Application of Derivatives: Numerical Problems Set (1)

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Tags Application Of Derivatives

Question

A spherical balloon is being inflated such that its radius is increasing at 0.5 cm/s0.5 \text{ cm/s}. Find the rate at which its volume is increasing when the radius is 5 cm5 \text{ cm}. Then, find the rate at which the surface area is increasing at the same instant.

Solution — Step by Step

Volume of a sphere:

V=43πr3V = \frac{4}{3}\pi r^3

Differentiate both sides with respect to time:

dVdt=4πr2drdt\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}

Given: r=5 cmr = 5 \text{ cm}, dr/dt=0.5 cm/sdr/dt = 0.5 \text{ cm/s}.

dVdt=4π(5)2(0.5)=50π cm3/s\frac{dV}{dt} = 4\pi (5)^2 (0.5) = 50\pi \text{ cm}^3/\text{s}

So the volume is increasing at 50π≈157.1 cm3/s50\pi \approx 157.1 \text{ cm}^3/\text{s}.

Surface area:

S=4πr2S = 4\pi r^2

Differentiate:

dSdt=8πrdrdt\frac{dS}{dt} = 8\pi r \frac{dr}{dt}

Plug in:

dSdt=8π(5)(0.5)=20π cm2/s\frac{dS}{dt} = 8\pi (5)(0.5) = 20\pi \text{ cm}^2/\text{s}

About 62.8 cm2/s62.8 \text{ cm}^2/\text{s}.

Why This Works

Related-rates problems all share the same structure: a geometric quantity changes over time, and we differentiate the formula relating it to other quantities (radius, height, side length) to link rates.

The chain rule does the heavy lifting: dV/dt=(dV/dr)(dr/dt)dV/dt = (dV/dr)(dr/dt). The trick is that we keep rr symbolic during differentiation and only plug in numbers at the end. If you plug in r=5r = 5 first and then differentiate, you’d lose the rr dependence and get the wrong answer.

Alternative Method

Linear approximation: in Δt\Delta t seconds, Δr≈0.5Δt\Delta r \approx 0.5 \Delta t, so ΔV≈4πr2Δr\Delta V \approx 4\pi r^2 \Delta r. Setting Δt=1 s\Delta t = 1 \text{ s}, ΔV≈4π(25)(0.5)=50π\Delta V \approx 4\pi (25)(0.5) = 50\pi. Same answer; this is just the rate written informally.

For any “radius increasing at rate” problem on a sphere, the volume rate scales as r2r^2 and the surface rate scales as rr. So as the balloon gets bigger, the volume rate accelerates faster than the surface rate — useful intuition for MCQ proportions.

Common Mistake

The killer mistake is plugging in r=5r = 5 before differentiating. If you write V=43π(5)3V = \frac{4}{3}\pi(5)^3 and try to differentiate that with respect to tt, you get 00 — because there’s no tt on the right. The order is: differentiate first (keeping rr as a function of tt), then substitute the instantaneous value of rr.

The other slip: forgetting that the formula S=4πr2S = 4\pi r^2 has a derivative with respect to rr of 8πr8\pi r, not 4πr4\pi r. Watch the chain.

Final answer: dV/dt=50π cm3/sdV/dt = 50\pi \text{ cm}^3/\text{s}, dS/dt=20π cm2/sdS/dt = 20\pi \text{ cm}^2/\text{s}.

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