16N.2.AHL.TZ0.H_6

pestleMathematicsAAHLPaper 216N· sl-5-3-differentiating-polynomials-n-e-zsource ↗

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.

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

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

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

Examiners’ report

[N/A]