# 3 Differentiation u Basic Rules of Differentiation u

- Slides: 111

3 Differentiation u Basic Rules of Differentiation u The Product and Quotient Rules u The Chain Rule u Marginal Functions in Economics u Higher Order Derivatives

3. 1 Basic Rules of Differentiation 1. 2. 3. 4. Derivative of a Constant The Power Rule Derivative of a Constant Multiple Function The Sum Rule

Four Basic Rules u We’ve learned that to find the rule for the derivative f ′of a function f, we first find the difference quotient u But this method is tedious and time consuming, even for relatively simple functions. u This chapter we will develop rules that will simplify the process of finding the derivative of a function.

Rule 1: Derivative of a Constant u We will use the notation u To mean “the derivative of f with respect to x at x. ” Rule 1: Derivative of a constant u The derivative of a constant function is equal to zero.

Rule 1: Derivative of a Constant u We can see geometrically why the derivative of a constant must be zero. u The graph of a constant function is a straight line parallel to the x axis. u Such a line has a slope that is constant with a value of zero. u Thus, the derivative of a constant must be zero as well. y f(x) = c x

Rule 1: Derivative of a Constant u We can use the definition of the derivative to demonstrate this:

Rule 2: The Power Rule u If n is any real number, then

Rule 2: The Power Rule u Lets verify this rule for the special case of n = 2. u If f(x) = x 2, then

Rule 2: The Power Rule Practice Examples: u If f(x) = x, then u If f(x) = x 8, then u If f(x) = x 5/2, then Example 2, page 159

Rule 2: The Power Rule Practice Examples: u Find the derivative of Example 3, page 159

Rule 2: The Power Rule Practice Examples: u Find the derivative of Example 3, page 159

Rule 3: Derivative of a Constant Multiple Function u If c is any constant real number, then

Rule 3: Derivative of a Constant Multiple Function Practice Examples: u Find the derivative of Example 4, page 160

Rule 3: Derivative of a Constant Multiple Function Practice Examples: u Find the derivative of Example 4, page 160

Rule 4: The Sum Rule

Rule 4: The Sum Rule Practice Examples: u Find the derivative of Example 5, page 161

Rule 4: The Sum Rule Practice Examples: u Find the derivative of Example 5, page 161

Applied Example: Conservation of a Species u A group of marine biologists at the Neptune Institute of Oceanography recommended that a series of conservation measures be carried out over the next decade to save a certain species of whale from extinction. u After implementing the conservation measure, the population of this species is expected to be where N(t) denotes the population at the end of year t. u Find the rate of growth of the whale population when t = 2 and t = 6. u How large will the whale population be 8 years after implementing the conservation measures? Applied Example 7, page 162

Applied Example: Conservation of a Species Solution u The rate of growth of the whale population at any time t is given by u In particular, for t = 2, we have u And for t = 6, we have u Thus, the whale population’s rate of growth will be 34 whales per year after 2 years and 338 per year after 6 years. Applied Example 7, page 162

Applied Example: Conservation of a Species Solution u The whale population at the end of the eighth year will be Applied Example 7, page 162

3. 2 The Product and Quotient Rules

Rule 5: The Product Rule u The derivative of the product of two differentiable functions is given by

Rule 5: The Product Rule Practice Examples: u Find the derivative of Example 1, page 172

Rule 5: The Product Rule Practice Examples: u Find the derivative of Example 2, page 172

Rule 6: The Quotient Rule u The derivative of the quotient of two differentiable functions is given by

Rule 6: The Quotient Rule Practice Examples: u Find the derivative of Example 3, page 173

Rule 6: The Quotient Rule Practice Examples: u Find the derivative of Example 4, page 173

Applied Example: Rate of Change of DVD Sales u The sales ( in millions of dollars) of DVDs of a hit movie t years from the date of release is given by u Find the rate at which the sales are changing at time t. u How fast are the sales changing at: ✦ The time the DVDs are released (t = 0)? ✦ And two years from the date of release (t = 2)? Applied Example 6, page 174

Applied Example: Rate of Change of DVD Sales Solution u The rate of change at which the sales are changing at time t is given by Applied Example 6, page 174

Applied Example: Rate of Change of DVD Sales Solution u The rate of change at which the sales are changing when the DVDs are released (t = 0) is That is, sales are increasing by $5 million per year. Applied Example 6, page 174

Applied Example: Rate of Change of DVD Sales Solution u The rate of change two years after the DVDs are released (t = 2) is That is, sales are decreasing by $600, 000 per year. Applied Example 6, page 174

3. 3 The Chain Rule

Deriving Composite Functions u Consider the function u To compute h′(x), we can first expand h(x) and then derive the resulting polynomial u But how should we derive a function like H(x)?

Deriving Composite Functions Note that is a composite function: u H(x) is composed of two simpler functions u So that u We can use this to find the derivative of H(x).

Deriving Composite Functions To find the derivative of the composite function H(x): u We let u = f(x) = x 2 + x + 1 and y = g(u) = u 100. u Then we find the derivatives of each of these functions u The ratios of these derivatives suggest that u Substituting x 2 + x + 1 for u we get

Rule 7: The Chain Rule u If h(x) = g[f(x)], then u Equivalently, if we write y = h(x) = g(u), where u = f(x), then

The Chain Rule for Power Functions u Many composite functions have the special form h(x) = g[f(x)] where g is defined by the rule g(x) = xn (n, a real number) so that h(x) = [f(x)]n u In other words, the function h is given by the power of a function f. u Examples:

The General Power Rule u If the function f is differentiable and h(x) = [f(x)]n (n, a real number), then

The General Power Rule Practice Examples: u Find the derivative of Solution u Rewrite as a power function: u Apply the general power rule: Example 2, page 184

The General Power Rule Practice Examples: u Find the derivative of Solution u Apply the product rule and the general power rule: Example 3, page 185

The General Power Rule Practice Examples: u Find the derivative of Solution u Rewrite as a power function: u Apply the general power rule: Example 5, page 186

The General Power Rule Practice Examples: u Find the derivative of Solution u Apply the general power rule and the quotient rule: Example 6, page 186

Applied Problem: Arteriosclerosis u Arteriosclerosis begins during childhood when plaque forms in the arterial walls, blocking the flow of blood through the arteries and leading to heart attacks, stroke and gangrene. Applied Example 8, page 188

Applied Problem: Arteriosclerosis u Suppose the idealized cross section of the aorta is circular with radius a cm and by year t the thickness of the plaque is h = g(t) cm then the area of the opening is given by A = p (a – h)2 cm 2 u Further suppose the radius of an individual’s artery is 1 cm (a = 1) and the thickness of the plaque in year t is given by h = g(t) = 1 – 0. 01(10, 000 – t 2)1/2 cm Applied Example 8, page 188

Applied Problem: Arteriosclerosis u Then we can use these functions for h and A h = g(t) = 1 – 0. 01(10, 000 – t 2)1/2 A = f(h) = p (1 – h)2 to find a function that gives us the rate at which A is changing with respect to time by applying the chain rule: Applied Example 8, page 188

Applied Problem: Arteriosclerosis u For example, at age 50 (t = 50), u So that u That is, the area of the arterial opening is decreasing at the rate of 0. 03 cm 2 per year for a typical 50 year old. Applied Example 8, page 188

3. 4 Marginal Functions in Economics

Marginal Analysis u Marginal analysis is the study of the rate of change of economic quantities. u These may have to do with the behavior of costs, revenues, profit, output, demand, etc. u In this section we will discuss the marginal analysis of various functions related to: ✦ Cost ✦ Average Cost ✦ Revenue ✦ Profit ✦ Elasticity of Demand

Applied Example: Rate of Change of Cost Functions u Suppose the total cost in dollars incurred each week by Polaraire for manufacturing x refrigerators is given by the total cost function C(x) = 8000 + 200 x – 0. 2 x 2 (0 x 400) a. What is the actual cost incurred for manufacturing the 251 st refrigerator? b. Find the rate of change of the total cost function with respect to x when x = 250. c. Compare the results obtained in parts (a) and (b). Applied Example 1, page 194

Applied Example: Rate of Change of Cost Functions Solution a. The cost incurred in producing the 251 st refrigerator is C(251) – C(250) = [8000 + 200(251) – 0. 2(251)2] – [8000 + 200(250) – 0. 2(250)2] = 45, 599. 8 – 45, 500 = 99. 80 or $99. 80. Applied Example 1, page 194

Applied Example: Rate of Change of Cost Functions Solution b. The rate of change of the total cost function C(x) = 8000 + 200 x – 0. 2 x 2 with respect to x is given by C´(x) = 200 – 0. 4 x So, when production is 250 refrigerators, the rate of change of the total cost with respect to x is C´(x) = 200 – 0. 4(250) = 100 or $100. Applied Example 1, page 194

Applied Example: Rate of Change of Cost Functions Solution c. Comparing the results from (a) and (b) we can see they are very similar: $99. 80 versus $100. ✦ This is because (a) measures the average rate of change over the interval [250, 251], while (b) measures the instantaneous rate of change at exactly x = 250. ✦ The smaller the interval used, the closer the average rate of change becomes to the instantaneous rate of change. Applied Example 1, page 194

Applied Example: Rate of Change of Cost Functions Solution u The actual cost incurred in producing an additional unit of a good is called the marginal cost. u As we just saw, the marginal cost is approximated by the rate of change of the total cost function. u For this reason, economists define the marginal cost function as the derivative of the total cost function. Applied Example 1, page 194

Applied Example: Marginal Cost Functions u A subsidiary of Elektra Electronics manufactures a portable music player. u Management determined that the daily total cost of producing these players (in dollars) is C(x) = 0. 0001 x 3 – 0. 08 x 2 + 40 x + 5000 where x stands for the number of players produced. a. Find the marginal cost function. b. Find the marginal cost for x = 200, 300, 400, and 600. c. Interpret your results. Applied Example 2, page 195

Applied Example: Marginal Cost Functions Solution a. If the total cost function is: C(x) = 0. 0001 x 3 – 0. 08 x 2 + 40 x + 5000 then, its derivative is the marginal cost function: C´(x) = 0. 0003 x 2 – 0. 16 x + 40 Applied Example 2, page 195

Applied Example: Marginal Cost Functions Solution b. The marginal cost for x = 200, 300, 400, and 600 is: C´(200) = 0. 0003(200)2 – 0. 16(200) + 40 = 20 C´(300) = 0. 0003(300)2 – 0. 16(300) + 40 = 19 C´(400) = 0. 0003(400)2 – 0. 16(400) + 40 = 24 C´(600) = 0. 0003(600)2 – 0. 16(600) + 40 = 52 or $20/unit, $19/unit, $24/unit, and $52/unit, respectively. Applied Example 2, page 195

Applied Example: Marginal Cost Functions Solution c. From part (b) we learn that at first the marginal cost is decreasing, but as output increases, the marginal cost increases as well. This is a common phenomenon that occurs because of several factors, such as excessive costs due to overtime and high maintenance costs for keeping the plant running at such a fast rate. Applied Example 2, page 195

Applied Example: Marginal Revenue Functions u Suppose the relationship between the unit price p in dollars and the quantity demanded x of the Acrosonic model F loudspeaker system is given by the equation p = – 0. 02 x + 400 (0 x 20, 000) a. Find the revenue function R. b. Find the marginal revenue function R′. c. Compute R′(2000) and interpret your result. Applied Example 5, page 199

Applied Example: Marginal Revenue Functions Solution a. The revenue function is given by R (x ) = px = (– 0. 02 x + 400)x = – 0. 02 x 2 + 400 x Applied Example 5, page 199 (0 x 20, 000)

Applied Example: Marginal Revenue Functions Solution b. Given the revenue function R(x) = – 0. 02 x 2 + 400 x We find its derivative to obtain the marginal revenue function: R′(x) = – 0. 04 x + 400 Applied Example 5, page 199

Applied Example: Marginal Revenue Functions Solution c. When quantity demanded is 2000, the marginal revenue will be: R′(2000) = – 0. 04(2000) + 400 = 320 Thus, the actual revenue realized from the sale of the 2001 st loudspeaker system is approximately $320. Applied Example 5, page 199

Applied Example: Marginal Profit Function u Continuing with the last example, suppose the total cost (in dollars) of producing x units of the Acrosonic model F loudspeaker system is C (x ) = 100 x + 200, 000 a. Find the profit function P. b. Find the marginal profit function P′. c. Compute P′ (2000) and interpret the result. Applied Example 6, page 199

Applied Example: Marginal Profit Function Solution a. From last example we know that the revenue function is R(x) = – 0. 02 x 2 + 400 x ✦ Profit is the difference between total revenue and total cost, so the profit function is P (x ) = R (x ) – C (x ) = (– 0. 02 x 2 + 400 x) – (100 x + 200, 000) = – 0. 02 x 2 + 300 x – 200, 000 Applied Example 6, page 199

Applied Example: Marginal Profit Function Solution b. Given the profit function P(x) = – 0. 02 x 2 + 300 x – 200, 000 we find its derivative to obtain the marginal profit function: P′(x)= – 0. 04 x + 300 Applied Example 6, page 199

Applied Example: Marginal Profit Function Solution c. When producing x = 2000, the marginal profit is P′(2000) = – 0. 04(2000) + 300 = 220 Thus, the profit to be made from producing the 2001 st loudspeaker is $220. Applied Example 6, page 199

Elasticity of Demand u Economists are frequently concerned with how strongly do changes in prices cause quantity demanded to change. u The measure of the strength of this reaction is called the elasticity of demand, which is given by Note: Since the ratio is negative, economists use the negative of the ratio, to make the elasticity be a positive number.

Elasticity of Demand u Suppose the price of a good increases by h dollars from p to (p + h) dollars. u The percentage change of the price is Percentage change in price u The percentage change in quantity demanded is Percentage change in quantity demanded Change in quantity demanded Quantity demanded at price p

Elasticity of Demand u One good way to measure the effect that a percentage change in price has on the percentage change in the quantity demanded is to look at the ratio of the latter to the former. We find Percentage change in quantity demanded Percentage change in price

Elasticity of Demand u We have u If f is differentiable at p, then, when h is small, u Therefore, if h is small, the ratio is approximately equal to u Economists call the negative of this quantity the elasticity of demand.

Elasticity of Demand u If f is a differentiable demand function defined by x = f(p) , then the elasticity of demand at price p is given by Note: Since the ratio is negative, economists use the negative of the ratio, to make the elasticity be a positive number.

Applied Example: Elasticity of Demand u Consider the demand equation for the Acrosonic model F loudspeaker system: p = – 0. 02 x + 400 (0 x 20, 000) a. Find the elasticity of demand E(p). b. Compute E(100) and interpret your result. c. Compute E(300) and interpret your result. Applied Example 7, page 201

Applied Example: Elasticity of Demand Solution a. Solving the demand equation for x in terms of p, we get x = f(p) = – 50 p + 20, 000 From which we see that f ′(p) = – 50 Therefore, Applied Example 7, page 201

Applied Example: Elasticity of Demand Solution b. When p = 100 the elasticity of demand is ✦ This means that for every 1% increase in price we can expect to see a 1/3% decrease in quantity demanded. ✦ Because the response (change in quantity demanded) is less than the action (change in price), we say demand is inelastic. ✦ Demand is said to be inelastic whenever E(p) < 1. Applied Example 7, page 201

Applied Example: Elasticity of Demand Solution c. When p = 300 the elasticity of demand is ✦ This means that for every 1% increase in price we can expect to see a 3% decrease in quantity demanded. ✦ Because the response (change in quantity demanded) is greater than the action (change in price), we say demand is elastic. ✦ Demand is said to be elastic whenever E(p) > 1. ✦ Finally, demand is said to be unitary whenever E(p) = 1. Applied Example 7, page 201

3. 5 Higher Order Derivatives

Higher-Order Derivatives u The derivative f ′ of a function f is also a function. u As such, f ′ may also be differentiated. u Thus, the function f ′ has a derivative f ″ at a point x in the domain of f if the limit of the quotient exists as h approaches zero. u The function f ″ obtained in this manner is called the second derivative of the function f, just as the derivative f ′ of f is often called the first derivative of f. u By the same token, you may consider the third, fourth, fifth, etc. derivatives of a function f.

Higher-Order Derivatives Practice Examples: u Find the third derivative of the function f(x) = x 2/3 and determine its domain. Solution u We have and u So the required derivative is u The domain of the third derivative is the set of all real numbers except x = 0. Example 1, page 208

Higher-Order Derivatives Practice Examples: u Find the second derivative of the function f(x) = (2 x 2 +3)3/2 Solution u Using the general power rule we get the first derivative: Example 2, page 209

Higher-Order Derivatives Practice Examples: u Find the second derivative of the function f(x) = (2 x 2 +3)3/2 Solution u Using the product rule we get the second derivative: Example 2, page 209

Applied Example: Acceleration of a Maglev u The distance s (in feet) covered by a maglev moving along a straight track t seconds after starting from rest is given by the function s = 4 t 2 (0 t 10) u What is the maglev’s acceleration after 30 seconds? Solution u The velocity of the maglev t seconds from rest is given by u The acceleration of the maglev t seconds from rest is given by the rate of change of the velocity of t, given by or 8 feet per second (ft/sec 2). Applied Example 4, page 209

3. 6 Implicit Differentiation and Related Rates

Differentiating Implicitly u Up to now we have dealt with functions in the form u u u y = f (x ) That is, the dependent variable y has been expressed explicitly in terms of the independent variable x. However, not all functions are expressed explicitly. For example, consider x 2 y + y – x 2 + 1 = 0 This equation expresses y implicitly as a function of x. Solving for y in terms of x we get which expresses y explicitly.

Differentiating Implicitly u Now, consider the equation u u y 4 – y 3 – y + 2 x 3 – x = 8 With certain restrictions placed on y and x, this equation defines y as a function of x. But in this case it is difficult to solve for y in order to express the function explicitly. How do we compute dy/dx in this case? The chain rule gives us a way to do this.

Differentiating Implicitly u Consider the equation y 2 = x. u To find dy/dx, we differentiate both sides of the equation: u Since y is a function of x, we can rewrite y = f(x) and find: Example 1, page 216 Using chain rule

Differentiating Implicitly u Therefore the above equation is equivalent to: u Solving for dy/dx yields: Example 1, page 216

Steps for Differentiating Implicitly u To find dy/dx by implicit differentiation: 1. Differentiate both sides of the equation with respect to x. (Make sure that the derivative of any term involving y includes the factor dy/dx) 2. Solve the resulting equation for dy/dx in terms of x and y.

Differentiating Implicitly Examples u Find dy/dx for the equation Solution u Differentiating both sides and solving for dy/dx we get Example 2, page 216

Differentiating Implicitly Examples u Find dy/dx for the equation u Then, find the value of dy/dx when y = 2 and x = 1. Solution Example 4, page 217

Differentiating Implicitly Examples u Find dy/dx for the equation u Then, find the value of dy/dx when y = 2 and x = 1. Solution u Substituting y = 2 and x = 1 we find: Example 4, page 217

Differentiating Implicitly Examples u Find dy/dx for the equation Solution Example 5, page 219

Related Rates u Implicit differentiation is a useful technique for solving a class of problems known as related-rate problems. u Here are some guidelines to solve related-rate problems: 1. Assign a variable to each quantity. 2. Write the given values of the variables and their rate of change with respect to t. 3. Find an equation giving the relationship between the variables. 4. Differentiate both sides of the equation implicitly with respect to t. 5. Replace the variables and their derivatives by the numerical data found in step 2 and solve the equation for the required rate of change.

Applied Example: Rate of Change of Housing Starts u A study prepared for the National Association of Realtors estimates that the number of housing starts in the southwest, N(t) (in millions), over the next 5 years is related to the mortgage rate r(t) (percent per year) by the equation 9 n 2 + r = 36 u What is the rate of change of the number of housing starts with respect to time when the mortgage rate is 11% per year and is increasing at the rate of 1. 5% per year? Applied Example 6, page 220

Applied Example: Rate of Change of Housing Starts Solution u We are given that r = 11% and dr/dt = 1. 5 at a certain instant in time, and we are required to find d. N/dt. 1. Substitute r = 11 into the given equation: (rejecting the negative root) Applied Example 6, page 220

Applied Example: Rate of Change of Housing Starts Solution u We are given that r = 11% and dr/dt = 1. 5 at a certain instant in time, and we are required to find d. N/dt. 2. Differentiate the given equation implicitly on both sides with respect to t: Applied Example 6, page 220

Applied Example: Rate of Change of Housing Starts Solution u We are given that r = 11% and dr/dt = 1. 5 at a certain instant in time, and we are required to find d. N/dt. 3. Substitute N = 5/3 and dr/dt = 1. 5 into this equation and solve for d. N/dt: u Thus, at the time under consideration, the number of housing starts is decreasing at rate of 50, 000 units per year. Applied Example 6, page 220

Applied Example: Watching a Rocket Launch u At a distance of 4000 feet from the launch site, a spectator is observing a rocket being launched. u If the rocket lifts off vertically and is rising at a speed of 600 feet per second when it is at an altitude of 3000 feet, how fast is the distance between the rocket and the spectator changing at that instant? Rocket x Spectator Launch Pad 4000 ft Applied Example 8, page 221 y

Applied Example: Watching a Rocket Launch Solution 1. Let y = altitude of the rocket x = distance between the rocket and the spectator at any time t. 2. We are told that at a certain instant in time and are asked to find dx/dt at that instant. Applied Example 8, page 221

Applied Example: Watching a Rocket Launch Solution 3. Apply the Pythagorean theorem to the right triangle we find that Therefore, when y = 3000, Rocket x Spectator Launch Pad 4000 ft Applied Example 8, page 221 y

Applied Example: Watching a Rocket Launch Solution 4. Differentiate with respect to t, obtaining 5. Substitute x = 5000, y = 3000, and dy/dt = 600, to find Therefore, the distance between the rocket and the spectator is changing at a rate of 360 feet per second. Applied Example 8, page 221

3. 7 Differentials

Increments u Let x denote a variable quantity and suppose x changes from x 1 to x 2. u This change in x is called the increment in x and is denoted by the symbol Dx (read “delta x”). u Thus, Dx = x 2 – x 1 Examples: u Find the increment in x as x changes from 3 to 3. 2. Solution u Here, x 1 = 3 and x 2 = 3. 2, so Dx = x 2 – x 1 = 3. 2 – 3 = 0. 2 Example 1, page 227

Increments u Let x denote a variable quantity and suppose x changes from x 1 to x 2. u This change in x is called the increment in x and is denoted by the symbol Dx (read “delta x”). u Thus, Dx = x 2 – x 1 Examples: u Find the increment in x as x changes from 3 to 2. 7. Solution u Here, x 1 = 3 and x 2 = 2. 7, so Dx = x 2 – x 1 = 2. 7 – 3 = – 0. 3 Example 1, page 227

Increments u Now, suppose two quantities, x and y, are related by an equation y = f(x), where f is a function. u If x changes from x to x + Dx, then the corresponding change in y is called the increment in y. u It is denoted Dy and is defined by D y = f (x + D x ) – f (x ) y f (x + D x ) Dy f (x ) x + Dx x Example 1, page 227 Dx x

Example u Let y = x 3. u Find Dx and Dy when x changes a. from 2 to 2. 01, and b. from 2 to 1. 98. Solution a. Here, Dx = 2. 01 – 2 = 0. 01 Next, b. Here, Dx = 1. 98 – 2 = – 0. 02 Next, Example 2, page 228

Differentials u We can obtain a relatively quick and simple way of approximating Dy, the change in y due to small change Dx. u Observe below that near the point of tangency P, the tangent line T is close to the graph of f. u Thus, if Dx is small, then dy is a good approximation of Dy. y T f (x + D x ) f (x ) Dy dy P x + Dx x

Differentials u Notice that the slope of T is given by dy/Dx (rise over run). u But the slope of T is given by f ′(x), so we have dy/Dx = f ′(x) or dy = f ′(x) Dx u Thus, we have the approximation Dy ≈ dy = f ′(x)Dx u The quantity dy is called the differential of y. y T f (x + D x ) f (x ) Dy dy P x + Dx x

The Differential u Let y = f(x) define a differentiable function x. Then 1. The differential dx of the independent variable x is dx = Dx 2. The differential dy of the dependent variable y is dy = f ′(x)Dx = f ′(x)dx

Example u Approximate the value of using differentials. Solution u Let’s consider the function y = f(x) =. u Since 25 is the number nearest 26. 5 whose square root is readily recognized, let’s take x = 25. u We want to know the change in y, Dy, as x changes from x = 25 to x = 26. 5, an increase of Dx = 1. 5. u So we find u Therefore, Example 4, page 229

Applied Example: Effect of Speed on Vehicular Operating u The total cost incurred in operating a certain type of truck on a 500 -mile trip, traveling at an average speed of v mph, is estimated to be dollars. u Find the approximate change in the total operating cost when the average speed is increased from 55 to 58 mph. Applied Example 5, page 230

Applied Example: Effect of Speed on Vehicular Operating Solution u Total operating cost is given by u With v = 55 an Dv = dv = 3, we find so the total operating cost is found to decrease by $1. 46. u This might explain why so many independent truckers often exceed the 55 mph speed limit. Applied Example 5, page 230

End of Chapter