Problem 53 - Essential Calculus

Question

Find the points of horizontal and vertical tangency (if any) to the polar curve. $$ r=1-\sin \theta $$

Step-by-Step Solution

Answer

The horizontal tangents occur at (r, \(\theta\)) coordinates where \(r = 1 - \sin(\theta)\), \(\theta = \frac{\pi}{2} + k\pi\) and k is an integer. The vertical tangents occur at (r, \(\theta\)) points where r = 0, \(\theta = \frac{\pi}{2} + 2k\pi\) and k is an integer.

Step 1 - Differentiate the Polar curve with respect to \(\theta\)

Differentiation of \(r = 1 - \sin(\theta)\) with respect to \(\theta\) gives: \[ \frac{dr}{d\theta} = - \cos(\theta) \] This derivative represents the rate of change of the radius \(r\) with respect to the angle \(\theta\).

Step 2 - Find the points where derivative equals zero

We solve the equation \(- \cos(\theta) = 0\).From trigonometry, this occurs when \(\theta = \frac{\pi}{2} + k\pi\), where \(k\) is an integer. These are the angles at which horizontal tangents occur. Note, to get the corresponding points (r, \(\theta\)) on the curve, substitute \(\theta\) into \(r = 1 - \sin(\theta)\).

Step 3 - Set r = 0

As vertical tangents occur at points where \(r = 0\), set \(r = 1 - \sin(\theta) = 0\). Solving the equation gives \(\sin(\theta) = 1\). From trigonometry, we know this happens at \(\theta = \frac{\pi}{2} + 2k\pi\), where k is an integer. These are the angles at which vertical tangents occur. We get the points (r ,\(\theta\)) by substituting \(\theta\) into \(r = 1 - \sin(\theta)\). Please note that set r=0 only gives us \(\theta\), as the corresponding r is zero.