Q.A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is $10 \text{ cm}$.

The volume of a sphere is $V = \dfrac{4}{3}\pi r^3$. The question asks for $\dfrac{dV}{dr}$ directly (no time variable involved).

Step 1: Differentiate: $\dfrac{dV}{dr} = \dfrac{4}{3}\pi\cdot 3r^2 = 4\pi r^2$. …

Reading the question

The balloon "always remains spherical," so at any instant its volume is given by the sphere-volume formula in terms of its radius $r$. The question asks for the rate at which the volume increases with the radius — that is the rate of change of $V$ with respect to $r$ directly, $\dfrac{dV}{dr}$, evaluated at $r = 10$ cm. There is no time variable anywhere in this question, so this is a direct rate-of-change computation.

Step 1 — Write the volume formula

Step 2 — Differentiate $V$ with respect to $r$

$\frac{dV}{dr} = \frac{4}{3}\pi \cdot 3r^2 = 4\pi r^2.$ …

Method: Direct Derivative for "Rate With Respect To" (No Time Variable)

This method applies whenever a question asks for the rate of change of one quantity with respect to another spatial quantity (like the radius), not with respect to time — the telltale phrase is "rate ... with the radius" or "with respect to $x$", never "per second" or "with time".

Steps

Step 1: Recognise there is no time variable here

Unlike a related-rates problem, only two quantities appear — no clock is running. So you do NOT need the chain rule through $t$; you differentiate one variable directly with respect to the other.

Step 2: Write the quantity to be differentiated as a function of the given variable

For a sphere, $V = \frac{4}{3}\pi r^3$ expresses volume purely as a function of radius $r$.

Step 3: Differentiate directly with respect to that variable

This is now a formula valid for any radius, giving volume change per unit change in radius (not per unit time).

Step 4: Substitute the given value of the variable …

Common Mistakes

Mistake 1: Treating this as a time-based related-rates problem and introducing $\frac{dr}{dt}$

Why it's wrong: the question asks for the rate "with the radius", not "with time" — there is no clock in this problem, so writing $\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$ and then hunting for a missing $\frac{dr}{dt}$ value is solving the wrong problem entirely. Correct approach: differentiate $V$ directly with respect to $r$ to get $\frac{dV}{dr}=4\pi r^2$ — no chain rule through time is needed here.

Mistake 2: Misreading the units of the final answer …