Standard +0.3 This is a straightforward related rates problem requiring differentiation of the sphere volume formula with respect to time, substitution of given values, and solving for dr/dt. While it involves implicit differentiation and chain rule application, it follows a standard template taught in C3/C4 with no conceptual surprises or multi-step reasoning—slightly easier than average.
8. The volume \(V\) of a spherical balloon is increasing at a constant rate of \(250 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\). Find the rate of increase of the radius of the balloon, in \(\mathrm { cm } \mathrm { s } ^ { - 1 }\), at the instant when the volume of the balloon is \(12000 \mathrm {~cm} ^ { 3 }\).
Give your answer to 2 significant figures. [0pt]
[You may assume that the volume \(V\) of a sphere of radius \(r\) is given by the formula \(V = \frac { 4 } { 3 } \pi r ^ { 3 }\).]
Applies \(12000 = \frac{4}{3}\pi r^3\) and rearranges to find \(r\) using division then cube root with accurate algebra. May state \(r = \sqrt[3]{\frac{3V}{4\pi}}\) then substitute \(V=12000\) later
awrt 0.099. Units may be ignored. If this answer is seen, award A1 and isw. Premature approximation usually results in all marks being earned prior to this one
[5] Total: 5
## Question 8:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dV}{dt} = 250$ | | |
| $\left\{V = \frac{4}{3}\pi r^3 \Rightarrow\right\}\ \frac{dV}{dr} = 4\pi r^2$ | B1 | May be stated or used, need not be simplified |
| $V = 12000 \Rightarrow 12000 = \frac{4}{3}\pi r^3 \Rightarrow r = \sqrt[3]{\frac{9000}{\pi}}\ (=14.202480...)$ | B1 | Applies $12000 = \frac{4}{3}\pi r^3$ and rearranges to find $r$ using division then cube root with accurate algebra. May state $r = \sqrt[3]{\frac{3V}{4\pi}}$ then substitute $V=12000$ later |
| $\frac{dr}{dt} \left\{= \frac{dr}{dV} \times \frac{dV}{dt}\right\} = \frac{1}{4\pi r^2} \times 250$ | M1 | Uses chain rule correctly so $\frac{1}{\left(\text{their } \frac{dV}{dr}\right)} \times 250$ |
| When $r = \sqrt[3]{\frac{9000}{\pi}}$, $\quad \frac{dr}{dt} = \frac{250}{4\pi\left(\sqrt[3]{\frac{9000}{\pi}}\right)^2}$ | dM1 | Substitutes their $r$ correctly into their equation for $\frac{dr}{dt}$. Depends on previous method mark |
| $\frac{dr}{dt} = 0.0986283...\,(\text{cms}^{-1})$ awrt 0.099 | A1 | awrt 0.099. Units may be ignored. If this answer is seen, award A1 and isw. Premature approximation usually results in all marks being earned prior to this one |
| **[5] Total: 5** | | |
8. The volume $V$ of a spherical balloon is increasing at a constant rate of $250 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }$. Find the rate of increase of the radius of the balloon, in $\mathrm { cm } \mathrm { s } ^ { - 1 }$, at the instant when the volume of the balloon is $12000 \mathrm {~cm} ^ { 3 }$.\\
Give your answer to 2 significant figures.\\[0pt]
[You may assume that the volume $V$ of a sphere of radius $r$ is given by the formula $V = \frac { 4 } { 3 } \pi r ^ { 3 }$.]
\hfill \mbox{\textit{Edexcel C34 2014 Q8 [5]}}