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 \((0,\pm2)\) and slowest near \((\pm3,0)\) — invisible in Cartesian form.
• Number of traversals. Extending to \(t \le 4\pi\) 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 \((d^2y/dt^2)/(d^2x/dt^2) = 6t/2 = 6\), 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 \(\pm 2\sin t\cos t\) 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 \(r^2\) and \(r\cos\theta\), 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: \(r = 2a\cos\theta\) is a circle of radius \(a\) centered at \((a, 0)\), and \(r = 2a\sin\theta\) is centered at \((0, a)\). Note the whole circle is traced as \(\theta\) runs from \(0\) to \(\pi\), not \(2\pi\) — integrating over the full turn would count it twice.

The full cardioid is traced once as \(\theta\) goes \(0\) to \(2\pi\):

Take the three pieces separately:

• \(\int_0^{2\pi} 1\,d\theta = 2\pi\)
• \(\int_0^{2\pi} 2\cos\theta\,d\theta = 0\)  (full period of a cosine)
• \(\int_0^{2\pi} \cos^2\theta\,d\theta = \pi\)  (average value \(\tfrac12\) over length \(2\pi\) — the same fact behind RMS in Unit 2)

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

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

Half-angle: \(\cos^2(2\theta) = (1 + \cos 4\theta)/2\).

At both limits \(\sin 4\theta = \sin(\pm\pi) = 0\), so only the \(\theta\) terms survive:

Total for all four petals: \(2\pi\). Note this rose has four petals because \(n = 2\) is even — the whole curve needs \(\theta\) from \(0\) to \(2\pi\). 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 \(i^2 = -1\): 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: \(|2+3i| = \sqrt{13}\), \(\arg = 56.3^\circ\). \(|1-2i| = \sqrt5\), \(\arg = -63.4^\circ\). Quotient modulus \(= \sqrt{13/5} = 1.61\) ✓, argument \(= 56.3 - (-63.4) = 119.7^\circ\) ✓. 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 \(\pi\):

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 \(z^4 + 16\) has real coefficients. In circuit terms, complex poles always appear in conjugate pairs, which is what keeps the time-domain response real.

Angular frequency: \(\omega = 2\pi(500) = 3142\) rad/s.

Series \(\Rightarrow\) 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. ✓

\(\phi = 0^\circ\): \(y = x\) exactly. The trace is a straight line of slope \(+1\), swept back and forth between \((-5,-5)\) and \((5,5)\).

\(\phi = 180^\circ\): \(\cos(\omega t + \pi) = -\cos(\omega t)\), so \(y = -x\) — a straight line of slope \(-1\).

\(\phi = 90^\circ\): \(\cos(\omega t + \pi/2) = -\sin(\omega t)\), 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 \(|\sin\phi|\). This was the standard bench technique for measuring phase before dual-trace scopes, and it still works.