Guides / Parametric and polar

Parametric and polar curves: dy/dx as a ratio of rates, and areas from ½r²

A parametric curve gives a point's position at each moment, x(t) and y(t); a polar curve gives its distance from the origin in each direction, r(θ). The calculus is the same as for y = f(x), with the chain rule translating between variables. The mistakes come from forgetting that translation: a second derivative taken only in t, a speed without its absolute value, and the area between polar curves set up like the area between graphs.

Why it works

Along the curve, y depends on x through t, so the chain rule gives dy/dt = (dy/dx)(dx/dt), and dividing gives the slope. The same rule turns any t-derivative into an x-derivative: d/dx = (1/(dx/dt)) d/dt. That's why the second derivative needs one more division, since the slope is itself a function of t. In a short time dt the point moves x′dt across and y′dt up, so Pythagoras gives the distance √(x′² + y′²) dt, and the arc length adds those up.

In polar coordinates, a thin wedge with angle dθ and radius r is nearly a sector of a circle, with area ½r² dθ. Adding the wedges from α to β gives the area swept out. A region between two curves is the outer wedges minus the inner ones, and because a wedge's area grows with the square of its radius, the squares are what get subtracted.

Example 1: slope, tangent line, and horizontal and vertical tangents

For x = t³ − 3t, y = t², find the tangent line at t = 2, and the points where the tangent is horizontal or vertical.

1. dydx = 2t3t² − 3dy/dt over dx/dt.
2. t = 2: point (2, 4), slope 49, so y = 49x + 289x = 8 − 6 = 2 and y = 4. Point-slope form: y − 4 = (4/9)(x − 2).
3. horizontal: 2t = 0, t = 0, at (0, 0)dx/dt = −3 there, so the slope really is 0.
4. vertical: 3t² − 3 = 0, t = ±1, at (−2, 1) and (2, 1)dy/dt = ±2 there, not 0. The curve also crosses itself at (0, 3), where t = ±√3, with two different slopes, ±√3/3.

Example 2: the second derivative

For the same curve, find d²y/dx² at t = 2.

1. ddt(dydx) = 2 · (3t² − 3) − 2t · 6t(3t² − 3)² = −2(t² + 1)3(t² − 1)²The quotient rule, in t. The top is −6t² − 6.
2. d²ydx² = −2(t² + 1)3(t² − 1)² ÷ 3(t² − 1) = −2(t² + 1)9(t² − 1)³Divide by dx/dt = 3t² − 3 to turn the t-derivative into an x-derivative.
3. t = 2: d²ydx² = −2 · 59 · 27 = −10243Negative, so the curve is concave down at (2, 4).

Example 3: arc length with an absolute value

Find the length of the astroid x = cos³ t, y = sin³ t, 0 ≤ t ≤ 2π.

1. x′ = −3cos² t sin t, y′ = 3sin² t cos tThe chain rule on each cube.
2. x′² + y′² = 9sin² t cos² t(cos² t + sin² t) = 9sin² t cos² tFactor out the common part, then use the Pythagorean identity.
3. speed = 3|sin t cos t|√(a²) = |a|, and sin t cos t is negative in the second and fourth quadrants.
4. L = 4∫₀^π/2 3sin t cos t dt = 4 · 32 = 6The four arcs have the same length, and on [0, π/2] the absolute value can go.

Example 4: the area of a polar rose

Find the area enclosed by r = sin 2θ.

1. one petal: 0 ≤ θ ≤ π2r = 0 at both ends and r ≥ 0 in between. The curve has four petals.
2. 12∫₀^π/2 sin² 2θ dθ = 12 · π4 = π8sin² 2θ = (1 − cos 4θ)/2, and the cosine part integrates to 0 over this interval.
3. area = 4 · π8 = π2Integrating r² from 0 to 2π gives the same π/2, because each petal is traced exactly once in a full turn.

Example 5: the area between two polar curves

Find the area inside the circle r = 1 and outside the cardioid r = 1 − cos θ.

1. 1 = 1 − cos θ gives cos θ = 0, θ = ±π2Where the curves cross. Between them cos θ > 0, so the circle is outside.
2. A = 12∫_−π/2^π/2 (1² − (1 − cos θ)²) dθ = 12∫_−π/2^π/2 (2cos θ − cos² θ) dθOuter radius squared minus inner radius squared.
3. A = 12(4 − π2) = 2 − π4 ≈ 1.215∫ 2cos θ over the interval is 4, and ∫ cos² θ is π/2. Check: it is less than the half disk on the right, π/2 ≈ 1.571.

Common mistakes, and the exact line they happen on

1. Stopping at d/dt of dy/dx

The curve from Example 2.

1. ddt(dydx) = −2(t² + 1)3(t² − 1)²Right so far.
2. d²ydx² = −1027 at t = 2This is the mistake. That is the rate of change of the slope per unit of t, not per unit of x. At t = 2, x is changing 9 times as fast as t (dx/dt = 9), so dividing by it gives −10/243, a ninth of this.

2. Dropping the absolute value in the speed

The astroid from Example 3.

1. x′² + y′² = 9sin² t cos² tRight so far.
2. L = ∫₀^2π 3sin t cos t dt = 0This is the mistake. √(9 sin² t cos² t) is 3|sin t cos t|, and without the bars the negative stretches cancel the positive ones. A closed curve has a positive length; with the absolute value, L = 6.

3. Squaring the difference of the radii

The region from Example 5.

1. r_out = 1, r_in = 1 − cos θ, −π2 ≤ θ ≤ π2Right so far.
2. A = 12∫_−π/2^π/2 (r_out − r_in)² dθ = 12∫_−π/2^π/2 cos² θ dθ = π4 ≈ 0.785This is the mistake. That is the area of wedges of radius r_out − r_in starting at the origin, a smaller region of a different shape. The region between the curves is the big wedges minus the small ones, ½∫(r_out² − r_in²) dθ = 2 − π/4 ≈ 1.215.

Practice

1. Find the tangent line to x = t², y = 4t at t = 1.

   Show answer

   dy/dx = 42t = 2 at the point (1, 4), so y = 2x + 2.

2. Identify the curve x = 1 + 2cos t, y = −1 + 2sin t, 0 ≤ t ≤ 2π, and find its length with the arc length integral.

   Show answer

   (x − 1)² + (y + 1)² = 4: a circle of radius 2 centered at (1, −1). The speed is √(4sin² t + 4cos² t) = 2, so L = 4π, its circumference.

3. Find the length of x = 3t², y = 2t³, 0 ≤ t ≤ 1.

   Show answer

   The speed is √(36t² + 36t⁴) = 6t√(1 + t²), so L = 2[(1 + t²)^3/2]₀¹ = 4√2 − 2 ≈ 3.657.

4. Find the area enclosed by the limaçon r = 2 + cos θ.

   Show answer

   r > 0 throughout, so A = 12∫₀^2π (4 + 4cos θ + cos² θ) dθ = 12(8π + π) = 9π2.

5. Find the slope of the spiral r = θ at θ = π2.

   Show answer

   dy/dθ = r′sin θ + r cos θ = 1 and dx/dθ = r′cos θ − r sin θ = −π2, so the slope is −2π ≈ −0.637, at the point (0, π2).