Moderate -0.3 This is a standard related rates problem requiring students to differentiate V = πr²h with respect to time, substitute given values (converting 3mm to 0.3cm), and solve for dr/dt. While it involves multiple steps and unit conversion, it follows a well-established template taught in C4 with no novel problem-solving required, making it slightly easier than average.
6. Oil is leaking from a storage container onto a flat section of concrete at a rate of \(0.48 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\). The leaking oil spreads to form a pool with an increasing circular cross-section. The pool has a constant uniform thickness of 3 mm .
Find the rate at which the radius \(r\) of the pool of oil is increasing at the instant when \(r = 5 \mathrm {~cm}\). Give your answer, in \(\mathrm { cm } \mathrm { s } ^ { - 1 }\), to 3 significant figures.
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| Answer/Working | Marks | Guidance |
|---|---|---|
| From question, $\frac{dV}{dt}=0.48$ | | |
| $V=\pi r^2(0.3)$ | B1 oe | $V=0.3\pi r^2$ (Can be implied.) |
| $\frac{dV}{dr}=0.6\pi r$ | B1 ft | |
| $\left\{\frac{dV}{dt}=\frac{dV}{dr}\times\frac{dr}{dt}\right\}\Rightarrow(0.6\pi r)\frac{dr}{dt}=0.48$ | M1; oe | |
| $\left\{\frac{dr}{dt}=(0.48)\frac{1}{0.6\pi r}=\left\{-\frac{4}{5\pi r}\right\}\right\}$ | | or $0.48\div\text{Candidate's}\frac{dV}{dr}$ |
| When $r=5$cm, $\frac{dr}{dt}=\frac{0.48}{0.6\pi(5)}=\left\{-\frac{5\pi(5)}\right\}$ | dM1 | Substitutes $r=5$ into an equation containing $\frac{dr}{dt}$ |
| Hence, $\frac{dr}{dt}=0.05092958179...$ (cm s$^{-1}$) | A1 | anything that rounds to $0.0509$ |
| | | [5] |
**Notes**
B1: $V=\pi r^2(0.3)$ or equivalent.
B1ft: Correct follow through differentiation of their $V$ or their $A$ with respect to $r$.
M1: $\left(\text{Candidate's}\frac{dV}{dr}\right)\times\frac{dr}{dt}=0.48$ or $0.48\div\text{Candidate's}\frac{dV}{dr}$
dM1: (dependent on the previous method mark) Substitutes $r=5$ into an equation containing $\frac{dr}{dt}$.
A1: anything that rounds to $0.0509$
**Example 1: Using thickness = 3 (cm) and not 0.3 (cm)**
$V=3\pi r^2 \Rightarrow \frac{dV}{dr}=6\pi r$ leading to $\frac{dr}{dt}=\frac{0.48}{6\pi(5)}=0.005092958179...$ gets B0B1ftM1M1A0.
**Example 2: Using thickness = 0.03 (cm) and not 0.3 (cm)**
$V=0.03\pi r^2 \Rightarrow \frac{dV}{dr}=0.06\pi r$ leading to $\frac{dr}{dt}r_{|r=5}=\frac{0.48}{0.06\pi(5)}=0.50929581...79...$ gets B0B1ftM1M1A0.
**Alternative method 1 First 3 marks**
6. Oil is leaking from a storage container onto a flat section of concrete at a rate of $0.48 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }$. The leaking oil spreads to form a pool with an increasing circular cross-section. The pool has a constant uniform thickness of 3 mm .
Find the rate at which the radius $r$ of the pool of oil is increasing at the instant when $r = 5 \mathrm {~cm}$. Give your answer, in $\mathrm { cm } \mathrm { s } ^ { - 1 }$, to 3 significant figures.
\includegraphics[max width=\textwidth, alt={}, center]{245bbe52-3a14-4494-af17-7711caf79b22-19_104_95_2617_1786}\\
\hfill \mbox{\textit{Edexcel C4 2014 Q6 [5]}}