| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2018 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Related rates |
| Difficulty | Standard +0.3 This is a structured, multi-part related rates question with clear scaffolding. Part (a) is routine differentiation, part (b) applies the chain rule with guidance ('show that'), part (c) is separation of variables with given constants, and parts (d-e) are straightforward substitution. While it requires multiple techniques (chain rule, separation of variables, integration), each step is standard and well-signposted, making it slightly easier than the average A-level question. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates4.10c Integrating factor: first order equations |
| END |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dV}{dr} = 4\pi r^2\) | B1 | cao; condone \(= \frac{4}{3} \times 3\pi r^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dr}{dt} = \frac{dV}{dt} \div \frac{dV}{dr} = \frac{9000\pi}{(t+81)^{\frac{5}{4}}} \times \frac{1}{4\pi r^2}\) | M1 | Correct use of chain rule; allow slips |
| \(= \frac{2250}{r^2(t+81)^{\frac{5}{4}}}\) | A1 | cao; condone \(r \leftrightarrow R\), \(t \leftrightarrow T\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dr}{dt} = \frac{k}{r^n(t+81)^{\frac{5}{4}}}\), so \(\int r^n\, dr = \int \frac{k}{(t+81)^{\frac{5}{4}}}\, dt\) | B1 | Correct separation of variables; dr and dt must be present as numerators |
| \(\frac{r^3}{3} = \frac{2250}{-1/4}(t+81)^{-\frac{1}{4}}(+c)\) | M1 A1ft | M1: correct integration method both sides following through on \(n\); A1ft: correct integration of both sides ft their \(k\) and \(n\) |
| When \(t=0\), \(r=3\) so \(c = 9 + 9000 \times (81)^{-\frac{1}{4}}\) | M1 | Uses \(t=0\), \(r=3\) to find \(c\) |
| \(\frac{r^3}{3} = -9000(t+81)^{-\frac{1}{4}} + 3009\) | dM1 | Dependent on previous M; proceeds to \(r =\) using correct order of operations |
| \(r = \left[9027 - 27000(t+81)^{-\frac{1}{4}}\right]^{\frac{1}{3}}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Uses \(t = 175\) to give \(r = 13.2\) | B1 | accept awrt 13.2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Uses \(t = 175\) and \(r = 13.2\) into \(\frac{dr}{dt}\) to give \(\frac{dr}{dt} = 0.0127\) or \(0.0126\) | M1 A1 | M1: substitutes \(t=175\) and \(r=13.2\) into \(\frac{dr}{dt}\); A1: awrt 0.0126 or 0.0127 |
## Question 14:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dV}{dr} = 4\pi r^2$ | B1 | cao; condone $= \frac{4}{3} \times 3\pi r^2$ |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dr}{dt} = \frac{dV}{dt} \div \frac{dV}{dr} = \frac{9000\pi}{(t+81)^{\frac{5}{4}}} \times \frac{1}{4\pi r^2}$ | M1 | Correct use of chain rule; allow slips |
| $= \frac{2250}{r^2(t+81)^{\frac{5}{4}}}$ | A1 | cao; condone $r \leftrightarrow R$, $t \leftrightarrow T$ |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dr}{dt} = \frac{k}{r^n(t+81)^{\frac{5}{4}}}$, so $\int r^n\, dr = \int \frac{k}{(t+81)^{\frac{5}{4}}}\, dt$ | B1 | Correct separation of variables; dr and dt must be present as numerators |
| $\frac{r^3}{3} = \frac{2250}{-1/4}(t+81)^{-\frac{1}{4}}(+c)$ | M1 A1ft | M1: correct integration method both sides following through on $n$; A1ft: correct integration of both sides ft their $k$ and $n$ |
| When $t=0$, $r=3$ so $c = 9 + 9000 \times (81)^{-\frac{1}{4}}$ | M1 | Uses $t=0$, $r=3$ to find $c$ |
| $\frac{r^3}{3} = -9000(t+81)^{-\frac{1}{4}} + 3009$ | dM1 | Dependent on previous M; proceeds to $r =$ using correct order of operations |
| $r = \left[9027 - 27000(t+81)^{-\frac{1}{4}}\right]^{\frac{1}{3}}$ | A1 | |
### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Uses $t = 175$ to give $r = 13.2$ | B1 | accept awrt 13.2 |
### Part (e):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Uses $t = 175$ and $r = 13.2$ into $\frac{dr}{dt}$ to give $\frac{dr}{dt} = 0.0127$ or $0.0126$ | M1 A1 | M1: substitutes $t=175$ and $r=13.2$ into $\frac{dr}{dt}$; A1: awrt 0.0126 or 0.0127 |14. The volume of a spherical balloon of radius $r \mathrm {~cm}$ is $V \mathrm {~cm} ^ { 3 }$, where $V = \frac { 4 } { 3 } \pi r ^ { 3 }$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } V } { \mathrm {~d} r }$
The volume of the balloon increases with time $t$ seconds according to the formula
$$\frac { \mathrm { d } V } { \mathrm {~d} t } = \frac { 9000 \pi } { ( t + 81 ) ^ { \frac { 5 } { 4 } } } \quad t \geqslant 0$$
\item Using the chain rule, or otherwise, show that
$$\frac { \mathrm { d } r } { \mathrm {~d} t } = \frac { k } { r ^ { n } ( t + 81 ) ^ { \frac { 5 } { 4 } } } \quad t \geqslant 0$$
where $k$ and $n$ are constants to be found.
Initially, the radius of the balloon is 3 cm .
\item Using the values of $k$ and $n$ found in part (b), solve the differential equation
$$\frac { \mathrm { d } r } { \mathrm {~d} t } = \frac { k } { r ^ { n } ( t + 81 ) ^ { \frac { 5 } { 4 } } } \quad t \geqslant 0$$
to obtain a formula for $r$ in terms of $t$.
\item Hence find the radius of the balloon when $t = 175$, giving your answer to 3 significant figures.\\
(1)
\item Find the rate of increase of the radius of the balloon when $t = 175$. Give your answer to 3 significant figures.
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\hfill \mbox{\textit{Edexcel C34 2018 Q14 [12]}}