Problem 30 - Essential Calculus

Question

Find all points (if any) of horizontal and vertical tangency to the curve. Use a graphing utility to confirm your results. $$ x=\cos \theta, \quad y=2 \sin 2 \theta $$

Step-by-Step Solution

Answer

The horizontal points of tangency are at \( \theta = \pm \pi/2, \pm 3\pi/2 \) and the vertical points of tangency are at \( \theta = 0, \pi, 2\pi \). A graphing utility can be used to visually confirm these points on the curve.

Step 1: Derive parametric equations

Derive the given parametric equations with respect to \( \theta \). \nDerivative of x with respect to \( \theta \) is \( dx/d\theta = -\sin(\theta) \)\nDerivative of y with respect to \( \theta \) is \( dy/d\theta = 4\cos(2\theta) \)

Step 2: Find the derivative of y with respect to x

Calculate dy/dx which is the derivative of y with respect to x. Using the Chain rule, it can be computed as \( dy/d\theta \) divided by \( dx/d\theta \), i.e.\n \( dy/dx = (4\cos(2\theta))/(-\sin(\theta)) \)

Step 3: Find points of horizontal tangency

Set dy/dx to zero to find points of horizontal tangency i.e. when \( \theta = \pm \pi/2, \pm 3\pi/2 \), since cosines of these values are zero.

Step 4: Find points of vertical tangency

Vertical tangency will occur when dy/dx is undefined, i.e. \( \theta = 0, \pi, 2\pi \), since sin(\theta) of these values will result in denominator becoming zero. This results in values of x and y as x = 1, x = -1 and x = 1 with corresponding y values as 0, 0 and 0 respectively.