A balloon radius increases at 2 cm/s — rate of volume increase when r=5

Question

A spherical balloon is being inflated. The radius increases at a rate of 2 cm/s. Find the rate at which the volume is increasing when the radius is 5 cm.

Solution — Step by Step

Given: Rate of change of radius with time: $\frac{dr}{dt} = 2$ cm/s

Find: Rate of change of volume with time: $\frac{dV}{dt}$ when $r = 5$ cm

This is a related rates problem — two quantities (V and r) are both changing with time, and we know one rate, so we find the other using differentiation.

Volume of a sphere:

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

Both $V$ and $r$ are functions of time $t$ .

Apply the chain rule:

\frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right) = \frac{4}{3}\pi \cdot 3r^2 \cdot \frac{dr}{dt} = 4\pi r^2 \cdot \frac{dr}{dt}

This makes intuitive sense — $4\pi r^2$ is the surface area of the sphere. The rate at which volume increases equals surface area times rate of radius increase.

At $r = 5$ cm and $\frac{dr}{dt} = 2$ cm/s:

\frac{dV}{dt} = 4\pi (5)^2 \times 2 = 4\pi \times 25 \times 2 = 200\pi \text{ cm}^3/\text{s}

\boxed{\frac{dV}{dt} = 200\pi \approx 628.3 \text{ cm}^3/\text{s}}

The volume is increasing at a rate of $200\pi$ cm³/s when the radius is 5 cm.

Why This Works

The chain rule tells us how to differentiate composite functions. Since $V$ is a function of $r$ , and $r$ is a function of $t$ , we have:

\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt}

Here, $\frac{dV}{dr} = 4\pi r^2$ (derivative of volume with respect to radius = surface area). Multiplying by $\frac{dr}{dt}$ gives the rate of volume change.

The beautiful fact that $\frac{dV}{dr} =$ surface area has a geometric meaning: if you increase the radius by a tiny amount $\delta r$ , the additional volume is approximately a thin shell of surface area $4\pi r^2$ and thickness $\delta r$ , giving $\delta V \approx 4\pi r^2 \cdot \delta r$ .

Alternative Method — Direct Substitution Check

We can express $V$ as a function of $t$ conceptually: if $r = r_0 + 2t$ (radius grows at 2 cm/s starting from some $r_0$ ), then $V = \frac{4}{3}\pi(r_0 + 2t)^3$ . Differentiating: $\frac{dV}{dt} = \frac{4}{3}\pi \cdot 3(r_0 + 2t)^2 \cdot 2 = 4\pi r^2 \cdot 2$ . At $r = 5$ : $4\pi(25)(2) = 200\pi$ . Same answer.

Common Mistake

The most common error is forgetting to apply the chain rule and writing $\frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right) = 4\pi r^2$ — missing the $\frac{dr}{dt}$ factor. The chain rule is essential here: you must multiply $\frac{dV}{dr}$ by $\frac{dr}{dt}$ .

Another slip: substituting $r = 5$ before differentiating. Always differentiate the general expression first, then substitute values. Substituting a specific value before differentiating destroys the rate information.

In JEE and CBSE, related rates questions almost always use the chain rule: $\frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt}$ (for area problems) or $\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt}$ (for volume problems). Write this structure first, then fill in the derivatives.

Question

Solution — Step by Step

Why This Works

Alternative Method — Direct Substitution Check

Common Mistake

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