C4 January 2009 Q1
1. A curve \(C\) has the equation \(y^2 - 3y = x^3 + 8\).
| Scheme | Marks |
|---|---|
| \(C\colon\ y^2 - 3y = x^3 + 8\) | |
| \(\left\{\xcancel{\tfrac{\mathrm{d}y}{\mathrm{d}x}\times}\right\}\quad 2y\dfrac{\mathrm{d}y}{\mathrm{d}x} - 3\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2\) | M1 A1 |
| \((2y - 3)\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2\) | M1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3x^2}{2y - 3}\) | A1 oe |
| (4) |
Notes
M1: Differentiates implicitly to include either \(\pm ky\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(\pm 3\dfrac{\mathrm{d}y}{\mathrm{d}x}\). (Ignore \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \right)\).)
A1: Correct equation.
M1: A correct (condoning sign error) attempt to combine or factorise their ‘\(2y\dfrac{\mathrm{d}y}{\mathrm{d}x} - 3\dfrac{\mathrm{d}y}{\mathrm{d}x}\)’. Can be implied.
A1 oe: \(\dfrac{3x^2}{2y - 3}\)
Aliter 1. (a) Way 2
| Scheme | Marks |
|---|---|
| \(C\colon\ y^2 - 3y = x^3 + 8\) | |
| \(\left\{\xcancel{\tfrac{\mathrm{d}x}{\mathrm{d}y}\times}\right\}\quad 2y - 3 = 3x^2\dfrac{\mathrm{d}x}{\mathrm{d}y}\) | M1 A1 |
| \(2y - 3 = 3x^2\dfrac{1}{\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)}\) | dM1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3x^2}{2y - 3}\) | A1 oe |
| (4) |
M1: Differentiates implicitly to include either \(\pm kx^2\dfrac{\mathrm{d}x}{\mathrm{d}y}\). (Ignore \(\left(\dfrac{\mathrm{d}x}{\mathrm{d}y} = \right)\).) A1: Correct equation.
dM1: Applies \(\dfrac{\mathrm{d}x}{\mathrm{d}y} = \dfrac{1}{\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)}\) A1 oe: \(\underline{\dfrac{3x^2}{2y - 3}}\)
Aliter 1. (a) Way 3
| Scheme | Marks |
|---|---|
| \(C\colon\ y^2 - 3y = x^3 + 8\) | |
| gives \(x^3 = y^2 - 3y - 8\) \(\quad \Rightarrow x = \left(y^2 - 3y - 8\right)^{\frac{1}{3}}\) | |
| \(\dfrac{\mathrm{d}x}{\mathrm{d}y} = \dfrac{1}{3}\left(y^2 - 3y - 8\right)^{-\frac{2}{3}}\left(2y - 3\right)\) | M1 A1 |
| \(\dfrac{\mathrm{d}x}{\mathrm{d}y} = \dfrac{2y - 3}{3\left(y^2 - 3y - 8\right)^{\frac{2}{3}}}\) | |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3\left(y^2 - 3y - 8\right)^{\frac{2}{3}}}{2y - 3}\) | dM1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \underline{\dfrac{3\left(x^3\right)^{\frac{2}{3}}}{2y - 3}} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \underline{\dfrac{3x^2}{2y - 3}}\) | A1 oe |
| (4) |
M1: Differentiates in the form \(\tfrac{1}{3}\left(\mathrm{f}(y)\right)^{-\frac{2}{3}}\left(\mathrm{f}^{\prime}(y)\right)\). A1: Correct differentiation.
dM1: Applies \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{\left(\frac{\mathrm{d}x}{\mathrm{d}y}\right)}\) A1 oe: \(\underline{\dfrac{3\left(x^3\right)^{\frac{2}{3}}}{2y - 3}}\) or \(\underline{\dfrac{3x^2}{2y - 3}}\)
| Scheme | Marks |
|---|---|
| \(y = 3 \Rightarrow 9 - 3(3) = x^3 + 8\) | M1 |
| \(x^3 = -8 \Rightarrow \underline{x = -2}\) | A1 |
| \((-2, 3) \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3(4)}{6 - 3} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = 4\) | A1ft |
| (3) | |
| (7 marks) |
Notes
M1: Substitutes \(y = 3\) into \(C\).
A1: Only \(\underline{x = -2}\)
A1ft: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 4\) from correct working. Also can be ft using their ‘\(x\)’ value and \(y = 3\) in the correct part (a) of \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3x^2}{2y - 3}\)
1(b) final A1ft. Note if the candidate inserts their \(x\) value and \(y = 3\) into \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3x^2}{2y - 3}\), then an answer of \(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\) their \(x^2\), may indicate a correct follow through.