Parametric, Polar & Complex

Isolate the trig functions, then use the Pythagorean identity:

An ellipse with semi-axes 3 (horizontal) and 2 (vertical).

What is lost:

• Direction. The parametrization starts at (3,0) and moves counterclockwise; the Cartesian equation has no orientation.
• Starting point. t = 0 puts you at (3,0), which the equation cannot express.
• Speed and timing. The point moves fastest near (0,±2) and slowest near (±3,0) — invisible in Cartesian form.
• Number of traversals. Extending to t ≤ 4π traces the same ellipse twice, with an identical Cartesian equation.

Second derivative. Differentiate dy/dx with respect to t, then divide by dx/dt. Rewrite first to make it easy:

At t = 2 that is 3/2 + 3/8 = 15/8. Now divide by dx/dt = 4:

The classic wrong answer is (d²y/dt²)/(d²x/dt²) = 6t/2 = 6, which is not a second derivative of anything. There is no shortcut here; you must differentiate the ratio.

Product rule on each:

Now square and add — the cross terms cancel, which is the whole point:

The ±2 sin t cos t cancellation is engineered into the problem. If your cross terms do not cancel in a spiral problem, re-check the product rule before doing anything else.

Multiply both sides by r — the standard trick, because it creates r² and r cos θ, both of which convert directly:

Complete the square in x:

A circle of radius 2 centered at (2, 0) — passing through the origin, tangent to the y-axis.

Worth knowing as a pattern: r = 2a cos θ is a circle of radius a centered at (a, 0), and r = 2a sin θ is centered at (0, a). Note the whole circle is traced as θ runs from 0 to π, not 2π — integrating over the full turn would count it twice.

The full cardioid is traced once as θ goes 0 to 2π:

Take the three pieces separately:

• ∫₀^2π 1 dθ = 2π
• ∫₀^2π 2 cos θ dθ = 0  (full period of a cosine)
• ∫₀^2π cos²θ dθ = π  (average value ¹⁄₂ over length 2π — the same fact behind RMS in Unit 2)

Sanity check: the cardioid fits inside a circle of radius 2 (area 4π ≈ 12.6) and contains a circle of radius about 1 (area π ≈ 3.1). 4.71 sits sensibly between. ✓

Limits. A petal begins and ends where r = 0:

Half-angle: cos²(2θ) = (1 + cos 4θ)/2.

At both limits sin 4θ = sin(±π) = 0, so only the θ terms survive:

Total for all four petals: 2π. Note this rose has four petals because n = 2 is even — the whole curve needs θ from 0 to 2π. Integrating over the full turn and calling it one petal is the standard error here.

Multiply numerator and denominator by the conjugate of the denominator:

Using i² = −1: numerator = 2 + 7i − 6 = −4 + 7i; denominator = 1 + 4 = 5.

Modulus:

Argument. The point is in quadrant II (negative real, positive imaginary), so add 180° to the raw arctangent:

Verify with the exponential form: |2+3i| = √13, arg = 56.3°. |1−2i| = √5, arg = −63.4°. Quotient modulus = √(13/5) = 1.61 ✓, argument = 56.3 − (−63.4) = 119.7° ✓. Two routes agreeing is worth the thirty seconds.

Put −16 in exponential form. It lies on the negative real axis, so modulus 16 and argument π:

Sketch: four points on a circle of radius 2, spaced 90° apart, at 45°, 135°, 225°, 315° — the corners of a square tilted 45°.

Check k = 0:

Note: the roots come in conjugate pairs, as they must — the polynomial z⁴ + 16 has real coefficients. In circuit terms, complex poles always appear in conjugate pairs, which is what keeps the time-domain response real.

Angular frequency: ω = 2π(500) = 3142 rad/s.

Series ⇒ add:

Polar form:

Current phasor, by Ohm’s law with complex numbers:

Current leads voltage by 57.9° — capacitive, as expected. Dividing in polar form is where the work disappears: divide the moduli, subtract the arguments.

Sanity check: the magnitude 188 Ω must exceed the resistance alone (100 Ω) but be less than the arithmetic sum 259 Ω, because the two impedances add as perpendicular vectors, not as scalars. ✓

φ = 0°: y = x exactly. The trace is a straight line of slope +1, swept back and forth between (−5,−5) and (5,5).

φ = 180°: cos(ωt + π) = −cos(ωt), so y = −x — a straight line of slope −1.

φ = 90°: cos(ωt + π/2) = −sin(ωt), so

A circle of radius 5, traced clockwise.

Path length in one period, by the parametric arc-length formula:

Which is the circumference of a radius-5 circle — the correct answer, arrived at the long way, and a good check that the machinery works.

Why this is genuinely useful: the shape of the ellipse tells you the phase difference at a glance. A line means in phase (or 180° out); a circle means 90°; anything between is a tilted ellipse whose openness scales with |sin φ|. This was the standard bench technique for measuring phase before dual-trace scopes, and it still works.