| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2017 |
| Session | October |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Substitution method |
| Difficulty | Standard +0.8 This is a multi-step differential equations problem requiring a guided substitution to find an integral (non-trivial with the square root term), then applying it to solve a separable DE with numerical limits. The substitution itself requires careful manipulation, and part (b) involves solving a transcendental equation numerically. More demanding than standard C3/C4 questions but the substitution is given, making it accessible with systematic work. |
| Spec | 1.08h Integration by substitution1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{du}{dh} = -\frac{1}{2}h^{-\frac{1}{2}}\) OR \(\frac{dh}{du} = -2(5-u)\) or \(du = -\frac{1}{2}h^{-\frac{1}{2}}dh\) | B1 | |
| \(\int \frac{dh}{5-\sqrt{h}} = \int \frac{-2(5-u)du}{u} = \int\left(\frac{-10}{u}+2\right)du\) | M1dM1A1 | Rewrite in terms of \(u\); must see both \(dh\) and \(5-\sqrt{h}\) in terms of \(u\) but \(dh \neq du\); divide by \(u\) to reach \(\int\left(\frac{A}{u}+B\right)du\) |
| \(= -10\ln u + 2u + c\) | ||
| \(= -10\ln(5-\sqrt{h}) + 2(5-\sqrt{h}) + c\) | M1 | Back-substitute |
| \(= -10\ln(5-\sqrt{h}) - 2\sqrt{h} + k\) | A1* | CSO; must have constant at integration stage and evidence \(2(5-\sqrt{h})+c \to -2\sqrt{h}+k\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dh}{dt} = \frac{t^{0.2}(5-\sqrt{h})}{5} \Rightarrow \int\frac{dh}{(5-\sqrt{h})} = \int\frac{t^{0.2}}{5}dt\) | B1 | Separates variables; accept even without integral signs |
| \(-10\ln(5-\sqrt{h}) - 2\sqrt{h} + k = \frac{t^{1.2}}{6}\) or equivalent | M1A1 | Must see \(\int\frac{dh}{(5-\sqrt{h})} \to A\ln(5-\sqrt{h})+B\sqrt{h}\) and \(\int t^{0.2}dt \to Ct^{1.2}\) |
| Substitute \(t=0, h=2 \Rightarrow k = 10\ln(5-\sqrt{2})+2\sqrt{2}\) (awrt 15.6) | M1 | |
| Substitute \(h=15\): \(\frac{t^{1.2}}{6} = -10\ln(5-\sqrt{15})-2\sqrt{15}+10\ln(5-\sqrt{2})+2\sqrt{2}\) | dM1 | Dependent on previous M1 |
| \(t^{1.2} = 39.94 \Rightarrow t = 21.6\) (or 21.7) | dM1A1 | All previous M marks must have been scored |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dh}{dt} = \frac{21.6^{0.2}(5-\sqrt{15})}{5} = 0.42\) cm per year | B1 | 42 only |
# Question 11:
## Part (a):
| $\frac{du}{dh} = -\frac{1}{2}h^{-\frac{1}{2}}$ OR $\frac{dh}{du} = -2(5-u)$ or $du = -\frac{1}{2}h^{-\frac{1}{2}}dh$ | B1 | |
| $\int \frac{dh}{5-\sqrt{h}} = \int \frac{-2(5-u)du}{u} = \int\left(\frac{-10}{u}+2\right)du$ | M1dM1A1 | Rewrite in terms of $u$; must see both $dh$ and $5-\sqrt{h}$ in terms of $u$ but $dh \neq du$; divide by $u$ to reach $\int\left(\frac{A}{u}+B\right)du$ |
| $= -10\ln u + 2u + c$ | | |
| $= -10\ln(5-\sqrt{h}) + 2(5-\sqrt{h}) + c$ | M1 | Back-substitute |
| $= -10\ln(5-\sqrt{h}) - 2\sqrt{h} + k$ | A1* | CSO; must have constant at integration stage and evidence $2(5-\sqrt{h})+c \to -2\sqrt{h}+k$ |
## Part (b):
| $\frac{dh}{dt} = \frac{t^{0.2}(5-\sqrt{h})}{5} \Rightarrow \int\frac{dh}{(5-\sqrt{h})} = \int\frac{t^{0.2}}{5}dt$ | B1 | Separates variables; accept even without integral signs |
| $-10\ln(5-\sqrt{h}) - 2\sqrt{h} + k = \frac{t^{1.2}}{6}$ or equivalent | M1A1 | Must see $\int\frac{dh}{(5-\sqrt{h})} \to A\ln(5-\sqrt{h})+B\sqrt{h}$ and $\int t^{0.2}dt \to Ct^{1.2}$ |
| Substitute $t=0, h=2 \Rightarrow k = 10\ln(5-\sqrt{2})+2\sqrt{2}$ (awrt 15.6) | M1 | |
| Substitute $h=15$: $\frac{t^{1.2}}{6} = -10\ln(5-\sqrt{15})-2\sqrt{15}+10\ln(5-\sqrt{2})+2\sqrt{2}$ | dM1 | Dependent on previous M1 |
| $t^{1.2} = 39.94 \Rightarrow t = 21.6$ (or 21.7) | dM1A1 | All previous M marks must have been scored |
## Part (c):
| $\frac{dh}{dt} = \frac{21.6^{0.2}(5-\sqrt{15})}{5} = 0.42$ cm per year | B1 | 42 only |
---\begin{enumerate}
\item (a) Given $0 \leqslant h < 25$, use the substitution $u = 5 - \sqrt { h }$ to show that
\end{enumerate}
$$\int \frac { \mathrm { d } h } { 5 - \sqrt { h } } = - 10 \ln ( 5 - \sqrt { h } ) - 2 \sqrt { h } + k$$
where $k$ is a constant.\\
(6)
A team of scientists is studying a species of tree.\\
The rate of change in height of a tree of this species is modelled by the differential equation
$$\frac { \mathrm { d } h } { \mathrm {~d} t } = \frac { t ^ { 0.2 } ( 5 - \sqrt { h } ) } { 5 }$$
where $h$ is the height of the tree in metres and $t$ is the time in years after the tree is planted.\\
One of these trees is 2 metres high when it is planted.\\
(b) Use integration to calculate the time it would take for this tree to reach a height of 15 metres, giving your answer to one decimal place.\\
(c) Hence calculate the rate of change in height of this tree when its height is 15 metres. Write your answer in centimetres per year to the nearest centimetre.
\hfill \mbox{\textit{Edexcel C34 2017 Q11 [14]}}