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Pure · Coordinate geometry

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Parametric curves

Defining a curve by parametric equations x = f(t), y = g(t), finding the coordinates of points on the curve, finding the parameter value at a given point, and reading off the range of x- and y-values from the domain of t.

Pure · Coordinate geometry

Parametric curves

Defining a curve by parametric equations x = f(t), y = g(t), finding the coordinates of points on the curve, finding the parameter value at a given point, and reading off the range of x- and y-values from the domain of t.

Why it works

So far every curve has been a single equation linking xx and yy directly — a Cartesian equation like y=x2y = x^2. A parametric equation describes the same sort of curve a different way: both coordinates are given separately, each as a function of a third variable called the parameter, usually tt: x=f(t),y=g(t).x = f(t), \qquad y = g(t).

Think of tt as time. At each instant tt the two formulas hand you an xx-coordinate and a yy-coordinate, so they hand you a point. Let tt run through its values and the moving point traces out the curve. This is exactly how you would describe the path of a thrown ball — its horizontal and vertical positions are each their own function of time — which is why parametric equations are the natural language for motion (and why this chapter ends with modelling).

Finding a point. Substitute the value of tt into both formulas. If x=3tx = 3t and y=t2+1y = t^2 + 1, then at t=2t = 2 the point is (3×2, 22+1)=(6,5)(3\times 2,\ 2^2+1) = (6, 5).

Going backwards — finding tt at a point. You are given (x,y)(x, y) and must find the parameter. Use one coordinate to solve for tt, then check it in the other: the value must work for both, or the point is not on the curve. For x=3t, y=t2+1x = 3t,\ y = t^2+1 and the point (12,17)(12, 17): from x=12x = 12 we get t=4t = 4, and then y=42+1=17y = 4^2 + 1 = 17 matches — so t=4t = 4.

The domain of tt controls the curve. If the parameter is restricted (say −1≤t≤5-1 \le t \le 5, or t>0t > 0), only part of the curve is drawn. The set of xx-values it covers — the range of xx — comes from putting that domain through x=f(t)x = f(t); likewise for yy. Treat each coordinate as a function of tt and find its greatest and least values over the domain, watching for a turning point inside the interval, not just the endpoints.

The picture below is the curve x=t2, y=2tx = t^2,\ y = 2t — a parabola lying on its side, opening to the right. Notice the same curve a Cartesian equation would write as x=14y2x = \tfrac14 y^2.2468-6-4-2246xy