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Exercise 15

Avg difficulty 2.5

Let f(v)f(v) be the gas consumption (in liters/km) of a car going at velocity vv (in km/hour). In other words, f(v)f(v) tells you how many liters of gas the car uses to go one kilometer if it is traveling at vv kilometers per hour. In addition, suppose that f(80)=0.05f(80)=0.05 and f′(80)=0.0004f'(80) = 0.0004.

  1. Let g(v)g(v) be the distance the same car goes on one liter of gas at velocity vv. What is the relationship between f(v)f(v) and g(v)g(v)? Hence find g(80)g(80) and g′(80)g'(80).

  2. Let h(v)h(v) be the gas consumption in liters per hour of a car going at velocity vv. In other words, h(v)h(v) tells you how many liters of gas the car uses in one hour if it is going at velocity vv. What is the algebraic relationship between h(v)h(v) and f(v)f(v)? Hence find h(80)h(80) and h′(80)h'(80).

  3. How would you explain the practical meaning of these function and derivative values to a driver who knows no calculus? Include units on each of the function and derivative values you discuss in your response.

Boelkins, Active Calculus, §2.3 The product and quotient rules, Exercise 15

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Adapted from Active Calculus © 2012–2025 Matthew Boelkins. Licensed under CC BY-SA 4.0. Source: Active Calculus. Imported at revision dc86ba3.