22. Parametric curves and their derivatives

Every curve in this course so far has been y=f(x) — one y for every x, which immediately rules out circles, loops, and anything that doubles back on itself. A parametric curve frees both x and y to depend on a third variable, a parameter t, and in doing so can trace shapes no single function y=f(x) could ever describe.

What a parametric curve is

x=f(t),\qquad y=g(t)

As t sweeps through its domain, the point (f(t),g(t)) traces a path in the plane. Think of t as time and (x,y) as the position of a moving particle — the parametrization carries more information than the path alone, since it also says when the particle is where, and it naturally allows the path to cross itself (different times, same location) or reverse direction, neither of which y=f(x) permits.

A circle, impossible to write as one y=f(x) over its full range, is immediate parametrically:

x=r\cos t,\qquad y=r\sin t,\qquad0\le t<2\pi

Eliminating t recovers the familiar implicit equation: x^2+y^2=r^2\cos^2t+r^2\sin^2t=r^2(\cos^2t+\sin^2t)=r^2 — the Pythagorean identity from §0.4 doing the work.

The derivative of a parametric curve

\frac{dy}{dx} still means "how fast does y change as x changes" — but now both x and y are only directly related to t. The chain rule (§2.4) connects them:

\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}\quad\Longrightarrow\quad\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\qquad\left(\frac{dx}{dt}\ne0\right)

This is the parametric analogue of §2.9's implicit differentiation: instead of solving for y explicitly, you divide two derivatives that are each easy to compute directly.

Horizontal tangents occur where \frac{dy}{dt}=0 (but \frac{dx}{dt}\ne0) — the curve momentarily stops rising or falling. Vertical tangents occur where \frac{dx}{dt}=0 (but \frac{dy}{dt}\ne0) — the curve momentarily stops moving left-right. Points where both derivatives vanish simultaneously need closer inspection (often a cusp, like the cycloid in §6.1).

The second derivative

A common mistake is writing \frac{d^2y}{dx^2}=\frac{d^2y/dt^2}{d^2x/dt^2} — this is wrong. The correct route treats \frac{dy}{dx} itself as a new function of t, then differentiates that with respect to x using the chain rule again:

\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{dx/dt}

Two divisions by dx/dt, not one — first to get \frac{dy}{dx}, then again to convert its t-derivative into an x-derivative. Skipping the second division is the single most common parametric-calculus error.

Doing it in Python

\frac{dy}{dx} and \frac{d^2y}{dx^2} for x=t^2, y=t^3-3t — a curve that loops back on itself, impossible to write as y=f(x) over its full range:

import sympy as sp

t = sp.Symbol('t')
x = t**2
y = t**3 - 3*t

dx_dt = sp.diff(x, t)
dy_dt = sp.diff(y, t)
dy_dx = sp.simplify(dy_dt / dx_dt)
print(f"dx/dt = {dx_dt}")
print(f"dy/dt = {dy_dt}")
print(f"dy/dx = {dy_dx}")

d2y_dx2 = sp.simplify(sp.diff(dy_dx, t) / dx_dt)
print(f"d2y/dx2 = {d2y_dx2}")

Finding horizontal and vertical tangents for the same curve:

import sympy as sp

t = sp.Symbol('t')
x = t**2
y = t**3 - 3*t
dx_dt = sp.diff(x, t)
dy_dt = sp.diff(y, t)

horiz = sp.solve(sp.Eq(dy_dt, 0), t)
vert = sp.solve(sp.Eq(dx_dt, 0), t)
print(f"horizontal tangents at t = {horiz}")
print(f"vertical tangents at t   = {vert}")

for tv in horiz:
    print(f"  t={tv}: point = ({x.subs(t, tv)}, {y.subs(t, tv)})")
for tv in vert:
    print(f"  t={tv}: point = ({x.subs(t, tv)}, {y.subs(t, tv)})")

Confirming the circle's implicit equation by eliminating t:

import sympy as sp

t, r = sp.symbols('t r', positive=True)
x = r * sp.cos(t)
y = r * sp.sin(t)

check = sp.simplify(x**2 + y**2 - r**2)
print(f"x^2 + y^2 - r^2 = {check}   (should be 0, confirming a circle)")

Worked example

Find \frac{dy}{dx} for x=t^2, y=t^3-3t, and identify all horizontal and vertical tangents.

\frac{dx}{dt}=2t,\qquad\frac{dy}{dt}=3t^2-3

\frac{dy}{dx}=\frac{3t^2-3}{2t}=\frac{3(t^2-1)}{2t}

Horizontal tangents: 3t^2-3=0\Rightarrow t=\pm1 (and dx/dt\ne0 there, since 2(\pm1)=\pm2\ne0). At t=1: (x,y)=(1,-2). At t=-1: (x,y)=(1,2).

Vertical tangent: 2t=0\Rightarrow t=0 (and dy/dt=-3\ne0 there). At t=0: (x,y)=(0,0).

Sanity check. Both horizontal tangents share x=1 — meaning the curve passes through x=1 twice, once rising and once falling, which is exactly the self-intersecting loop this curve is known for; a function y=f(x) could never do this, since it can only take one y value per x. The vertical tangent at the origin is the loop's leftmost point, consistent with x=t^2\ge0 always — the curve never crosses into x<0. ✓

Your turn

1. For x=t^3, y=t^2, find \frac{dy}{dx} in terms of t, and note where it's undefined.

2. For the circle x=3\cos t, y=3\sin t, find \frac{dy}{dx} and confirm it matches the slope of the tangent line to x^2+y^2=9 found via implicit differentiation (§2.9).

3. True or false: a point where both \frac{dx}{dt}=0 and \frac{dy}{dt}=0 simultaneously is guaranteed to be a smooth point of the curve.

Solutions

1.

\frac{dx}{dt}=3t^2,\qquad\frac{dy}{dt}=2t,\qquad\frac{dy}{dx}=\frac{2t}{3t^2}=\boxed{\frac2{3t}}

Undefined at t=0 — exactly where \frac{dx}{dt}=3t^2=0 too, so both derivatives vanish together there. (This point, (0,0), is in fact a cusp — foreshadowing problem 3.)

2.

\frac{dx}{dt}=-3\sin t,\qquad\frac{dy}{dt}=3\cos t,\qquad\frac{dy}{dx}=\frac{3\cos t}{-3\sin t}=-\cot t=-\frac{\cos t}{\sin t}=-\frac xy

Implicit differentiation on x^2+y^2=9: $2x+2y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac xy$ — identical. Two different derivations of the same tangent line, which must agree since they describe the same circle.

3. False. When both derivatives vanish at once, \frac{dy}{dx}=\frac00 is indeterminate — the formula simply fails to apply, and such points are frequently cusps (sharp corners, as in problem 1's (0,0), or the cycloid's cusps in §6.1) rather than smooth points. The condition \frac{dx}{dt}\ne0 or \frac{dy}{dt}\ne0 (at least one nonzero) is precisely what smoothness requires.

Check yourself in code

For x=t^2, y=t^3-3t, compute \frac{dy}{dx} symbolically, then evaluate it at t=2.

Print exactly this:

dy/dx = 3*(t**2 - 1)/(2*t)
dy/dx at t=2 = 9/4

import sympy as sp

t = sp.Symbol('t')
x = t**2
y = t**3 - 3*t

dx_dt = sp.diff(x, t)
dy_dt = sp.diff(y, t)
dy_dx = sp.simplify(dy_dt / dx_dt)
print("dy/dx = ...")
print("dy/dx at t=2 = ...")

import sympy as sp

t = sp.Symbol('t')
x = t**2
y = t**3 - 3*t

dx_dt = sp.diff(x, t)
dy_dt = sp.diff(y, t)
dy_dx = sp.simplify(dy_dt / dx_dt)
print(f"dy/dx = {dy_dx}")
print(f"dy/dx at t=2 = {dy_dx.subs(t, 2)}")

A parametric curve x=f(t), y=g(t) escapes the one-y-per-x restriction of ordinary functions, and its slope comes from dividing two ordinary derivatives via the chain rule: \frac{dy}{dx}=\frac{dy/dt}{dx/dt}. The second derivative needs that division applied twice — once for the slope, once more to convert the slope's own t-derivative into an x-derivative — and points where both \frac{dx}{dt} and \frac{dy}{dt} vanish together are exactly where the tangent-line formula breaks down, usually at a cusp.

Next: now that parametric curves have derivatives, they get the rest of this module's toolkit too — arc length and area, adapted from Cartesian form to handle a curve that isn't a function at all.