91Ó°ÊÓ

The polar equation of the cardioid is given as \( r = -1 + \sin \theta \). First, we need to convert this into a Cartesian equation. Using the polar-to-Cartesian conversions, \( x = r \cos \theta \) and \( y = r \sin \theta \), substitute \( r = -1 + \sin \theta \) into these equations. This gives: \( x = (-1 + \sin \theta) \cos \theta \) and \( y = (-1 + \sin \theta) \sin \theta \).

Now, evaluate the Cartesian coordinates for the given values of \( \theta \). For \( \theta = 0 \):- \( \sin 0 = 0 \) and \( \cos 0 = 1 \), so \( r = -1 \).- Thus, \( x = -1 \cdot 1 = -1 \) and \( y = -1 \cdot 0 = 0 \). So, the point is \((-1, 0)\).For \( \theta = \pi \):- \( \sin \pi = 0 \) and \( \cos \pi = -1 \), so \( r = -1 \).- Thus, \( x = -1 \cdot (-1) = 1 \) and \( y = -1 \cdot 0 = 0 \).So, the point is \((1, 0)\).

The slope of the tangent line to a polar curve is given by the formula \( \frac{dr}{d\theta} = \frac{x' \sin\theta + r\cos\theta} {x'\cos\theta - r\sin\theta} \) where \( x' = \frac{dx}{d\theta} \) and \( y' = \frac{dy}{d\theta} \). First, find \( x' = \frac{d}{d\theta}((-1 + \sin\theta )\cos\theta) \) and \( y' = \frac{d}{d\theta}((-1 + \sin\theta) \sin\theta) \).**For \( \theta = 0 \):**- Differentiate to get: \( x' = \cos^2\theta - (1-\sin\theta)\sin\theta = 1 \) and \( y' = \cos\theta\sin\theta = 0 \).- Therefore, the slope is \( m = \frac{0(-1) + (-1)1}{1(1) - (-1)0} = -1 \).**For \( \theta = \pi \):**- Differentiate to find: \( x' = -1 \) and \( y' = 0 \).- Slope here is \( m = \frac{0(1) + (-1)(-1)}{-1(-1) - (-1)0} = 1 \).

Sketch the cardioid with polar equation \( r = -1 + \sin\theta \). At \( \theta = 0 \), the point is \((-1, 0)\) with a slope of \(-1\). Draw the tangent line at this point with this slope.At \( \theta = \pi \), the point is \((1, 0)\) with a slope of \(1\). Likewise, draw the tangent line at this point. This sketch helps visualize how the cardioid and its tangent lines interact at these points.