Pure 3.3 - Parametric equations and Cartesian form

What a parameter does

Can the same circle equation describe a whole loop and only half a loop? Consider $x=2+5\cos t$, $y=-4+5\sin t$: at $t=0$ you get $(7, -4)$, while at $t=\pi$ you get $(-3, -4)$. The allowed inputs decide how far around you travel. We will make that distinction precise.

Begin without trigonometry. For $x=5t$, $y=3t^2$, the input $t=-2$ gives $(-10, 12)$, $t=0$ gives $(0, 0)$, and $t=2$ gives $(10, 12)$. Both coordinates use the same value of $t$.

A table shows separate coordinates; plotting each pair turns them into a curve.

[DIAGRAM: asset_slug: p1_3_3_understand_and_use_the_parametric_equations_of_curves_and_conversion_between_cartesian_and_parametric_forms__diagram_01; asset_name: One parameter generates two coordinates; caption: Increasing the same parameter traces successive points of a parabola; description: Parameter inputs minus 2, minus 1, 0, 1 and 2 beside the points generated by x equals 5t and y equals 3t squared, with direction arrows]

For $x=t-3$, $y=2t^2+1$, the values $t=1$ and $t=-2$ give $(-2, 3)$ and $(-5, 9)$. Combining the first point's $x$ with the second point's $y$ has no justification: one input must generate the whole pair.

There is a useful surprise: $x=5t$, $y=3t^2$ and $x=5u^3$, $y=3u^6$, with real parameters, trace the same parabola. Since $u^3$ takes every real value, writing $t=u^3$ changes the labelling and spacing of inputs, not the set of points. A curve does not have just one possible parametrisation.

Check that distinction before eliminating any parameter.

Eliminating a non-trigonometric parameter

To obtain a Cartesian equation, remove the parameter while keeping the relationship between the two coordinates. For $x=5t$, $y=3t^2$, write $t=x/5$, then $y=3(x/5)^2$, giving $25y=3x^2$.

Usually isolate the parameter in whichever equation makes it easiest. All conversions in this part are derived by substitution; no conversion recipe is supplied in the formula booklet.

Change only the $y$ rule by adding a reciprocal term. The excluded parameter value must now be carried into the answer.

The same substitution handles more algebraic expressions. For the useful form $x=3t+2$, $y=6t-1+4/t$, $t\ne0$, take $t=(x-2)/3$. Then

For the simple reciprocal pair $x=5t$, $y=5/t$, multiplication gives $xy=25$, with $x\ne0$.

Change just the reciprocal coefficient in the second example.

Trigonometric parameters and circles

For $x=3\cos t$, $y=3\sin t$, the familiar identity $\cos^2t+\sin^2t=1$ says $x^2+y^2=9$. The point stays exactly $3$ units from the origin.

Subtract each centre coordinate before using the identity. The translation matters just as much as the radius.

Change only the parameter interval, to half a turn. The Cartesian relationship stays the same, but its allowed points change.

A different translation works the same way: $x=-1+3\cos\theta$, $y=4+3\sin\theta$, for a full turn, gives $(x+1)^2+(y-4)^2=9$, centre $(-1, 4)$, radius $3$.

Change only the interval again. Use the sign of the appropriate trigonometric function.

Creating parametric forms

You can introduce a parameter as well as remove one. For a graph $y=f(x)$, choosing $x=t$ gives $y=f(t)$. Thus $y=2x^2+1$ can be written $x=t$, $y=2t^2+1$, $t\in\mathbb R$.

A different choice $x=3t$ gives $y=18t^2+1$. Both cover the whole parabola because both choices let $x$ take every real value. A valid parametrisation must satisfy the equation and reach the intended points; satisfaction alone is not enough.

Change only the curve's coefficient from $3$ to $4$. Keep the same parameter scale.

Other familiar shapes suggest other choices. For $xy=25$, take $x=5t$, $y=5/t$, $t\ne0$: their product is $25$, and each allowed $x$ has $t=x/5$. For $(x-2)^2+(y+4)^2=25$, take $x=2+5\cos t$, $y=-4+5\sin t$, for one full turn. No new formula is supplied in the booklet: these are constructions from the Cartesian equations.

Change the coefficient once more and check coverage of the whole curve.

The parameter domain matters

A curve's relationship and its allowed section are separate pieces of information. Return to the opening circle: one full turn reaches every point, while half a turn reaches only the upper semicircle.

[DIAGRAM: asset_slug: p1_3_3_understand_and_use_the_parametric_equations_of_curves_and_conversion_between_cartesian_and_parametric_forms__diagram_02; asset_name: Same equation different parameter domains; caption: A full turn gives a circle and a half turn gives its upper semicircle; description: Two equally scaled panels for x equals 2 plus 5 cos t and y equals minus 4 plus 5 sin t, comparing 0 to 2 pi with 0 to pi and labelled endpoints]

For a non-trigonometric example, $x=5t$, $y=3t^2$, $0\le t\le2$, gives $25y=3x^2$ only for $0\le x\le10$. Its height rises from $0$ to $12$. If instead $-1\le t\le2$, the correct horizontal restriction is $-5\le x\le10$. The unrestricted parabola would include many extra points.

Change only whether the upper parameter endpoint is included. That tiny symbol changes one endpoint of the arc.

Do not assume every coordinate is monotonic in the parameter. For $x=t^2$, $y=t$, $-1\le t\le2$, the $x$-range is $0\le x\le4$, not $1\le x\le4$: the minimum occurs at $t=0$. The exact curve is $x=y^2$, $-1\le y\le2$.

Change the lower endpoint in the decreasing-coordinate example; retain the strict upper endpoint.

Quick Recap

• Use the same parameter value in both coordinate rules.
• Eliminate by substitution or an appropriate identity, retaining all restrictions.
• A parametrisation must reach the intended points as well as satisfy their equation.
• Endpoints, exclusions and non-monotonic coordinate rules need explicit checks.

The opening equations give the whole circle for a full turn, but only $y\ge-4$ on that circle when $0\le t\le\pi$.