June 2025 Paper 1 Q13
13 A curve \(C\) has parametric equations
\[\begin{gathered}x = 4(4t + 1)^2\\ y = \mathrm{e}^{-4t}\end{gathered}\]for \(-\dfrac{1}{4} \leqslant t \leqslant 0\)
(a) Find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(t\) [3 marks]
(b) Find an equation of the tangent to \(C\) at the point where \(t = 0\) [3 marks]
(c) Find a Cartesian equation for \(C\) in the form \(y = \mathrm{f}(x)\)
Fully justify your answer.
[3 marks]| Scheme | Marks | AO |
|---|---|---|
| Obtains \(32(4t + 1)\) Can be unsimplified OE | B1 | 1.1b |
| Obtains \(-4\mathrm{e}^{-4t}\) | B1 | 1.1b |
| Uses their \(\dfrac{\mathrm{d}y}{\mathrm{d}t} \div\) their \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\) to obtain an expression for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) Can be unsimplified ISW ACF | B1F | 3.1a |
| (3) |
Typical solution
\[\frac{\mathrm{d}x}{\mathrm{d}t} = 32(4t + 1)\]\[\frac{\mathrm{d}y}{\mathrm{d}t} = -4\mathrm{e}^{-4t}\]\[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-4\mathrm{e}^{-4t}}{32(4t + 1)} = \frac{-\mathrm{e}^{-4t}}{8(4t + 1)}\]| Scheme | Marks | AO |
|---|---|---|
| Substitutes \(t = 0\) to obtain values for \(x\), \(y\) and their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) PI by correct values | M1 | 3.1a |
| Obtains at least two of \(x = 4\), \(y = 1\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{1}{8}\) | M1 | 1.1a |
| Obtains \(y - 1 = -\dfrac{1}{8}(x - 4)\) ACF ISW | A1 | 1.1b |
| (3) |
Typical solution
\[t = 0\]\[x = 4\]\[y = 1\]\[\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{1}{8}\]\[y - 1 = -\frac{1}{8}(x - 4)\]| Scheme | Marks | AO |
|---|---|---|
| Eliminates \(t\) correctly to form a Cartesian equation. May use positive or negative square root or both. Condone a sign error when rearranging either parametric equation. | M1 | 3.1a |
| Obtains \(y = \mathrm{e}^{1 - \frac{\sqrt{x}}{2}}\) or \(y = \mathrm{e}^{1 + \frac{\sqrt{x}}{2}}\) or \(y = \mathrm{e}^{1 \pm \frac{\sqrt{x}}{2}}\) OE May be unsimplified. | M1 | 1.1a |
| Gives a valid explanation for selecting the appropriate square root for example \(-\dfrac{1}{4} \leqslant t\) or \(y \leqslant \mathrm{e}\) AND Obtains \(y = \mathrm{e}^{1 - \frac{\sqrt{x}}{2}}\) OE | R1 | 2.4 |
| (3) | ||
| (9 marks) |
Typical solution
\[4t + 1 = \frac{\pm\sqrt{x}}{2}\]The positive square root is used since \(-\dfrac{1}{4} \leqslant t \leqslant 0\)
\[4t = \frac{\sqrt{x} - 2}{2}\]\[y = \mathrm{e}^{-\frac{\sqrt{x} - 2}{2}}\]