Rates of Change & Connected Rates

A derivative is a rate of change, and “connected rates” questions link a rate you know to a rate you want through the chain rule:

$\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt}$

The single most important habit: write the chain symbolically before any numbers. That line is the method mark, and it forces clarity about which derivative comes from where.

The standard question, dissected

The radius of a circular oil patch grows at 0.5 cm/s. Find the rate at which the area grows when $r = 12$ cm. Chain: $\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt}$ (M, the plan, stated) Geometry: $A = \pi r^2 \to \frac{dA}{dr} = 2\pi r$ (M, differentiate the link formula) Substitute at the instant: $\frac{dA}{dt} = 2\pi(12) \times 0.5 =$ $12\pi \text{ cm}^2/\text{s}$ (A)

Three sources feed the chain: the given rate ($\frac{dr}{dt}$, from the words “grows at 0.5 cm/s”), the link formula (geometry: $A = \pi r^2$, $V = \frac{4}{3}\pi r^3$, similar-triangle relations), and the instant ($r = 12$, substituted after differentiating, never before). Differentiating after substituting numbers is the fatal reversal: a constant differentiates to zero and the question evaporates.

Decreasing quantities: negative rates

“The volume decreases at $8 \text{ cm}^3/\text{s}$” means $\frac{dV}{dt} = -8$. Carry the sign through the chain; a negative final answer then means something, and the interpretation sentence (“the radius is decreasing at …”) is often the closing mark. Watch units throughout, $\text{cm}^2/\text{s}$ for areas, $\text{cm}^3/\text{s}$ for volumes; stating them is part of the answer.

Rearranged chains and small changes

Sometimes the unknown sits in the middle: given $\frac{dV}{dt}$ and wanting $\frac{dr}{dt}$, the chain rearranges to $\frac{dr}{dt} = \frac{dV/dt}{dV/dr}$, write the chain first and the algebra sorts itself. The cousin topic, small changes, uses $\delta y \approx \frac{dy}{dx} \times \delta x$ for “find the approximate increase in $y$ when $x$ increases from 4 by 0.02”, same logic, one instant, no time variable.

Common mistakes

• Numbers substituted before differentiating
• The chain never written, so a wrong derivative slots in unnoticed
• Decrease rates entered as positive
• The link formula misdifferentiated. $V = \frac{4}{3}\pi r^3$ gives $\frac{dV}{dr} = 4\pi r^2$, not $\frac{4}{3}\pi r^2$ (the power must multiply down)
• Units missing from the final answer

Full topic context: Calculus notes.