Differentiation & Applications: Rate of change — RI 2025 H2 Math Prelim Paper 1
What this question tests
Question
The point \(P\) travels along the curve \(C\) with equation \(y = x\sin^{-1}x\), \(-1 < x < 1\). Let the gradient of the curve \(C\) at the point \(P\) be \(m\).
If the \(x\)-coordinate of \(P\) is increasing at the rate of 9 units per second when \(x = \dfrac{1}{2}\), find the exact value of the rate at which \(m\) is changing at this instant.
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The point \(P\) travels along \(y = x\sin^{-1}x\), \(-1<x<1\). Let \(m\) be the gradient of the curve at \(P\). When \(x = \frac{1}{2}\), \(\frac{\mathrm{d}x}{\mathrm{d}t} = 9\). Find the exact rate of change of \(m\).
\[ m = \frac{\mathrm{d}y}{\mathrm{d}x} = \sin^{-1}x + \frac{x}{\sqrt{1-x^2}} \]Differentiating with respect to \(x\) again: \[\begin{aligned} \frac{\mathrm{d}m}{\mathrm{d}x} &= \frac{1}{\sqrt{1-x^2}} + \frac{\sqrt{1-x^2} - x\cdot\dfrac{-2x}{2\sqrt{1-x^2}}}{1-x^2}\\[6pt] &= \frac{1}{\sqrt{1-x^2}} + \frac{(1-x^2)+x^2}{\sqrt{1-x^2}(1-x^2)} = \frac{2-x^2}{\sqrt{(1-x^2)^3}} \end{aligned}\]
When \(x = \dfrac{1}{2}\): \[ \frac{\mathrm{d}m}{\mathrm{d}t} = \frac{\mathrm{d}m}{\mathrm{d}x}\cdot\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{2 - \tfrac{1}{4}}{\sqrt{\!\left(1-\tfrac{1}{4}\right)^{\!3}}}\times 9 = \frac{\tfrac{7}{4}}{\sqrt{\!\left(\tfrac{3}{4}\right)^{\!3}}}\times 9 = 14\sqrt{3} \]