Compute length when radius and direction both change

Try

A point moves outward while turning around the origin. Its distance traveled includes both radial movement and angular movement. Predict why integrating only the radius change cannot measure the length of the spiral r=θ. At θ=0 the radius is zero, but the point can still have nonzero velocity.

Predict

For a constant radius R, the length over an angle interval should reduce to R times angular change. Predict a formula involving both r and r′ that gives this familiar answer when r′=0. Compare it with the polar area integrand r²/2; they have different dimensions and purposes.

Experiment

For r=1+cos θ, calculate the speed algebraically before integrating.

Quantity	Expression
Radial derivative	−sin θ
Squared Cartesian speed	(1+cos θ)²+sin²θ=2+2cos θ
Speed	2∣cos(θ/2)∣
Total length on [0,2π]	8

Sketch the sign of cos(θ/2) over this interval. It changes at π, so removing the absolute value without splitting would incorrectly cancel positive length contributions. Draw the cardioid and mark the origin reached at θ=π; the zero speed there is compatible with a cusp and finite total length.

Understand

Start from x=r(θ)cos θ and y=r(θ)sin θ. Differentiation gives x′=r′cos θ−r sin θ and y′=r′sin θ+r cos θ. Squaring and adding cancels the mixed products and uses sin²θ+cos²θ=1, leaving (r′)²+r². Therefore the length integral is ∫√[r²+(r′)²]dθ over the stated traversal interval, under appropriate piecewise smoothness conditions.

The derivation explains both contributions. Even if r stays constant, rotating the direction moves the point. Even if r=0 at one parameter value, a nonzero r′ can give nonzero speed there. For r=θ, the speed is √(θ²+1), equal to 1 at zero. Direct Cartesian differentiation gives velocity (1,0) at that point, confirming that the origin is not automatically a singularity of the curve.

For the cardioid, the length is ∫₀^2π2|cos(θ/2)|dθ. The integral from zero to π is 4, and symmetry gives another 4 from π to 2π. Alternatively split and reverse the sign of the cosine on the second interval. Both methods preserve nonnegative speed. A square-root simplification √(z²)=z without checking z’s sign would produce the wrong result.

As with parametric length, repeated tracing counts repeatedly. If a polar equation traverses the same loop twice over the chosen interval, the integral gives twice that traversal length even though its image is unchanged. Negative radius does not make length negative; the Cartesian derivative derivation remains the guide, while the tracing interval determines multiplicity.

Some polar length integrals have no elementary primitive. An exact definite integral still specifies the mathematical length. Numerical quadrature then needs appropriate error control and possible splitting at nonsmooth points. A finite polygonal approximation can be useful evidence, but it does not by itself prove an error tolerance for an arbitrary curve.

Apply

In the practice bank, derive the formula rather than recalling it without explanation. For the cardioid, retain absolute values until the interval is analyzed. For r=θ at the origin, compute a Cartesian velocity to separate a coordinate ambiguity from an actual failure of differentiability.

Continue

Next revisit graph length as a special parameterized curve. This connects the vector formula to the familiar expression involving a graph’s derivative and clarifies the role of a chosen coordinate direction.