Calculus (Texas A&M University) | Source: MATH 308 Lecture Notes, Ch. 8.1, 10.1–10.2
Tags: parametric equations, parametric curves, arc length, slope of parametric curve, surface area of revolution, parameter elimination, Gabriel's horn, MATH 308
Parametric curves describe motion in the plane using a pair of functions x(t) and y(t), where t usually represents time. You can sometimes eliminate t to get a Cartesian equation. Calculus on parametric curves lets you find slopes (dy/dx = (dy/dt)/(dx/dt)), arc lengths, and surface areas of revolution, all by working through the parameter.
Parametric curve
A curve defined by the set of all points (x(t), y(t)) as the parameter t varies over some interval. The parameter t is usually interpreted as time, with the point tracing out a path in the x-y plane.
Parameter elimination
The process of removing t from a pair of parametric equations to obtain a single Cartesian equation involving only x and y. Sometimes possible, sometimes not.
Arc length (ordinary curve)
For a curve y = f(x) from x = a to x = b: L = ∫ₐᵇ √(1 + (dy/dx)²) dx. When x = g(y), the analogous formula integrates with respect to y.
Arc length (parametric curve)
For a curve x = f(t), y = g(t) from t = a to t = b: L = ∫ₐᵇ √((dx/dt)² + (dy/dt)²) dt.
Surface area of revolution
The area of the surface generated by rotating a parametric curve about an axis. About the x-axis: A = ∫ 2πy ds. About the y-axis: A = ∫ 2πx ds, where ds = √((dx/dt)² + (dy/dt)²) dt.
Gabriel's horn
The surface formed by rotating y = 1/x (for x ≥ 1) about the x-axis. It has finite volume (π) but infinite surface area, a classic example in improper integrals.
A parametric curve consists of all points (x(t), y(t)) as t ranges over some interval. Think of the point as moving through the plane over time.
Different parameterisations can trace the same geometric curve at different speeds or over different portions
(t, t²) traces out the parabola y = x²
(t³, t⁶) traces the same parabola but moves through it at a different rate
(t², t⁴) traces only the right half (x ≥ 0)
(sin t, sin² t) traces only the segment where –1 ≤ x ≤ 1
Example (a): x = t + 1, y = t³. From the first equation, t = x – 1. Substitute: y = (x – 1)³. This is a shifted cubic.
Example (b): x = t², y = t – 2, for –2 ≤ t ≤ 4. From the second equation, t = y + 2. Substitute: x = (y + 2)². This is a sideways parabola. The restriction on t limits which portion is traced.
Example (c): x = √t, y = 2 – t. Since x = √t, we have t = x² (and x ≥ 0). So y = 2 – x², the right half of a downward parabola.
Example (d): x = 2 cos θ, y = 2 sin θ. Use the identity cos²θ + sin²θ = 1: x² + y² = 4. This is a circle of radius 2.
Example (e): x = 3 + sin t, y = 1 + cos t. Rearrange: (x – 3) = sin t, (y – 1) = cos t. Square and add: (x – 3)² + (y – 1)² = 1. A circle of radius 1 centred at (3, 1).
Example (f): x = –4 cos θ, y = 5 sin θ, for 0 ≤ θ ≤ π. Rearrange: cos θ = –x/4, sin θ = y/5. Square and add: x²/16 + y²/25 = 1. This is the upper half of an ellipse (since 0 ≤ θ ≤ π keeps y ≥ 0).
Spiral: (t cos t, t sin t) for t ≥ 0 traces a spiral that expands outward from the origin.
The slope of a parametric curve at a point is:
dy/dx = (dy/dt) / (dx/dt), provided dx/dt ≠ 0.
A useful mnemonic: treat dy/dx as a fraction of differentials, with dt cancelling.
Tangent line behaviour:
Horizontal tangent when dy/dt = 0 (and dx/dt ≠ 0)
Vertical tangent when dx/dt = 0 (and dy/dt ≠ 0)
Worked example: (x(t), y(t)) = (t², t³ – 3t).
dx/dt = 2t, dy/dt = 3t² – 3
dy/dx = (3t² – 3) / (2t)
Horizontal tangent: 3t² – 3 = 0, so t = ±1. Points: (1, –2) and (1, 2)
Vertical tangent: 2t = 0, so t = 0. Point: (0, 0)
At the point (3, 0): solve t² = 3 and t³ – 3t = 0. From the second equation, t(t² – 3) = 0, so t = 0 or t = ±√3. Since t² = 3, we need t = ±√3. These give two different slopes, meaning the curve crosses itself at (3, 0) with two distinct tangent lines
For y = f(x) from x = a to x = b:
L = ∫ₐᵇ √(1 + (f'(x))²) dx
When the curve is given as x = g(y):
L = ∫ₐᵇ √((g'(y))² + 1) dy
Example: y = ln(sec x), from x = 0 to x = π/4. y' = sec x · tan x / sec x = tan x. L = ∫₀^(π/4) √(1 + tan²x) dx = ∫₀^(π/4) sec x dx = ln|sec x + tan x| evaluated from 0 to π/4.
For x = f(t), y = g(t) from t = a to t = b:
L = ∫ₐᵇ √((dx/dt)² + (dy/dt)²) dt
Deriving the circumference of a circle: Parameterise as x = r cos t, y = r sin t, for 0 ≤ t ≤ 2π. dx/dt = –r sin t, dy/dt = r cos t. (dx/dt)² + (dy/dt)² = r²sin²t + r²cos²t = r². L = ∫₀^(2π) r dt = 2πr. Exactly as expected.
Example: x = t² + 4, y = t³ + 1, from (4, 1) to (8, 9). At (4, 1): t² + 4 = 4 and t³ + 1 = 1, so t = 0. At (8, 9): t² + 4 = 8 and t³ + 1 = 9, so t = 2. dx/dt = 2t, dy/dt = 3t². L = ∫₀² √(4t² + 9t⁴) dt = ∫₀² t√(4 + 9t²) dt.
When a parametric curve is rotated about an axis, the surface area uses ds = √((dx/dt)² + (dy/dt)²) dt.
Rotation about the x-axis (radius = y):
A = ∫ₐᵇ 2πy · √((dx/dt)² + (dy/dt)²) dt
Rotation about the y-axis (radius = x):
A = ∫ₐᵇ 2πx · √((dx/dt)² + (dy/dt)²) dt
Example (astroid): x = cos³θ, y = sin³θ, for 0 ≤ θ ≤ π/2, rotated about the x-axis. dx/dθ = –3cos²θ sin θ, dy/dθ = 3sin²θ cos θ. ds = √(9cos⁴θ sin²θ + 9sin⁴θ cos²θ) dθ = 3 sin θ cos θ dθ. A = ∫₀^(π/2) 2π sin³θ · 3 sin θ cos θ dθ.
Let R be bounded by y = 1/x, y = 0, and x = 1 (extending to infinity).
Area of R: ∫₁^∞ (1/x) dx = ln x |₁^∞ = ∞ (infinite)
Volume when R is rotated about the x-axis: π∫₁^∞ (1/x²) dx = π (finite)
Surface area of the surface of revolution: infinite (can be shown via comparison)
This is the famous paradox: you can fill Gabriel's horn with paint, but you cannot paint its surface.
Slope of parametric curve: dy/dx = (dy/dt) / (dx/dt)
Arc length (Cartesian): L = ∫ₐᵇ √(1 + (dy/dx)²) dx
Arc length (parametric): L = ∫ₐᵇ √((dx/dt)² + (dy/dt)²) dt
Surface area, rotation about x-axis: A = ∫ₐᵇ 2πy ds
Surface area, rotation about y-axis: A = ∫ₐᵇ 2πx ds
Differential arc length: ds = √((dx/dt)² + (dy/dt)²) dt
⚠️ When eliminating the parameter, pay attention to the range of t. The Cartesian equation may describe a larger curve than what the parametric equations actually trace.
⚠️ Different parameterisations can trace the same geometric curve but produce different parametric curves (different speeds, different portions, different directions).
⚠️ Horizontal tangent requires dy/dt = 0 with dx/dt ≠ 0. If both are zero simultaneously, the situation is more subtle and needs further analysis.
⚠️ For arc length problems, check the limits of integration carefully. The parameter values, not the x or y values, serve as the bounds.
⚠️ Gabriel's horn is a classic exam topic. Remember: finite volume, infinite surface area.
Q: Given x = t + 1, y = t³, eliminate the parameter to find a Cartesian equation.
A: t = x – 1, so y = (x – 1)³.
Q: For the parametric curve (t², t³ – 3t), at what values of t is the tangent line horizontal?
A: dy/dt = 3t² – 3 = 0 gives t = ±1. At t = 1 the point is (1, –2); at t = –1 it is (1, 2).
Q: Write the arc length integral for x = r cos t, y = r sin t over 0 ≤ t ≤ 2π. What does it simplify to?
A: L = ∫₀^(2π) √(r²sin²t + r²cos²t) dt = ∫₀^(2π) r dt = 2πr.
Q: What is dy/dx for the parametric curve x = 3 + sin t, y = 1 + cos t?
A: dy/dx = (–sin t) / (cos t) = –tan t.
Q: Is the volume of Gabriel's horn finite or infinite? What about its surface area?
A: The volume is finite (equal to π). The surface area is infinite.
parametric equations, parametric curves, parameter elimination, arc length, curve length, surface area of revolution, slope of parametric curve, tangent to parametric curve, horizontal tangent, vertical tangent, Gabriel's horn, astroid, spiral, ellipse parameterisation, circle parameterisation, MATH 308, Texas A&M, calculus