# Question #6cb9f

Aug 20, 2016

The length of subtangent is $a \left(1 - \cos t\right) \cdot \sin \frac{t}{1 + \cos t}$

#### Explanation: The subtangent is a segment $S T$ on the X-axis from point $S$, an abscissa of some point $P$ on a curve, to point $T$, intersection of a tangent to a curve at point $P$ with the X-axis.

Knowing parametric expressions for $x = f \left(t\right)$ and $y = g \left(t\right)$, for any parameter $t$ we know abscissa and ordinate of a point $P \left(t\right)$ on a curve.
In this case we will use ordinate $y = a \left(1 - \cos t\right)$ and will determine $\tan \left(\Psi\right)$. That will be sufficient to find $S T$.

To calculate $\tan \left(\Psi\right) = \frac{\mathrm{dy}}{\mathrm{dx}}$ we will use the property of parametric curve:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

Using this approach,
$\tan \left(\Psi\right) = \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{a \left(\sin t\right)}{a \left(1 + \cos t\right)} = \sin \frac{t}{1 + \cos t}$

Now we can calculate $S T$:
$S T = P S \cdot \tan \left(\Psi\right) = y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = a \left(1 - \cos t\right) \cdot \sin \frac{t}{1 + \cos t}$

This does not resemble any of the answers provided.
However, if $x = a \left(t - \sin t\right)$ then $\frac{\mathrm{dx}}{\mathrm{dt}} = 1 - \cos t$ and answer $a \sin t$ would be correct.