Differentiation & Applications: Optimisation — SAJC 2025 H2 Math Prelim Paper 1
What this question tests
Question
A curve has equation \((x - y)^2 = 2\mathrm{e}^{xy}\).
(i) Show that \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{x - y - y\mathrm{e}^{xy}}{x - y + x\mathrm{e}^{xy}}\).
(ii) Given that the curve cuts the positive \(y\)-axis at point \(A\), find the equation of the normal to the curve at \(A\).
(iii) The normal to the curve at \(A\) meets the curve at another point \(B\). Find the coordinates of \(B\).
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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 \(A\)).
