2010 H2 Mathematics Paper 2 Question 3

Differentiation II: Maxima, Minima, Rates of Change

Answers

{\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{3x+4}{2\sqrt{x+2}}}

{x=-\frac{4}{3}}

{\pm \sqrt{2}}

Out of syllabus

Out of syllabus

\begin{align*} y &= x \sqrt{x+2} \\ y^2 &= x^2 (x+2) \\ \end{align*}

Differentiating implicitly w.r.t.

{x,}

\begin{align*} 2y \frac{\mathrm{d}y}{\mathrm{d}x} &= 2x(x+2) + x^2 \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{3x^2+4x}{2y} \\ &= \frac{3x^2+4x}{2x\sqrt{x+2}} \\ &= \frac{3x+4}{2\sqrt{x+2}} \; \blacksquare \end{align*}

At turning point,

{\frac{\mathrm{d}y}{\mathrm{d}x}=0}

\begin{align*} \frac{3x+4}{2\sqrt{x+2}} &= 0 \\ 3x+4 &= 0 \\ x &= -\frac{4}{3} \; \blacksquare \end{align*}

Hence there is only one value of

{x}

for which the curve has a turning point

{\blacksquare}

(iia)

For

{y=x\sqrt{x+2}, }

when

{x=0,}

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{3(0)+4}{2\sqrt{0+2}} \\ &= \frac{2}{\sqrt{2}} \\ &= \sqrt{2} \end{align*}

Hence the possible values of the gradient for

{y^2=x^2(x+2)}

are

{\pm \sqrt{2} \; \blacksquare}

(iib)

Out of syllabus

Out of syllabus