| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2014 |
| Session | June |
| Marks | 11 |
| 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 requiring similar triangles to relate r and h, differentiation of volume, separation of variables, and integration of h². While it involves multiple steps, each technique is routine for C4 level with clear signposting through the parts. The geometric setup is straightforward and the algebra uncomplicated, making it slightly easier than average. |
| Spec | 1.07t Construct differential equations: in context1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{dV}{dt} = \pm 0.01\) | B1 | |
| by similar triangles, \(\dfrac{h}{4.5} = \dfrac{r}{3}\) | B1 | may be implied by \(r = \dfrac{2h}{3}\) oe |
| \(\dfrac{dV}{dh} = \dfrac{4}{9}\pi h^2\) oe | B1 | |
| \(\dfrac{dh}{dt} = \pm 0.01 \times\) their \(\dfrac{dh}{dV}\) oe | M1 | use of Chain rule |
| \(-0.01 = \left(\dfrac{4}{9}\pi h^2\right) \times \dfrac{dh}{dt}\) | A1 | completion to given result |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int h^2\,dh = \int\dfrac{-9}{400\pi}\,dt\) oe soi | M1 | separation of variables |
| \(\dfrac{h^3}{3} = \dfrac{-9}{400\pi}t\,(+c)\) | A1 | |
| substitution of \(t=0\) and \(h=4.5\) in their expression following integration | M1 | expression must include \(c\) and powers must be correct on each side |
| \(h = \sqrt[3]{\dfrac{729}{8} - \dfrac{27t}{400\pi}}\) oe isw | A1 | allow \(-0.0215\) or \(-0.02148591\ldots\) r.o.t to 4 sf or more and similarly \(91.125\) |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| set \(h=0\) and solve to obtain positive \(t\) | M1 | or \((t=)\dfrac{1}{3}\pi \times 3^2 \times 4.5 \div 0.01\) (= \(1350\pi\)) |
| 71 minutes cao | A1 | |
| [2] |
# Question 10(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{dV}{dt} = \pm 0.01$ | B1 | |
| by similar triangles, $\dfrac{h}{4.5} = \dfrac{r}{3}$ | B1 | may be implied by $r = \dfrac{2h}{3}$ oe |
| $\dfrac{dV}{dh} = \dfrac{4}{9}\pi h^2$ oe | B1 | |
| $\dfrac{dh}{dt} = \pm 0.01 \times$ their $\dfrac{dh}{dV}$ oe | M1 | use of Chain rule | may follow from incorrect differentiation: expressions must be a function of either $r$ or $h$ or both |
| $-0.01 = \left(\dfrac{4}{9}\pi h^2\right) \times \dfrac{dh}{dt}$ | A1 | completion to given result | $h^2\dfrac{dh}{dt} = \dfrac{-0.09}{4\pi} = \dfrac{-9}{400\pi}$ |
| **[5]** | | |
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# Question 10(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int h^2\,dh = \int\dfrac{-9}{400\pi}\,dt$ oe soi | M1 | separation of variables | if no subsequent work, integral signs needed, but allow omission of $dh$ or $dt$, but must be correctly placed if present |
| $\dfrac{h^3}{3} = \dfrac{-9}{400\pi}t\,(+c)$ | A1 | |
| substitution of $t=0$ and $h=4.5$ in their expression following integration | M1 | expression must include $c$ and powers must be correct on each side |
| $h = \sqrt[3]{\dfrac{729}{8} - \dfrac{27t}{400\pi}}$ oe isw | A1 | allow $-0.0215$ or $-0.02148591\ldots$ r.o.t to 4 sf or more and similarly $91.125$ | $91.125 = \dfrac{729}{8}$ |
| **[4]** | | |
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# Question 10(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| set $h=0$ and solve to obtain positive $t$ | M1 | or $(t=)\dfrac{1}{3}\pi \times 3^2 \times 4.5 \div 0.01$ (= $1350\pi$) | NB $1350\pi = 4241.150082\ldots$ |
| 71 minutes cao | A1 | |
| **[2]** | | |10 A container in the shape of an inverted cone of radius 3 metres and vertical height 4.5 metres is initially filled with liquid fertiliser. This fertiliser is released through a hole in the bottom of the container at a rate of $0.01 \mathrm {~m} ^ { 3 }$ per second. At time $t$ seconds the fertiliser remaining in the container forms an inverted cone of height $h$ metres.\\[0pt]
[The volume of a cone is $V = \frac { 1 } { 3 } \pi r ^ { 2 } h$.]\\
(i) Show that $h ^ { 2 } \frac { \mathrm {~d} h } { \mathrm {~d} t } = - \frac { 9 } { 400 \pi }$.\\
(ii) Express $h$ in terms of $t$.\\
(iii) Find the time it takes to empty the container, giving your answer to the nearest minute.
\hfill \mbox{\textit{OCR C4 2014 Q10 [11]}}