Parametric Calculus

Keep the interval with the equations

The equations

describe an arc from $(0,0)$ to $(1,1)$ in the first quadrant. Eliminating the parameter gives $y^2=x^3$, which by itself includes another branch and extends beyond the specified segment.

The interval, endpoints and direction are part of the parametrisation. Evaluate representative points and check for retracing before integrating. Parametric Equations provides the coordinate foundation.

The derivative vector $(x’,y’)$ describes change with respect to the parameter. When the parameter is not time, calling this velocity does not automatically give it a physical interpretation.

Substitute into an oriented integral

When $x(t)$ increases and the curve is a graph over the corresponding interval,

For the first example,

Because $x$ increases and $y\ge0$, this signed integral equals the area under the arc. Integrating $y(t)$ without the factor $x'(t)$ would answer a different question.

Direction affects the sign

The upper semicircle of radius two has parametrisation

It moves right to left, so

The geometric area is $2\pi$, while the oriented integral is negative. Reversing traversal reverses the signed integral.

Taking an absolute value at the end is not a universal area method. Folding, axis crossings and retracing can cause cancellation. Split into pieces whose geometric roles are understood. The parametric-calculus reference develops the substitution framework.

Derive the arc-length formula

A short chord has approximate components $x'(t)\Delta t$ and $y'(t)\Delta t$. Its length is approximately $\sqrt{x’^2+y’^2}|\Delta t|$. For a continuously differentiable curve,

This nonnegative integrand is speed with respect to the increasing parameter. Piecewise-smooth curves can be handled by summing lengths. A zero derivative at an isolated parameter does not automatically invalidate the length integral, although formulas that divide by speed need additional care.

Calculate a nonconstant-speed length

For $x=t^2$, $y=t^3$ on $[0,1]$,

The simplification uses $t\ge0$. On an interval containing negative values, $\sqrt{t^2}=|t|$ must be retained. Set $u=4+9t^2$, giving $du=18t\,dt$:

This exceeds the endpoint distance $\sqrt2$. A chord does not measure the length of a bent path. For the semicircle, speed is constantly two and length is $2\pi$, independent of traversal direction.

Reparameterise without changing a once-traced path

Replace $t$ by $u^2$, with $0\le u\le1$. Then $x=u^4$ and $y=u^6$ trace the same first-quadrant arc in the same order, at a different speed.

The oriented integral becomes $\int_0^1u^6(4u^3)\,du=2/5$. The length is

A suitable monotone change of parameter preserves length. Reversing orientation reverses $\int y\,dx$, but not length. The absolute change-of-parameter factor in a length calculation must not be discarded.

Distinguish reparameterisation from repeated tracing

The equations $x=\cos t$, $y=\sin t$ on $[0,4\pi]$ trace the unit circle twice. Their length integral is $4\pi$, not its geometric circumference $2\pi$.

A path that travels out and back along a segment has nonzero length but zero displacement and potentially zero net oriented integral. Repeated tracing is particularly easy to miss when a polar description is converted into coordinate functions.

Before reporting a geometric area or length, identify how many times the parameter interval covers each relevant portion. A correct integral can answer a different tracing question from the one intended.

Investigate exceptional parameter values

Where $x’\ne0$, the chain rule gives $dy/dx=y’/x’$. A vertical tangent can occur with $x’=0$, $y’\ne0$. If both vanish, the curve might have a cusp, a stationary tracing of a smooth path or another special behaviour.

For $x=t^2$, $y=t^3$, the velocity vanishes at zero. Yet the one-sided tangent on the positive arc can be examined from $y/x=t$. An undefined quotient at one parameter is a prompt for local analysis, not proof that every geometric quantity is meaningless.

The companion lesson develops slopes and concavity without overloading the integral calculations here.

Choose parametric form for the information it retains

A closed path, vertical tangent or spatial motion can be simpler parametrically than as $y=f(x)$. The representation treats coordinates symmetrically and preserves chronology.

When using a parametrisation, state the interval, preserve restrictions, check orientation and distinguish tracing from geometric size. These decisions come before numerical quadrature just as they come before finding an exact antiderivative.

Related topics

Derivatives of Parametric Curves covers slopes and singular parameters. Polar Calculus and Vector-Valued Functions extend this framework to radial and spatial descriptions.