P4 3.1 - Parametric equations

P4 3.1 - Parametric equations

Describe a curve by giving both coordinates in terms of a third variable. Learn to move between parametric and Cartesian descriptions while keeping the allowed points, excluded values and direction of travel.

Coordinates from a parameter

Can a curve give two different heights at the same horizontal position? Let x=t2x=t^2 and y=ty=t. The values t=−1t=-1 and t=1t=1 give (1,−1)(1, -1) and (1,1)(1, 1): one horizontal coordinate, two points.

Parameter

A parameter is a third variable used to specify coordinates together. In x=f(t),y=g(t)x=f(t), y=g(t), each allowed tt gives the point (f(t),g(t))(f(t), g(t)). The parameter need not represent time, but increasing tt gives an order in which points are traced.

Substitute the same value into both coordinate equations. The coordinate pair is (x,y)(x, y), never (x,t)(x, t). A few points guide a sketch; the equations determine the intervening curve.

Worked example 1: paired coordinates

Question. Find the points on x=t+1,y=t2x=t+1, y=t^2 at t=−1,0,1t=-1, 0, 1, and describe their order as tt increases.

Plan. Evaluate both coordinates at each parameter value.

t−101x012y101\begin{array}{c|rrr} t&-1&0&1\\\hline x&0&1&2\\ y&1&0&1 \end{array}

Answer. The points are (0,1),(1,0),(2,1)(0, 1), (1, 0), (2, 1), in that order. Horizontal position increases throughout.

Check. Each pair satisfies y=(x−1)2y=(x-1)^2; this also checks that tt was not accidentally used as the horizontal coordinate.

Change only the power in the second coordinate, from t2t^2 to t3t^3. The horizontal positions stay fixed but one height changes sign.

Worked example 2: a changed vertical coordinate

Question. Find the points on x=t+1,y=t3x=t+1, y=t^3 at t=−1,0,1t=-1, 0, 1, in increasing parameter order.

Plan. Reuse the horizontal coordinates and calculate the new cubes.

t−101x012y−101\begin{array}{c|rrr} t&-1&0&1\\\hline x&0&1&2\\ y&-1&0&1 \end{array}

Answer. (0,−1),(1,0),(2,1)(0, -1), (1, 0), (2, 1), in that order.

Check. Substituting t=x−1t=x-1 gives y=(x−1)3y=(x-1)^3, satisfied by all three pairs.

Direction belongs to the parameter description: changing tt to −t-t in both coordinate equations can trace the same curve in reverse.

Watch out

Calculate a coordinate pair using one shared value of tt. Choosing t=1t=1 for xx and t=2t=2 for yy generally produces a point off the curve.

Use each supplied parameter value in both coordinate equations.

Eliminating the parameter

A Cartesian equation relates xx and yy directly. To eliminate a parameter, often rearrange the simpler coordinate equation to find tt, then substitute into the other.

Every rearrangement has conditions. A denominator introduced while solving for tt can identify a horizontal coordinate the curve never reaches. Keep that restriction alongside the final equation.

Worked example 3: a linear coordinate

Question. Eliminate tt from x=2t+1,y=1−t2x=2t+1, y=1-t^2, where tt is any real number.

Plan. Make tt the subject of the linear equation.

t=x−12,y=1−(x−12)2.t=\frac{x-1}{2}, \qquad y=1-\left(\frac{x-1}{2}\right)^2.

Every real xx gives an allowed real tt.

Answer. y=1−(x−1)2/4y=1-(x-1)^2/4, for all real xx.

Check. At t=2t=2, the original gives (5,−3)(5, -3). The Cartesian equation gives 1−16/4=−31-16/4=-3.

Change only the equation for xx to a rational expression. Collect all the terms containing tt before dividing.

Worked example 4: a rational coordinate

Question. Eliminate tt from x=(2+t)/(1−t),y=1−t2x=(2+t)/(1-t), y=1-t^2, where t≠1t\ne1.

Plan. Solve the rational equation for tt, then simplify yy to one fraction.

x(1−t)=2+tclear the denominatorx−2=t(x+1)collect parameter termst=x−2x+1so x≠−1.\begin{aligned} x(1-t)&=2+t &&\text{clear the denominator}\\ x-2&=t(x+1) &&\text{collect parameter terms}\\ t&=\frac{x-2}{x+1} &&\text{so }x\ne-1. \end{aligned}

Therefore

y=1−(x−2)2(x+1)2=(x+1)2−(x−2)2(x+1)2=6x−3(x+1)2.y=1-\frac{(x-2)^2}{(x+1)^2} =\frac{(x+1)^2-(x-2)^2}{(x+1)^2} =\frac{6x-3}{(x+1)^2}.

The value x=−1x=-1 is impossible in the original equation: it would require −1+t=2+t-1+t=2+t. For every other xx, the recovered tt is never 11.

Answer. y=(6x−3)/(x+1)2y=(6x-3)/(x+1)^2, with x≠−1x\ne-1.

Check. t=0t=0 gives (2,1)(2, 1); substituting x=2x=2 gives (12−3)/9=1(12-3)/9=1.

Finding the formula and finding its domain are connected parts of the conversion.

Watch out

An equation with an omitted restriction may describe extra points. Check both the original parameter range and any values excluded during rearrangement.

Retain the excluded horizontal value when writing the simplified equation.

Choosing a parameterisation

You can also build parametric equations from a Cartesian relation. There is usually more than one correct choice.

For a graph y=x2+2y=x^2+2, the simple choice x=t,y=t2+2x=t, y=t^2+2 works for all real tt. A circle or ellipse suggests cos⁡2t+sin⁡2t=1\cos^2t+\sin^2t=1. Scaling cosine and sine supplies the different horizontal and vertical radii.

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