CAIE P3 2009 November — Question 10 10 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2009
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDifferential equations
TypeSpherical geometry differential equations
DifficultyStandard +0.3 This is a straightforward differential equations question requiring standard techniques: translating a word problem into a DE using given formulas (part i), solving a separable DE by integration (part ii), and interpreting the domain from the solution (part iii). While it involves multiple steps, each is routine for P3 level with no novel insight required, making it slightly easier than average.
Spec1.07t Construct differential equations: in context1.08k Separable differential equations: dy/dx = f(x)g(y)

10 In a model of the expansion of a sphere of radius \(r \mathrm {~cm}\), it is assumed that, at time \(t\) seconds after the start, the rate of increase of the surface area of the sphere is proportional to its volume. When \(t = 0\), \(r = 5\) and \(\frac { \mathrm { d } r } { \mathrm {~d} t } = 2\).
  1. Show that \(r\) satisfies the differential equation $$\frac { \mathrm { d } r } { \mathrm {~d} t } = 0.08 r ^ { 2 }$$ [The surface area \(A\) and volume \(V\) of a sphere of radius \(r\) are given by the formulae \(A = 4 \pi r ^ { 2 }\), \(V = \frac { 4 } { 3 } \pi r ^ { 3 }\).]
  2. Solve this differential equation, obtaining an expression for \(r\) in terms of \(t\).
  3. Deduce from your answer to part (ii) the set of values that \(t\) can take, according to this model.

(i)
AnswerMarks
State or imply \(\frac{dA}{dt} = kY\)M1*
Obtain equation in \(r\) and \(\frac{dr}{dt}\), e.g. \(8\pi r\frac{dr}{dt} = k\frac{4}{3}\pi r^3\)A1
Use \(\frac{dr}{dt} = 2, r = 5\) to evaluate \(k\)M1(dep*)
Obtain given answerA1
[4]
(ii)
AnswerMarks
Separate variables correctly and integrate both sidesM1
Obtain terms \(-\frac{1}{r}\) and \(0.08t\), or equivalentA1 + A1
Evaluate a constant or use limits \(t = 0, r = 5\) with a solution containing terms of the form \(\frac{a}{r}\) and \(bt\)M1
Obtain solution \(r = \frac{5}{(1-0.4t)}\), or equivalentA1
[5]
(iii)
AnswerMarks
State the set of values \(0 \leq t < 2.5\), or equivalentB1
[Allow \(t < 2.5\) and \(0 < t < 2.5\) to earn B1.][1]
**(i)**

| State or imply $\frac{dA}{dt} = kY$ | M1* |
| Obtain equation in $r$ and $\frac{dr}{dt}$, e.g. $8\pi r\frac{dr}{dt} = k\frac{4}{3}\pi r^3$ | A1 |
| Use $\frac{dr}{dt} = 2, r = 5$ to evaluate $k$ | M1(dep*) |
| Obtain given answer | A1 |
| [4] |

**(ii)**

| Separate variables correctly and integrate both sides | M1 |
| Obtain terms $-\frac{1}{r}$ and $0.08t$, or equivalent | A1 + A1 |
| Evaluate a constant or use limits $t = 0, r = 5$ with a solution containing terms of the form $\frac{a}{r}$ and $bt$ | M1 |
| Obtain solution $r = \frac{5}{(1-0.4t)}$, or equivalent | A1 |
| [5] |

**(iii)**

| State the set of values $0 \leq t < 2.5$, or equivalent | B1 |
| [Allow $t < 2.5$ and $0 < t < 2.5$ to earn B1.] | [1] |
10 In a model of the expansion of a sphere of radius $r \mathrm {~cm}$, it is assumed that, at time $t$ seconds after the start, the rate of increase of the surface area of the sphere is proportional to its volume. When $t = 0$, $r = 5$ and $\frac { \mathrm { d } r } { \mathrm {~d} t } = 2$.\\
(i) Show that $r$ satisfies the differential equation

$$\frac { \mathrm { d } r } { \mathrm {~d} t } = 0.08 r ^ { 2 }$$

[The surface area $A$ and volume $V$ of a sphere of radius $r$ are given by the formulae $A = 4 \pi r ^ { 2 }$, $V = \frac { 4 } { 3 } \pi r ^ { 3 }$.]\\
(ii) Solve this differential equation, obtaining an expression for $r$ in terms of $t$.\\
(iii) Deduce from your answer to part (ii) the set of values that $t$ can take, according to this model.

\hfill \mbox{\textit{CAIE P3 2009 Q10 [10]}}