Lecture 9 Notes: Arc Length and Polar Coordinates
Context and Course Logistics
- Announcement: Week 6 has no workshops; mid-semester exam on Monday; quiz extended to Thursday due to the Monday exam.
- Topic focus: Arc length and polar coordinates; polar curves are closely related to parametric curves and can be viewed as a special type of parametric curve.
Parametric Curves: Key Concepts Recap
- How to plot parametric curves: identify tangents, horizontal/vertical tangents, multiple tangents, and test for concavity using the second-derivative test.
- Arc length for parametric curves: the arc length is the integral of speed along the curve.
- Speed interpretation: treat the parametric path as the trajectory of a particle; at time t the velocity vector is
\mathbf{v}(t) = (x'(t), y'(t)) and its speed is the magnitude |\mathbf{v}(t)| = \sqrt{(x'(t))^2 + (y'(t))^2}. - Arc length formula: for a parameter t in ([\alpha, \beta]),
L = \int_{\alpha}^{\beta} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt. - Example approach: compute x'(t) and y'(t), set the limits (\alpha, \beta) (e.g., 0 to 2), and form the integral; some integrals are not elementary and may require numerical methods.
- Numerical methods: Riemann sums, left/right endpoints, trapezoid rule; software (e.g., Wolfram Alpha) can provide numerical approximations (e.g., decimal results).
Example: Arc Length for a Parametric Curve
- Given a parametric curve with
x'(t) = 3t^2 - 3,
\quad y'(t) = 2t,\qquad t \in [0,2],
the arc length is
L = \int{0}^{2} \sqrt{(3t^2-3)^2 + (2t)^2} \, dt = \int{0}^{2} \sqrt{9t^4 - 14t^2 + 9} \, dt. - This integral is typically hard to evaluate in elementary closed form; numerical evaluation yields approximately
L \approx 7.0461. - Conceptual takeaway: arc length is the accumulated speed along the path; if the integral is difficult, numerical methods are standard in practice.
The Cycloid: Parametrization, Arc Length, and Area
- Physical picture: A wheel of radius r rolls without slipping; a fixed point on the rim traces a cycloid.
- Standard parametric equations for one arch (θ from 0 to 2π):
x(\theta) = r(\theta - \sin\theta),\qquad y(\theta) = r(1 - \cos\theta). - Derivation idea: the horizontal translation equals the arc length rolled (proportional to θ), while the vertical motion comes from the rotation of the wheel.
- One arch corresponds to θ ∈ [0, 2π].
- Area under one arch (parametric area): for a region under a parametric curve, the area can be computed via
A = \int y \, dx = \int y(t)\, x'(t) \, dt,
where x and y are given parametrically by x = f(t), y = g(t). - For the cycloid, with the above x and y, we have
x'(\theta) = r(1 - \cos\theta),\quad y(\theta) = r(1 - \cos\theta). - Therefore, the area under one arch is
A = \int{0}^{2\pi} y(\theta) x'(\theta) \, d\theta = \int{0}^{2\pi} r^2(1 - \cos\theta)^2 \, d\theta. - Expanding and integrating:
A = r^2 \int{0}^{2\pi} \left(1 - 2\cos\theta + \cos^2\theta\right) d\theta
= r^2 \left[ \int{0}^{2\pi} 1 \, d\theta - 2\int{0}^{2\pi} \cos\theta \, d\theta + \int{0}^{2\pi} \cos^2\theta \, d\theta \right].
Using (\cos^2\theta = \frac{1 + \cos 2\theta}{2}) and the fact that the integrals of (\cos\theta) and (\cos 2\theta) over one full period vanish, we get
A = r^2 \left( 2\pi - 0 + \pi \right) = 3\pi r^2. - Key takeaway: area under one arch of a cycloid is (A = 3\pi r^2).
Parametric Curves in Higher Dimensions: Helix and Coordinate Curves
- Helix example: a circle in the xy-plane with a linear rise in z.
- Parametrization: x(t) = \cos t,\quad y(t) = \sin t,\quad z(t) = t, with t typically real.
- For three full turns, t goes from 0 to (6\pi).
- Velocity: \mathbf{v}(t) = (-\sin t, \cos t, 1), so the speed is
|\mathbf{v}(t)| = \sqrt{(-\sin t)^2 + (\cos t)^2 + 1^2} = \sqrt{2}. - Arc length for three turns:
L = \int_{0}^{6\pi} \sqrt{2} \, dt = 6\pi\sqrt{2}.
- Coordinate curves on a paraboloid (a 3D surface): consider a paraboloid with equation e.g. z = x^2 + y^2.
- A coordinate curve can be generated by fixing a direction in the xy-plane, described by a unit vector (\mathbf{u} \in \mathbb{R}^2) and letting t run along that direction while z follows the surface.
- If (\mathbf{u} = (ux, uy)) with (ux^2 + uy^2 = 1), a coordinate curve is parameterized by
x(t) = t ux,\quad y(t) = t uy,\quad z(t) = x(t)^2 + y(t)^2 = t^2. - Example: take (\mathbf{u} = \left( \tfrac{1}{\sqrt{3}}, \tfrac{\sqrt{2}}{3} \right)) (which is unit-length). Then
x(t) = \frac{t}{\sqrt{3}},\quad y(t) = t\cdot \frac{\sqrt{2}}{3},\quad z(t) = t^2,
and t runs over the real numbers.
- Practical note: these coordinate curves help build geometric intuition in higher dimensions; partial derivatives along these curves give insight into surface behavior.
Polar Coordinates: Basics and Why They Help
- Polar coordinates describe a point in the plane using radius r and angle θ: the same point satisfies
- Cartesian relations: x = r\cos\theta,\quad y = r\sin\theta,\quad r^2 = x^2 + y^2.
- Inverse relations: \theta = \operatorname{arctan}\left(\frac{y}{x}\right) (with quadrant awareness).
- Relationship between the two coordinate systems: rotating by π changes the sign of r, i.e., (−r, θ) is the same as (r, θ+π).
- Conversion back and forth relies on basic trigonometry and the identity \sin^2\theta + \cos^2\theta = 1.
- Why polar can be simpler: radial or symmetric shapes are often easier to describe in polar form; many curves are straightforward in polar form but more involved in Cartesian form, and vice versa for some lines and circles.
Examples in Polar Coordinates: From Polar to Cartesian and Back
- Circle via r = 2: this is a circle of radius 2 centered at the origin; in Cartesian form it’s simply x^2 + y^2 = 4.
- Line through the origin with a fixed angle: θ = φ describes a straight line through the origin making angle φ with the positive x-axis.
- Cardioid: r = 1 + \sin\theta
- Plot strategy: first plot r as a function of θ (a 1D plot), then interpret the polar points ((r, \theta)) as Cartesian points via x = r cos θ, y = r sin θ.
- Key points: at θ = 0, r = 1 ⇒ (x,y) = (1,0); at θ = π/2, r = 2 ⇒ (x,y) = (0,2); at θ = π, r = 1 ⇒ (x,y) = (−1,0); at θ = 3π/2, r = 0 ⇒ (0,0) (the origin).
- The plotted path forms a heart-shaped cardioid as θ goes from 0 to 2π.
- Rose curves: r = 2 cos(2θ) yields a four-leaved rose.
- Explanation: replacing θ by 2θ increases the angular frequency, producing more oscillations (petals) within 0 to 2π; the amplitude is doubled, so petals reach out to r = 2 and cross the origin multiple times.
- Plotting approach mirrors the cardioid example, but with more frequent crossings of the origin due to the factor 2 in the cosine argument.
Plotting Polar Curves: Practical Tips
- Plot the function r = f(θ) first to understand how far out the curve extends for given angles.
- Convert to Cartesian when helpful to recognize the shape (e.g., circles, lines, ellipses) by using x = r cos θ and y = r sin θ together with r^2 = x^2 + y^2.
- Be mindful of the origin crossings: r = 0 at certain θ values means the curve passes through the origin; count how often this occurs to understand the shape (e.g., in roses).
- Polar plots can reveal symmetry and radial structure that are not immediately obvious from Cartesian plots.
Connections and Implications
- Relationship to calculus in higher dimensions: polar coordinates provide a bridge to analyzing curves and surfaces with symmetry; they complement the parametric perspective by offering alternative descriptions that can simplify differentiation, integration, and geometric interpretation.
- Real-world relevance: arc length calculations model actual distances traveled; area under curves in parametric form extends to surfaces and higher-dimensional analogues via appropriate coordinate systems.
- Conceptual takeaways: drawing and geometric intuition are emphasized for higher-dimensional thinking; parameterizations expose how motion and geometry intertwine (velocity, speed, and area under parametric graphs).
- Arc length for a parametric curve:
L = \int_{\alpha}^{\beta} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt. - Velocity vector and speed:
\mathbf{v}(t) = \bigl(x'(t), y'(t)\bigr),\quad |\mathbf{v}(t)| = \sqrt{(x'(t))^2 + (y'(t))^2}. - Cycloid parametrics (one arch):
x(\theta) = r(\theta - \sin\theta),\quad y(\theta) = r(1 - \cos\theta),\qquad \theta \in [0, 2\pi]. - Area under one arch of cycloid:
A = \int{0}^{2\pi} y(\theta) x'(\theta) \, d\theta = \int{0}^{2\pi} r^2(1 - \cos\theta)^2 \, d\theta = 3\pi r^2. - Helix parametrization and arc length for three turns:
x(t) = \cos t,\quad y(t) = \sin t,\quad z(t) = t,
|\mathbf{v}(t)| = \sqrt{(-\sin t)^2 + (\cos t)^2 + 1^2} = \sqrt{2},
\quad L = \int_{0}^{6\pi} \sqrt{2} \, dt = 6\pi\sqrt{2}.$$