The curve \(y=\mathrm{f}(x)\) passes through the point \((0,69)\) and has gradient given by \(\frac{\mathrm dy}{\mathrm dx}=\left(\frac13y-15\right)^{\frac13}\).
(i) Let \(v=\frac13y-15\). Then \(3v^{-1/3}\,\mathrm dv=\mathrm dx\).
\(\frac92v^{2/3}=x+C\). At \((0,69)\), \(v=8\), so \(C=18\).
\(v^{2/3}=4+\frac{2x}{9}\Rightarrow\mathrm{f}(x)=45+3\left(4+\frac{2x}{9}\right)^{3/2}\) on the branch through \((0,69)\).
(ii) \(\frac{\mathrm dy}{\mathrm dx}=4\Rightarrow v=4^3=64\Rightarrow y=3(64+15)=237\).
\(16=4+\frac{2x}{9}\Rightarrow x=54\).
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