Problem 59 - Essential Calculus

Question

Use a graphing utility to graph the polar equation and find all points of horizontal tangency. $$ r=2 \csc \theta+5 $$

Step-by-Step Solution

Answer

The points of horizontal tangency for the given polar equation occur at \( \theta = \frac{\pi}{2} + k\pi \) where k is an integer.

Step 1: Graph the Polar Equation

First the given polar equation should be graphed using a graphing utility. \(r = 2 \csc(\theta) + 5\). This will give a general understanding of the shape and where tangent lines might be.

Step 2: Derive the equation

Next, differential calculus is performed on the polar equation to find the derivative. This requires knowledge on how to derive csc function. The derivative of \( \csc(\theta) \) is \( -\csc(\theta) \cot(\theta) \). Therefore, after applying the derivative, the equation becomes \( r' = -2\csc(\theta) \cot(\theta) \)

Step 3: Look for Horizontal Tangency

In a polar coordinate graph, horizontal tangency occurs when r' = 0. So, setting \( -2\csc(\theta) \cot(\theta) \) equal to 0 will give the points of horizontal tangency.

Step 4: Solve for Tangency Points

In order for \( -2\csc(\theta) \cot(\theta) \) to be 0, either \( \csc(\theta) \) or \( \cot(\theta) \) needs to be 0. However, csc function is never 0, hence we set \( \cot(\theta) = 0 \) which gives \( \theta = \frac{\pi}{2} + k\pi \) where k is an integer. These are the all the angles for which horizontal tangency occurs. To find the corresponding 'r' value we substitute \(\theta = \frac{\pi}{2} + k\pi \) into original equation \( r = 2 \csc(\theta) + 5 \)