Rates of change

If a quantity $y$ depends on time $t$, either directly or through another variable $x$, then $\dfrac{dy}{dt}$ is its instantaneous rate of change — positive if $y$ is increasing, negative if it is decreasing. Most physical situations do not hand you $y$ as an explicit function of $t$; instead they hand you a relationship between two changing quantities, such as a geometric formula, and tell you how fast one of them is changing. The bridge between the two is the chain rule. If $y = f(x)$ and both $x$ and $y$ vary with $t$, then

where: $x$ and $y$ are the two related quantities, $f$ is the function connecting them (so $y=f(x)$ at every instant), and $\dfrac{dx}{dt}$, $\dfrac{dy}{dt}$ are their respective rates of change with time. Every related-rates problem follows the same three moves: identify the relating equation between the quantities involved (an area, volume, or Pythagoras-type formula), differentiate that equation with respect to $t$ — attaching a chain-rule rate factor to every time-varying quantity, not just one — and only then substitute the known numerical values to solve for the unknown rate. Two things make this technique so useful in practice. First, you almost never need to know $x(t)$ or $y(t)$ explicitly — real problems only ever hand you a snapshot rate and a snapshot value, not full functions of time, and related rates works with exactly that limited information. Second, the technique reduces every new scenario to the same short checklist: name the changing quantities, write the geometric or physical law that ties them together, differentiate once with respect to $t$, and substitute. Once that pattern is internalised, circles, spheres, cones and right-triangle setups all become the same three lines of algebra with a different relating equation plugged in.

The goal is to derive the general working rule used in every related-rates problem: given any relating equation linking two quantities such as $y=f(x)$, produce the equation that links their rates $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$, so that a single differentiation step turns a static geometric formula into a dynamic one usable for word problems where only rates and one instant are known, never the full functions of time.

Step 1. Start from the relating equation. Suppose two quantities are connected by $y = f(x)$ — this could be $A = \pi r^2$ for a circle's area and radius, or $V = \tfrac{4}{3}\pi r^3$ for a sphere's volume and radius, or any similar geometric or physical formula. Both $x$ and $y$ are understood to be functions of time $t$, even though neither is written out explicitly.

Step 2. Differentiate both sides with respect to $t$. Because $y$ is a function of $x$, and $x$ is itself a function of $t$, the left side needs the chain rule: $\dfrac{d}{dt}\big[f(x)\big] = f'(x)\cdot\dfrac{dx}{dt}$. So differentiating $y=f(x)$ throughout with respect to $t$ gives $\dfrac{dy}{dt} = f'(x)\cdot\dfrac{dx}{dt}$.

Step 3. Read off the working rule. Every term that involves a changing quantity picks up a rate factor for that quantity; nothing is ever differentiated with respect to $x$ alone. This is why the method is called related rates: the equation now expresses one rate directly in terms of the other rate and the instantaneous value of $x$, which is exactly the form needed to substitute numbers and solve for whichever rate is unknown.

Step 4. Confirm with the circle. For $A=\pi r^2$, $f(r)=\pi r^2$ so $f'(r)=2\pi r$, giving $\dfrac{dA}{dt} = 2\pi r\cdot\dfrac{dr}{dt}$ — precisely the formula used below, and the pattern every related-rates setup in this syllabus (circles, spheres, cones, right triangles) follows.

The radius of a circular oil spill is increasing at a rate of $0.5$ m/s. Find the rate at which the area of the spill is increasing when the radius is $6$ m.

Step 1. Write the relating equation: $A = \pi r^2$.

Step 2. Differentiate with respect to $t$, applying the chain rule since $r=r(t)$: $\dfrac{dA}{dt} = 2\pi r\cdot\dfrac{dr}{dt}$.

Step 3. Substitute the known values, $\dfrac{dr}{dt}=0.5$ m/s and $r=6$ m: $\dfrac{dA}{dt} = 2\pi(6)(0.5) = 6\pi$.

Step 4. State the result with units: $\dfrac{dA}{dt} = 6\pi \approx 18.85\ \text{m}^2/\text{s}$.

The area of the oil spill is increasing at $6\pi\ \text{m}^2/\text{s} \approx 18.85\ \text{m}^2/\text{s}$ when $r=6$ m.

Competitive stretch

A $5$ m ladder leans against a wall; its foot is pulled away from the wall at $0.3$ m/s. Let $x$ be the foot's distance from the wall and $y$ the height of the top on the wall, so $x^2+y^2=25$. Differentiating gives $2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=0$. At $x=3$, Pythagoras gives $y=4$. Substituting $\dfrac{dx}{dt}=0.3$: $2(3)(0.3)+2(4)\dfrac{dy}{dt}=0 \Rightarrow \dfrac{dy}{dt} = \dfrac{-1.8}{8} = -0.225$. The top slides down at $0.225$ m/s — the negative sign confirming $y$ is decreasing — when the foot is $3$ m from the wall. Both examples above share the same shape: one geometric fact fixed for all time (the circle's area formula, or the ladder's constant length via Pythagoras), one given rate, and one instant at which the unknown rate is wanted. Recognising that shape — rather than memorising each scenario separately — is what lets the same three-step method handle spheres filling with water, cones draining, shadows lengthening, and any other pair of quantities tied together by a fixed relationship.

• Differentiating the relating equation with respect to $x$ instead of $t$ — every time-varying quantity needs a chain-rule rate factor like $\dfrac{dx}{dt}$, not a bare derivative.
• Substituting numerical values before differentiating — this destroys the variable you still need to differentiate against; always differentiate the general relation first, then plug in numbers.
• Dropping or misreading the sign of a rate — a decreasing quantity must appear as a negative rate, as in the sliding-ladder example.
• Forgetting to find an intermediate value (like $y$ from Pythagoras) needed to evaluate the final rate at the specific instant asked about.

Ask the tutor: Got a cone-shaped tank draining or a similar related-rates setup you are stuck on — bring your relating equation to the AI tutor and work through the differentiation together.