CALCULUS POLAR CURVES NEED HELP ASAP PLS

Question

Find the slope of the tangent line to the polar curve r = 2 sin θ at the point where  . (4 points)

ANSWER CHOICES

	Undefined
	0
	1
	-1

Answer 1

If you think about what this curve looks like (r has a maximum at θ = π/2 and the function will be symmetric about the y-axis), we should expect a slope of 0.

Note that, at θ = π/2, r = 2 sin θ, so r = 2, x = r cos θ and y = r sin θ, so (x, y) = (0, 2)

Not the most elegant solution, but let's re-express the equation in Cartesian coordinates, and it will be easy to show dy/dx at x = 0

As y = r sin θ, sin θ =y/r

Then,

r = 2 sin θ

r = 2 y/r

r² = 2y

x² + y² = 2y

We could simply note that we can rewrite this equation as

x² + y² - 2y = 0, or, completing the square,

x² + y² - 2y + 1 = 1

x² + (y - 1)² = 1

This is a circle with center (0, 1) and radius 1, so at (0, 2), we have a horizontal tangent.

However, if we start from x² + y² - 2y = 0, we can use implicit differentiation to solve for the slope.

2x dx + (2y - 2) dy = 0

dy/dx = -2x/(2y - 2) = -x/(y - 1)

At (0, 2), this equals -0/(2 - 1) = 0