| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Connected Rates of Change |
| Type | Cone: related rates of dimensions |
| Difficulty | Standard +0.3 This is a straightforward connected rates problem requiring Pythagoras to express radius in terms of height, substitution into the volume formula (part i is a 'show that'), then standard differentiation and second derivative test. All steps are routine A-level techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.07d Second derivatives: d^2y/dx^2 notation1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use of Pythagoras \(\rightarrow r^2 = 15^2 - h^2\) | M1 | |
| \(V = \frac{1}{3}\pi(225 - h^2) \times h \rightarrow \frac{1}{3}\pi(225h - h^3)\) | A1 | AG; WWW e.g. sight of \(r = 15 - h\) gets A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{dV}{dh} = \frac{\pi}{3}(225 - 3h^2)\) | B1 | |
| Their \(\frac{dV}{dh} = 0\) | M1 | Differentiates, sets their differential to 0 and attempts to solve at least as far as \(h^2 \neq 0\) |
| \((h =)\ \sqrt{75},\ 5\sqrt{3}\) or AWRT \(8.66\) | A1 | Ignore \(-\sqrt{75}\) OE and ISW for both A marks |
| \(\frac{d^2V}{dh^2} = \frac{\pi}{3}(-6h)\ (\rightarrow -\text{ve})\) | M1 | Differentiates for a second time and considers the sign of the second differential or any other valid complete method |
| \(\rightarrow\) Maximum | A1FT | Correct conclusion from correct 2nd differential, value for \(h\) not required, or any other valid complete method. FT for their \(h\), if used, as long as it is positive. SC Omission of \(\pi\) or \(\frac{\pi}{3}\) throughout can score B0M1A1M1A0 |
## Question 5(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use of Pythagoras $\rightarrow r^2 = 15^2 - h^2$ | M1 | |
| $V = \frac{1}{3}\pi(225 - h^2) \times h \rightarrow \frac{1}{3}\pi(225h - h^3)$ | A1 | AG; WWW e.g. sight of $r = 15 - h$ gets A0 |
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## Question 5(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dV}{dh} = \frac{\pi}{3}(225 - 3h^2)$ | B1 | |
| Their $\frac{dV}{dh} = 0$ | M1 | Differentiates, sets their differential to 0 and attempts to solve at least as far as $h^2 \neq 0$ |
| $(h =)\ \sqrt{75},\ 5\sqrt{3}$ or AWRT $8.66$ | A1 | Ignore $-\sqrt{75}$ OE and ISW for both A marks |
| $\frac{d^2V}{dh^2} = \frac{\pi}{3}(-6h)\ (\rightarrow -\text{ve})$ | M1 | Differentiates for a second time and considers the sign of the second differential or any other valid complete method |
| $\rightarrow$ Maximum | A1FT | Correct conclusion from correct 2nd differential, value for $h$ not required, or any other valid complete method. FT for their $h$, if used, as long as it is positive. **SC** Omission of $\pi$ or $\frac{\pi}{3}$ throughout can score B0M1A1M1A0 |
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\includegraphics[max width=\textwidth, alt={}, center]{567c3d72-c633-4ae0-8605-f63f93d718c4-08_512_460_258_772}\\
\includegraphics[max width=\textwidth, alt={}, center]{567c3d72-c633-4ae0-8605-f63f93d718c4-08_462_85_260_1279}
The diagram shows a solid cone which has a slant height of 15 cm and a vertical height of $h \mathrm {~cm}$.\\
(i) Show that the volume, $V \mathrm {~cm} ^ { 3 }$, of the cone is given by $V = \frac { 1 } { 3 } \pi \left( 225 h - h ^ { 3 } \right)$.\\[0pt]
[The volume of a cone of radius $r$ and vertical height $h$ is $\frac { 1 } { 3 } \pi r ^ { 2 } h$.]\\
(ii) Given that $h$ can vary, find the value of $h$ for which $V$ has a stationary value. Determine, showing all necessary working, the nature of this stationary value.\\
\hfill \mbox{\textit{CAIE P1 2019 Q5 [7]}}