Core Concept
What is Rate of Change?
If a quantity y depends on another quantity x, then the derivative dy/dx represents how fast y changes as x changes. It is the instantaneous rate of change of y with respect to x.
Instantaneous Rate of Change
dy/dx = limΔx→0 Δy/Δx
Key Points
- 1dy/dx at x = a gives the rate of change of y at that specific point x = a.
- 2Positive dy/dx → y is increasing as x increases.
- 3Negative dy/dx → y is decreasing as x increases.
- 4If both x and y are functions of time t, use the chain rule: dy/dt = (dy/dx)·(dx/dt).
- 5Units of rate of change = units of y / units of x.
Remember: dy/dx is not a fraction — it's a limit. But in chain rule, you can treat it like one.
Useful Results
Volume of Sphere
V = (4/3)πr³
dV/dr = 4πr²
dV/dr = 4πr²
Area of Circle
A = πr²
dA/dr = 2πr
dA/dr = 2πr
Volume of Cube
V = x³
dV/dx = 3x²
dV/dx = 3x²
Surface Area of Sphere
S = 4πr²
dS/dr = 8πr
dS/dr = 8πr
Solved Examples
Example 01
Rate of change of area of circle
Q. The radius of a circle is increasing at 3 cm/s. Find the rate at which area is increasing when r = 5 cm.
Given: dr/dt = 3 cm/s, r = 5 cm
A = πr² → dA/dt = 2πr · (dr/dt)
dA/dt = 2π(5)(3) = 30π
∴ Rate of increase of area = 30π cm²/s
Example 02
Rate of change of volume of sphere
Q. The volume of a sphere is increasing at 8 cm³/s. Find the rate of increase of radius when r = 2 cm.
V = (4/3)πr³ → dV/dt = 4πr² · (dr/dt)
8 = 4π(2)²· (dr/dt)
8 = 16π · (dr/dt) → dr/dt = 8/(16π) = 1/(2π)
∴ Rate of increase of radius = 1/(2π) cm/s
Example 03
Rate of change on a curve
Q. Find the rate of change of y = x² + 2x with respect to x at x = 3.
y = x² + 2x → dy/dx = 2x + 2
At x = 3: dy/dx = 2(3) + 2 = 8
∴ Rate of change = 8 (y increases 8 units per unit increase in x)