| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2007 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Deriving the differential equation |
| Difficulty | Moderate -0.8 This is a straightforward multi-part differential equations question requiring standard techniques. Part (i) is simple substitution of given values to find a constant of proportionality. Parts (ii)-(iv) involve routine separation of variables with a substitution, followed by basic algebraic manipulation to answer contextual questions. All steps are mechanical applications of standard A-level methods with no novel insight required. |
| Spec | 1.07t Construct differential equations: in context1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| (i) State \(\frac{dh}{dt} = k(9 - h)^{\mathbf{\top}}\) | B1 | |
| Show that \(k = 0.1\) | B1 | Total: 2 marks |
| (ii) Separate variables correctly and attempt integration of at least one side | M1 | |
| Obtain terms \(-\frac{2}{3}(9-h)^{\frac{3}{2}}\) and \(0.1t\), or equivalent | A1 + A1 | |
| Evaluate a constant, or use limits \(t = 0, h = 1\) with a solution containing terms of the form \(a(9-h)^p\) and \(bt\), where \(p > 0\) | M1* | |
| Obtain solution in any form, e.g. \(-\frac{2}{3}(9-h)^{\frac{3}{2}} = 0.1t - 6\) | A1 | |
| Rearrange and make \(h\) the subject | M1(dep*) | |
| Obtain answer \(h = 9 - (4 - \frac{1}{15}t)^{\frac{3}{2}}\), or equivalent | A1 | Total: 7 marks |
| (iii) State that the maximum height is \(h = 9\) | B1 | |
| State that the time taken is 60 years | B1 | Total: 2 marks |
| (iv) Substitute \(h = 9/2\) and obtain \(t = 19.1\) (accept 19, 19.0 and 19.2) | B1 | Total: 1 mark |
(i) State $\frac{dh}{dt} = k(9 - h)^{\mathbf{\top}}$ | B1 |
Show that $k = 0.1$ | B1 | **Total: 2 marks**
(ii) Separate variables correctly and attempt integration of at least one side | M1 |
Obtain terms $-\frac{2}{3}(9-h)^{\frac{3}{2}}$ and $0.1t$, or equivalent | A1 + A1 |
Evaluate a constant, or use limits $t = 0, h = 1$ with a solution containing terms of the form $a(9-h)^p$ and $bt$, where $p > 0$ | M1* |
Obtain solution in any form, e.g. $-\frac{2}{3}(9-h)^{\frac{3}{2}} = 0.1t - 6$ | A1 |
Rearrange and make $h$ the subject | M1(dep*) |
Obtain answer $h = 9 - (4 - \frac{1}{15}t)^{\frac{3}{2}}$, or equivalent | A1 | **Total: 7 marks**
(iii) State that the maximum height is $h = 9$ | B1 |
State that the time taken is 60 years | B1 | **Total: 2 marks**
(iv) Substitute $h = 9/2$ and obtain $t = 19.1$ (accept 19, 19.0 and 19.2) | B1 | **Total: 1 mark**10 A model for the height, $h$ metres, of a certain type of tree at time $t$ years after being planted assumes that, while the tree is growing, the rate of increase in height is proportional to $( 9 - h ) ^ { \frac { 1 } { 3 } }$. It is given that, when $t = 0 , h = 1$ and $\frac { \mathrm { d } h } { \mathrm {~d} t } = 0.2$.\\
(i) Show that $h$ and $t$ satisfy the differential equation
$$\frac { \mathrm { d } h } { \mathrm {~d} t } = 0.1 ( 9 - h ) ^ { \frac { 1 } { 3 } } .$$
(ii) Solve this differential equation, and obtain an expression for $h$ in terms of $t$.\\
(iii) Find the maximum height of the tree and the time taken to reach this height after planting.\\
(iv) Calculate the time taken to reach half the maximum height.
\hfill \mbox{\textit{CAIE P3 2007 Q10 [12]}}