P7.5Implicit and parametric differentiation

• M1 for attempting \(\frac{\frac{\mathrm{d}y}{\mathrm{d}t}}{\frac{\mathrm{d}x}{\mathrm{d}t}}\) with both derivatives of the correct form
• A1 for \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1 - 3t^{2}}{2t}\) oe

Worked solution: \(\frac{\mathrm{d}x}{\mathrm{d}t} = 2t\) and \(\frac{\mathrm{d}y}{\mathrm{d}t} = 1 - 3t^{2}\), so \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{\mathrm{d}y}{\mathrm{d}t}}{\frac{\mathrm{d}x}{\mathrm{d}t}} = \frac{1 - 3t^{2}}{2t}\).

• M1 for finding the point \((-3, 0)\)
• M1 for finding the gradient \(-1\) and using a correct straight-line method
• A1 for \(y = -x - 3\)

Worked solution: At \(t = 1\): \(x = -3\), \(y = 0\) and \(\frac{\mathrm{d}y}{\mathrm{d}x} = -1\).
\(y = -(x + 3)\), so \(y = -x - 3\).

• M1 for \(\frac{\mathrm{d}x}{\mathrm{d}t} = -4\sin t\) and \(\frac{\mathrm{d}y}{\mathrm{d}t} = 6\cos t\)
• M1 for \(\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{6\cos t}{4\sin t}\) and substituting \(t = \frac{\pi}{4}\)
• A1 for the gradient \(-\frac{3}{2}\)
• M1 for the point \((2\sqrt{2}, 3\sqrt{2})\) and a correct straight-line method
• A1* for \(6x + 4y = 24\sqrt{2}\) (cso)

Worked solution: \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{6\cos t}{-4\sin t}\), so at \(t = \frac{\pi}{4}\) the gradient is \(-\frac{3}{2}\), and the point is \((2\sqrt{2}, 3\sqrt{2})\).
\(y - 3\sqrt{2} = -\frac{3}{2}\left(x - 2\sqrt{2}\right)\), which rearranges to \(6x + 4y = 24\sqrt{2}\).

• M1 for finding \(A\) and \(B\) from the equation of the tangent
• A1 for \(OA = 4\sqrt{2}\) and \(OB = 6\sqrt{2}\)
• A1 for area \(24\)

Worked solution: \(y = 0\): \(x = 4\sqrt{2}\); \(x = 0\): \(y = 6\sqrt{2}\).
Area \(= \frac{1}{2} \times 4\sqrt{2} \times 6\sqrt{2} = 24\).

• M1 for implicit differentiation, including the product rule on \(xy\)
• A1 for \(2x + y + x\frac{\mathrm{d}y}{\mathrm{d}x} + 2y\frac{\mathrm{d}y}{\mathrm{d}x} = 0\)
• A1* for rearranging to the given result

Worked solution: \(2x + \left(y + x\dfrac{\mathrm{d}y}{\mathrm{d}x}\right) + 2y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0 \Rightarrow (x + 2y)\dfrac{\mathrm{d}y}{\mathrm{d}x} = -(2x + y) \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{2x + y}{x + 2y}\).

• M1 for \(2x + y = 0\), i.e. \(y = -2x\)
• M1 for substituting: \(x^2 - 2x^2 + 4x^2 = 12\)
• A1 for \((2, -4)\) and \((-2, 4)\)

Worked solution: \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0 \Rightarrow y = -2x\). Then \(x^2 - 2x^2 + 4x^2 = 3x^2 = 12\), \(x = \pm 2\): the points are \((2, -4)\) and \((-2, 4)\).

• M1 for writing \(y^2 + xy + (x^2 - 12) = 0\)
• M1 for requiring discriminant \(\ge 0\): \(x^2 - 4(x^2 - 12) \ge 0\)
• A1* for \(x^2 \le 16\), so \(-4 \le x \le 4\), with a conclusion

Worked solution: For a given \(x\), a point on \(C\) has a real \(y\) satisfying \(y^2 + xy + (x^2 - 12) = 0\). So the discriminant is non-negative: \(x^2 - 4(x^2 - 12) = 48 - 3x^2 \ge 0\), giving \(x^2 \le 16\), i.e. \(-4 \le x \le 4\).

• M1 for \(x + 2y = 0\) (or using \(x = \pm 4\) from part (c), where the discriminant is zero)
• A1 for \((4, -2)\) and \((-4, 2)\)

Worked solution: The tangent is vertical where \(x + 2y = 0\), i.e. \(x = -2y\): \(4y^2 - 2y^2 + y^2 = 3y^2 = 12\), \(y = \pm 2\). The points are \((-4, 2)\) and \((4, -2)\) — the extreme values of \(x\) found in part (c).