AP Calculus BC 4.1 Interpreting the Meaning of the Derivative in Context Study Notes - New Syllabus
AP Calculus BC 4.1 Interpreting the Meaning of the Derivative in Context Study Notes- New syllabus
AP Calculus BC 4.1 Interpreting the Meaning of the Derivative in Context Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- Derivatives allow us to solve real-world problems involving rates of change.
Key Concepts:
- Derivative as Instantaneous Rate of Change
- Using Derivatives in Applied Contexts
- Units in Derivatives
Derivative as Instantaneous Rate of Change
Derivative as Instantaneous Rate of Change
The derivative of a function at a point gives the instantaneous rate of change of the function with respect to its independent variable at that specific point.
In simpler terms, it tells us how fast the function is changing right now at a particular input value, not over an interval. Graphically, this is the slope of the tangent line at that point on the curve.
This is different from average rate of change, which considers the slope between two points on the graph.
Instantaneous Rate of Change
The instantaneous rate of change at a point \( P = (x_1, f(x_1)) \) on the graph of the function \( f(x) \) is how much the graph is changing at this particular point. Let the tangent line to \( f(x) \) at \( x = x_1 \) be given by
\( y = ax + b \)
Then, the instantaneous rate of change at this point is equal to the slope of the tangent. Therefore,
Instantaneous rate of change = \( a \)
For a function \( f(x) \) and a point \( P \) on the graph, you can choose two points \( (x_1, y_1) \) and \( (x_2, y_2) \) on the tangent at \( P \). Then
Instantaneous rate of change = \( \dfrac{\text{change in } y}{\text{change in } x} = \dfrac{y_2 – y_1}{x_2 – x_1} \)
If \( (x_1, y_1) = P \), then:
\( \dfrac{y_2 – y_1}{x_2 – x_1} = \dfrac{y_2 – f(x_1)}{x_2 – x_1} \)
Example:
The height \( h(t) \) in meters of a ball thrown upward is given by:
\( h(t) = -5t^2 + 20t + 1 \)
Find the instantaneous rate of change of height at \( t = 2 \) seconds.
▶️Answer/Explanation
- Differentiate: \( h'(t) = \frac{d}{dt}(-5t^2 + 20t + 1) = -10t + 20 \)
- Substitute \( t = 2 \): \( h'(2) = -10(2) + 20 = -20 + 20 = 0 \)
Final Answer: At \( t = 2 \), the instantaneous rate of change is 0 m/s — the ball is momentarily at its highest point.
Example:
The graph below shows the function \( f(x) \).
Estimate the instantaneous rate of change at \( x = 1 \) by finding the slope of the tangent line at that point.
▶️Answer/Explanation
The tangent line at \( x = 1 \) appears to pass through the points \( (0, 1) \) and \( (2, 5) \).
- Find slope: \( m = \frac{5 – 1}{2 – 0} = \frac{4}{2} = 2 \)
Final Answer: The instantaneous rate of change at \( x = 1 \) is approximately 2.
Using Derivatives in Applied Contexts
Using Derivatives in Applied Contexts
The derivative of a function represents the rate of change of a quantity with respect to another.
In real-world (applied) contexts, it is used to describe things like:
- The rate at which a car’s position changes — velocity.
- The rate at which water drains from a tank — flow rate.
- The rate at which a population grows — growth rate.
- The rate at which a cost function changes — marginal cost.
Example: A balloon is being inflated, and its volume (in cm³) at time \( t \) seconds is given by:
\( V(t) = 4t^2 + 2t \)
Find the rate at which the volume is increasing when \( t = 3 \) seconds.
▶️Answer/Explanation
- Differentiate: \( V'(t) = \frac{d}{dt}(4t^2 + 2t) = 8t + 2 \)
- Substitute \( t = 3 \): \( V'(3) = 8(3) + 2 = 24 + 2 = 26 \)
Final Answer: The volume is increasing at a rate of 26 cm³/sec at \( t = 3 \) seconds.
Example:
A car’s position is modeled by \( s(t) = t^3 – 6t^2 + 9t \), where \( s \) is in meters and \( t \) is in seconds.
Explain the difference between the average rate of change from \( t = 1 \) to \( t = 3 \) and the instantaneous rate at \( t = 2 \).
▶️Answer/Explanation
- Average Rate: \( \frac{s(3) – s(1)}{3 – 1} = \frac{(27 – 54 + 27) – (1 – 6 + 9)}{2} = \frac{0 – 4}{2} = -2 \, \text{m/s} \)
- Instantaneous Rate: First, differentiate: \( s'(t) = 3t^2 – 12t + 9 \). Then evaluate: \( s'(2) = 3(4) – 12(2) + 9 = 12 – 24 + 9 = -3 \, \text{m/s} \)
Conclusion: The average rate over the interval is -2 m/s, but at exactly \( t = 2 \), the car’s speed is -3 m/s — showing the car is slowing down more rapidly at that instant.
Units in Derivatives
Understanding Units in Derivatives
The derivative \( f'(x) \) represents the rate of change of a function \( f(x) \) with respect to the variable \( x \). Therefore:
The unit of \( f'(x) \) is:
$\dfrac{\mathrm{Unit ~~of ~~( f(x)}}{ \mathrm{ Unit ~~of ~~( x )}}$
This is because derivatives measure how much \( f \) changes per unit of \( x \).
Example:
A car travels a distance \( s(t) \), where \( s \) is measured in meters (m) and time \( t \) is measured in seconds (s). The function is:
\( s(t) = 10t^2 \)
Then the derivative \( s'(t) \) gives the velocity.
▶️Answer/Explanation
- \( s'(t) = \frac{d}{dt}(10t^2) = 20t \)
- Units: \( \text{meters} \div \text{seconds} = \text{m/s} \)
Final Answer: The unit of \( s'(t) \) is m/s (meters per second).
Example:
A company’s cost function is defined as \( C(x) \), where \( x \) is the number of items produced, and \( C \) is the cost in dollars ($\$$).
Suppose the cost function is given by:
\( C(x) = 1000 + 5x + 0.01x^2 \)
This function gives the total cost to produce \( x \) items. The marginal cost is the derivative \( C'(x) \), which represents the cost to produce one more item.
▶️Answer/Explanation
- \( C'(x) = \frac{d}{dx}(1000 + 5x + 0.01x^2) = 5 + 0.02x \)
- Units of \( C(x) \) = dollars ($), units of \( x \) = items
- Units of \( C'(x) \) = dollars per item ($\$/$item)
Interpretation: At any value of \( x \), the derivative tells how many more dollars it will cost to produce one additional item.
Example: If the company is producing 100 items, then:
- \( C'(100) = 5 + 0.02(100) = 7 \)
Final Answer: The marginal cost is $\$7$ per item — meaning it will cost about $\$7$ to produce the 101st item.