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2017 Paper 2 Q3
D: 1600.0
B: 1500.0
implicit equations
implicit differentiation
curve sketching
trigonometric
second derivative
symmetry
sin y = sin x
- Sketch, on \(x\)-\(y\) axes, the set of all points satisfying \(\sin y = \sin x\), for \(-\pi \le x \le \pi\) and \(-\pi \le y \le \pi\). You should give the equations of all the lines on your sketch.
- Given that \[ \sin y = \tfrac12 \sin x \] obtain an expression, in terms of \(x\), for \(y'\) when \(0\le x \le \frac12 \pi\) and \(0\le y \le \frac12 \pi\), and show that \[ y'' = - \frac {3\sin x}{(4-\sin^2 x)^{\frac32}} \;. \] Use these results to sketch the set of all points satisfying \(\sin y = \tfrac12 \sin x\) for \(0 \le x \le \frac12 \pi\) and \(0 \le y \le \frac12 \pi\). Hence sketch the set of all points satisfying \(\sin y = \tfrac12 \sin x\) for \(-\pi\! \le \! x \! \le \! \pi\) and \mbox{\( -\pi \, \le\, y\, \le\, \pi\,\)}.
- Without further calculation, sketch the set of all points satisfying \(\cos y = \tfrac12 \sin x\) for \(- \pi \le x \le \pi\) and \( -\pi \le y \le \pi\).
Solution:
- \(\,\)
- \(\,\) \begin{align*}
&& \sin y &= \tfrac12 \sin x \\
\Rightarrow && \frac{\d y}{\d x} \cos y &= \tfrac12 \cos x \\
\Rightarrow && \frac{\d y}{\d x} &= \frac{\cos x}{2 \cos y} \\
&&&= \frac{\cos x}{2 \sqrt{1-\sin^2 y}} \\
&&&= \frac{\cos x}{2 \sqrt{1-\frac14 \sin^2 x}} \\
&&&= \frac{\cos x}{\sqrt{4-\sin^2 x}} \\
\\
&& y'' &= \frac{-\sin x \cdot (4-\sin^2 x)^{\frac12} - \cos x \cdot (4-\sin^2 x)^{-\frac12} \cdot 2 \sin x \cos x}{(4-\sin^2 x)} \\
&&&= \frac{-\sin x \cdot (4-\sin^2 x) - \cos x \cdot 2 \sin x \cos x}{(4-\sin^2x)^{\frac32}} \\
&&&= \frac{-\sin x \cdot (4-\sin^2 x) - \sin x (1-\sin^2x)}{(4-\sin^2x)^{\frac32}} \\
&&&= \frac{-3\sin x }{(4-\sin^2x)^{\frac32}} \\
\end{align*}
- \(\,\)
