Math · Calculus I · Worked example

Find how fast a balloon’s radius grows

Air is pumped into a spherical balloon at 100 cm³/s. How fast is the radius increasing when the radius is 5 cm?

V = \frac{4}{3} π r^{3}, \frac{d V}{d t} = 100

Link the volume and the radius

The volume of a sphere is V = (4/3)πr³. Both V and r change with time.

V = \frac{4}{3} π r^{3}

Differentiate with respect to t

The derivative of r³ with respect to t is 3r²·dr/dt, so the 4/3 and the 3 combine to 4.

\frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}

Substitute and solve

Put dV/dt = 100 and r = 5: 100 = 4π(25)·dr/dt = 100π·dr/dt, so dr/dt = 1/π.

100 = 100 π \frac{d r}{d t} ⟹ \frac{d r}{d t} = \frac{1}{π}

\begin{array}{c} 100 = 100 π \frac{d r}{d t} \\ \frac{d r}{d t} = \frac{1}{π} \end{array}

Interpret the answer

dr/dt = 1/π ≈ 0.318 cm/s. The radius grows more slowly as the balloon gets bigger, because the same new volume is spread over a larger surface: 4πr² is the sphere’s surface area.

Result

dr/dt = 1/π ≈ 0.318 cm/s when the radius is 5 cm.

Your turn

A circular oil slick spreads so that its radius grows at 2 m/min. How fast is its area growing when the radius is 30 m?

Show the answer and explanation

120π ≈ 377 m²/min.

A = πr², so dA/dt = 2πr·dr/dt = 2π(30)(2) = 120π ≈ 377 m²/min.

\frac{d A}{d t} = 2 π r \frac{d r}{d t} = 2 π (30) (2) = 120 π

\begin{array}{c} \frac{d A}{d t} = 2 π r \frac{d r}{d t} \\ = 2 π (30) (2) = 120 π \end{array}

Keep exploring

Open the rows in Math and add V′(10): it is 400π, so at r = 10 the radius grows four times more slowly, at 1/(4π) ≈ 0.080 cm/s.

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