A curve has equation $x^{\frac{1}{2}} + y^{\frac{1}{2}} = 3$, for $x 0$ and \(y 0

Question

A curve has equation $x^{\frac{1}{2}} + y^{\frac{1}{2}} = 3$, for $x > 0$ and $y > 0$. Show that $\frac{\mathrm{d}y}{\mathrm{d}x} = -\left(\frac{y}{x}\right)^{\frac{1}{2}}$.[2] Find the equation of the normal to the curve at the point where $x = 1$ and determine the acute angle that the normal makes with the $x$-axis.[5]

Accepted Answer

(a) $\frac{\mathrm dy}{\mathrm dx}=-\sqrt{\frac{y}{x}}$ (b) $y=\frac{1}{2}x+\frac{7}{2}$, acute angle $0.464\text{ rad}$ (3 s.f.)

N2021 P1 Q3 [Modified]

N2021 P1 Q3 [Modified]

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