An earth satellite moves in a path that can be described by the curve $72.5{x^2} + 71.5{y^2} = 1$ where $x = x(t)$ and $y = y(t)$ are in thousands of kilometres and $t$ is time in seconds.

Given that $\frac{{{\text{d}}x}}{{{\text{d}}t}} = 7.75 \times {10^{ - 5}}$ when $x = 3.2 \times {10^{ - 3}}$, find the possible values of $\frac{{{\text{d}}y}}{{{\text{d}}t}}$.

Give your answers in standard form.

METHOD 1

substituting for $x$ and attempting to solve for $y$ (or vice versa)     (M1)

$y = ( \pm )0.11821 \ldots$    (A1)

EITHER

$145x + 143y\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0{\text{ }}\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}} =  - \frac{{145x}}{{143y}}} \right)$    M1A1

OR

$145x\frac{{{\text{d}}x}}{{{\text{d}}t}} + 143y\frac{{{\text{d}}y}}{{{\text{d}}t}} = 0$    M1A1

THEN

attempting to find $\frac{{{\text{d}}x}}{{{\text{d}}t}}{\text{ }}\left( {\frac{{{\text{d}}y}}{{{\text{d}}t}} =  - \frac{{145(3.2 \times {{10}^{ - 3}})}}{{143\left( {( \pm )0.11821 \ldots } \right)}} \times (7.75 \times {{10}^{ - 5}})} \right)$     (M1)

$\frac{{{\text{d}}y}}{{{\text{d}}t}} =  \pm 2.13 \times {10^{ - 6}}$    A1

Note: Award all marks except the final A1 to candidates who do not consider ±.

METHOD 2

$y = ( \pm )\sqrt {\frac{{1 - 72.5{x^2}}}{{71.5}}}$    M1A1

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = ( \pm )0.0274 \ldots$    (M1)(A1)

$\frac{{{\text{d}}y}}{{{\text{d}}t}} = ( \pm )0.0274 \ldots  \times 7.75 \times {10^{ - 5}}$    (M1)

$\frac{{{\text{d}}y}}{{{\text{d}}t}} =  \pm 2.13 \times {10^{ - 6}}$    A1

Note: Award all marks except the final A1 to candidates who do not consider ±.

[6 marks]