June 2023 Paper 2 Q7
7.
In this question you must show all stages of your working.
Solutions relying on calculator technology are not acceptable.
A curve has equation\[x^3 + 2xy + 3y^2 = 47\]
The point \(P(-2,\ 5)\) lies on the curve.
| Scheme | Marks | AO |
|---|---|---|
| \(x^3 \rightarrow \ldots x^2\) and \(3y^2 \rightarrow \ldots y\dfrac{\mathrm{d}y}{\mathrm{d}x}\) | M1 | 1.1b |
| \(2xy \rightarrow 2y + 2x\dfrac{\mathrm{d}y}{\mathrm{d}x}\) | B1 | 1.1b |
| \(3x^2 + 2x\dfrac{\mathrm{d}y}{\mathrm{d}x} + 2y + 6y\dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots\) | M1 | 2.1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{2y + 3x^2}{2x + 6y}\) | A1 | 1.1b |
| (4) |
Notes
(a) Allow equivalent notation for the \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) e.g. \(y^{\prime}\)
M1: Attempts to differentiate \(x^3 \rightarrow \ldots x^2\) and \(3y^2 \rightarrow \ldots y\dfrac{\mathrm{d}y}{\mathrm{d}x}\) where … are constants
B1: Correct application of the product rule on \(2xy\): \(2xy \rightarrow 2x\dfrac{\mathrm{d}y}{\mathrm{d}x} + 2y\)
Note that some candidates have a spurious \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots\) at the start (as their intention to differentiate) and this can be ignored for the first 2 marks
M1: For a valid attempt to make \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) the subject, with exactly 2 different terms in \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) coming from \(3y^2\) and \(2xy\). Look for \((\ldots \pm \ldots)\dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots\) which may be implied by their working.
Condone slips provided the intention is clear.
For those candidates who had a spurious \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots\) at the start, they may incorporate this in their rearrangement in which case they will have 3 terms in \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) and so score M0.
If they ignore it, then this mark is available for the condition as described above.
A1: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{2y + 3x^2}{2x + 6y}\) oe e.g. \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{-2y - 3x^2}{2x + 6y},\ \dfrac{2y + 3x^2}{-2x - 6y}\) Isw once a correct expression is seen.
Note that it is sometimes unclear if the minus sign(s) is/are correctly placed and you may have to use your judgement. Evidence may be available in part (b) to help you decide if they have the correct expression.
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{2(5) + 3(-2)^2}{2(-2) + 6(5)}\) or e.g. \(3(-2)^2 + 2(-2)\dfrac{\mathrm{d}y}{\mathrm{d}x} + 2 \times 5 + 6 \times 5\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0 \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots\ \left(-\dfrac{11}{13}\right)\) | M1 | 1.1b |
| \(y - 5 = \text{``}\dfrac{13}{11}\text{''}(x + 2)\) | dM1 | 1.1b |
| \(13x - 11y + 81 = 0\) | A1 | 2.2a |
| (3) | ||
| (7 marks) |
Notes
M1: Substitutes \(x = -2\) and \(y = 5\) into \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \text{``}{-}\dfrac{2y + 3x^2}{2x + 6y}\text{''}\)
They must have \(x\)’s and \(y\)’s in their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) but condone slips in substitution provided the intention is clear.
As a minimum look for at least one \(x\) and at least one \(y\) substituted correctly.
Note that this mark may be implied by their value for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) and may be implied if, for example, they find the negative reciprocal or the reciprocal of \(\text{``}{-}\dfrac{2y + 3x^2}{2x + 6y}\text{''}\) and then substitute \(x = -2\) and \(y = 5\)
Alternatively, substitutes \(x = -2\) and \(y = 5\) into their attempt to differentiate and then rearranges to find a value or numerical expression for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
dM1: Attempts to find the equation of the normal using their gradient of the tangent and \(x = -2\) and \(y = 5\) correctly placed. Score for an expression of the form \((y - 5) = \text{``}\dfrac{13}{11}\text{''}(x + 2)\) or if they use \(y = mx + c\) they must proceed as far as \(c = \ldots\) Must be using the negative reciprocal of the tangent gradient.
Note that \(y - 5 = \dfrac{2x + 6y}{2y + 3x^2}(x + 2)\) is not a correct method unless the gradient is evaluated first before expanding.
A1: \(13x - 11y + 81 = 0\) or any integer multiple of this equation including the “= 0”, not just a, b, c given.
e.g., \(26x - 22y + 162 = 0\) is likely if they don’t cancel down their gradient.