15 The diagram shows part of a spiral curve.
The point \(P\) has polar coordinates \(( r , \theta )\) where \(0 \leq \theta \leq \frac { \pi } { 2 }\)
The points \(T\) and \(S\) lie on the initial line and \(O\) is the pole.
\(T P Q\) is the tangent to the curve at \(P\).
\includegraphics[max width=\textwidth, alt={}, center]{68359582-cd8b-4807-9127-eaf8fd339746-26_624_730_653_653}
15
- Show that the gradient of \(T P Q\) is equal to
$$\frac { \frac { \mathrm { d } r } { \mathrm {~d} \theta } \sin \theta + r \cos \theta } { \frac { \mathrm {~d} r } { \mathrm {~d} \theta } \cos \theta - r \sin \theta }$$
15
- The curve has polar equation
$$r = \mathrm { e } ^ { ( \cot b ) \theta }$$
where \(b\) is a constant such that \(0 < b < \frac { \pi } { 2 }\)
Use the result of part (a) to show that the angle between the line \(O P\) and the tangent TPQ does not depend on \(\theta\).
\includegraphics[max width=\textwidth, alt={}, center]{68359582-cd8b-4807-9127-eaf8fd339746-28_2488_1719_219_150}
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