Standard +0.3 This is a straightforward related rates problem requiring the chain rule (dV/dt = dV/dr × dr/dt) with a given formula. Students substitute r=8 into the differentiated volume formula and solve for dr/dt. It's slightly easier than average as it's a standard textbook exercise with clear setup and direct application of a single technique.
2 A spherical balloon of radius \(r \mathrm {~cm}\) has volume \(V \mathrm {~cm} ^ { 3 }\), where \(V = \frac { 4 } { 3 } \pi r ^ { 3 }\). The balloon is inflated at a constant rate of \(10 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\). Find the rate of increase of \(r\) when \(r = 8\).
2 A spherical balloon of radius $r \mathrm {~cm}$ has volume $V \mathrm {~cm} ^ { 3 }$, where $V = \frac { 4 } { 3 } \pi r ^ { 3 }$. The balloon is inflated at a constant rate of $10 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }$. Find the rate of increase of $r$ when $r = 8$.
\hfill \mbox{\textit{OCR MEI C3 Q2 [5]}}