Three ways to sketch a parametric curve

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What a parameter describes

A parametric curve gives both coordinates in terms of another variable:

x=f(t),y=g(t),t∈I.x=f(t),\qquad y=g(t),\qquad t\in I.

The parameter tt describes where the point is on the curve. Such a curve need not be the graph of a single-valued function y=h(x)y=h(x), and its parameter interval is part of its definition.

There are three useful approaches to sketching it:

  1. Eliminate tt, retaining the restrictions inherited from its interval.
  2. Differentiate the coordinates with respect to tt and track their changes.
  3. Recognize a familiar curve after a rotation or another transformation.

1. Eliminate the parameter

Problem

Sketch the path of P=(x,y)P=(x,y) when

x=12t−1,y=t2−3t+1,−1≤t≤1.x=\frac12t-1,\qquad y=t^2-3t+1,\qquad -1\le t\le1.

From the first equation t=2(x+1)t=2(x+1). Substituting into both the second equation and the parameter restriction gives

y=4(x+1)2−6(x+1)+1=4x2+2x−1,y=4(x+1)^2-6(x+1)+1=4x^2+2x-1, −1≤2(x+1)≤1⟺−32≤x≤−12.-1\le2(x+1)\le1 \quad\Longleftrightarrow\quad -\frac32\le x\le-\frac12.

Thus the answer is the indicated closed segment of a parabola, with endpoints (−3/2,5)(-3/2,5) and (−1/2,−1)(-1/2,-1) included.

The parabola y=4x squared+2x-1 restricted to -3/2 at most x at most -1/2, with both endpoints included.

The full parabola would be an incorrect answer. Elimination without the parameter range loses information. Conversely, every xx in the displayed interval corresponds to an allowed t=2(x+1)t=2(x+1), so the restriction is sufficient as well as necessary.

2. Follow derivatives with respect to the parameter

Elimination may produce an equation that is harder to read than the original parametrization. In that case, study x(t)x(t) and y(t)y(t) directly.

Problem

For all real tt, sketch

x=32t2−12t,y=12t2+32t.x=\frac{\sqrt3}{2}t^2-\frac12t, \qquad y=\frac12t^2+\frac{\sqrt3}{2}t.

The coordinate derivatives are

x′(t)=3t−12,y′(t)=t+32.x'(t)=\sqrt3t-\frac12,\qquad y'(t)=t+\frac{\sqrt3}{2}.

The important parameter values are t=−3/2t=-\sqrt3/2 and t=3/6t=\sqrt3/6:

  • Before −3/2-\sqrt3/2, both coordinates decrease.
  • Between −3/2-\sqrt3/2 and 3/6\sqrt3/6, xx decreases while yy increases.
  • After 3/6\sqrt3/6, both coordinates increase.

At these turning points,

P(−32)=(538,−38),P(36)=(−324,724).P\left(-\frac{\sqrt3}{2}\right)=\left(\frac{5\sqrt3}{8},-\frac38\right), \qquad P\left(\frac{\sqrt3}{6}\right)=\left(-\frac{\sqrt3}{24},\frac7{24}\right).

Also P(0)=(0,0)P(0)=(0,0), and both coordinates tend to +∞+\infty as t→±∞t\to\pm\infty. Where x′(t)≠0x'(t)\ne0,

dydx=t+3/23t−1/2,d2ydx2=−2(3t−1/2)3.\frac{dy}{dx}=\frac{t+\sqrt3/2}{\sqrt3t-1/2}, \qquad \frac{d^2y}{dx^2}=-\frac2{(\sqrt3t-1/2)^3}.

There is a horizontal tangent at the first turning point and a vertical tangent at the second; at each point the other coordinate derivative is nonzero. These observations give the following sketch.

A rotated parabola passing through the origin, with a horizontal tangent below the x-axis and a vertical tangent to the left of the y-axis.

This method is broadly useful, but a reliable sketch can also require checking self-intersections, endpoints, asymptotes, or points where both derivatives vanish. Coordinate monotonicity alone does not settle every possible curve.

3. Recognize a rotation

For the same example, group the coefficients as vectors:

(xy)=t2(3/21/2)+t(−1/23/2).\begin{pmatrix}x\\y\end{pmatrix} =t^2\begin{pmatrix}\sqrt3/2\\1/2\end{pmatrix} +t\begin{pmatrix}-1/2\\\sqrt3/2\end{pmatrix}.

The two coefficient vectors are the standard coordinate directions rotated counterclockwise through π/6\pi/6. Equivalently,

x+iy=(t2+it)(32+i2).x+iy=(t^2+it)\left(\frac{\sqrt3}{2}+\frac i2\right).

The point (t2,t)(t^2,t) describes the parabola X=Y2X=Y^2 for all real tt. Multiplication by cos⁡(π/6)+isin⁡(π/6)\cos(\pi/6)+i\sin(\pi/6) rotates it by π/6\pi/6 about the origin. Thus the curve above is exactly that rotated parabola, rather than merely a shape that resembles one.

In matrix notation the same statement is

(xy)=(cos⁡(π/6)−sin⁡(π/6)sin⁡(π/6)cos⁡(π/6))(t2t).\begin{pmatrix}x\\y\end{pmatrix} =\begin{pmatrix}\cos(\pi/6)&-\sin(\pi/6)\\\sin(\pi/6)&\cos(\pi/6)\end{pmatrix} \begin{pmatrix}t^2\\t\end{pmatrix}.

You can also recover the unrotated coordinates:

X=3x+y2,Y=−x+3y2,X=Y2.X=\frac{\sqrt3x+y}{2},\qquad Y=\frac{-x+\sqrt3y}{2},\qquad X=Y^2.

This verifies the geometry without relying on a visual guess. Recognizing a transformation is especially efficient when the coefficient vectors reveal it; otherwise, direct differentiation is often the more straightforward starting point.

For subsequent applications, see areas enclosed by parametric curves and volumes of their solids of revolution.

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About the author

Koki Saito

Physician and graduate of the University of Tokyo Faculty of Medicine. Drawing on his entrance-exam experience, he explains how to choose mathematical methods and construct complete solutions.

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