Q.Find the value of the following: The rate of change of the area of a circle with respect to its radius $r$ at $r = 6 \text{ cm}$ is (A) $10\pi$ (B) $12\pi$ (C) $8\pi$ (D) $11\pi$

"Rate of change of $A$ with respect to $r$" means the derivative $\frac{dA}{dr}$ — there is no time rate involved here.

Step 1 — Area formula. $A=\pi r^2$.

Step 2 — Differentiate with respect to $r$. $\dfrac{dA}{dr}=2\pi r$. …

The idea

The phrase "rate of change of the area with respect to the radius" is a direct instruction to differentiate the area with respect to $r$. This is a plain derivative evaluation, not a related-rates problem — no time is given, so there is no $\frac{dr}{dt}$ and no chain rule.

Set up

The area of a circle of radius $r$ is

Work the steps

1. Differentiate with respect to $r$ (treat $\pi$ as a constant):

Neatly, this equals the circumference — adding a thin ring of thickness $dr$ adds an area of about (circumference)$\times dr$.

2. Substitute $r=6$ cm:

$\frac{dA}{dr}\Big|_{r=6}=2\pi(6)=12\pi.$ …

Method: Recognising a Plain Derivative Question Disguised as a "Rate" Question

This method applies to MCQs phrased as "rate of change of [quantity] with respect to [variable]" where the second variable is a spatial/algebraic quantity (like the radius $r$) rather than time — a frequent one-mark trap testing whether a student can tell a plain derivative apart from a related-rates setup.

Steps

Step 1: Read the phrase carefully — "with respect to $r$", not "with respect to time"

No clock is mentioned and no rate like $\frac{dr}{dt}$ is given anywhere in the question — that is the signal that only a direct derivative is needed, not the related-rates chain-rule machinery.

Step 2: Write the quantity as a function of the stated variable

For a circle, $A = \pi r^2$ expresses area purely as a function of radius.

Step 3: Differentiate directly with respect to that variable

keeping this as a general formula in $r$ — do not substitute the given radius yet.

Step 4: Substitute the given value of the variable into this derivative …

Common Mistakes

Mistake 1: Substituting $r=6$ into the area formula before differentiating

Why it's wrong: computing $A=\pi(6)^2=36\pi$ first and treating THAT as the "rate" answers a completely different question (the area itself, not its rate of change) — it also makes $r$ a fixed number, so there is nothing left to differentiate. Correct approach: differentiate $A=\pi r^2$ with respect to $r$ FIRST to get $\frac{dA}{dr}=2\pi r$, and substitute $r=6$ only into this derivative.

Mistake 2: Mistaking this for a related-rates (time-based) problem …