| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2019 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Conical geometry differential equations |
| Difficulty | Standard +0.3 This is a standard related rates problem involving a cone, requiring similar triangles to relate r and h, differentiation of volume, and integration of a separable differential equation. While it has multiple parts, each step follows a well-established procedure taught in C3/C4 with no novel insights required, making it slightly easier than average. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{r}{h}=\frac{3}{5},\; r=\frac{3}{5}h\) | B1 | Any correct equation connecting \(r\) and \(h\) |
| \(V=\frac{1}{3}\pi r^2 h = \frac{1}{3}\pi\left(\frac{3}{5}h\right)^2 h \Rightarrow V=\frac{9}{75}\pi h^3\) | M1 | Obtains \(V=kh^3\) using their equation connecting \(h\) and \(r\) |
| \(\frac{dV}{dh}=\frac{27}{75}\pi h^2\) | dM1 | Attempts \(\frac{dV}{dh}\). Dependent on first M. |
| \(\frac{dV}{dh}=\frac{dV}{dt}\times\frac{dt}{dh} \Rightarrow \frac{27}{75}\pi h^2 = -0.02\frac{dt}{dh}\) | M1 | Uses chain rule with their \(\frac{dV}{dh}\) and \(\frac{dV}{dt}=\pm0.02\) |
| \(h^2\frac{dh}{dt}=-\frac{1}{18\pi}\) | A1 cso | Correct equation or states \(k=18\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{h^3}{3}=-\frac{1}{18\pi}t\,(+c)\) | M1 | \(ph^3 = qt\,(+c)\). Note "+c" is not required for this mark. |
| \(t=0,\; h=5 \Rightarrow c=\frac{125}{3}\) | M1 | Uses \(h=5\) and \(t=0\) to find \(c\). Must have constant of integration. |
| \(h=\sqrt[3]{125-\frac{t}{6\pi}}\) | A1 | Correct equation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(h=0 \Rightarrow 125-\frac{t}{6\pi}=0 \Rightarrow t=750\pi\) seconds | M1 | Puts \(h=0\) and solves for \(t\) |
| \(39\) (minutes) | A1 | Must be positive. Allow awrt 39 minutes. |
## Question 10:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{r}{h}=\frac{3}{5},\; r=\frac{3}{5}h$ | B1 | Any correct equation connecting $r$ and $h$ |
| $V=\frac{1}{3}\pi r^2 h = \frac{1}{3}\pi\left(\frac{3}{5}h\right)^2 h \Rightarrow V=\frac{9}{75}\pi h^3$ | M1 | Obtains $V=kh^3$ using their equation connecting $h$ and $r$ |
| $\frac{dV}{dh}=\frac{27}{75}\pi h^2$ | dM1 | Attempts $\frac{dV}{dh}$. Dependent on first M. |
| $\frac{dV}{dh}=\frac{dV}{dt}\times\frac{dt}{dh} \Rightarrow \frac{27}{75}\pi h^2 = -0.02\frac{dt}{dh}$ | M1 | Uses chain rule with their $\frac{dV}{dh}$ and $\frac{dV}{dt}=\pm0.02$ |
| $h^2\frac{dh}{dt}=-\frac{1}{18\pi}$ | A1 **cso** | Correct equation or states $k=18$ |
### Part (b) Way 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{h^3}{3}=-\frac{1}{18\pi}t\,(+c)$ | M1 | $ph^3 = qt\,(+c)$. Note "+c" is not required for this mark. |
| $t=0,\; h=5 \Rightarrow c=\frac{125}{3}$ | M1 | Uses $h=5$ and $t=0$ to find $c$. Must have constant of integration. |
| $h=\sqrt[3]{125-\frac{t}{6\pi}}$ | A1 | Correct equation |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $h=0 \Rightarrow 125-\frac{t}{6\pi}=0 \Rightarrow t=750\pi$ seconds | M1 | Puts $h=0$ and solves for $t$ |
| $39$ (minutes) | A1 | Must be positive. Allow awrt 39 minutes. |
---10.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ae871952-f525-44e6-8bac-09308aa1964f-38_570_671_310_680}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Diagram not drawn to scale
Figure 3 shows a container in the shape of an inverted right circular cone which contains some water.
The cone has an internal radius of 3 m and a vertical height of 5 m as shown in Figure 3.
At time $t$ seconds,the height of the water is $h$ metres,the volume of the water is $V \mathrm {~m} ^ { 3 }$ and water is leaking from a hole in the bottom of the container at a constant rate of $0.02 \mathrm {~m} ^ { 3 } \mathrm {~s} ^ { - 1 }$\\
[The volume of a cone of radius $r$ and height $h$ is $\frac { 1 } { 3 } \pi r ^ { 2 } h$ .]
\begin{enumerate}[label=(\alph*)]
\item Show that,while the water is leaking,
$$h ^ { 2 } \frac { \mathrm {~d} h } { \mathrm {~d} t } = - \frac { 1 } { \mathrm { k } \pi }$$
where $k$ is a constant to be found.
Given that the container is initially full of water,
\item express $h$ in terms of $t$ .
\item Find the time taken for the container to empty,giving your answer to the nearest minute.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2019 Q10 [10]}}