Please explain each step on how you get to the answer.

Polar Coordinates II Tangents to Polar Curves Doing calculus with curves given by polar equations is most often carried out by viewing the curve as given by parametric equations with the polar coordinate 0 acting as the parameter. For example, we have already learned how to find the slope of the line tangent at a point on a curve given parametrically. Now, suppose we wanted to find the slope of the line tangent to the 8-petal roser = f(0) = sin40 when 0 = 7. Thinking back to the formulas for converting polar coordinates to Cartesian coordinates, we see x = rcos0 = f(0) cose = sin 40 cos 0, and y = rsine = f(0) sine = sin 40 sin 0. That gives us parametric equations for the curve. Recalling the formula for the slope of the tangent line to a curve given parametric cally, we get dy de 4 cos 40 sin 0 + sin 40 cos 0 dx dx 4 cos 40 cos 0 - sin 40 sin 0 Plugging in the parameter value 0 = , we find the slope of the tangent line is 1 5V3 CO Lengths of Polar Curves The length of a polar curve r = f(0), a