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]