CAIE P3 2004 November — Question 10 10 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2004
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFirst order differential equations (integrating factor)
TypeTank/reservoir mixing problems
DifficultyStandard +0.3 This is a structured multi-part differential equations question with significant scaffolding. Part (i) requires setting up a rate equation from a word problem (straightforward with given information). Part (ii) involves separable variables after substitution is provided, and part (iii) is direct substitution. While it requires multiple techniques, the heavy guidance and standard methods place it slightly above average difficulty.
Spec1.08k Separable differential equations: dy/dx = f(x)g(y)

10 A rectangular reservoir has a horizontal base of area \(1000 \mathrm {~m} ^ { 2 }\). At time \(t = 0\), it is empty and water begins to flow into it at a constant rate of \(30 \mathrm {~m} ^ { 3 } \mathrm {~s} ^ { - 1 }\). At the same time, water begins to flow out at a rate proportional to \(\sqrt { } h\), where \(h \mathrm {~m}\) is the depth of the water at time \(t \mathrm {~s}\). When \(h = 1 , \frac { \mathrm {~d} h } { \mathrm {~d} t } = 0.02\).
  1. Show that \(h\) satisfies the differential equation $$\frac { \mathrm { d } h } { \mathrm {~d} t } = 0.01 ( 3 - \sqrt { } h )$$ It is given that, after making the substitution \(x = 3 - \sqrt { } h\), the equation in part (i) becomes $$( x - 3 ) \frac { \mathrm { d } x } { \mathrm {~d} t } = 0.005 x$$
  2. Using the fact that \(x = 3\) when \(t = 0\), solve this differential equation, obtaining an expression for \(t\) in terms of \(x\).
  3. Find the time at which the depth of water reaches 4 m .

AnswerMarks Guidance
(i) State or imply \(\frac{dV}{dt} = 1000\frac{dh}{dt}\)B1
State or imply \(\frac{dV}{dt} = 30 - k\sqrt{h}\) or \(\frac{dh}{dt} = 0.03 - m\sqrt{h}\)B1
Show that \(k = 10\) or \(m = 0.01\) and justify the given equationB1 [Allow the first B1 for the statement that \(0.03 = 30/1000\).]
Total: 3 marks
(ii) Separate variables and attempt integration of \(\frac{x-3}{x}\) with respect to \(x\)M1*
Obtain \(x - 3\ln x\), or equivalentA1
Obtain \(0.005t\), or equivalentA1
Use \(x = 3, t = 0\) in the evaluation of a constant or as limits in an answer involving \(\ln x\) and \(kt\)M1(dep*)
Obtain answer in any correct form e.g. \(t = 200(x-3-3\ln x+3\ln 3)\)A1 [To qualify for the first M mark, an attempt to solve the earlier differential equation in \(h\) and \(t\) must involve correct separation of variables, the use of a substitution such as \(\sqrt{h} = u\), and an attempt to integrate the resulting function of \(u\).]
Total: 5 marks
(iii) Substitute \(x = 1\) and calculate \(t\)M1
Obtain answer \(t = 259\) correctlyA1 Total: 2 marks
**(i)** State or imply $\frac{dV}{dt} = 1000\frac{dh}{dt}$ | B1 | |
State or imply $\frac{dV}{dt} = 30 - k\sqrt{h}$ or $\frac{dh}{dt} = 0.03 - m\sqrt{h}$ | B1 | |
Show that $k = 10$ or $m = 0.01$ and justify the given equation | B1 | [Allow the first B1 for the statement that $0.03 = 30/1000$.] |
| | | **Total: 3 marks** |

**(ii)** Separate variables and attempt integration of $\frac{x-3}{x}$ with respect to $x$ | M1* | |
Obtain $x - 3\ln x$, or equivalent | A1 | |
Obtain $0.005t$, or equivalent | A1 | |
Use $x = 3, t = 0$ in the evaluation of a constant or as limits in an answer involving $\ln x$ and $kt$ | M1(dep*) | |
Obtain answer in any correct form e.g. $t = 200(x-3-3\ln x+3\ln 3)$ | A1 | [To qualify for the first M mark, an attempt to solve the earlier differential equation in $h$ and $t$ must involve correct separation of variables, the use of a substitution such as $\sqrt{h} = u$, and an attempt to integrate the resulting function of $u$.] |
| | | **Total: 5 marks** |

**(iii)** Substitute $x = 1$ and calculate $t$ | M1 | |
Obtain answer $t = 259$ correctly | A1 | **Total: 2 marks** |
10 A rectangular reservoir has a horizontal base of area $1000 \mathrm {~m} ^ { 2 }$. At time $t = 0$, it is empty and water begins to flow into it at a constant rate of $30 \mathrm {~m} ^ { 3 } \mathrm {~s} ^ { - 1 }$. At the same time, water begins to flow out at a rate proportional to $\sqrt { } h$, where $h \mathrm {~m}$ is the depth of the water at time $t \mathrm {~s}$. When $h = 1 , \frac { \mathrm {~d} h } { \mathrm {~d} t } = 0.02$.\\
(i) Show that $h$ satisfies the differential equation

$$\frac { \mathrm { d } h } { \mathrm {~d} t } = 0.01 ( 3 - \sqrt { } h )$$

It is given that, after making the substitution $x = 3 - \sqrt { } h$, the equation in part (i) becomes

$$( x - 3 ) \frac { \mathrm { d } x } { \mathrm {~d} t } = 0.005 x$$

(ii) Using the fact that $x = 3$ when $t = 0$, solve this differential equation, obtaining an expression for $t$ in terms of $x$.\\
(iii) Find the time at which the depth of water reaches 4 m .

\hfill \mbox{\textit{CAIE P3 2004 Q10 [10]}}