C4 June 2013 Q4
4. A curve \(C\) has parametric equations \[x = 2\sin t, \quad y = 1 - \cos 2t, \quad -\frac{\pi}{2} \leqslant t \leqslant \frac{\pi}{2}\]
| Scheme | Marks |
|---|---|
| \(x = 2\sin t, \quad y = 1 - \cos 2t\ \left\{= 2\sin^2 t\right\}, \quad -\dfrac{\pi}{2} \leqslant t \leqslant \dfrac{\pi}{2}\) | |
| \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 2\cos t, \quad \dfrac{\mathrm{d}y}{\mathrm{d}t} = 2\sin 2t\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = 4\sin t\cos t\) At least one of \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) correct. Both \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) are correct. | B1 B1 |
| So, \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2\sin 2t}{2\cos t}\ \left\{= \dfrac{4\cos t\sin t}{2\cos t} = 2\sin t\right\}\) At \(t = \dfrac{\pi}{6}\), \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2\sin\left(\frac{2\pi}{6}\right)}{2\cos\left(\frac{\pi}{6}\right)};\ = 1\) Applies their \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) divided by their \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\) and substitutes \(t = \dfrac{\pi}{6}\) into their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\). Correct value for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) of 1 | M1; A1 cao cso |
| (4) |
Notes
B1: At least one of \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) correct. Note: that this mark can be implied from their working.
B1: Both \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) are correct. Note: that this mark can be implied from their working.
M1: Applies their \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) divided by their \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\) and attempts to substitute \(t = \dfrac{\pi}{6}\) into their expression for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\).
This mark may be implied by their final answer.
Ie. \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\sin 2t}{2\cos t}\) followed by an answer of \(\dfrac{1}{2}\) would be M1 (implied).
A1: For an answer of 1 by correct solution only.
Note: Don’t just look at the answer! A number of candidates are finding \(\boldsymbol{\dfrac{\mathrm{d}y}{\mathrm{d}x} = 1}\) from incorrect methods.
Note: Applying \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\) divided by their \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) is M0, even if they state \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}t} \div \dfrac{\mathrm{d}x}{\mathrm{d}t}\).
Special Case: Award SC: B0B0M1A1 for \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = -2\cos t\), \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = -2\sin 2t\) leading to \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{-2\sin 2t}{-2\cos t}\)
which after substitution of \(t = \dfrac{\pi}{6}\), yields \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 1\)
Note: It is possible for you to mark part(a), part (b) and part (c) together. Ignore labelling!
Aliter 4. (a) Way 2
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 2\cos t\), \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = 2\sin 2t\), So B1, B1. | |
| At \(t = \dfrac{\pi}{6}\), \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 2\cos\left(\dfrac{\pi}{6}\right) = \sqrt{3}\), \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = 2\sin\left(\dfrac{2\pi}{6}\right) = \sqrt{3}\) | |
| Hence \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 1\) So implied M1, A1. |
Aliter 4. (a) Way 3
| Scheme | Marks |
|---|---|
| \(y = \dfrac{1}{2}x^2 \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = x\) Correct differentiation of their Cartesian equation. Finds \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = x\), using the correct Cartesian equation only. | B1ft B1 |
| At \(t = \dfrac{\pi}{6}\), \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2\sin\left(\dfrac{\pi}{6}\right)\) Finds the value of “\(x\)” when \(t = \dfrac{\pi}{6}\) and substitutes this into their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) | M1 |
| \(= 1\) Correct value for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) of 1 | A1 |
| Scheme | Marks |
|---|---|
| \(y = 1 - \cos 2t = 1 - (1 - 2\sin^2 t)\) \(= 2\sin^2 t\) | M1 |
| So, \(y = 2\left(\dfrac{x}{2}\right)^2\) or \(y = \dfrac{x^2}{2}\) or \(y = 2 - 2\left(1 - \left(\dfrac{x}{2}\right)^2\right)\) \(y = \dfrac{x^2}{2}\) or equivalent. | A1 cso isw |
| Either \(k = 2\) or \(-2 \leqslant x \leqslant 2\) | B1 |
| (3) |
Notes
M1: Uses the correct double angle formula \(\cos 2t = 1 - 2\sin^2 t\) or \(\cos 2t = 2\cos^2 t - 1\) or
\(\cos 2t = \cos^2 t - \sin^2 t\) in an attempt to get \(y\) in terms of \(\sin^2 t\) or get \(y\) in terms of \(\cos^2 t\)
or get \(y\) in terms of \(\sin^2 t\) and \(\cos^2 t\). Writing down \(y = 2\sin^2 t\) is fine for M1.
A1: Achieves \(y = \dfrac{x^2}{2}\) or un-simplified equivalents in the form \(\boldsymbol{y = \mathrm{f}(x)}\). For example:
\(y = \dfrac{2x^2}{4}\) or \(y = 2\left(\dfrac{x}{2}\right)^2\) or \(y = 2 - 2\left(1 - \left(\dfrac{x}{2}\right)^2\right)\) or \(y = 1 - \dfrac{4 - x^2}{4} + \dfrac{x^2}{4}\)
and you can ignore subsequent working if a candidate states a correct version of the Cartesian equation.
IMPORTANT: Please check working as this result can be fluked from an incorrect method.
Award A0 if there is a \(+c\) added to their answer.
B1: Either \(k = 2\) or a candidate writes down \(-2 \leqslant x \leqslant 2\). Note: \(-2 \leqslant k \leqslant 2\) unless \(k\) stated as 2 is B0.
Aliter 4. (b) Way 2
| Scheme | Marks |
|---|---|
| \(y = 1 - \cos 2t = 1 - (2\cos^2 t - 1)\) | M1 |
| \(y = 2 - 2\cos^2 t \Rightarrow \cos^2 t = \dfrac{2 - y}{2} \Rightarrow 1 - \sin^2 t = \dfrac{2 - y}{2}\) | |
| \(1 - \left(\dfrac{x}{2}\right)^2 = \dfrac{2 - y}{2}\) (Must be in the form \(y = \mathrm{f}(x)\)). | |
| \(y = 2 - 2\left(1 - \left(\dfrac{x}{2}\right)^2\right)\) | A1 |
Aliter 4. (b) Way 3
| Scheme | Marks |
|---|---|
| \(x = 2\sin t \Rightarrow t = \sin^{-1}\left(\dfrac{x}{2}\right)\) | |
| So, \(y = 1 - \cos\left(2\sin^{-1}\left(\dfrac{x}{2}\right)\right)\) Rearranges to make \(t\) the subject and substitutes the result into \(y\). \(y = 1 - \cos\left(2\sin^{-1}\left(\dfrac{x}{2}\right)\right)\) | M1 A1 oe |
Aliter 4. (b) Way 4
| Scheme | Marks |
|---|---|
| \(y = 1 - \cos 2t \Rightarrow \cos 2t = 1 - y \Rightarrow t = \dfrac{1}{2}\cos^{-1}(1 - y)\) | |
| So, \(x = \pm 2\sin\left(\dfrac{1}{2}\cos^{-1}(1 - y)\right)\) Rearranges to make \(t\) the subject and substitutes the result into \(y\). | M1 |
| So, \(y = 1 - \cos\left(2\sin^{-1}\left(\dfrac{x}{2}\right)\right)\) \(y = 1 - \cos\left(2\sin^{-1}\left(\dfrac{x}{2}\right)\right)\) | A1 oe |
Aliter 4. (b) Way 5
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2\sin t = x \Rightarrow y = \dfrac{1}{2}x^2 + c\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = x \Rightarrow y = \dfrac{1}{2}x^2 + c\) | M1 |
| Eg: when eg: \(t = 0\) (nb: \(-\frac{\pi}{2} \leqslant t \leqslant \frac{\pi}{2}\)), \(x = 0,\ y = 1 - 1 = 0 \Rightarrow c = 0 \Rightarrow y = \dfrac{1}{2}x^2\) Full method of finding \(y = \dfrac{1}{2}x^2\) using a value of \(t\): \(-\frac{\pi}{2} \leqslant t \leqslant \frac{\pi}{2}\) | A1 |
Note: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2\sin t = x \Rightarrow y = \dfrac{1}{2}x^2\), with no attempt to find \(c\) is M1A0.
| Scheme | Marks |
|---|---|
| Range: \(0 \leqslant \mathrm{f}(x) \leqslant 2\) or \(0 \leqslant y \leqslant 2\) or \(0 \leqslant \mathrm{f} \leqslant 2\) See notes | B1 B1 |
| (2) | |
| (9 marks) |
Notes
Note: The values of 0 and/or 2 need to be evaluated in this part
B1: Achieves an inclusive upper or lower limit, using acceptable notation. Eg: \(\mathrm{f}(x) \geqslant 0\) or \(\mathrm{f}(x) \leqslant 2\)
B1: \(0 \leqslant \mathrm{f}(x) \leqslant 2\) or \(0 \leqslant y \leqslant 2\) or \(0 \leqslant \mathrm{f} \leqslant 2\)
Special Case: SC: B1B0 for either \(0 < \mathrm{f}(x) < 2\) or \(0 < \mathrm{f} < 2\) or \(0 < y < 2\) or \((0, 2)\)
Special Case: SC: B1B0 for \(0 \leqslant x \leqslant 2\).
IMPORTANT: Note that: Therefore candidates can use either \(y\) or f in place of \(\mathrm{f}(x)\)
Examples:
| \(0 \leqslant x \leqslant 2\) is SC: B1B0 | \(0 < x < 2\) is B0B0 |
| \(x \geqslant 0\) is B0B0 | \(x \leqslant 2\) is B0B0 |
| \(\mathrm{f}(x) > 0\) is B0B0 | \(\mathrm{f}(x) < 2\) is B0B0 |
| \(x > 0\) is B0B0 | \(x < 2\) is B0B0 |
| \(0 \geqslant \mathrm{f}(x) \geqslant 2\) is B0B0 | \(0 < \mathrm{f}(x) \leqslant 2\) is B1B0 |
| \(0 \leqslant \mathrm{f}(x) < 2\) is B1B0. | \(\mathrm{f}(x) \geqslant 0\) is B1B0 |
| \(\mathrm{f}(x) \leqslant 2\) is B1B0 | \(\mathrm{f}(x) \geqslant 0\) and \(\mathrm{f}(x) \leqslant 2\) is B1B1. Must state AND {or} \(\cap\) |
| \(2 \leqslant \mathrm{f}(x) \leqslant 2\) is B0B0 | \(\mathrm{f}(x) \geqslant 0\) or \(\mathrm{f}(x) \leqslant 2\) is B1B0. |
| \(\left|\mathrm{f}(x)\right| \leqslant 2\) is B1B0 | \(\left|\mathrm{f}(x)\right| \geqslant 2\) is B0B0 |
| \(1 \leqslant \mathrm{f}(x) \leqslant 2\) is B1B0 | \(1 < \mathrm{f}(x) < 2\) is B0B0 |
| \(0 \leqslant \mathrm{f}(x) \leqslant 4\) is B1B0 | \(0 < \mathrm{f}(x) < 4\) is B0B0 |
| \(0 \leqslant \text{Range} \leqslant 2\) is B1B0 | Range is in between 0 and 2 is B1B0 |
| \(0 < \text{Range} < 2\) is B0B0. | \(\text{Range} \geqslant 0\) is B1B0 |
| \(\text{Range} \leqslant 2\) is B1B0 | \(\text{Range} \geqslant 0\) and \(\text{Range} \leqslant 2\) is B1B0. |
| \(\left[0, 2\right]\) is B1B1 | \((0, 2)\) is SC B1B0 |