| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Related rates with cones, hemispheres, and bowls (variable depth) |
| Difficulty | Standard +0.3 This is a standard related rates problem requiring the cone volume formula, similar triangles to relate radius to height, differentiation, and substitution. While it involves multiple steps, each is routine for C3 level—establishing the volume formula using tan(30°), then applying dV/dt = (dV/dh)(dh/dt). The calculations are straightforward with no novel insight required, making it slightly easier than average. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates3.02b Kinematic graphs: displacement-time and velocity-time |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| let radius \(= r\), \(\tan 30° = \frac{1}{\sqrt{3}} = \frac{r}{h}\), \(r = \frac{h}{\sqrt{3}}\) | M1 | |
| \(V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi h \times \frac{h^2}{3} = \frac{1}{9}\pi h^3\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dV}{dt} = 120\), \(\frac{dV}{dh} = \frac{1}{3}\pi h^2\) | B1 | |
| \(\frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt}\), \(\quad 120 = \frac{1}{3}\pi h^2 \frac{dh}{dt}\), \(\quad \frac{dh}{dt} = \frac{360}{\pi h^2}\) | M1 A1 | |
| when \(h = 6\), \(\frac{dh}{dt} = 3.18\) cm s\(^{-1}\) (2dp) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(V = 8 \times 120 = 960 = \frac{1}{9}\pi h^3\) \(\therefore h = \sqrt[3]{\frac{9 \times 960}{\pi}} = 14.011\) | M1 | |
| \(\therefore \frac{dh}{dt} = 0.58\) cm s\(^{-1}\) (2dp) | A1 | (9) |
# Question 6:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| let radius $= r$, $\tan 30° = \frac{1}{\sqrt{3}} = \frac{r}{h}$, $r = \frac{h}{\sqrt{3}}$ | M1 | |
| $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi h \times \frac{h^2}{3} = \frac{1}{9}\pi h^3$ | A1 | |
## Part (b)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dV}{dt} = 120$, $\frac{dV}{dh} = \frac{1}{3}\pi h^2$ | B1 | |
| $\frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt}$, $\quad 120 = \frac{1}{3}\pi h^2 \frac{dh}{dt}$, $\quad \frac{dh}{dt} = \frac{360}{\pi h^2}$ | M1 A1 | |
| when $h = 6$, $\frac{dh}{dt} = 3.18$ cm s$^{-1}$ (2dp) | M1 A1 | |
## Part (b)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $V = 8 \times 120 = 960 = \frac{1}{9}\pi h^3$ $\therefore h = \sqrt[3]{\frac{9 \times 960}{\pi}} = 14.011$ | M1 | |
| $\therefore \frac{dh}{dt} = 0.58$ cm s$^{-1}$ (2dp) | A1 | **(9)** |
---6.\\
\includegraphics[max width=\textwidth, alt={}, center]{14a2477a-c40e-4b4b-bc39-7100d1df9b4d-2_397_488_1299_632}
The diagram shows a vertical cross-section through a vase.\\
The inside of the vase is in the shape of a right-circular cone with the angle between the sides in the cross-section being $60 ^ { \circ }$. When the depth of water in the vase is $h \mathrm {~cm}$, the volume of water in the vase is $V \mathrm {~cm} ^ { 3 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that $V = \frac { 1 } { 9 } \pi h ^ { 3 }$.
The vase is initially empty and water is poured in at a constant rate of $120 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }$.
\item Find, to 2 decimal places, the rate at which $h$ is increasing
\begin{enumerate}[label=(\roman*)]
\item when $h = 6$,
\item after water has been poured in for 8 seconds.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR C3 Q6 [9]}}