomes
βΊ Differentiating on both sides gives
βΊ Rearrange it to yield
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π =1
3π
β
5
2
β =1
75πβ3
ππ
ππ‘=
π
753β2
πβ
ππ‘=πβ2
25βπβ
ππ‘
πβ
ππ‘=
25
πβ2βππ
ππ‘
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βΊ Given β = 20 and
βΊ Substituting these to the equation yields
βΊ So, the water level is rising at a rate of about 0.20 in/s.
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ππ
ππ‘= 10
πβ
ππ‘=
25
π(20)2β 10 =
5
8π
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βΊ 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.
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βΊ 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.
βΊ Solution: Consider the situation
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βΊ From the diagram we can deduce the relationship between the quantities π₯ and π¦:
βΊ Differentiating this expression implicitly yields
βΊ Rearranging this we obtain the following
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π₯2 + π¦2 = 262
2π₯ππ₯
ππ‘+ 2π¦
ππ¦
ππ‘= 0
ππ¦
ππ‘= β
π₯
π¦βππ₯
ππ‘
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βΊ To find ππ¦/ππ‘, we need π₯, π¦, and ππ₯/ππ‘. The second and third quantities are already given in the question; namely
βΊ The value of π₯ is not provided, but it can be calculated using the Pythagorean theorem:
βΊ Hence,
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π¦ = 24ππ₯
ππ‘= 6
π₯ = 262 β π¦2 = 262 β 242 = 10
ππ¦
ππ‘= β
10
24β 6 = β2.5 in/s
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βΊ 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?
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βΊ 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 manβs shadow on the wall decreasing when he is 3 m from the wall?
βΊ Solution: The story can be represented by the following diagram.
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βΊ Note that the quantities π₯ and π¦ are not related by the Pythagorean theorem! Instead, they are related by similar triangles:
βΊ This gives
βΊ Differentiating both sides with respect to π‘ yields
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2
π₯=
π¦
10
π¦ =20
π₯
ππ¦
ππ‘= β
20
π₯2βππ₯
ππ‘
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βΊ It is given
βΊ Note that when the man is 3 m from the wall, π₯ is not 3 but π₯ = 10 β 3 = 7. Hence,
βΊ The shadow is decreasing at a rate of about 0.49 m/s.
βΊ Think carefully: Can we solve this question using Pythagorean theorem?
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ππ₯
ππ‘= 1.2
ππ¦
ππ‘= β
20
721.2 = β
24
49
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βΊ 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 skaterβs shadow lengthening when she is 25 ft from her starting position?
(b) How fast is the top of her shadowβs head moving when she is 25 ft from her starting position?
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βΊ 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 cameraβs angle of elevation change so that it is still aimed at the rocket?
βΊ Solution: We first sketch a diagram describing the motion of the rocket.
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βΊ Note that this question requires us to find ππ/ππ‘. The angle π and the height of the rocket, π¦, are related by
βΊ Differentiating this equation with respect to π‘ gives
βΊ Thus
βΊ To find ππ/ππ‘, we need π and ππ¦/ππ‘ when π¦ = 4000.
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π¦ = 3000 tan π
ππ¦
ππ‘= 3000 sec2 π β
ππ
ππ‘
ππ
ππ‘=
1
3000 sec2 πβππ¦
ππ‘=cos2 π
3000βππ¦
ππ‘
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βΊ When π¦ = 4000,
βΊ This gives
βΊ Since ππ¦/ππ‘ is constantly 880 ft/s, we have
βΊ Question: What is the unit of the answer?
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tan π =4000
3000
cos π =3000
5000= 0.6
ππ
ππ‘=
0.62
3000880 =
66
625
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βΊ 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 observerβs 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.
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4. Optimization Problems
βΊ Another common type of applications of differentiation is the determination of maximum or minimum values.
βΊ We have already seen this in PREC 11 indeed. Consider the following case.
βΊ 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?
βΊ This question can be solved without using calculus.
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βΊ The equation calculating the area is:
βΊ Rewriting it into vertex form:
βΊ Hence, the largest area can be obtained when π€ = 25 m. The resulting area will be 625 m2.
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π΄ = ππ€
= 50 β π€ π€
= βπ€2 + 50π€
π΄ = β π€ β 25 2 + 625
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βΊ 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.
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βΊ For a function that is continuous in a closed interval π, π , it would have both minimum and maximum on the interval.
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βΊ Depending on where the extrema appear on a closed or open interval, they are called absolute extrema or relative extrema.
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βΊ In general, a relative extremum occur on a βhillβ (maximum) or a βvalleyβ (minimum) for a smooth and differentiable function. At this point, the slope of the tangent line is zero.
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βΊ Therefore, to find a relative maximum or minimum of a function π(π₯), we differentiate it and set it equal to zero:
βΊ The solution(s) of this equation, called critical value(s), will correspond to the extrema of π(π₯).
βΊ Note that the converse of this statement is not guaranteed!
βΊ The identities of these values can be verified by back substitution to π(π₯) or by the first derivative test (which will be discussed later in this chapter).
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πβ² π₯ = 0
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βΊ Example: Find the relative extremum for each of the following functions.
βΊ (a) π π₯ =9 π₯2β3
π₯3
βΊ (b) π π₯ = π₯
βΊ (c) π π₯ = sin π₯
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βΊ We can make a more accurate description of the relationship between relative extrema and critical values:
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βΊ To find an extremum on a closed interval, we use the following guidelines.
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βΊ Example: Find the extrema of π π₯ = 3π₯4 β 4π₯3 on the interval β1, 2 .
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βΊ Example: Find the extrema of π π₯ = 2π₯ β 3π₯2/3 on the interval β1, 3 .
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βΊ Example: Find the extrema of π π₯ = 2 sin π₯ β cos 2π₯ on the interval 0, 2π .
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βΊ The extremum theorem states that both maximum and minimum exist for π(π₯) if it is continuous on a closed interval π, π , but these points may occur at the endpoints π₯ = π or π₯ = π.
βΊ There exists a theorem, called Rolleβs theorem, that ensures the existence of such points in the interior of a closed interval if the following condition is fulfilled:
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βΊ The Rolleβs theorem implies the following:
βΊ Note that such argument is not warranted if π(π₯) is not differentiable on π, π .
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βΊ We now have all the tools necessary for us to solve problems about optimization.
βΊ Consider the following example:
βΊ 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?
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βΊ The volume of the box is:
βΊ This is the primary equation that will be optimized.
βΊ In order to perform the optimization, one of the variables on the right hand side has to be eliminated. Using the condition about area:
βΊ We can write:
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π = π₯2β
π = π₯2 + 4π₯β = 108
β =108 β π₯2
4π₯
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βΊ Back substitution to the primary equation yields:
βΊ Now we can find the maximum value of π by differentiation:
βΊ Solving this gives:
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π = π₯2108 β π₯2
4π₯= 27π₯ β
π₯3
4
ππ
ππ₯= 27 β
3
4π₯2 = 0
π₯ = Β±6
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βΊ From the question, we know that the domain of π₯ is between 0 and 108. Therefore, we only take the positive answer 6.
βΊ To confirm the answer, we check the endpoints:
βΊ Hence, when π₯ = 6, the volume is largest. The dimensions are 6 Γ 6 Γ 3 inches.
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π 0 = 02 Γ 108 = 0
π 6 = 62 Γ 3 = 108
π 108 = 1082Γ 0 = 0
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βΊ Example: Which points on the graph of π¦ = 4 β π₯2 are closest to the point 0, 2 ?
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βΊ 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?
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βΊ 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?
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βΊ Example: Find the volume of the largest right circular cone that can be inscribed in a sphere of radius π.
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5. Economic applications
βΊ 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.
βΊ 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?
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βΊ The cost that a company incurs in producing items of a commodity is called a cost function, denoted by πΆ(π₯), in which π₯ is the number of units of production.
βΊ Cost functions may take different forms. The following is the most general one:
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βΊ There are two quantities associated with πΆ(π₯):
(1) Average cost
βΊ This is the cost per unit when π₯ units are produced.
(2) Marginal cost
βΊ This is the rate of change of cost with respect to π₯.
βΊ These two values are equal if the average cost is minimum.
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π π₯ =πΆ(π₯)
π₯
πΆβ² π₯ =ππΆ(π₯)
ππ₯
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βΊ Example: A company estimates that the cost of producing π₯ items is πΆ π₯ = 2600 + 2π₯ + 0.001π₯2.
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?
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βΊ Now, we change our focus to marketing. Let π(π₯) be the price per unit that the company can charge if it sells π₯units. This function is called the price function.
βΊ If π₯ items are sold and the price per unit is π(π₯), then the total revenue will be
βΊ The function π (π₯) is called the revenue function.
βΊ The rate of change of revenue with respect to π₯ is called the marginal revenue:
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π π₯ = π₯π(π₯)
π β² π₯ =ππ (π₯)
ππ₯= π π₯ + π₯πβ²(π₯)
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βΊ Hence, the total profit earned by selling π₯ units is given by:
βΊ The function π(π₯) is called the profit function.
βΊ The marginal profit is defined as
βΊ When the profit is maximum, πβ² π₯ = 0, and
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πβ² π₯ = π β² π₯ β πΆβ²(π₯)
π π₯ = π π₯ β πΆ π₯ = π₯π π₯ β πΆ(π₯)
π β² π₯ = πΆβ²(π₯)
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βΊ Example: Determine the production level that will maximize the profit for a company with cost and price functions:
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πΆ π₯ = 84 + 1.26π₯ β 0.01π₯2 + 0.00007π₯3
π π₯ = 3.5 β 0.01π₯
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