2.2 – Derivatives in Context (Motion & Rates of Change)
Learning Objectives
By the end of this section, you should be able to:
- Interpret derivatives as rates of change in real-world contexts
- Apply derivatives to motion problems (position, velocity, acceleration)
- Solve problems involving related rates
- Connect graphical representations with physical interpretations
- Analyze units of measurement for derivatives in applied contexts
Derivatives as Rates of Change
The derivative \( f'(x) \) represents the instantaneous rate of change of \( f(x) \) with respect to \( x \). In applied contexts, this concept becomes powerful for modeling real-world phenomena.
If \( y = f(x) \), then \( \frac{dy}{dx} \) represents how much \( y \) changes per unit change in \( x \).
Motion: Position, Velocity, and Acceleration
For an object moving along a straight line with position function \( s(t) \):
- Velocity: \( v(t) = s'(t) \) (rate of change of position)
- Acceleration: \( a(t) = v'(t) = s''(t) \) (rate of change of velocity)
| Function | Derivative | Physical Meaning | Typical Units |
|---|---|---|---|
| \( s(t) \) - Position | \( s'(t) = v(t) \) | Instantaneous velocity | m/s, ft/s |
| \( v(t) \) - Velocity | \( v'(t) = a(t) \) | Instantaneous acceleration | m/s², ft/s² |
| \( a(t) \) - Acceleration | \( a'(t) = j(t) \) | Jerk (rate of acceleration change) | m/s³, ft/s³ |
Worked Examples: Motion Problems
Example 1: Position to Velocity and Acceleration
Motion AnalysisProblem: A particle moves along a straight line with position function \( s(t) = t^3 - 6t^2 + 9t + 2 \) meters, where \( t \) is in seconds.
Find: (a) The velocity function (b) The acceleration function (c) When the particle is at rest
Find the velocity function:
\[ v(t) = s'(t) = \frac{d}{dt}[t^3 - 6t^2 + 9t + 2] \]
\[ v(t) = 3t^2 - 12t + 9 \]
Find the acceleration function:
\[ a(t) = v'(t) = \frac{d}{dt}[3t^2 - 12t + 9] \]
\[ a(t) = 6t - 12 \]
Find when the particle is at rest:
The particle is at rest when velocity is zero:
\[ v(t) = 3t^2 - 12t + 9 = 0 \]
\[ 3(t^2 - 4t + 3) = 0 \]
\[ 3(t-1)(t-3) = 0 \]
\[ t = 1 \text{ or } t = 3 \text{ seconds} \]
Interpretation: The particle is at rest at \( t = 1 \)s and \( t = 3 \)s, meaning it changes direction at these times.
Example 2: Related Rates Problem
Geometry ApplicationProblem: A spherical balloon is being inflated at a rate of 10 cm³/s. How fast is the radius increasing when the radius is 5 cm?
Identify known values and what we need to find:
Volume of sphere: \( V = \frac{4}{3}\pi r^3 \)
Given: \( \frac{dV}{dt} = 10 \) cm³/s
Find: \( \frac{dr}{dt} \) when \( r = 5 \) cm
Differentiate both sides with respect to time:
\[ \frac{d}{dt}[V] = \frac{d}{dt}\left[\frac{4}{3}\pi r^3\right] \]
\[ \frac{dV}{dt} = \frac{4}{3}\pi \cdot 3r^2 \frac{dr}{dt} \]
\[ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \]
Solve for \( \frac{dr}{dt} \):
\[ \frac{dr}{dt} = \frac{dV}{dt} \cdot \frac{1}{4\pi r^2} \]
Substitute known values:
\[ \frac{dr}{dt} = 10 \cdot \frac{1}{4\pi (5)^2} \]
\[ \frac{dr}{dt} = \frac{10}{100\pi} = \frac{1}{10\pi} \text{ cm/s} \]
\[ \frac{dr}{dt} \approx 0.0318 \text{ cm/s} \]
Key Insight: Related rates problems involve finding how different quantities change together over time using the chain rule.
Example 3: Economic Application
Marginal AnalysisProblem: A company's cost function is \( C(x) = 0.01x^3 - 0.3x^2 + 5x + 100 \), where \( x \) is the number of units produced.
Find the marginal cost when producing 10 units and interpret the result.
Find the marginal cost function:
Marginal cost is the derivative of the cost function:
\[ C'(x) = \frac{d}{dx}[0.01x^3 - 0.3x^2 + 5x + 100] \]
\[ C'(x) = 0.03x^2 - 0.6x + 5 \]
Evaluate at \( x = 10 \):
\[ C'(10) = 0.03(10)^2 - 0.6(10) + 5 \]
\[ C'(10) = 0.03(100) - 6 + 5 \]
\[ C'(10) = 3 - 6 + 5 = 2 \]
Interpret the result:
The marginal cost at 10 units is $2 per additional unit.
This means producing the 11th unit would cost approximately $2 more than producing the 10th unit.
Economic Insight: Marginal cost represents the approximate cost of producing one additional unit, which helps businesses make production decisions.
Real-World Applications of Derivatives
| Field | Application | Derivative Represents |
|---|---|---|
| Physics | Motion analysis | Velocity, acceleration |
| Economics | Cost analysis | Marginal cost, revenue |
| Biology | Population growth | Growth rate |
| Engineering | Structural design | Rate of stress change |
| Medicine | Drug concentration | Rate of absorption |
Problem-Solving Strategy: When solving applied derivative problems:
- Identify what you know and what you need to find
- Write the relevant equation(s)
- Differentiate with respect to the appropriate variable
- Substitute known values
- Solve for the unknown rate
- Interpret your answer in context
AP Exam Tip
On the AP Calculus exam, applied derivative problems often appear in the free-response section. Remember to:
- Include units in your final answer
- Show all steps of your work
- Interpret your results in the context of the problem
- Use proper notation (\( \frac{dy}{dx} \), \( f'(x) \), etc.)
- Check that your answer makes sense physically
Check Yourself
1. A car's position is given by \( s(t) = 2t^2 + 3t + 5 \) meters. Find its velocity at \( t = 4 \) seconds.
\[ v(t) = s'(t) = 4t + 3 \]
\[ v(4) = 4(4) + 3 = 16 + 3 = 19 \text{ m/s} \]
2. The area of a circle is increasing at 5 cm²/s. How fast is the radius increasing when the radius is 3 cm?
\[ A = \pi r^2 \]
\[ \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \]
\[ 5 = 2\pi (3) \frac{dr}{dt} \]
\[ \frac{dr}{dt} = \frac{5}{6\pi} \approx 0.265 \text{ cm/s} \]
3. A company's revenue function is \( R(x) = 50x - 0.1x^2 \). Find the marginal revenue when selling 100 units.
\[ R'(x) = 50 - 0.2x \]
\[ R'(100) = 50 - 0.2(100) = 50 - 20 = 30 \]
Marginal revenue is $30 per additional unit.