| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2012 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Connected Rates of Change |
| Type | Container filling: find depth rate |
| Difficulty | Standard +0.3 This is a straightforward connected rates of change question requiring chain rule application. Part (i) involves routine differentiation of a composite function using the chain rule, and part (ii) applies the standard formula dV/dt = (dV/dh)(dh/dt) with given values. While it requires careful calculation, it follows a standard template with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Attempt use of chain rule | *M1 | To obtain derivative of form \(kh(3h^2+4)^n\), any non-zero constants \(k\) and \(n\); condone retention of \(-8\) |
| Obtain \(9h(3h^2+4)^{\frac{1}{2}}\) | A1 | Or (unsimplified) equiv; no \(-8\) here |
| Substitute 0.6 in attempt at first derivative | M1 | dep *M; condone retention of \(-8\); implied by value following wrong derivative if no working seen |
| Obtain 12.17 | A1 | Or greater accuracy |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| State or imply \(\frac{dh}{dt} = -0.015\) or \(0.015\) | B1 | Implied by use in calculation with part (i) answer |
| Carry out multiplication of \((\pm)0.015\) and answer from part (i) | M1 | |
| Obtain \(0.18\) or \(-0.18\) (whatever this value is claimed to be) | A1 | Or greater accuracy; condone absence or misuse of negative signs throughout; ignore units; allow for answer rounding to 0.18 following slight inaccuracy due to use of 12.18 or 12.2 or … |
# Question 6(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt use of chain rule | *M1 | To obtain derivative of form $kh(3h^2+4)^n$, any non-zero constants $k$ and $n$; condone retention of $-8$ |
| Obtain $9h(3h^2+4)^{\frac{1}{2}}$ | A1 | Or (unsimplified) equiv; no $-8$ here |
| Substitute 0.6 in attempt at first derivative | M1 | dep *M; condone retention of $-8$; implied by value following wrong derivative if no working seen |
| Obtain 12.17 | A1 | Or greater accuracy |
---
# Question 6(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| State or imply $\frac{dh}{dt} = -0.015$ or $0.015$ | B1 | Implied by use in calculation with part (i) answer |
| Carry out multiplication of $(\pm)0.015$ and answer from part (i) | M1 | |
| Obtain $0.18$ or $-0.18$ (whatever this value is claimed to be) | A1 | Or greater accuracy; condone absence or misuse of negative signs throughout; ignore units; allow for answer rounding to 0.18 following slight inaccuracy due to use of 12.18 or 12.2 or … |
---6 The volume, $V \mathrm {~m} ^ { 3 }$, of liquid in a container is given by
$$V = \left( 3 h ^ { 2 } + 4 \right) ^ { \frac { 3 } { 2 } } - 8 ,$$
where $h \mathrm {~m}$ is the depth of the liquid.\\
(i) Find the value of $\frac { \mathrm { d } V } { \mathrm {~d} h }$ when $h = 0.6$, giving your answer correct to 2 decimal places.\\
(ii) Liquid is leaking from the container. It is observed that, when the depth of the liquid is 0.6 m , the depth is decreasing at a rate of 0.015 m per hour. Find the rate at which the volume of liquid in the container is decreasing at the instant when the depth is 0.6 m .
\hfill \mbox{\textit{OCR C3 2012 Q6 [7]}}