Differentiation – Rate of Change

Rate of change measures how fast one quantity changes when another quantity changes.

Examples of physical meaning:

• \(\dfrac{ds}{dt}\) → rate of change of distance (speed)
• \(\dfrac{dV}{dt}\) → rate of change of volume
• \(\dfrac{dA}{dt}\) → rate of change of area

To find a rate of change:

1. Differentiate to find \(\frac{dy}{dx}\).
2. Substitute the given value of \(x\).
3. State correct units (if context given).

The volume of a cube is \(V = x^3\). Find the rate of change of volume when \(x = 4\text{ cm}\).

Step 1: \[ V = x^3 \Rightarrow \frac{dV}{dx} = 3x^2 \] Step 2: \[ \frac{dV}{dx}\bigg|_{x=4} = 3(4)^2 = 48 \] Interpretation: Volume increases at \(48 \text{ cm}^3/\text{cm}\).

The area of a circle is \(A = \pi r^2\). Find the rate of change of area when \(r = 5\text{ cm}\).

Step 1: \[ \frac{dA}{dr} = 2\pi r \] Step 2: \[ \frac{dA}{dr}\bigg|_{r=5} = 2\pi(5) = 10\pi \] Interpretation: Area increases at \(10\pi \text{ cm}^2/\text{cm}\).

A quantity is given by \(y = \frac{4}{x^2}\). Find the rate of change of \(y\) when \(x = 2\).

Rewrite: \[ y = 4x^{-2} \] Differentiate: \[ \frac{dy}{dx} = -8x^{-3} \] Substitute: \[ \frac{dy}{dx}\bigg|_{x=2} = -8(2)^{-3} = -8\left(\frac{1}{8}\right) = -1 \] Interpretation: The quantity is decreasing at a rate of \(1\) per unit of \(x\).

Sometimes the variable depends on time. Use:

Example: If \(A = x^2\) and \(\frac{dx}{dt} = 3\), find \(\frac{dA}{dt}\) when \(x=2\).

\[ \frac{dA}{dx} = 2x \Rightarrow \frac{dA}{dt} = 2x \cdot 3 = 6x = 12 \]

• Rewrite roots and fractions as powers before differentiating.
• State units when context is given.
• Use the chain rule when rate of change involves time.
• Only differentiate functions that can be written as \(kx^n\) (not \(x^{-1}\)).
• Always evaluate at the specified value.