Differentiation & Applications: Optimisation — SAJC 2025 H2 Math Prelim Paper 1

The curve is \((x-y)^2 = 2\mathrm{e}^{xy}\).

(i) Differentiate with respect to \(x\): \[ 2(x-y)\!\left(1-\frac{\mathrm{d}y}{\mathrm{d}x}\right) = 2\mathrm{e}^{xy}\!\left(x\frac{\mathrm{d}y}{\mathrm{d}x}+y\right) \] \[ (x-y) - (x-y)\frac{\mathrm{d}y}{\mathrm{d}x} = x\mathrm{e}^{xy}\frac{\mathrm{d}y}{\mathrm{d}x} + y\mathrm{e}^{xy} \] \[ \frac{\mathrm{d}y}{\mathrm{d}x}\bigl[-(x-y) - x\mathrm{e}^{xy}\bigr] = y\mathrm{e}^{xy} - (x-y) \]

(ii) When \(x=0\): \[ (0-y)^2 = 2\mathrm{e}^{0} \implies y^2 = 2 \implies y = \pm\sqrt{2} \] Since \(y > 0\), \(y = \sqrt{2}\).

At \((0,\sqrt{2})\): \[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{0-\sqrt{2}-\sqrt{2}\,\mathrm{e}^{0(\sqrt{2})}} {0-\sqrt{2}+0\cdot\mathrm{e}^{0(\sqrt{2})}} = \frac{-2\sqrt{2}}{-\sqrt{2}} = 2 \]

Gradient of normal \(= -\dfrac{1}{2}\).

Equation of the normal to the curve at \((0,\sqrt{2})\): \[ y - \sqrt{2} = -\tfrac{1}{2}(x - 0) \implies y = -\tfrac{1}{2}x + \sqrt{2} \]

(iii) Substitute \(y = -\dfrac{1}{2}x + \sqrt{2}\) into \((x-y)^2 = 2\mathrm{e}^{xy}\): \[ \left(x - \left(-\tfrac{1}{2}x + \sqrt{2}\right)\right)^2 = 2\mathrm{e}^{x\left(-\frac{1}{2}x+\sqrt{2}\right)} \] \[ \left(\tfrac{3}{2}x - \sqrt{2}\right)^2 = 2\mathrm{e}^{-\frac{1}{2}x^2+\sqrt{2}\,x} \]

Using G.C., \(x = 2.2488\) or \(x = 0\) (reject: it's point \(A\)).

When \(x = 2.2488\): \[ y = -\tfrac{1}{2}(2.2488) + \sqrt{2} = 0.28980\text{ (5 s.f.)} \]