The equation of a curve $C$ is given by ${{y}^{3}}+xy-{{x}^{2}}=1$. Find the $x$- coordinate o

Question

The equation of a curve $C$ is given by ${{y}^{3}}+xy-{{x}^{2}}=1$. Find the $x$- coordinate of the point at which the tangent is parallel to the $x$-axis.[4] Show that $\left( 3{{y}^{2}}+x \right)\frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}}+6y{{\left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)}^{2}}+2\frac{\mathrm{d}y}{\mathrm{d}x}-2=0$. Given that the value of $\frac{{{\mathrm{d}}^{3}}y}{\mathrm{d}{{x}^{3}}}$ when $x=0$ is $\frac{20}{27}$, find the Maclaurin series for $y$, up to and including the term in ${{x}^{3}}$.[3]

Accepted Answer

(i) $x=0.462$ to ($3$ s.f.) (ii) $y=1-\frac13x+\frac13x^2+\frac{10}{81}x^3+\cdots$

2020 DHS P1 Q2

2020 DHS P1 Q2

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