• The derivative, or the rate of change, is: $$\frac{dv}{dt} = t^2$$

This means $$v$$ changes at a rate of $$t^2$$. Here, the rate of change depends on $$t$$.

• At $$t = 2$$, the rate of change is: $$2^2 = 4$$
• At $$t = 5$$, the rate of change is: $$5^2 = 25$$

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.

Interpreting rate of change in exam questions

• Read the scenario description carefully.
• Determine if the function describes an amount or a rate of change.

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: $$f(t) = 6 \sin \left( \frac{2t}{5} - 2 \right) + 16 \quad \text{for} \quad 0 \le t < 24$$ where $$t$$ 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 $$\sin\left( \frac{2t}{5} - 2 \right)$$ is 1. Thus, the maximum depth is: $$6(1) + 16 = 22 \text{ feet}$$

(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 $$f'(t)$$: $$f'(t) = 6 \cos \left( \frac{2t}{5} - 2 \right) \cdot \frac{2}{5} = \frac{12}{5} \cos \left( \frac{2t}{5} - 2 \right)$$

At 6 am ($t = 6$): $$f'(6) = \frac{12}{5} \cos \left( \frac{2(6)}{5} - 2 \right) \approx 2.211 \text{ feet per hour}$$

(c) Interpretation of given derivatives

Question: Explain the meaning of $$f'(12) = -2.261$$ and $$f''(12) = -0.322$$ in the context of the model.

Answer:

• $$f'(12) = -2.261$$ means that at noon, the depth of water is decreasing at a rate of 2.261 feet per hour.
• $$f''(12) = -0.322$$ 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 $$r(t)$$, where $$t$$ is measured in minutes.

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

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

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

(b) Interpretation of $$r(t)$$ being positive and $$r'(t)$$ negative

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

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

(c) Units of $$\int r(t) , dt$$

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

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

Practice Questions

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.