Parametric Curves & Calculus

Slope dy/dx, tangents, and arc length of a curve traced by x(t), y(t).

The idea

A parametric curve describes a path in the plane by giving both coordinates as functions of a parameter $t$, often thought of as time: $x = x(t), \qquad y = y(t).$ As $t$ runs over an interval, the moving point $(x(t), y(t))$ traces the curve. A graph $y = f(x)$ allows only one height above each $x$; a parametric curve has no such restriction, so it can be a circle, a spiral, or any path that doubles back on itself.

Over a short interval of the parameter, the point moves $dx = (dx/dt)\,dt$ across and $dy = (dy/dt)\,dt$ up. Dividing the second by the first cancels the common $dt$ and leaves the slope of the tangent, $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} \qquad (dx/dt \neq 0),$ so the tangent is horizontal where $dy/dt = 0$ and vertical where $dx/dt = 0$.

The same small step gives the arc length. Its components $dx$ and $dy$ are the legs of a right triangle, so the point travels the hypotenuse, $\sqrt{(dx/dt)^{2} + (dy/dt)^{2}}\,dt$ — its speed times the elapsed time. Adding these distances from $t = a$ to $t = b$ gives $L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^{2} + \left(\frac{dy}{dt}\right)^{2}}\,dt.$

Ways to work on it

• Walkthrough. Points, the tangent slope dy/dx, horizontal tangents, and the arc-length integrand.
• Practice. Compute the tangent slope dy/dx at a parameter value.
• Hardest. Exact parametric arc length of an exponential spiral.

Not sure where to start? Take the ten-question placement test.