Polar Coordinates and Curves

Polar Coordinates #

A point named by a distance and an angle.

A point can be named by how far it lies from the origin and in which direction

Instead of an x-coordinate and a y-coordinate, name the point by its distance from the origin and an angle.

To reach Cartesian form, use trigonometry: x and y are the two legs of the triangle.

Going back, Pythagoras gives r and an inverse tangent gives $\theta$θ. Then check the quadrant.

So r = 2 at $\theta =\frac{\pi }{3}$θ = π/3 is the point $\left(1,\sqrt{3}\right)$(1, √3), and $\left(1,\sqrt{3}\right)$(1, √3) returns r = 2.

Sketching Polar Curves #

Curves traced by r as the angle turns.

Letting r depend on the angle traces curves no function of x could draw

Hold r at 2 and let $\theta$θ make a full turn: every point is 2 from the origin, so the curve is a circle.

Let r fall as the angle grows: $r=2\cos \theta$r = 2 cos θ is a circle that passes through the origin.

$r=1+\cos \theta$r = 1 + cos θ reaches 2 at $\theta =0$θ = 0 and falls to 0 at $\theta =\pi$θ = π: this curve is a cardioid.

$r=2\cos 2\theta$r = 2 cos 2θ returns to the origin four times in a turn, so it draws four petals.

To sketch one, read r at the quarter turns, mark those points, and see where r reaches 0.

The Rate of Change of a Polar Function #

How fast r moves toward or away from the pole.

Differentiating r with respect to theta says how fast the curve moves toward or away from the pole

On $r=1+\cos \theta$r = 1 + cos θ the distance from the pole changes as the angle turns.

Plot r against $\theta$θ instead: r starts at 2, falls to 0 at $\theta =\pi$θ = π, and rises back to 2 at $\theta =2\pi$θ = 2π.

Differentiate r with respect to $\theta$θ, exactly as for any other function.

Between 0 and $\pi$π the derivative is negative, so the curve moves in toward the pole.

Where the derivative is zero the distance stops changing: r is at a maximum or a minimum.

For an average rate, divide the change in r by the change in $\theta$θ across that stretch.

The Area of a Polar Region #

Half the integral of r squared.

A polar region is swept from the pole by thin sectors, each contributing half r squared d theta

A polar region is swept out from the pole, not built up from the x-axis.

One thin slice is a fraction $\frac{d\theta }{2\pi }$dθ/2π of a disc of radius r, so its area is ½$r^{2}d\theta$r² dθ.

Add the slices and the area is half the integral of r squared, between the two angles.

Check it on a quarter disc: ½ times 4 times $\frac{\pi }{2}$π/2 is $\pi$π, and a quarter of $4\pi$4π is $\pi$π.

For a ring between two curves, subtract the squares — each sweep starts at the pole.

Sweep out to the outer curve, then take away the sweep out to the inner one.