A2 June 2025 Paper 1 Q17
17 A researcher is modelling the height of a particular type of tree over its lifetime.
Data suggests that the maximum possible height of this type of tree over its lifetime is double the height of the tree 5 years after planting.
It is given that, \(t\) years after planting a seed for this type of tree, the corresponding height of the tree is \(h\) m, and that \(h = 0\) when \(t = 0\).
\(\dfrac{\mathrm{d}^2h}{\mathrm{d}t^2} + 0.3\dfrac{\mathrm{d}h}{\mathrm{d}t} + 0.02h = 0.4\)
Further research determines that the initial rate of growth of this type of tree is 2.9 metres per year.
| Scheme | Marks | AO |
|---|---|---|
| (i) \(\dfrac{\mathrm{d}h}{\mathrm{d}t} = 0.2(20 - h)\) | B1 | 3.3 |
| [1] | ||
| (ii) \(\dfrac{\mathrm{d}h}{\mathrm{d}t} = 0 \Rightarrow h = 20\) | B1 | 3.4 |
| [1] | ||
| (iii) \(\int \frac{\mathrm{d}h}{20 - h} = \int 0.2\,\mathrm{d}t\) or \(h\mathrm{e}^{0.2t} = \int 4\mathrm{e}^{0.2t}\,\mathrm{d}t\) | M1* | 2.1 |
| \(-\ln(20 - h) = 0.2t \; [+ c]\) \(20 - h = A\mathrm{e}^{-0.2t}\) | A1 | 1.1 |
| when \(t = 0\), \(h = 0 \Rightarrow A = 20\) | M1dep | 3.4 |
| \(\Rightarrow h = 20\left(1 - \mathrm{e}^{-0.2t}\right)\) | A1 | 2.1 |
| [4] | ||
| (iv) When \(t = 5\), \(h = 20\left(1 - \frac{1}{\mathrm{e}}\right) = 12.64\) | M1 | 3.4 |
| 25.28… [\(\neq 20\)], so model does not fit the data well | A1 | 3.5a |
| [2] |
Notes
(a)(i)
B1: or any rearrangement. ISW.
(a)(ii)
B1: or converse. Could be in words. Do not condone \(h \to 20\).
Accept their expression for \(\frac{\mathrm{d}h}{\mathrm{d}t} = 0\) from (a)(i) provided it is correct or of the form \(k(20 - h)\).
(a)(iii)
M1*: correctly separating variables or using integrating factor for their DE, leading to an integral soi. If a CF/PI approach used award for a correct method leading to the CF and a value for the PI.
A1: or \(h\mathrm{e}^{0.2t} = 20\mathrm{e}^{0.2t}[+c]\) or \(h = B\mathrm{e}^{-0.2t} + 20\). Condone missing brackets if recovered.
M1dep: substituting initial conditions into their equation leading to a value for \(A\), \(B\) or \(c\)
A1: AG www, condone missing brackets earlier if recovered but do not condone \(\frac{\mathrm{d}h}{\mathrm{d}t}\) or \(\frac{\mathrm{d}}{\mathrm{d}x}\) used for \(\frac{\mathrm{d}}{\mathrm{d}t}\)
(a)(iv)
M1: substituting \(t = 5\) into given expression for \(h\)
A1: or \(12.64\ldots \neq 10\) soi but 10 oe must be seen, e.g. \(\frac{20}{2}\) or “half the maximum height”. Comment of inconsistency required (must not say that the value is consistent).
| Scheme | Marks | AO |
|---|---|---|
| (i) \(\lambda^2 + 0.3\lambda + 0.02 = 0\) \(\Rightarrow \lambda = -0.1\) or \(-0.2\) | M1 | 1.1 |
| CF: \([h =]\, A\mathrm{e}^{-0.1t} + B\mathrm{e}^{-0.2t}\) | A1 | 1.1 |
| PI: \([h =]\, k \Rightarrow 0.02k = 0.4 \quad [k = 20]\) | M1 | 2.1 |
| \(\Rightarrow h = A\mathrm{e}^{-0.1t} + B\mathrm{e}^{-0.2t} + 20\) | A1 | 1.1 |
| [4] | ||
| (ii) When \(t \to \infty\), \(\mathrm{e}^{-kt} \to 0\) so \(h \to 20\) [so final height is 20 m] | B1 | 3.4 |
| [1] | ||
| (iii) \(t = 0\): \(A + B + 20 = 0\) | M1FT | 3.4 |
| \(\dfrac{\mathrm{d}h}{\mathrm{d}t} = -0.1A\mathrm{e}^{-0.1t} - 0.2B\mathrm{e}^{-0.2t}\) | M1FT | 3.1b |
| \(\dfrac{\mathrm{d}h}{\mathrm{d}t} = 2.9: -0.1A - 0.2B = 2.9\) | M1FT | 1.1 |
| \(A = -11, B = -9 \quad \left[h = 20 - 11\mathrm{e}^{-0.1t} - 9\mathrm{e}^{-0.2t}\right]\) | A1 | 1.1 |
| When \(t = 5\), \(h = 10.017\ldots \approx 10\) so model does fit the data well | A1 | 3.5a |
| [5] |
Notes
(b)(i)
M1: \(= 0\) must be seen unless implied by correct roots
M1: correct form of particular integral (any polynomial with a constant term) and substituting into DE correctly, soi by correct \(k\)
A1: must see \(h =\)
(b)(ii)
B1: must state that each exponential term (or their sum) tends to 0. Accept general term with negative exponent, e.g. \(\mathrm{e}^{-kt}\), but not \(\mathrm{e}^{-t}\). Condone \(h = 20\).
(b)(iii)
M1FT: substituting \(t = 0\), \(h = 0\) correctly into their formula for \(h\) from part (b)(i) to form an equation in \(A\) and \(B\); must be seen
M1FT: differentiate their \(h\) from part (b)(i) correctly, soi by correct second equation in \(A\) and \(B\)
M1FT: substituting \(t = 0\) and \(\frac{\mathrm{d}h}{\mathrm{d}t} = 2.9\) to form (a second) equation in \(A\) and \(B\); must be seen
A1: or \(10.017\ldots \times 2 = 20.03\ldots\) so does fit the data well. Accept “half the maximum height” oe for 10.