Rate of change of quantities

If $y = f(x)$, the derivative $\frac{dy}{dx}$ is the rate of change of $y$ with respect to $x$ at a point. In most physical situations, both $x$ and $y$ are themselves changing with time $t$, so the natural question is how fast $y$ changes with $t$. The chain rule answers this directly:

where: $\frac{dy}{dt}$ is the rate of change of $y$ with time, $\frac{dx}{dt}$ is the rate of change of $x$ with time, and $\frac{dy}{dx}$ is the instantaneous rate of change of $y$ with respect to $x$ from the functional relation $y=f(x)$. A related-rates problem goes one step further: several quantities are connected not by $y=f(x)$ directly but by a shared geometric formula — area of a circle, volume of a sphere, Pythagoras' theorem for a ladder against a wall, and so on. Four moves solve every such problem, in this order and no other: First, write the relation connecting the quantities using symbols only — no numbers yet. Second, differentiate both sides with respect to $t$, applying the chain rule to every variable that itself depends on $t$. Third, substitute the specific numerical values given — for the variables and for any known rates — but only after differentiating. Fourth, solve the resulting equation for the one unknown rate. The sign of a rate carries meaning. A positive $\frac{dy}{dt}$ means $y$ is increasing; a negative value means it is decreasing. When a question asks "how fast is $y$ decreasing," it usually wants the positive magnitude of $-\frac{dy}{dt}$, so read the wording carefully before reporting a sign.

A stone dropped into a pond creates a circular ripple whose radius increases at the constant rate $\frac{dr}{dt} = 4\text{ cm/s}$. Find the rate at which the area of the circle is increasing at the instant the radius is $10\text{ cm}$.

Step 1. Identify the relation: $A = \pi r^2$, with both $A$ and $r$ functions of time.

Step 2. Differentiate with respect to $t$ using the result derived above: $\frac{dA}{dt} = 2\pi r\,\frac{dr}{dt}$.

Step 3. Substitute the given numbers — $r = 10\text{ cm}$ and $\frac{dr}{dt} = 4\text{ cm/s}$ — only now, after differentiating: $\frac{dA}{dt} = 2\pi(10)(4) = 80\pi\text{ cm}^2/\text{s}$.

Step 4. Evaluate numerically: $80\pi \approx 251.3\text{ cm}^2/\text{s}$. So the area is increasing at $80\pi \approx 251.3\text{ cm}^2/\text{s}$ at that instant.

Competitive stretch

A ladder $5\text{ m}$ long leans against a vertical wall; its foot is pulled away from the wall along the ground at $2\text{ m/s}$. How fast is the top sliding down when the foot is $3\text{ m}$ from the wall? Let $x$ be the foot's distance from the wall and $y$ the height of the top, both functions of $t$, related by Pythagoras' theorem since the ladder length is fixed: $x^2+y^2=25$. Differentiating with respect to $t$ gives $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$, i.e. $x\frac{dx}{dt}+y\frac{dy}{dt}=0$. When $x=3$, $y^2=25-9=16$ so $y=4$. With $\frac{dx}{dt}=2\text{ m/s}$: $3(2)+4\frac{dy}{dt}=0 \Rightarrow \frac{dy}{dt}=-1.5\text{ m/s}$. The negative sign confirms $y$ is decreasing — the top slides down at $1.5\text{ m/s}$ at that instant.

• Substituting numbers before differentiating. Plugging in the given values into the relation first freezes the variables and produces a wrong (often zero) derivative. Always differentiate the general relation, then substitute.
• Forgetting the chain rule on an inner variable. Differentiating $r^2$ with respect to $t$ must give $2r\frac{dr}{dt}$, not merely $2r$ — $r$ itself depends on $t$.
• Sign confusion. A decreasing quantity has a negative rate. If asked "how fast is it decreasing," report the positive magnitude but remember the derivative itself is negative.
• Mixing units. Convert all lengths, areas, and times to one consistent system (e.g. all in metres and seconds) before differentiating and substituting numbers.

Ask the tutor: Want to try a related-rates problem with a cone-shaped tank draining, or a shadow lengthening as someone walks under a streetlight — ask the AI tutor to work through the setup with you step by step.