October 2021 Paper 1 Q9
9 A particle moves in the \(x\)-\(y\) plane so that at time \(t\) seconds, where \(t \geqslant 0\), its coordinates are given by
\(x = \mathrm{e}^{2t} - 4\mathrm{e}^t + 3\), \(y = 2\mathrm{e}^{-3t}\).
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{e}^{kt} \gt 0\) for all \(t\), so \(y \gt 0\) for \(t \geqslant 0\) (or for all \(t\)) hence never crosses \(x\)-axis | B1 | 2.4 |
| [1] |
Notes
B1: Or show that \(2\mathrm{e}^{-3t} = 0\) has no solutions
Need to see \(y \neq 0\) or \(y \gt 0\) and reason relating to exponential (or logarithmic) function
Must clearly be referring to \(y\) or \(2\mathrm{e}^{-3t}\)
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{e}^{2t} - 4\mathrm{e}^t + 3 = 0\) \((\mathrm{e}^t - 1)(\mathrm{e}^t - 3) = 0\) \(\mathrm{e}^t = 1\), \(\mathrm{e}^t = 3\) | M1 | 3.1a |
| \(t = 0\), \(t = \ln 3\) | A1 | 1.1 |
| [2] |
Notes
M1: Equate to 0 and attempt to solve disguised quadratic
‘determine’ so some evidence of method needed
A1: Obtain both correct values
A0 for \(\ln 1\) and not 0
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 2\mathrm{e}^{2t} - 4\mathrm{e}^t\) | B1 | 1.1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = -6\mathrm{e}^{-3t}\) | B1 | 1.1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{-6\mathrm{e}^{-3t}}{2\mathrm{e}^{2t} - 4\mathrm{e}^t}\) | M1 | 2.4 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{-3\mathrm{e}^{-3t}}{\mathrm{e}^{2t} - 2\mathrm{e}^t} = \dfrac{3}{\mathrm{e}^{3t}(2\mathrm{e}^t - \mathrm{e}^{2t})} = \dfrac{3}{2\mathrm{e}^{4t} - \mathrm{e}^{5t}}\) A.G. | A1 | 2.1 |
| [4] |
Notes
B1: Correct \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\)
Mark derivative and condone no/wrong label
B1: Correct \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\)
Mark derivative and condone no/wrong label
M1: Attempt correct method to combine derivatives
Combine their derivatives correctly
A1: Show manipulation to given answer
Need to see some evidence of how \(\mathrm{e}^{-3t}\) is dealt with
AG so method must be fully correct
| Scheme | Marks | AO |
|---|---|---|
| \(2\mathrm{e}^{4t} - \mathrm{e}^{5t} = 0\) | M1 | 3.1a |
| \(\mathrm{e}^{4t}(2 - \mathrm{e}^t) = 0\) \(t = \ln 2\) | A1 | 1.1 |
| \(\left(-1,\ \frac{1}{4}\right)\) | A1 | 1.1 |
| [3] |
Notes
M1: Equate denominator to 0
Or \(\dfrac{\mathrm{d}x}{\mathrm{d}y} = 0\) or \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 0\)
A1: Solve for \(t\) to obtain \(t = \ln 2\)
No need to see \(\mathrm{e}^{4t} = 0\) discounted
A1: Obtain correct coordinate
Or \(x = -1\), \(y = \frac{1}{4}\)