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Parametric & Polar Calculus

Useful preparation: Sequences, Convergence & Taylor Series

For a long equation, swipe sideways inside its formula box to see every term.

Meaning & notation

Parametric equations describe both coordinates through a parameter; polar coordinates use angle and signed radius. Calculus follows that geometry: slopes compare coordinate rates, lengths accumulate speed, and polar areas accumulate sectors.

Parametric curve
Both coordinates depend on a parameter. Its interval determines the portion traced, direction, and repeated traversal.
Regular point
A parameter value with nonzero velocity. Vertical tangents are possible even when the slope quotient is unavailable.
Polar coordinates
An angle and signed radial coordinate describing a point. Negative radius places the point opposite the angle's forward ray, and representations are not unique.

Conditions before calculation

  • The quotient formula for slope requires a nonzero derivative of the horizontal coordinate. If both coordinate derivatives vanish, analyze limits or the curve itself.
  • Use continuously differentiable or piecewise smooth parameterizations for the length formulas; repeated traversal contributes repeatedly to the computed length.
  • For ordinary enclosed polar area, choose intervals that cover the intended region once. Signed radii and repeated points require checking the actual tracing.

Why the method works

1. Compare rates using the parameter

Divide the coordinate rates to obtain slope. For the second derivative, differentiate the slope in t, then divide again by the horizontal rate.

dydx=y′(t)x′(t),d2ydx2=ddt(y′(t)/x′(t))x′(t)\frac{dy}{dx}=\frac{y'(t)}{x'(t)},\qquad \frac{d^2y}{dx^2}=\frac{\dfrac{d}{dt}\left(y'(t)/x'(t)\right)}{x'(t)}

2. Accumulate speed for length

Pythagoras combines coordinate changes into displacement length. Integrating the velocity magnitude therefore accumulates nonnegative length.

L=∫abx′(t)2+y′(t)2 dtL=\int_a^b\sqrt{x'(t)^2+y'(t)^2}\,dt

3. Convert polar geometry carefully

Coordinate conversion gives polar speed. A narrow sector contributes approximately half its squared radius times angle width, explaining the area formula.

x=rcos⁡θ,y=rsin⁡θ;A=12∫αβr2 dθ,L=∫αβr2+(drdθ)2 dθx=r\cos\theta,\quad y=r\sin\theta;\qquad A=\frac12\int_\alpha^\beta r^2\,d\theta,\qquad L=\int_\alpha^\beta\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}\,d\theta

Worked examples

EXAMPLE 1

Find the quarter-circle slope at the stated parameter and its arc length.

  1. The parameter interval traces the upper-right quarter once, moving counterclockwise. Differentiate both coordinates.

    x(t)=3cos⁡t,y(t)=3sin⁡t,0≤t≤π2;x′=−3sin⁡t,y′=3cos⁡tx(t)=3\cos t,\quad y(t)=3\sin t,\quad 0\le t\le\frac\pi2;\qquad x'=-3\sin t,\quad y'=3\cos t

  2. At the chosen parameter the horizontal rate is nonzero, so division is legitimate.

    dydx∣t=π/6=3cos⁡(π/6)−3sin⁡(π/6)=−3\left.\frac{dy}{dx}\right|_{t=\pi/6}=\frac{3\cos(\pi/6)}{-3\sin(\pi/6)}=-\sqrt3

  3. The velocity magnitude is constant even though its components vary. Integrating this speed gives the arc length.

    9sin⁡2t+9cos⁡2t=3,L=∫0π/23 dt=3π2\sqrt{9\sin^2t+9\cos^2t}=3,\qquad L=\int_0^{\pi/2}3\,dt=\frac{3\pi}{2}

Conclusion: The slope is negative square root of 3; arc length is three pi over two.

EXAMPLE 2

Find the area enclosed by the polar curve and explain why a full angle interval would overcount it.

  1. Convert to Cartesian form to identify the geometry. On the chosen interval, the radius is nonnegative and the circle is traced once.

    r=2cos⁡θ,−π2≤θ≤π2;x2+y2=2x  ⟺  (x−1)2+y2=1r=2\cos\theta,\quad -\frac\pi2\le\theta\le\frac\pi2;\qquad x^2+y^2=2x\iff(x-1)^2+y^2=1

  2. Integrate the sector areas on this single traversal. The double-angle identity evaluates the squared cosine.

    A=12∫−π/2π/24cos⁡2θ dθ=[θ+12sin⁡(2θ)]−π/2π/2=πA=\frac12\int_{-\pi/2}^{\pi/2}4\cos^2\theta\,d\theta=\left[\theta+\frac12\sin(2\theta)\right]_{-\pi/2}^{\pi/2}=\pi

  3. Increasing the angle by pi reverses both the radius sign and direction, producing the same point. A full turn therefore traces the circle twice.

    r(θ+π)=−r(θ),(x(θ+π),y(θ+π))=(x(θ),y(θ))r(\theta+\pi)=-r(\theta),\qquad (x(\theta+\pi),y(\theta+\pi))=(x(\theta),y(\theta))

Conclusion: The enclosed area is pi; using an interval of length two pi would give twice that area.

Pitfalls & counterexamples

Tempting claim: A zero horizontal rate means the curve has no tangent.

Correction: With nonzero vertical rate it gives a vertical tangent; if both rates vanish, the quotient is inconclusive.

Test the claim: At the starting point of the quarter-circle example, the horizontal rate is zero and the vertical rate is 3, so the tangent is vertical.

Check your understanding

Does tracing the same circle twice double its geometric circumference?

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Compare your reasoning

No, but it doubles the parameterized path length.

The geometric set is unchanged, while the length integral counts every traversal.

Connect this idea

Parametric speed becomes the arc-length element in line integrals. Polar area anticipates the radial Jacobian factor in double integrals.

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Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.

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