Find the slope of the tangent line to the polar curve r = 2 - sin(θ) at the point specified byθ = π/3?

where [tex] \sf dy/d\theta [/tex] and [tex] \sf dx/d\theta [/tex] are the derivatives of the Cartesian coordinates [tex] \sf y [/tex] and [tex] \sf x [/tex] concerning [tex] \sf \theta [/tex], which are given by:

[tex] \sf x = r(\theta) \cos(\theta) \quad \text{and} \quad y = r(\theta) \sin(\theta) [/tex]

For the polar curve [tex] \sf r = 2 - \sin(\theta) [/tex], we compute the derivatives:

[tex] \sf \frac{dx}{d\theta} = \frac{d}{d\theta}[(2 - \sin(\theta)) \cos(\theta)] = (2 - \sin(\theta))(-\sin(\theta)) + \cos(\theta)(-\cos(\theta)) [/tex]

[tex] \sf \frac{dy}{d\theta} = \frac{d}{d\theta}[(2 - \sin(\theta)) \sin(\theta)] = (2 - \sin(\theta))\cos(\theta) + \cos(\theta)(-\sin(\theta)) [/tex]

At [tex] \sf \theta = \frac{\pi}{3} [/tex], substituting the values of [tex] \sf \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2} [/tex] and [tex] \sf \cos(\frac{\pi}{3}) = \frac{1}{2} [/tex]:

[tex] \sf \frac{dx}{d\theta} = (2 - \frac{\sqrt{3}}{2})(-\frac{\sqrt{3}}{2}) + \frac{1}{2}(-\frac{1}{2}) = -\frac{3\sqrt{3}}{4} - \frac{1}{4} [/tex]

[tex] \sf \frac{dy}{d\theta} = (2 - \frac{\sqrt{3}}{2})\frac{1}{2} + \frac{1}{2}(-\frac{\sqrt{3}}{2}) = \frac{4 - \sqrt{3}}{4} - \frac{\sqrt{3}}{4} = 1 - \frac{\sqrt{3}}{2} [/tex]

Now, calculate the slope:

[tex] \sf \text{slope} = \frac{dy/d\theta}{dx/d\theta} = \frac{1 - \frac{\sqrt{3}}{2}}{-\frac{3\sqrt{3}}{4} - \frac{1}{4}} = \sqrt{3} [/tex]

This calculation is derived from applying the product and chain rules of calculus to the expressions for [tex] \sf x [/tex] and [tex] \sf y [/tex] in terms of [tex] \sf r [/tex] and [tex] \sf \theta [/tex], and then evaluating them at [tex] \sf \theta = \frac{\pi}{3} [/tex]. The result, [tex] \sf \sqrt{3} [/tex], indicates the steepness and direction of the tangent at that specific point on the curve.