Polar Calculus

A polar curve specifies radius and angle, but a point’s velocity is not just the derivative of its radius. Both distance from the origin and direction can change. Converting to coordinate functions makes their contributions explicit.

The same conversion explains why negative radii, zero radii and repeated tracing require care. Geometry comes before substitution into a slope or length formula.

On this page

Convert to parametric coordinates

For $r=r(\theta)$ , write

x=r\cos\theta,\qquad y=r\sin\theta.

Using radians, the product rule gives

x’=r’\cos\theta-r\sin\theta,\qquad y’=r’\sin\theta+r\cos\theta.

Primes here mean derivatives with respect to angle. For constant nonzero radius, $r’=0$ but the coordinate derivative is not zero. Constant radius describes circular movement, not a stationary point.

If angle is not physical time, these derivatives do not yet give physical velocity. A time-dependent angle supplies another chain-rule factor.

Derive a tangent slope

Where the denominator is nonzero,

\frac{dy}{dx}=\frac{r’\sin\theta+r\cos\theta}{r’\cos\theta-r\sin\theta}.

A horizontal tangent needs a zero numerator and nonzero denominator. A vertical tangent needs a zero denominator and nonzero numerator. If both vanish, analyse the local curve instead of assigning a slope to $0/0$ .

The radial ray is not generally the tangent line. The formula follows from parametric derivatives, treating both coordinates symmetrically.

Find a cardioid tangent

For $r=1+\cos\theta$ , $r’=-\sin\theta$ . At $\theta=\pi/2$ , the radius is one and the point is $(0,1)$ . Both coordinate derivatives are $-1$ , so the slope is one and the tangent is

y=x+1.

As angle increases through that point, the tracing moves down and left despite its positive tangent slope. A geometric line and direction of movement are different descriptions.

Tangent geometry on a cardioid

Equal coordinate units; the complete traced length is eight.

Diagram explanation and exact data

At θ=π/2 both coordinate derivatives are −1. At θ=π a local expansion is needed instead of the 0/0 slope quotient.

Tangent geometry on a cardioid: exact reference data
θ	r	point
0	2	(2,0)
π/2	1	(0,1)
π	0	(0,0)
3π/2	1	(0,−1)

Inspect the cusp at the pole

At $\theta=\pi$ , both $r$ and $r’$ vanish. Let $u=\theta-\pi$ . The leading terms are

x=-\frac{u^2}{2}+O(u^4),\qquad y=-\frac{u^3}{2}+O(u^5).

For nearby nonzero $u$ , slope behaves like $3u/2$ , tending to zero. The cusp therefore has a horizontal limiting tangent, although its angular parametrisation is not regular there.

A zero radius does not always cause a cusp. If $r=0$ but $r’\ne0$ , the coordinate derivative is $r'(\cos\theta,\sin\theta)$ and is nonzero. The curve can pass regularly through the pole.

Derive the polar arc-length formula

Squaring and adding the coordinate derivatives cancels the cross terms:

x’^2+y’^2=r’^2+r^2.

For a continuously differentiable radial function on the interval,

L=\int_\alpha^\beta\sqrt{r^2+r’^2}\,d\theta.

This counts length with tracing multiplicity. It is the geometric length only when the interval traces the curve once, apart from shared endpoints or isolated intersections. The polar arc-length reference develops the formula.

Squaring the radius does not remove the importance of its sign when determining the path.

Calculate the cardioid length

For $r=1+\cos\theta$ ,

r^2+r’^2=2+2\cos\theta=4\cos^2(\theta/2).

Thus speed with respect to angle is $2|\cos(\theta/2)|$ . Splitting the interval at $\pi$ gives

L=\int_0^\pi2\cos(\theta/2)\,d\theta-\int_\pi^{2\pi}2\cos(\theta/2)\,d\theta=8.

Dropping the absolute value would yield zero through cancellation, an impossible length. The interval traces this cardioid once.

Recognise negative-radius tracing

The representations $(r,\theta)$ and $(-r,\theta+\pi)$ give the same Cartesian point. A negative radius places the point opposite the ray at angle $\theta$ .

For $r=2\cos\theta$ ,

x=1+\cos2\theta,\qquad y=\sin2\theta.

This is a unit circle centred at $(1,0)$ . An interval of angular length $\pi$ changes $2\theta$ by a full turn, tracing the circle once. The interval from zero to $2\pi$ traces it twice.

The length integrand is constantly two. It gives $2\pi$ over $[0,\pi]$ and $4\pi$ over $[0,2\pi]$ . Both are valid traced lengths, but only the first is the circle’s circumference.

Distinguish angular and time derivatives

If $\theta=\theta(t)$ , physical speed becomes

v(t)=\sqrt{r^2+r’^2}\left|\frac{d\theta}{dt}\right|,

where $r’$ still means differentiation with respect to angle. If radius and angle are instead given independently as time functions,

v=\sqrt{\dot r^{\,2}+r^2\dot\theta^{\,2}}.

Leaving the differentiated variable unstated can hide a missing angular-speed factor. Units distinguish length per radian from length per second.

Prepare area calculations by checking the path

Identify zeros, sign changes, symmetries and a once-traced interval before using an area formula. Evaluate representative Cartesian points when a negative radius is confusing.

Two curves can reach the pole at different angles. Solving $r_1(\theta)=r_2(\theta)$ at a common angle is useful but may not find every geometric intersection. The next lesson handles those decisions before integration.

Practice questions

Find the full length of $r=3$ on $[0,2\pi]$ .
State the cardioid point and tangent slope at $\theta=\pi/2$ .
Why does the length integral for $r=2\cos\theta$ over $[0,2\pi]$ count twice the circumference?

Check your answers

With $r’=0$ , the integrand is three and the length is $6\pi$ .
The point is $(0,1)$ and slope is one because both coordinate derivatives equal $-1$ .
The Cartesian formula depends on $2\theta$ , which makes two full turns on that interval.