Question

Find all points (if any) of horizontal and vertical tangency to the curve. Use a graphing utility to confirm your results. $$ x=3 \cos \theta, \quad y=3 \sin \theta $$

Short answer

The points of horizontal tangency to the curve are at (0,3) and (0,-3) while the points of vertical tangency to the curve are at (3,0) and (-3,0).

Step-by-step solution

Find Derivatives

Compute the derivatives of x and y with respect to \( \theta \).\n\( dx/d\theta = -3 \sin \theta \)\n\( dy/d\theta = 3 \cos \theta \).

Find Values for Horizontal Tangency

For horizontal tangency, we need to find the values of \( \theta \) when \( dy/d\theta = 0 \).\nSolving \( dy/d\theta = 3 \cos \theta = 0 \) gives \( \theta = \pi/2 \), or \( 3\pi/2 \).

Find Points of Horizontal Tangency

Substitute these \( \theta \) values into the original equations to find the x and y coordinates of these points.\nFor \( \theta = \pi/2 \), \( x = 3 \cos(\pi/2) = 0 \), and \( y = 3 \sin(\pi/2) = 3 \), hence the point is (0,3).\nFor \( \theta = 3\pi/2 \), \( x = 3 \cos(3\pi/2) = 0 \) and \( y = 3 \sin(3\pi/2) = -3 \), the point is (0,-3).

Find Values for Vertical Tangency

For vertical tangency, we need to find the values of \( \theta \) when \( dx/d\theta = 0 \).\nSolving \( dx/d\theta = -3 \sin \theta = 0 \) gives \( \theta = 0 \), or \( \pi \).

Find Points of Vertical Tangency

Substitute these \( \theta \) values into the original equations to find the x and y coordinates of these points.\nFor \( \theta = 0 \), \( x = 3 \cos(0) = 3 \), and \( y = 3 \sin(0) = 0 \), hence the point is (3,0).\nFor \( \theta = \pi \), \( x = 3 \cos(\pi) = -3 \) and \( y = 3 \sin(\pi) = 0 \), the point is (-3,0).