If the rate of change of volume of a sphere is equal to the rate of change of its radius, then find its radius

3.2 · Q3

Q. If the rate of change of volume of a sphere is equal to the rate of change of its radius, then find its radius. Also find its surface area.

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Concept understanding — Related Rates

Related Rates: When Two Things Change Together

Imagine you're blowing up a balloon. Your lungs push air in at a certain rate — say, 3 cubic centimetres per second. As the balloon grows, its radius increases. The question is: how fast is the radius increasing at the exact moment when the radius is 5 cm?

That's a related rates problem. Two quantities — the volume $V$ and the radius $r$ — are linked by a geometric formula ( $V = \frac{4}{3}\pi r^3$ ). You know how fast one is changing ( $dV/dt = 3$ ), and you want to find how fast the other is changing ( $dr/dt$ ) at a specific instant.

The core idea is simple: if two quantities are connected by an equation, their rates of change are also connected — by the derivative of that equation.

The Precise Statement

Let $x$ and $y$ be two quantities that both depend on time $t$ , and suppose they satisfy some equation $F(x, y) = 0$ (or $y = f(x)$ , etc.). Then:

Differentiate both sides of the equation with respect to time $t$ .
Use the chain rule wherever you see a variable that depends on $t$ .
The result is an equation linking $dx/dt$ and $dy/dt$ .

That's it. The "related rates" are $dx/dt$ and $dy/dt$ , and the chain rule is the tool that connects them.

\frac{d}{dt} \big[ \text{equation linking variables} \big] \quad\Longrightarrow\quad \text{equation linking rates}

The Chain Rule in Action

In the balloon example, the equation is $V = \frac{4}{3}\pi r^3$ . Differentiate both sides with respect to $t$ :

\frac{dV}{dt} = \frac{d}{dt}\left( \frac{4}{3}\pi r^3 \right)

The right side is a function of $r$ , and $r$ itself depends on $t$ . By the chain rule:

\frac{dV}{dt} = \frac{4}{3}\pi \cdot 3r^2 \cdot \frac{dr}{dt} = 4\pi r^2 \frac{dr}{dt}

Now plug in what you know: $dV/dt = 3$ and $r = 5$ :

3 = 4\pi (5)^2 \frac{dr}{dt} \quad\Longrightarrow\quad 3 = 100\pi \frac{dr}{dt}

So:

\frac{dr}{dt} = \frac{3}{100\pi} \text{ cm/s}

That's the answer. The radius is growing at about 0.0095 cm/s when the balloon's radius is 5 cm.

Watch out

A common mistake is to plug in known values before differentiating. Don't. If you substitute $r = 5$ into $V = \frac{4}{3}\pi r^3$ first, you get a constant — and its derivative is zero. You lose the relationship between the rates. Always differentiate first, then substitute.

The General Recipe

For any related rates problem, follow these steps:

Identify all changing quantities and assign them variables. Note which rates you know and which you need.
Write an equation that relates the quantities at any time $t$ (not just at the instant of interest).
Differentiate implicitly with respect to $t$ . Use the chain rule for every variable that depends on $t$ .
Substitute the known values (including the specific instant's measurements) into the rate equation.
Solve for the unknown rate.

A Second Example: The Ladder Problem

A 10-metre ladder leans against a wall. The bottom slides away from the wall at 1 m/s. How fast is the top sliding down when the bottom is 6 m from the wall?

Let $x$ be the distance from the wall to the ladder's bottom, and $y$ be the height of the ladder's top. By Pythagoras:

x^2 + y^2 = 10^2 = 100

Differentiate with respect to $t$ :

2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0

Divide by 2:

x \frac{dx}{dt} + y \frac{dy}{dt} = 0

You know $dx/dt = 1$ (positive because $x$ increases). At the instant $x = 6$ , find $y$ from the original equation: $6^2 + y^2 = 100 \Rightarrow y = 8$ . Substitute:

(6)(1) + (8)\frac{dy}{dt} = 0 \quad\Longrightarrow\quad 8\frac{dy}{dt} = -6

\frac{dy}{dt} = -\frac{6}{8} = -0.75 \text{ m/s}

The negative sign means $y$ is decreasing — the top slides down at 0.75 m/s.

Tip

The sign of a rate tells you direction. Positive means increasing, negative means decreasing. Always include the sign in your final answer.

Why This Matters

Related rates are everywhere in physics, engineering, and economics. Whenever two quantities move together — a piston compressing gas, a shadow lengthening as you walk away from a lamp, the surface area of a melting snowball shrinking — the same idea applies: differentiate the relationship, and the rates reveal themselves.

The key is to see the geometry or the physics first. Draw a picture. Label everything. Then let the chain rule do the work.

Equate the rate of change of volume $4\pi r^2\frac{dr}{dt}$ to $\frac{dr}{dt}$ , giving $4\pi r^2=1\Rightarrow r=\frac{1}{2\sqrt{\pi}}$ ; then the surface area $S=4\pi r^2=1$ square unit.

V=\dfrac{4}{3}\pi r^3

Differentiate the volume w.r.t. time $t$ :

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

Apply the given condition $\dfrac{dV}{dt}=\dfrac{dr}{dt}$ :

4\pi r^2\frac{dr}{dt}=\frac{dr}{dt}.

Cancel $\dfrac{dr}{dt}$ (non-zero):

4\pi r^2=1\;\Rightarrow\; r^2=\frac{1}{4\pi}\;\Rightarrow\; r=\frac{1}{2\sqrt{\pi}}.

Find the surface area:

S=4\pi r^2=4\pi\cdot\frac{1}{4\pi}=1.

✓Final answer

r=\dfrac{1}{2\sqrt{\pi}}

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Q.If the rate of change of volume of a sphere is equal to the rate of change of its radius, then find its radius. Also find its surface area.

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