| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 5 |
| 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 of V = (π/12)x³, and chain rule application. Part (i) is routine geometry, part (ii) is straightforward implicit differentiation with dV/dt = 14. Slightly easier than average as it's a textbook application of related rates with clear scaffolding. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use \(\alpha\) (possibly implicitly) to state that radius of 'base' is \(\frac{1}{2}x\) | *B1 | Or to obtain equiv such as \(2r=x\) or \(\frac{r}{x}=\frac{1}{2}\) or \(\frac{x}{r}=2\) |
| Substitute into formula to obtain \(\frac{1}{3}\pi(\frac{1}{2}x)^2 x\) or \(\frac{1}{3}\pi\frac{1}{4}x^2 x\) and obtain \(\frac{1}{12}\pi x^3\) | B1 | Dep *B; AG; necessary detail needed. Note: comparing formulae \(\frac{1}{3}\pi r^2 h\) and \(\frac{1}{12}\pi x^3\) to 'deduce' is B0B0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Differentiate to obtain \(\frac{1}{4}\pi x^2\) or equiv | B1 | Whatever they call it |
| Attempt division involving 14 and their value of derivative when \(x=8\) | M1 | i.e. \(14 \div \text{deriv}\) or \(\text{deriv} \div 14\) with \(x=8\) |
| Obtain 0.28 | A1 | Allow 0.279 but not greater accuracy. Alternatives: 1. \(14t = \frac{1}{12}\pi x^3\) Obtain \(\frac{dt}{dx}=\frac{1}{56}\pi x^2\) B1 Sub 8 and invert M1 Ans A1; 2. \(x^3=\frac{168t}{\pi}\) Obtain \(3x^2\frac{dx}{dt}=\frac{168}{\pi}\) B1 Sub 8 M1 Ans A1 |
## Question 3:
### Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use $\alpha$ (possibly implicitly) to state that radius of 'base' is $\frac{1}{2}x$ | *B1 | Or to obtain equiv such as $2r=x$ or $\frac{r}{x}=\frac{1}{2}$ or $\frac{x}{r}=2$ |
| Substitute into formula to obtain $\frac{1}{3}\pi(\frac{1}{2}x)^2 x$ or $\frac{1}{3}\pi\frac{1}{4}x^2 x$ and obtain $\frac{1}{12}\pi x^3$ | B1 | Dep *B; AG; necessary detail needed. Note: comparing formulae $\frac{1}{3}\pi r^2 h$ and $\frac{1}{12}\pi x^3$ to 'deduce' is B0B0 |
### Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Differentiate to obtain $\frac{1}{4}\pi x^2$ or equiv | B1 | Whatever they call it |
| Attempt division involving 14 and their value of derivative when $x=8$ | M1 | i.e. $14 \div \text{deriv}$ or $\text{deriv} \div 14$ with $x=8$ |
| Obtain 0.28 | A1 | Allow 0.279 but not greater accuracy. Alternatives: 1. $14t = \frac{1}{12}\pi x^3$ Obtain $\frac{dt}{dx}=\frac{1}{56}\pi x^2$ B1 Sub 8 and invert M1 Ans A1; 2. $x^3=\frac{168t}{\pi}$ Obtain $3x^2\frac{dx}{dt}=\frac{168}{\pi}$ B1 Sub 8 M1 Ans A1 |
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\includegraphics[max width=\textwidth, alt={}, center]{71e01d8f-d0ed-4f17-b7cd-6f5a93bbe329-2_435_472_932_794}
The diagram shows a container in the form of a right circular cone. The angle between the axis and the slant height is $\alpha$, where $\alpha = \tan ^ { - 1 } \left( \frac { 1 } { 2 } \right)$. Initially the container is empty, and then liquid is added at the rate of $14 \mathrm {~cm} ^ { 3 }$ per minute. The depth of liquid in the container at time $t$ minutes is $x \mathrm {~cm}$.\\
(i) Show that the volume, $V \mathrm {~cm} ^ { 3 }$, of liquid in the container when the depth is $x \mathrm {~cm}$ is given by
$$V = \frac { 1 } { 12 } \pi x ^ { 3 } .$$
[The volume of a cone is $\frac { 1 } { 3 } \pi r ^ { 2 } h$.]\\
(ii) Find the rate at which the depth of the liquid is increasing at the instant when the depth is 8 cm . Give your answer in cm per minute correct to 2 decimal places.
\hfill \mbox{\textit{OCR C3 2013 Q3 [5]}}