Q.If the area of a circle increases at a uniform rate, then prove that the perimeter varies inversely as the radius.

Idea: "Uniform rate" means the area grows at a constant rate, $\frac{dA}{dt}=k$. Link this to the perimeter through the common variable $r$.

For a circle, $A=\pi r^2$ and perimeter $P=2\pi r$.

Differentiate the area with respect to time:

Now differentiate the perimeter:

Since $k$ is constant, $\frac{dP}{dt}\propto\frac{1}{r}$ — the perimeter's rate of change varies inversely as the radius.

The intuition

We are told the area is increasing at a uniform (constant) rate. Both the area and the perimeter of a circle depend on the same radius $r$, and $r$ itself is changing with time. So this is a related-rates situation: knowing how fast the area grows tells us how fast the radius grows, and that in turn tells us how fast the perimeter changes.

Set up the formulas

For a circle of radius $r$:

• Area: $A=\pi r^2$
• Perimeter (circumference): $P=2\pi r$

"Area increases at a uniform rate" translates to

Work the steps

1. Find $\frac{dr}{dt}$ from the area. Differentiate $A=\pi r^2$ with respect to $t$ (chain rule):

Set this equal to $k$:

2. Differentiate the perimeter.

3. Substitute $\frac{dr}{dt}$.

Conclude

Because $k$ is a fixed constant, $\frac{dP}{dt}=\frac{k}{r}$ is a constant divided by $r$. Hence the rate of change of the perimeter is inversely proportional to the radius, which is exactly what we set out to prove.

Method: Proving an Inverse-Variation Relationship Between Two Related Rates

Many related-rates "prove that ___ varies as ___" questions ask you to show one rate is proportional (or inversely proportional) to another quantity, without the problem ever giving you numbers — the proof runs entirely on the constant rate itself.

Steps

Step 1: Identify the two changing quantities and the constant rate.

Read the problem for a phrase like "increases/changes at a uniform (constant) rate" — this tells you one derivative, say $\frac{dQ_1}{dt}=k$, is a fixed constant. It does NOT mean $Q_1$ itself stays constant.

Step 2: Express both quantities in terms of a single common variable.

Find the variable (often a length such as radius or side) that both $Q_1$ and $Q_2$ depend on, and write $Q_1=f(x)$, $Q_2=g(x)$ using the standard geometric formulas.

Step 3: Differentiate the first relation and solve for the common variable's own rate.

Step 4: Differentiate the second relation and substitute.

Simplify the resulting expression — since $k$ is a fixed constant, whichever variable is left over (in the numerator or denominator) tells you the type of variation.

Step 5: State the conclusion.

If $x$ ends up only in the denominator, $\frac{dQ_2}{dt}$ varies inversely as $x$; if only in the numerator, it varies directly as $x$. This is the complete proof — no specific numerical values are ever needed.

Common Mistakes

Mistake 1: Substituting a numeric value for $r$ before differentiating

Why it's wrong: once a specific number replaces $r$ in $A=\pi r^2$, $r$ is frozen and $\dfrac{dr}{dt}$ disappears from the equation entirely, so the link between $\dfrac{dP}{dt}$ and $\dfrac{dA}{dt}$ can never be recovered. Correct approach: differentiate the symbolic relation $A=\pi r^2$ with respect to $t$ first to get $\dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt}$, and only plug in numbers afterward (here there are none to plug in, which is itself a clue the answer must stay in terms of $r$).

Mistake 2: Confusing "perimeter varies inversely as radius" with "rate of change of perimeter varies inversely as radius"

Why it's wrong: since $P=2\pi r$, the perimeter itself is directly proportional to $r$, not inversely — students sometimes conclude $P\propto\frac{1}{r}$, contradicting the basic circle formula. Correct approach: keep clear that it is only $\dfrac{dP}{dt}=\dfrac{k}{r}$, the rate, that is inversely proportional to $r$; the perimeter itself still grows as $r$ grows.

Mistake 3: Treating $\frac{dr}{dt}$ as constant because $\frac{dA}{dt}$ is constant

Why it's wrong: the problem states the area grows uniformly ($\frac{dA}{dt}=k$), but $\frac{dr}{dt}=\frac{k}{2\pi r}$ actually changes as $r$ changes — it is not itself constant. Correct approach: keep $\frac{dr}{dt}$ as an expression in $r$ throughout, and only observe that the final quantity $\frac{dP}{dt}$ simplifies to a form independent of the growing-but-non-constant $\frac{dr}{dt}$'s numeric value.