AP®MathsCollege BoardCalculus BCStudy GuidesUnit 4: Contextual Applications of DifferentiationRates of Change & Related RatesMeaning of a Derivative in Context

Meaning of a Derivative in Context (College Board AP® Calculus BC): Study Guide

Jamie Wood

Written by: Jamie Wood

Reviewed by: Dan Finlay

Updated on

College Board Calculus BCCollege Board Calculus AB

Meaning of a derivative in context

What does the derivative mean?

  • The derivative of a function is the rate of change of that function

  • The rate of change describes how the dependent variable changes as the independent variable changes

  • Consider a simple example of y equals 3 x

    • The derivative, or the rate of change, is fraction numerator d y over denominator d x end fraction equals 3

    • This means for every 1 unit that x increases by, y increases by 3 units

    • In this case, this is true at every point on the graph of y against x

      • the rate of change is always 3

  • For a more complicated example, consider v equals 1 third t cubed

    • The derivative, or the rate of change, is fraction numerator d v over denominator d t end fraction equals t squared

    • This means that v is changing at a rate of t squared

    • In this case, the rate of change is dependent on t

      • Therefore every point on the graph of v against t will have a different rate of change

        • At the point where t equals 2, the rate of change is 2 squared equals 4

        • At the point where t equals 5, the rate of change is 5 squared equals 25

    • The rate of change at a particular point is the instantaneous rate of change

  • The derivative (rate of change) at a point, is equal to the slope of the tangent at that point

What are the units for the rate of change?

  • The units for fraction numerator d y over denominator d x end fraction will be the units for y, divided by the units for x

    • E.g. The rate at which the volume of water in a tank changes as it is filled could be described by fraction numerator d v over denominator d t end fraction where v is in liters and t is in seconds

    • The units for fraction numerator d v over denominator d t end fraction would be liters per second

How do I interpret a rate of change given in an exam question?

  • Read the description of the scenario carefully

  • Is the function describing an amount, or a rate of change?

  • For example:

  • "The volume of gasoline pumped is described by f open parentheses t close parentheses"

    • This means that f open parentheses t close parentheses represents the volume (amount), most likely measured in gallons

    • f to the power of apostrophe open parentheses t close parentheses would then be describing the rate of change of volume, most likely measured in gallons per second

  • "The rate of flow of gasoline is described by f open parentheses t close parentheses"

    • This means that f open parentheses t close parentheses represents a rate, most likely measured in gallons per second

    • f to the power of apostrophe open parentheses t close parentheses would then be describing the rate of change of the flow rate

      • Most likely measured in gallons per second, per second (or gallons per second squared)

      • It is describing how the rate of flow is changing: is it flowing faster or slower than before?

    • To find a function for the volume (amount) of gasoline pumped in this case,

      • you would need to integrate f open parentheses t close parentheses

  • If you are not sure if something is a rate or an amount, considering the stated units is usually helpful

Worked Example

The depth of the water in a harbor, measured in feet, is modeled by the function f open parentheses t close parentheses. The variable t represents the number of hours after midnight.

f open parentheses t close parentheses equals 6 sin open parentheses fraction numerator 2 t over denominator 5 end fraction minus 2 close parentheses plus 16 space space space space space space space space space space space space space space 0 less or equal than t less than 24

(a) State the maximum depth of the water in the harbor according to the model.

Answer:

f open parentheses t close parentheses models the depth of the water, so we need to find the maximum value of f open parentheses t close parentheses

The maximum of sin open parentheses fraction numerator 2 t over denominator 5 end fraction minus 2 close parentheses will be 1

Use this to find the maximum of the function

6 open parentheses 1 close parentheses plus 16 equals 22

Maximum depth = 22 feet

(b) Find the rate at which the depth of the water in the harbor is changing at 6 am. State appropriate units for your answer.

Answer:

The rate of change of the depth will be given by f to the power of apostrophe open parentheses t close parentheses

Differentiate f open parentheses t close parentheses, using the chain rule for 6 sin open parentheses fraction numerator 2 t over denominator 5 end fraction minus 2 close parentheses

f to the power of apostrophe open parentheses t close parentheses equals 6 cos open parentheses fraction numerator 2 t over denominator 5 end fraction minus 2 close parentheses times 2 over 5 equals 12 over 5 cos open parentheses fraction numerator 2 t over denominator 5 end fraction minus 2 close parentheses

6 am is 6 hours after midnight, so substitute in t equals 6

Make sure your calculator is set to use radians as the angle measure

f to the power of apostrophe open parentheses 6 close parentheses equals 12 over 5 cos open parentheses fraction numerator 2 open parentheses 6 close parentheses over denominator 5 end fraction minus 2 close parentheses equals 2.210546386...

Depth is in feet, and time is in hours, so the units will be feet per hour

2.211 feet per hour (to 3 decimal places)

(c) It is given that f to the power of apostrophe open parentheses 12 close parentheses equals negative 2.261 and f to the power of apostrophe apostrophe end exponent open parentheses 12 close parentheses equals negative 0.322.

Explain the meaning of these two values in the context of the model.

Answer:

t equals 12 is 12 hours after midnight, so noon

f to the power of apostrophe open parentheses t close parentheses is the rate of change of the depth

f to the power of apostrophe open parentheses 12 close parentheses equals negative 2.261 means that at noon, the depth of water in the harbor is decreasing at a rate of 2.261 feet per hour.

f to the power of apostrophe apostrophe end exponent open parentheses t close parentheses is the rate of change of the rate of change of the depth

f to the power of apostrophe apostrophe end exponent open parentheses 12 close parentheses equals negative 0.322 means that at noon, the rate at which the depth of water in the harbor is changing, is decreasing at rate of 0.322 feet per hour, per hour.

Worked Example

The rate of change of the volume of water in a container is modeled by the function r open parentheses t close parentheses.

r open parentheses t close parentheses is measured in gallons per minute and t is measured in minutes.

(a) Explain the meaning of r open parentheses 0.1 close parentheses equals 2 in the context of the model.

Answer:

Note that in this problem, r open parentheses t close parentheses is modelling a rate, rather than an amount

The volume of water in the container at t=0.1 minutes (6 seconds) is increasing at a rate of 2 gallons per minute.

(b) At a particular time, r open parentheses t close parentheses is positive and r to the power of apostrophe open parentheses t close parentheses is negative. Explain what this means in the context of the model.

Answer:

r open parentheses t close parentheses models the rate of change of volume, while r to the power of apostrophe open parentheses t close parentheses models the rate of change of the rate of change (how fast it is increasing or decreasing)

The volume of water in the container is increasing, but at a decreasing rate.

(c) State the units for the quantity found by calculatingintegral r open parentheses t close parentheses space d t.

Answer:

Integrating the rate of change of a quantity will produce an expression for the change in the quantity

I.e. integral fraction numerator d y over denominator d x end fraction d x equals y plus C

So in this context we are integrating gallons per minute, with respect to minutes

The units of integral r open parentheses t close parentheses space d t will be gallons.

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    Author
    Reviewer
    Jamie Wood

    Author: Jamie Wood

    Expertise: Maths Content Creator

    Jamie graduated in 2014 from the University of Bristol with a degree in Electronic and Communications Engineering. He has worked as a teacher for 8 years, in secondary schools and in further education; teaching GCSE and A Level. He is passionate about helping students fulfil their potential through easy-to-use resources and high-quality questions and solutions.

    Dan Finlay

    Reviewer: Dan Finlay

    Expertise: Maths Subject Lead

    Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.