C.8 · Trigonometric phase and parametric curves

For y=A cos(ωx−φ) with A>0 and ω>0, amplitude is A, period is 2π/ω, and horizontal shift is φ/ω. The phase angle φ is measured inside the cosine argument; it is not itself the horizontal shift unless ω=1. Start with peak-to-peak spacing for the period and midline-to-peak distance for amplitude. To find phase, substitute a known point and check the direction of motion there. Equivalent phases differ by 2π.

For y=−2 cos(3x), rewrite the model with positive amplitude as 2 cos(3x−π). Its amplitude is 2, period 2π/3, and phase π; the equivalent right shift is π/3. Peaks occur where 3x−π is a multiple of 2π. A graph point at x=0 and y=0 alone cannot determine phase uniquely: slopes or another point distinguish the possible angles. Units and angle conventions matter; ordinary calculus trigonometric derivatives use radians.

A parametric curve specifies x=x(t), y=y(t). Eliminating t describes a point set but can lose restrictions and direction. For x=cos³t, y=sin³t, real cube roots give |x|^(2/3)+|y|^(2/3)=1, an astroid with cusps on the axes. As t runs from 0 to 2π it starts at (1,0), passes (0,1) at π/2 and travels counterclockwise. Restricting t to [0,π/2] gives only the first-quadrant arc, not the entire implicit locus.

When dx/dt≠0, dy/dx=(dy/dt)/(dx/dt). A horizontal tangent generally requires dy/dt=0 and dx/dt≠0; a vertical tangent generally reverses those conditions. If both vanish, inspect a limit or the local expansion rather than taking 0/0 as a slope. For x=t²,y=t³ at t=0, the quotient for t≠0 is 3t/2 and tends to zero: the cusp has a horizontal tangent. A zero parameter velocity is not by itself a local maximum or minimum of y as a function of x.

For y=−2 cos(3x), minima include x=0 and maxima include x=π/3. The phase representation 2 cos(3x−π) gives the same values. For x=2 cos t,y=sin t, elimination gives x²/4+y²=1; at t=π/4, dx/dt=−√2 and dy/dt=√2/2, hence slope −1/2. The full parameter interval [0,2π] traces the ellipse once counterclockwise.

Phase angle and horizontal shift differ by the frequency factor. An implicit equation can add untraced parts of a curve. If both parameter derivatives vanish, use local reasoning instead of labelling every such point an extremum.

For $y=-3\cos(2x)$, state amplitude, period, a positive-amplitude phase representation and the corresponding horizontal shift. Parameterise the ellipse $x^2/9+y^2/4=1$ counterclockwise.

Worked solution

Amplitude is three and period $2\pi/2=\pi$. Write $y=3\cos(2x-\pi)$, with phase $\pi$ and rightward shift $\pi/2$. Phase and shift are not the same. One parametrisation is $x=3\cos t,y=2\sin t$, $0\le t\le2\pi$, which begins at $(3,0)$ and initially moves upward. Eliminating t checks the ellipse equation.

For $x=t^2,y=t^3$ on $-1\le t\le1$, determine the traced locus and tangent at t=0. Does vanishing parameter velocity prove there is no tangent? Does the implicit equation alone retain the whole parameter restriction?

Check after attempting

Elimination gives $y^2=x^3$, with $0\le x\le1$ and both signs of y. For t nonzero, $dy/dx=3t/2\to0$. The secant slope from the origin is $y/x=t\to0$ too, so the cusp has horizontal tangent. Both parameter derivatives vanish at zero, making the raw quotient $0/0$ inconclusive, not proving absence of a tangent. The unrestricted implicit equation also has points with x>1; the interval restriction must be carried separately.