7.2 Calculus of Parametric Curves

Big idea

A curve described by a pair of coordinate functions still has tangent lines, still bends one way or the other, and still has a length. The question is how to compute those things when there is no formula for $y$ in terms of $x$ to differentiate. The answer is that you differentiate what you have, namely each coordinate with respect to the parameter, and then assemble the geometric quantity from those two rates.

The slope is the cleanest case. The chain rule says $dy/dt = (dy/dx)(dx/dt)$, so dividing by $dx/dt$ isolates the slope wherever that denominator is nonzero. Nothing has been assumed about solving for $y$; the formula extracts the slope from the motion. It also explains the two special cases immediately. If the vertical rate vanishes and the horizontal one does not, the tangent is horizontal. If the horizontal rate vanishes and the vertical one does not, the tangent is vertical, which is a situation an equation of the form $y = f(x)$ could never produce.

Concavity takes one extra step and one common error. The second derivative with respect to $x$ is not the ratio of the two second derivatives with respect to $t$. It is the derivative of the slope with respect to $x$, and since the slope is naturally a function of the parameter, you differentiate it with respect to $t$ and divide by $dx/dt$ again, exactly as the chain rule requires.

Length and surface area come from a different idea. Over a short interval of the parameter the point moves almost in a straight line, and the Pythagorean theorem turns the two coordinate changes into a distance. Adding those distances and passing to a limit gives an integral of the speed, which is the size of the velocity built from the two rates. That single integrand does double duty: multiply it by the circumference swept by a point at distance $y$ from the axis, and you get the surface area of a solid of revolution.

Decoder

The slope of a parametric curve is the rate of the vertical coordinate divided by the rate of the horizontal coordinate, and the arc length is the integral of the square root of the sum of their squares.

The division in the first half is doing the same job that cancelling $dt$ would do if these were fractions. They are not fractions, but the chain rule makes the cancellation legitimate. The condition attached is that the denominator is nonzero, and the condition is not a technicality: the points where $dx/dt = 0$ are precisely where the curve is turning around horizontally, which is where vertical tangents and cusps live.

The square root in the second half is a speed. If you read $t$ as time, then $dx/dt$ and $dy/dt$ are the components of velocity and the root is its magnitude. Distance is speed times time, so integrating the speed over the parameter interval gives the distance travelled. That phrasing also warns you about the standard trap: if the parametrization covers the curve twice, the integral returns twice the length, because the point really did travel that far.

The error worth naming is computing the second derivative as $\frac{d^2y/dt^2}{d^2x/dt^2}$. That quotient has no geometric meaning. The correct procedure differentiates the already-computed slope with respect to $t$ and then divides by $dx/dt$, so the denominator is the first derivative of $x$, never the second.

Definitions and results

Smooth curve. A parametrization is smooth on an interval if $x'(t)$ and $y'(t)$ are continuous there and never simultaneously zero. Smoothness is what rules out corners produced by the parametrization stalling.

First derivative. Where $x'(t) \neq 0$,

$$ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{y'(t)}{x'(t)} $$

Horizontal and vertical tangents. The tangent is horizontal at parameter values where $y'(t) = 0$ and $x'(t) \neq 0$, and vertical where $x'(t) = 0$ and $y'(t) \neq 0$. When both vanish the point needs separate examination; a cusp is the usual outcome.

Second derivative. Treat the slope as a function of $t$ and apply the same rule again:

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

Positive values mean the curve is concave up as a graph in the plane, negative values concave down.

Arc length. For a smooth curve traced once as $t$ runs from $a$ to $b$,

$$ L = \int_a^b \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2}\; dt $$

Consistency with the graph formula. Taking $x = t$ and $y = f(t)$ reduces the integrand to $\sqrt{1 + [f'(t)]^2}$, which is the arc length formula for an ordinary graph. The parametric version contains the familiar one as a special case.

Traced once. The formula measures distance travelled. If the parameter interval covers the curve more than once, restrict it to a single pass before integrating.

Surface area of revolution. Revolving a smooth curve with $y \ge 0$ about the $x$-axis produces a surface of area

$$ S = 2\pi \int_a^b y(t)\sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2}\; dt $$

Revolving about the $y$-axis instead, with $x \ge 0$, replaces the leading $y(t)$ by $x(t)$. In both cases the factor in front of the speed is the distance from the axis, which is what the circumference $2\pi r$ needs.

Worked examples

Horizontal and vertical tangents on a looping curve

Consider $x = t^2$, $y = t^3 - 3t$.

The two rates are $x'(t) = 2t$ and $y'(t) = 3t^2 - 3 = 3(t^2-1)$, so

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

Horizontal tangents need $y'(t) = 0$ with $x'(t) \neq 0$, which happens at $t = 1$ and $t = -1$. Both give $x = 1$, and the $y$ values are $1 - 3 = -2$ and $-1 + 3 = 2$. So the curve has horizontal tangents at $(1,-2)$ and $(1,2)$, one on each branch.

Vertical tangents need $x'(t) = 0$, which happens only at $t = 0$. There $y'(0) = -3$, which is nonzero, so the tangent at the origin is vertical. The curve reaches $(0,0)$, stops moving horizontally, and reverses: this is a cusp, and it is the kind of feature that exists only because both coordinates are free to have their own rates.

Concavity on the same curve

Write the slope as a function of $t$ in a form that is easy to differentiate:

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

Differentiate with respect to $t$: $\frac{3}{2} + \frac{3}{2}t^{-2}$. Now divide by $x'(t) = 2t$:

$$ \frac{d^2y}{dx^2} = \frac{\frac{3}{2} + \frac{3}{2t^2}}{2t} = \frac{3t^2 + 3}{4t^3} = \frac{3(t^2+1)}{4t^3} $$

The numerator is always positive, so the sign is the sign of $t^3$, that is the sign of $t$. The curve is concave up for $t > 0$ and concave down for $t < 0$.

Check this against the tangents just found. At $t = 1$ the second derivative is $6/4 = 3/2 > 0$, and a horizontal tangent with concave-up behaviour is a low point, matching $(1,-2)$ as the bottom of the lower branch. At $t = -1$ it is $-3/2 < 0$, and the horizontal tangent at $(1,2)$ is a high point. Both agree with the shape.

Arc length of an arc

Find the length of $x = t^2$, $y = t^3$ for $0 \le t \le 1$.

The rates are $x'(t) = 2t$ and $y'(t) = 3t^2$, so the speed is

$$ \sqrt{4t^2 + 9t^4} = \sqrt{t^2(4 + 9t^2)} = t\sqrt{4+9t^2} $$

using $t \ge 0$ to drop the absolute value. Substitute $u = 4 + 9t^2$, so $du = 18t\,dt$ and $t\,dt = du/18$:

$$ L = \int_0^1 t\sqrt{4+9t^2}\;dt = \frac{1}{18}\int_{4}^{13} \sqrt{u}\;du = \frac{1}{27}\Big[ u^{3/2} \Big]_4^{13} = \frac{13\sqrt{13} - 8}{27} $$

Numerically that is about $1.4397$. Sanity check the size: the curve runs from $(0,0)$ to $(1,1)$, and the straight-line distance between those points is $\sqrt{2} \approx 1.4142$. The arc is slightly longer than the chord, as it must be.

A second check on the formula itself: for $x = 5\cos t$, $y = 5\sin t$ on $[0, 2\pi]$ the speed is $\sqrt{25\sin^2 t + 25\cos^2 t} = 5$, and the integral gives $5(2\pi) = 10\pi$, the circumference of a circle of radius $5$.

Surface area of a sphere

Revolve the upper half of the circle of radius $2$ about the $x$-axis. Parametrize it as $x = 2\cos t$, $y = 2\sin t$ for $0 \le t \le \pi$, which keeps $y \ge 0$.

The rates are $x'(t) = -2\sin t$ and $y'(t) = 2\cos t$, so the speed is $\sqrt{4\sin^2 t + 4\cos^2 t} = 2$, constant. Then

$$ S = 2\pi \int_0^{\pi} (2\sin t)(2)\;dt = 8\pi \int_0^{\pi} \sin t\;dt = 8\pi \Big[ -\cos t \Big]_0^{\pi} = 8\pi (1 + 1) = 16\pi $$

The standard surface area of a sphere of radius $r$ is $4\pi r^2$, which for $r = 2$ is $16\pi$. The formula reproduces a known result, which is the strongest kind of check available.

Practice

Start with derivatives. These drills ask for $dy/dx$ at a parameter value, for the points with horizontal or vertical tangents, and for the second derivative and the concavity it reports.

Then lengths. Set up the speed first, simplify under the radical before integrating, and confirm that the parameter interval covers the curve exactly once.

Quiz

Five items on parametric slopes, second derivatives, tangent lines, and arc length.