Derivative as Rate of Change
Understanding derivatives through real-world rates of change
The Derivative as Rate of Change
Derivatives measure how fast things change. This is their most powerful real-world application!
Average vs. Instantaneous Rate of Change
Average rate of change (over an interval):
Instantaneous rate of change (at a point):
The derivative is the instantaneous rate of change!
Motion: Position, Velocity, Acceleration
This is where derivatives truly shine!
If is position (where you are):
- is velocity (how fast you're moving)
- is acceleration (how fast your speed is changing)
The derivative hierarchy:
Example: Free Fall
An object is dropped from a building. Its height is meters (where is in seconds).
Find the velocity:
Find velocity at seconds:
The negative sign means downward!
Find the acceleration:
This is constant (gravity)!
Sign of the Derivative
The sign tells you the direction of change:
| | Meaning | |---------|---------| | | is increasing | | | is decreasing | | | is stationary (not changing) |
Example: If , you're moving in the positive direction. If , you're moving in the negative direction.
Units Matter!
The derivative's units are:
Examples:
- Position in meters, time in seconds → Velocity in m/s
- Cost in dollars, quantity in items → Marginal cost in $/item
- Temperature in °C, time in hours → Rate of temperature change in °C/hr
Related Rates
When two quantities both depend on time, their rates of change are related through derivatives!
Example: A balloon is being inflated. The radius increases at 2 cm/s. How fast is the volume changing?
We know:
- cm/s (given)
- (volume formula)
Take the derivative with respect to time:
At cm:
Economics: Marginal Analysis
In economics, "marginal" means "derivative"!
Marginal cost: = cost to produce one more unit Marginal revenue: = revenue from selling one more unit Marginal profit:
Example: If dollars:
At units: C'(100) = 5 + 2 = \7$ per unit
Population Growth
If is population at time :
- : Population growing
- : Population shrinking
- : Rate of growth/decline
Example: (exponential growth)
The population is always growing (positive derivative)!
Optimization Preview
To maximize or minimize something:
- Find where (critical points)
- Test to see if it's a max or min
- Check endpoints if on a closed interval
We'll explore this deeply in later lessons!
The Power of Derivatives
Derivatives let us:
- ✓ Predict future behavior
- ✓ Find optimal solutions
- ✓ Understand dynamic systems
- ✓ Model real-world phenomena
- ✓ Make informed decisions
Calculus transforms static snapshots into dynamic understanding!
📚 Practice Problems
1Problem 1medium
❓ Question:
A particle moves along a line with position meters at time seconds. Find the velocity and acceleration at seconds.
💡 Show Solution
Step 1: Find velocity (first derivative)
At :
The particle is moving at 3 m/s in the negative direction.
Step 2: Find acceleration (second derivative)
At :
The acceleration is zero at - the particle is neither speeding up nor slowing down at that instant.
Summary:
- Velocity at : -3 m/s (moving backward)
- Acceleration at : 0 m/s² (constant velocity)
2Problem 2hard
❓ Question:
The temperature of coffee is given by degrees Celsius, where is in minutes. How fast is the temperature changing at minutes?
💡 Show Solution
We need to find to get the rate of temperature change.
Step 1: Find the derivative
The derivative of 20 is 0. Using the chain rule:
Step 2: Evaluate at t = 10
Answer: The temperature is decreasing at about 2.21 °C per minute at minutes.
The negative sign indicates the coffee is cooling down (as expected!).
3Problem 3easy
❓ Question:
If the radius of a circle is r = 5 cm and increasing at dr/dt = 2 cm/s, find how fast the area is changing.
💡 Show Solution
Step 1: Write area formula: A = πr²
Step 2: Find dA/dr: dA/dr = 2πr
Step 3: Use chain rule: dA/dt = (dA/dr)·(dr/dt)
Step 4: Substitute values: r = 5, dr/dt = 2 dA/dt = 2π(5)·2 = 20π
Step 5: Interpret: The area is increasing at 20π ≈ 62.83 cm²/s
Answer: 20π cm²/s
4Problem 4medium
❓ Question:
A car's position is s(t) = t³ - 6t² + 9t meters at time t seconds. Find the velocity and acceleration at t = 2.
💡 Show Solution
Step 1: Velocity is the derivative of position: v(t) = s'(t) = 3t² - 12t + 9
Step 2: Find velocity at t = 2: v(2) = 3(2)² - 12(2) + 9 = 12 - 24 + 9 = -3 m/s
Step 3: Acceleration is the derivative of velocity: a(t) = v'(t) = s''(t) = 6t - 12
Step 4: Find acceleration at t = 2: a(2) = 6(2) - 12 = 12 - 12 = 0 m/s²
Step 5: Interpretation: At t = 2, the car is moving backward at 3 m/s but the acceleration is 0 (instantaneously not speeding up or slowing down)
Answer: v(2) = -3 m/s, a(2) = 0 m/s²
5Problem 5hard
❓ Question:
The temperature T (in °F) of a cooling object after t minutes is T(t) = 80 + 120e^(-0.1t). Find the rate of cooling at t = 10 minutes.
💡 Show Solution
Step 1: Find dT/dt: T(t) = 80 + 120e^(-0.1t) T'(t) = 120·e^(-0.1t)·(-0.1) T'(t) = -12e^(-0.1t)
Step 2: Evaluate at t = 10: T'(10) = -12e^(-0.1·10) = -12e^(-1) = -12/e
Step 3: Calculate numerical value: -12/e ≈ -12/2.71828 ≈ -4.41 °F/min
Step 4: Interpret: The negative sign means temperature is decreasing At t = 10 minutes, cooling at about 4.41°F per minute
Answer: -12/e ≈ -4.41 °F/min (cooling rate)
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