OCR MEI C3 — Question 4 5 marks

Exam BoardOCR MEI
ModuleC3 (Core Mathematics 3)
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicConnected Rates of Change
TypeSphere: radius rate from volume rate
DifficultyModerate -0.3 This is a straightforward connected rates of change question requiring differentiation of the volume formula with respect to time and substitution of given values. While it involves the chain rule (dV/dt = dV/dr × dr/dt), the setup is direct with all information provided explicitly, making it slightly easier than average for A-level.
Spec1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates
4 The volume of a sphere, \(V \mathrm {~cm} ^ { 3 }\) is given by the formula \(V = \frac { 4 } { 3 } \pi r ^ { 3 }\) where \(r \mathrm {~cm}\) is the radius.
The radius of a sphere increases at a constant rate of 2 cm per second.
Find the rate of increase of \(V\) when \(r = 10 \mathrm {~cm}\).
Question 4:
AnswerMarks Guidance
AnswerMark Guidance
\(V = \frac{4}{3}\pi r^3\)M1
\(\frac{dV}{dr} = 4\pi r^2\)B1
\(\frac{dr}{dt} = 2\)B1
\(\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt} = 4\pi \times 10^2 \times 2\)A1
\(= 800\pi\)A1 B1 for number, B1 for units
The rate of increase of \(V\) is \(2500 \text{ cm}^3/\text{sec}\), to 2sf
Total: 5 
## Question 4:
| Answer | Mark | Guidance |
|--------|------|----------|
| $V = \frac{4}{3}\pi r^3$ | M1 | |
| $\frac{dV}{dr} = 4\pi r^2$ | B1 | |
| $\frac{dr}{dt} = 2$ | B1 | |
| $\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt} = 4\pi \times 10^2 \times 2$ | A1 | |
| $= 800\pi$ | A1 | B1 for number, B1 for units |
| The rate of increase of $V$ is $2500 \text{ cm}^3/\text{sec}$, to 2sf | | |
| | **Total: 5** | |

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4 The volume of a sphere, $V \mathrm {~cm} ^ { 3 }$ is given by the formula $V = \frac { 4 } { 3 } \pi r ^ { 3 }$ where $r \mathrm {~cm}$ is the radius.\\
The radius of a sphere increases at a constant rate of 2 cm per second.\\
Find the rate of increase of $V$ when $r = 10 \mathrm {~cm}$.

\hfill \mbox{\textit{OCR MEI C3  Q4 [5]}}
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