Q.Find the rate of change of the area of a circle per second with respect to its radius $r$ when $r = 5$ cm.

Idea: "Rate of change of area with respect to the radius" means the derivative $\dfrac{dA}{dr}$ — no time is involved, so we just differentiate and substitute.

The area of a circle is $A=\pi r^2$. Differentiate with respect to $r$:

At $r=5$ cm,

Read the question carefully

We are asked for the rate at which the area changes with respect to the radius — that is precisely the derivative $\dfrac{dA}{dr}$. This is a plain derivative evaluation, not a related-rates (time) problem: no rate $\dfrac{dr}{dt}$ is given, so we must not invent one.

Step 1 — Write the area formula

Step 2 — Differentiate with respect to $r$

Since $\pi$ is a constant,

Nicely, this is just the circumference of the circle: increasing the radius by a sliver $dr$ adds a thin ring of area $\approx 2\pi r\,dr$.

Step 3 — Substitute $r=5$ cm

Units

Area is in $\text{cm}^2$ and radius in $\text{cm}$, so $\dfrac{dA}{dr}$ is in $\text{cm}^2/\text{cm}$ — square centimetres of area per centimetre of radius. (The word "per second" in the question is loose textbook phrasing; nothing here depends on time.)

Method: Distinguishing a Plain Derivative from a Related-Rates (Time) Derivative

This method teaches how to read a rate-of-change question carefully to decide whether it is asking for a plain derivative with respect to a given variable, or a related-rates derivative with respect to time — the two require different information and different setups.

Steps

Step 1: Identify exactly what the question is differentiating with respect to what

Read the phrase carefully: "rate of change of $A$ with respect to $r$" means $\frac{dA}{dr}$ — a plain derivative, evaluated at a given value of $r$. It is different from "rate of change of $A$ with respect to time," which would be $\frac{dA}{dt}$ and would require a given value of $\frac{dr}{dt}$.

Step 2: Check whether a time-rate is actually given

If the problem never states how fast $r$ itself is changing (no $\frac{dr}{dt}$ or "increasing at ... cm/s" for $r$), then no time variable is genuinely in play, regardless of stray wording like "per second" — you cannot invent a rate that isn't given.

Step 3: Write the formula connecting the two quantities

Step 4: Differentiate directly with respect to the variable named in the question

Step 5: Substitute the given value and attach the correct units

Evaluate at the given $r$, and state units as (units of $A$) per (unit of $r$) — e.g. $\text{cm}^2/\text{cm}$ — never a per-second unit unless a genuine time-rate was computed.

This same read-the-question-first discipline applies whenever a problem's wording is ambiguous between a plain derivative and a related-rates derivative — always check what quantity is actually given a rate before setting up the differentiation.

Common Mistakes

Mistake 1: Treating this as a related-rates (time) problem because of the phrase "per second"

A student sets up $\frac{dA}{dt}=2\pi r\frac{dr}{dt}$ and either invents a value for $\frac{dr}{dt}$ or leaves it as an unexplained symbol. Why it's wrong: the question explicitly asks for the rate of change of area with respect to the radius, not with respect to time — no $\frac{dr}{dt}$ is given anywhere, so introducing one fabricates information that was never provided. Correct approach: differentiate $A=\pi r^2$ directly with respect to $r$ to get $\frac{dA}{dr}=2\pi r$, ignoring the loose "per second" phrasing.

Mistake 2: Attaching time-based units (like $\text{cm}^2/\text{s}$) to the final answer

Even a student who differentiates correctly may then write the answer with an "s" (seconds) unit out of habit. Why it's wrong: since no time-rate was computed, the answer's units must be (area unit) per (length unit) — $\text{cm}^2/\text{cm}$ — not a rate per second. Correct approach: match the units to exactly what was differentiated with respect to what.