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{\bf Question}

Show that the angle between the tangent and the radial line at any
point P on the curve $\ds r = ae^{\theta\cot \alpha} \, (\alpha
{\rm \ constant})$ is equal to $\alpha$

\vspace{.25in}

{\bf Answer}

$\ds r = ae^{\theta \cot \alpha}$

$\ds \cot \phi = \frac{1}{r} \frac{dr}{d\theta} = \frac{ae^{\theta
\cot \alpha}\cot \alpha}{ae^{\theta \cot \alpha}} = \cot \alpha$

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So $$\theta=\alpha$$

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This curve is an equiangular spiral

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