Porisma

Worked solution

Sketch two parameterised curves with their full allowed ranges

Sketch (a) x=t2x=t^2, y=2ty=2t, t∈Rt\in\mathbb R; (b) x=cos⁡2tx=\cos^2t, y=sin⁡3ty=\sin^3t, 0≤t≤π/20\le t\le\pi/2.

CJC · Lecture · Example 2

(a) Locate the intercept and both branches

At x=0x=0, t=0t=0 and y=0y=0; at y=0y=0, again t=0t=0. Eliminating t gives y2=4xy^2=4x, a right-opening parabola with vertex (0,0)(0,0) and both positive and negative y branches. As t increases, it runs along the lower branch to the origin and then along the upper branch. The source calculator example uses −2π≤t≤2π-2\pi\le t\le2\pi to display a finite representative portion, not the complete infinite parabola.

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(b) Preserve the quarter-interval endpoints

At x=0x=0, t=π/2t=\pi/2, giving (0,1)(0,1); at y=0y=0, t=0t=0, giving (1,0)(1,0). Since both sine and cosine are nonnegative on the interval, y=(1−x)3/2y=(1-x)^{3/2} for 0≤x≤10\le x\le1. The curve runs from (1,0)(1,0) to (0,1)(0,1) as t increases. It has horizontal tangent at (1,0)(1,0) and limiting slope −3/2-3/2 at (0,1)(0,1); the source is a schematic, not a claim of a vertical tangent there.

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Two exact plots: the full-parabola representative portion y squared=4x with both branches and increasing-t direction, and y=(1−x) to power 3/2 on 0≤x≤1 from (1,0) to (0,1), with both endpoints included.
Two exact plots: the full-parabola representative portion y squared=4x with both branches and increasing-t direction, and y=(1−x) to power 3/2 on 0≤x≤1 from (1,0) to (0,1), with both endpoints included.
Two exact plots: the full-parabola representative portion y squared=4x with both branches and increasing-t direction, and y=(1−x) to power 3/2 on 0≤x≤1 from (1,0) to (0,1), with both endpoints included.

Common mistakes

A finite calculator window cannot establish an unrestricted real-parameter curve is bounded.

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