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): f(b)f(a)ba\frac{f(b) - f(a)}{b - a}

Instantaneous rate of change (at a point): f(a)=limh0f(a+h)f(a)hf'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}

The derivative is the instantaneous rate of change!

Motion: Position, Velocity, Acceleration

This is where derivatives truly shine!

If s(t)s(t) is position (where you are):

  • s(t)=v(t)s'(t) = v(t) is velocity (how fast you're moving)
  • v(t)=s(t)=a(t)v'(t) = s''(t) = a(t) is acceleration (how fast your speed is changing)

The derivative hierarchy: PositionderivativeVelocityderivativeAcceleration\text{Position} \xrightarrow{\text{derivative}} \text{Velocity} \xrightarrow{\text{derivative}} \text{Acceleration}

Example: Free Fall

An object is dropped from a building. Its height is h(t)=1004.9t2h(t) = 100 - 4.9t^2 meters (where tt is in seconds).

Find the velocity: v(t)=h(t)=9.8t m/sv(t) = h'(t) = -9.8t \text{ m/s}

Find velocity at t=2t = 2 seconds: v(2)=9.8(2)=19.6 m/sv(2) = -9.8(2) = -19.6 \text{ m/s}

The negative sign means downward!

Find the acceleration: a(t)=v(t)=h(t)=9.8 m/s2a(t) = v'(t) = h''(t) = -9.8 \text{ m/s}^2

This is constant (gravity)!

Sign of the Derivative

The sign tells you the direction of change:

| f(x)f'(x) | Meaning | |---------|---------| | f(x)>0f'(x) > 0 | ff is increasing | | f(x)<0f'(x) < 0 | ff is decreasing | | f(x)=0f'(x) = 0 | ff is stationary (not changing) |

Example: If v(t)>0v(t) > 0, you're moving in the positive direction. If v(t)<0v(t) < 0, you're moving in the negative direction.

Units Matter!

The derivative's units are: units of outputunits of input\frac{\text{units of output}}{\text{units of input}}

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:

  • drdt=2\frac{dr}{dt} = 2 cm/s (given)
  • V=43πr3V = \frac{4}{3}\pi r^3 (volume formula)

Take the derivative with respect to time: dVdt=4πr2drdt\frac{dV}{dt} = 4\pi r^2 \cdot \frac{dr}{dt}

At r=5r = 5 cm: dVdt=4π(5)2(2)=200π cm3/s\frac{dV}{dt} = 4\pi(5)^2(2) = 200\pi \text{ cm}^3\text{/s}

Economics: Marginal Analysis

In economics, "marginal" means "derivative"!

Marginal cost: C(x)C'(x) = cost to produce one more unit Marginal revenue: R(x)R'(x) = revenue from selling one more unit Marginal profit: P(x)=R(x)C(x)P'(x) = R'(x) - C'(x)

Example: If C(x)=1000+5x+0.01x2C(x) = 1000 + 5x + 0.01x^2 dollars: C(x)=5+0.02xC'(x) = 5 + 0.02x

At x=100x = 100 units: C'(100) = 5 + 2 = \7$ per unit

Population Growth

If P(t)P(t) is population at time tt:

  • P(t)>0P'(t) > 0: Population growing
  • P(t)<0P'(t) < 0: Population shrinking
  • P(t)|P'(t)|: Rate of growth/decline

Example: P(t)=1000e0.02tP(t) = 1000e^{0.02t} (exponential growth) P(t)=20e0.02tP'(t) = 20e^{0.02t}

The population is always growing (positive derivative)!

Optimization Preview

To maximize or minimize something:

  1. Find where f(x)=0f'(x) = 0 (critical points)
  2. Test to see if it's a max or min
  3. 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 s(t)=t36t2+9ts(t) = t^3 - 6t^2 + 9t meters at time tt seconds. Find the velocity and acceleration at t=2t = 2 seconds.

💡 Show Solution

Step 1: Find velocity (first derivative)

v(t)=s(t)=3t212t+9v(t) = s'(t) = 3t^2 - 12t + 9

At t=2t = 2: v(2)=3(2)212(2)+9=1224+9=3 m/sv(2) = 3(2)^2 - 12(2) + 9 = 12 - 24 + 9 = -3 \text{ m/s}

The particle is moving at 3 m/s in the negative direction.

Step 2: Find acceleration (second derivative)

a(t)=v(t)=s(t)=6t12a(t) = v'(t) = s''(t) = 6t - 12

At t=2t = 2: a(2)=6(2)12=1212=0 m/s2a(2) = 6(2) - 12 = 12 - 12 = 0 \text{ m/s}^2

The acceleration is zero at t=2t = 2 - the particle is neither speeding up nor slowing down at that instant.

Summary:

  • Velocity at t=2t = 2: -3 m/s (moving backward)
  • Acceleration at t=2t = 2: 0 m/s² (constant velocity)

2Problem 2hard

Question:

The temperature of coffee is given by T(t)=20+60e0.1tT(t) = 20 + 60e^{-0.1t} degrees Celsius, where tt is in minutes. How fast is the temperature changing at t=10t = 10 minutes?

💡 Show Solution

We need to find T(10)T'(10) to get the rate of temperature change.

Step 1: Find the derivative

T(t)=ddt[20+60e0.1t]T'(t) = \frac{d}{dt}[20 + 60e^{-0.1t}]

The derivative of 20 is 0. Using the chain rule: ddt[e0.1t]=e0.1t(0.1)\frac{d}{dt}[e^{-0.1t}] = e^{-0.1t} \cdot (-0.1)

T(t)=60(0.1)e0.1t=6e0.1tT'(t) = 60 \cdot (-0.1)e^{-0.1t} = -6e^{-0.1t}

Step 2: Evaluate at t = 10

T(10)=6e0.1(10)=6e1=6e2.21T'(10) = -6e^{-0.1(10)} = -6e^{-1} = -\frac{6}{e} \approx -2.21

Answer: The temperature is decreasing at about 2.21 °C per minute at t=10t = 10 minutes.

The negative sign indicates the coffee is cooling down (as expected!).