C4 January 2008 Q5
5. A curve is described by the equation\[x^3 - 4y^2 = 12xy.\]
| Scheme | Marks |
|---|---|
| \(x^3 - 4y^2 = 12xy\) ( eqn \(*\) ) | |
| \(x = -8 \Rightarrow -512 - 4y^2 = 12(-8)y\) \(-512 - 4y^2 = -96y\) | M1 |
| \(4y^2 - 96y + 512 = 0\) \(y^2 - 24y + 128 = 0\) | |
| \((y - 16)(y - 8) = 0\) \(y = \dfrac{24 \pm \sqrt{576 - 4(128)}}{2}\) | dM1 |
| \(y = 16\) or \(y = 8\). | A1 |
| (3) |
Notes
M1: Substitutes \(x = -8\) (at least once) into \(*\) to obtain a three term quadratic in \(y\). Condone the loss of \(= 0\).
dM1: An attempt to solve the quadratic in \(y\) by either factorising or by the formula or by completing the square.
A1: Both \(\underline{y = 16}\) and \(\underline{y = 8}\). or \(\underline{(-8, 8)}\) and \(\underline{(-8, 16)}\).
| Scheme | Marks |
|---|---|
| \(\left\{\xcancel{\tfrac{\mathrm{d}y}{\mathrm{d}x}\times}\right\}\quad 3x^2 - 8y\dfrac{\mathrm{d}y}{\mathrm{d}x};\ = \underline{\left(12y + 12x\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)}\) | M1 A1; (B1) |
| \(\left\{\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3x^2 - 12y}{12x + 8y}\right\}\) not necessarily required. | |
| @ \((-8, 8)\), \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3(64) - 12(8)}{12(-8) + 8(8)} = \dfrac{96}{-32} = \underline{-3}\), | dM1 |
| @ \((-8, 16)\), \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3(64) - 12(16)}{12(-8) + 8(16)} = \dfrac{0}{32} = \underline{0}\). | A1 A1 cso |
| (6) | |
| (9 marks) |
Notes
M1: Differentiates implicitly to include either \(\pm ky\tfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(12x\tfrac{\mathrm{d}y}{\mathrm{d}x}\). Ignore \(\tfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots\)
A1: Correct LHS equation; (B1): Correct application of product rule
dM1: Substitutes \(x = -8\) and at least one of their \(y\)-values to attempt to find any one of \(\tfrac{\mathrm{d}y}{\mathrm{d}x}\).
A1: One gradient found. A1 cso: Both gradients of -3 and 0 correctly found.
Aliter 5. (b) Way 2
| Scheme | Marks |
|---|---|
| \(\left\{\xcancel{\tfrac{\mathrm{d}x}{\mathrm{d}y}\times}\right\}\quad 3x^2\dfrac{\mathrm{d}x}{\mathrm{d}y} - 8y;\ = \underline{\left(12y\dfrac{\mathrm{d}x}{\mathrm{d}y} + 12x\right)}\) | M1 A1; (B1) |
| \(\left\{\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3x^2 - 12y}{12x + 8y}\right\}\) not necessarily required. | |
| @ \((-8, 8)\), \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3(64) - 12(8)}{12(-8) + 8(8)} = \dfrac{96}{-32} = \underline{-3}\), | dM1 |
| @ \((-8, 16)\), \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3(64) - 12(16)}{12(-8) + 8(16)} = \dfrac{0}{32} = \underline{0}\). | A1 A1 cso |
| (6) |
M1: Differentiates implicitly to include either \(\pm kx^2\tfrac{\mathrm{d}x}{\mathrm{d}y}\) or \(12y\tfrac{\mathrm{d}x}{\mathrm{d}y}\). Ignore \(\tfrac{\mathrm{d}x}{\mathrm{d}y} = \ldots\) A1: Correct LHS equation (B1): Correct application of product rule
dM1: Substitutes \(x = -8\) and at least one of their \(y\)-values to attempt to find any one of \(\tfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(\tfrac{\mathrm{d}x}{\mathrm{d}y}\).
A1: One gradient found. A1 cso: Both gradients of -3 and 0 correctly found.
Aliter 5. (b) Way 3
| Scheme | Marks |
|---|---|
| \(x^3 - 4y^2 = 12xy\) ( eqn \(*\) ) | |
| \(4y^2 + 12xy - x^3 = 0\) | |
| \(y = \dfrac{-12x \pm \sqrt{144x^2 - 4(4)(-x^3)}}{8}\) | |
| \(y = \dfrac{-12x \pm \sqrt{144x^2 + 16x^3}}{8}\) | |
| \(y = \dfrac{-12x \pm 4\sqrt{9x^2 + x^3}}{8}\) | |
| \(y = -\tfrac{3}{2}x \pm \tfrac{1}{2}(9x^2 + x^3)^{\frac{1}{2}}\) | |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\tfrac{3}{2} \pm \tfrac{1}{2}(\tfrac{1}{2})(9x^2 + x^3)^{-\frac{1}{2}};(18x + 3x^2)\) | M1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{3}{2} \pm \dfrac{18x + 3x^2}{4(9x^2 + x^3)^{\frac{1}{2}}}\) | A1 A1 |
| @ \(x = -8\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{3}{2} \pm \dfrac{18(-8) + 3(64)}{4(9(64) + (-512))^{\frac{1}{2}}}\) | dM1 |
| \(= -\dfrac{3}{2} \pm \dfrac{48}{4\sqrt{(64)}} = -\dfrac{3}{2} \pm \dfrac{48}{32}\) | |
| \(\therefore \dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{3}{2} \pm \dfrac{3}{2} = \underline{-3}, \underline{0}\). | A1 A1 |
| (6) |
M1: A credible attempt to make \(y\) the subject and an attempt to differentiate either \(-\tfrac{3}{2}x\) or \(\tfrac{1}{2}(9x^2 + x^3)^{\frac{1}{2}}\).
A1: \(\tfrac{\mathrm{d}y}{\mathrm{d}x} = -\tfrac{3}{2} \pm k(9x^2 + x^3)^{-\frac{1}{2}}(\mathrm{g}(x))\) A1: \(\tfrac{\mathrm{d}y}{\mathrm{d}x} = -\tfrac{3}{2} \pm \tfrac{1}{2}(\tfrac{1}{2})(9x^2 + x^3)^{-\frac{1}{2}};(18x + 3x^2)\)
dM1: Substitutes \(x = -8\) find any one of \(\tfrac{\mathrm{d}y}{\mathrm{d}x}\). A1: One gradient correctly found. A1: Both gradients of -3 and 0 correctly found.