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Rates of Change & Related Rates

Emily Davis

Emily Davis

7 min read

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Study Guide Overview

This study guide covers the meaning of a derivative in context, focusing on its interpretation as a rate of change. It explains how to determine the units of the derivative, interpret it in exam questions, and distinguish between a function representing an amount versus a rate. The guide includes worked examples involving real-world scenarios like depth of water and volume change, along with practice questions and a glossary of key terms like derivative, instantaneous rate of change, and slope of the tangent.

Study Notes on Derivatives

Table of Contents

  1. Introduction to Derivatives
  2. Meaning of a Derivative
  3. Worked Examples
  4. Practice Questions
  5. Glossary
  6. Summary and Key Takeaways

Introduction to Derivatives

Key Concept

Derivatives are a fundamental concept in calculus that measure how a function changes as its input changes. They are essential for understanding rates of change and have numerous applications in various fields such as physics, engineering, and economics.


Meaning of a Derivative

What does the derivative mean?

  • The derivative of a function represents 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: y=3xy = 3x
  • The derivative, or the rate of change, is given by: dydx=3\frac{dy}{dx} = 3

This means for every 1 unit increase in xx, yy increases by 3 units. In this case, the rate of change is constant (3) at every point on the graph of yy against xx.

For a more complex example, consider: v=13t3v = \frac{1}{3}t^3
  • The derivative, or the rate of change, is: dvdt=t2\frac{dv}{dt} = t^2

This means vv changes at a rate of t2t^2. Here, the rate of change depends on tt.

  • At t=2t = 2, the rate of change is: 22=42^2 = 4
  • At t=5t = 5, the rate of change is: 52=255^2 = 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.

Units for the rate of change

  • The units for the rate of change are derived from the units of the dependent and independent variables.
For instance, if the volume of water in a tank (vv) is measured in liters and time (tt) in seconds: - The rate of change of volume, dvdt\frac{dv}{dt}, would be in liters per second.

Interpreting rate of change in exam questions

  • Read the scenario description carefully.
  • Determine if the function describes an amount or a rate of change.
"The volume of gasoline pumped is described by f(t)f(t)": - Here, f(t)f(t) represents the volume (amount), likely measured in gallons. - fโ€ฒ(t)f'(t) would describe the rate of change of volume, likely in gallons per second. "The rate of flow of gasoline is described by f(t)f(t)": - Here, f(t)f(t) represents a rate, likely in gallons per second. - fโ€ฒ(t)f'(t) describes the rate of change of the flow rate, likely in gallons per second squared.
Exam Tip

If unsure whether a function is a rate or an amount, checking the units can provide helpful clues.


Worked Examples

Example 1: Depth of Water in a Harbor

The depth of the water in a harbor, measured in feet, is modeled by the function: f(t)=6sinโก(2t5โˆ’2)+16for0โ‰คt<24f(t) = 6 \sin \left( \frac{2t}{5} - 2 \right) + 16 \quad \text{for} \quad 0 \le t < 24 where tt represents the number of hours after midnight.

(a) Maximum depth of water

Question: State the maximum depth of the water in the harbor according to the model.

Answer: The maximum value of sinโก(2t5โˆ’2)\sin\left( \frac{2t}{5} - 2 \right) is 1. Thus, the maximum depth is: 6(1)+16=22ย feet6(1) + 16 = 22 \text{ feet}

(b) Rate of change at 6 am

Question: 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 is given by fโ€ฒ(t)f'(t): fโ€ฒ(t)=6cosโก(2t5โˆ’2)โ‹…25=125cosโก(2t5โˆ’2)f'(t) = 6 \cos \left( \frac{2t}{5} - 2 \right) \cdot \frac{2}{5} = \frac{12}{5} \cos \left( \frac{2t}{5} - 2 \right)

At 6 am (t=6t = 6): fโ€ฒ(6)=125cosโก(2(6)5โˆ’2)โ‰ˆ2.211ย feetย perย hourf'(6) = \frac{12}{5} \cos \left( \frac{2(6)}{5} - 2 \right) \approx 2.211 \text{ feet per hour}

(c) Interpretation of given derivatives

Question: Explain the meaning of fโ€ฒ(12)=โˆ’2.261f'(12) = -2.261 and fโ€ฒโ€ฒ(12)=โˆ’0.322f''(12) = -0.322 in the context of the model.

Answer:

  • fโ€ฒ(12)=โˆ’2.261f'(12) = -2.261 means that at noon, the depth of water is decreasing at a rate of 2.261 feet per hour.
  • fโ€ฒโ€ฒ(12)=โˆ’0.322f''(12) = -0.322 means that at noon, the rate at which the depth is decreasing is itself decreasing by 0.322 feet per hour per hour.

Example 2: Volume of Water in a Container

The rate of change of the volume of water in a container is modeled by the function r(t)r(t), where tt is measured in minutes.

(a) Interpretation of r(0.1)=2r(0.1) = 2

Question: Explain the meaning of r(0.1)=2r(0.1) = 2 in the context of the model.

Answer: r(t)r(t) models the rate of change of volume. Thus, at t=0.1t = 0.1 minutes (6 seconds), the volume of water is increasing at a rate of 2 gallons per minute.

(b) Interpretation of r(t)r(t) being positive and rโ€ฒ(t)r'(t) negative

Question: Explain what it means if r(t)r(t) is positive and rโ€ฒ(t)r'(t) is negative.

Answer: The volume of water is increasing, but the rate of increase is slowing down.

(c) Units of โˆซr(t),dt\int r(t) , dt

Question: State the units for the quantity found by calculating โˆซr(t),dt\int r(t) , dt.

Answer: Integrating the rate of change of volume gives the total change in volume. Thus, the units of โˆซr(t),dt\int r(t) , dt will be gallons.


Practice Questions

Practice Question

Question 1: Find the derivative of f(x)=5x2+3xโˆ’7f(x) = 5x^2 + 3x - 7.

Practice Question

Question 2: If g(t)=4t3โˆ’2t+1g(t) = 4t^3 - 2t + 1, what is the rate of change of g(t)g(t) at t=2t = 2?

Practice Question

Question 3: The rate of change of the temperature in a room is given by $$ h(t) = 3 \sin(t) ). What is the maximum rate of change?


Glossary

  • Derivative: The rate of change of a function with respect to its independent variable.
  • Rate of Change: How a quantity changes with respect to another quantity.
  • Instantaneous Rate of Change: The rate of change at a specific point.
  • Slope of the Tangent: The slope of the line that just touches a curve at a given point, representing the derivative at that point.

Summary and Key Takeaways

  • The derivative of a function measures its rate of change.
  • The units of the rate of change depend on the units of the dependent and independent variables.
  • Interpreting a rate of change requires understanding whether the function describes an amount or a rate.
  • The slope of the tangent to a curve at a point gives the instantaneous rate of change at that point.
  • Practice interpreting and calculating derivatives to gain proficiency.

Key Takeaways

  • Understand that the derivative represents how a function changes as its input changes.
  • Be able to calculate derivatives and interpret their meanings in real-world contexts.
  • Recognize the units associated with rates of change and how they relate to the problem context.
  • Use derivative concepts to solve practical problems, such as finding rates of change and interpreting graphs.

By mastering these concepts, you'll be well-prepared to understand and apply derivatives in various contexts, both in exams and real-world scenarios.

Question 1 of 9

What does the derivative of a function, f(x)f(x), represent? ๐Ÿค”

The area under the curve of f(x)f(x)

The rate of change of f(x)f(x) with respect to xx

The inverse of f(x)f(x)

The y-intercept of f(x)f(x)