Exercise · ZW-RELATED-RATES-002 · ★★☆ Intermediate

The radius of a sphere grows at \(\frac{1}{2}\) cm/s. How fast is the volume growing when \(r = 3\) cm?

Choose one option.

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  • D. \(18 \pi\)

Explanation

From \(V=\dfrac{4}{3}\pi r^{3}\) we get \(\dfrac{dV}{dt}=4\pi r^{2}\dfrac{dr}{dt}=4\pi(3)^{2}\cdot\frac{1}{2}=18 \pi\) cm³/s.

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Review the concept: Related rates

In a related rates problem several quantities change with time \(t\) and are linked by an equation. The equation is differentiated with respect to \(t\) (chain rule) and the known values are substituted.

\[A=\pi r^{2}\ \Rightarrow\ \frac{dA}{dt}=2\pi r\,\frac{dr}{dt}\]

More practice problems: related rates

← ZW-RELATED-RATES-001 ZW-RELATED-RATES-003 →