From Parameter Values to a Useful Sketch
A parametric curve is described by coordinate functions \(x=f(t)\) and \(y=g(t)\). For each allowed value of the parameter \(t\), the point \((x(t),y(t))\) lies on the curve. To sketch the curve, it is often more useful to choose a small set of informative parameter values than to calculate many arbitrary ones.
Good choices include values that make the coordinates easy to evaluate, values where a coordinate is zero, and values where a coordinate derivative is zero. For trigonometric parametrizations, standard angles often produce exact coordinates. The resulting points provide landmarks for a sketch; they are not, by themselves, a proof of every detail of the curve.
A Practical Point-Selection Method
First note the parameter interval. Then inspect both coordinate functions and select values that make them straightforward to compute. For a polynomial, try zeroes, simple integers, and values that make a factor vanish. For sine and cosine, use standard angles such as \(0\), \(\frac{\pi}{2}\), and \(\pi\). Include interval endpoints when they are part of the specified domain.
For each chosen value, calculate both coordinates and record the ordered pair. A table helps prevent a common mix-up: a parameter value is not itself a point on the curve. For instance, \(t=2\) produces the point \((x(2),y(2))\), not the point \((2,2)\) unless both coordinates happen to equal \(2\).
Record which parameter values are allowed, including whether interval endpoints are included.
Look for simple inputs, zeroes, symmetric values, trigonometric angles, and possible coordinate-derivative zeroes.
Evaluate \(x(t)\) and \(y(t)\) separately at every selected value.
Plot the points and use the formulas and any symmetry to guide a sketch between them.
Worked Example: A Sideways Parabola
Worked Example: Selecting Points for \(x=t^2,\ y=t\)
Sketch the curve for \(-2\leq t\leq 2\). The input values \(-2,-1,0,1,2\) make both coordinates easy to evaluate. Since \(x(t)=t^2\) is even and \(y(t)=t\) is odd, the points for opposite parameter values have the same \(x\)-coordinate and opposite \(y\)-coordinates.
For example, at \(t=-2\), \(x(-2)=(-2)^2=4\) and \(y(-2)=-2\), giving \((4,-2)\). At \(t=0\), both coordinates are zero. The other rows follow by substituting their listed values. The derivatives \(x'(t)=2t\) and \(y'(t)=1\) are also easy to evaluate; at \(t=0\), the first coordinate derivative is zero, so that parameter value is a particularly useful landmark.
Plot the five points. The negative and positive parameter values give reflected points across the \(x\)-axis, and the point \((0,0)\) lies on that axis. The coordinate formulas specify a sideways parabola over the stated parameter interval; the plotted points provide a useful sketch of that portion.
Check the entries by substitution: the point \((1,1)\) comes from \(t=1\), since \(1^2=1\) and \(y=1\); the point \((4,-2)\) comes from \(t=-2\), since \((-2)^2=4\) and \(y=-2\). The same substitutions in each row confirm all five pairs.
Why Symmetric Parameter Choices Help
Symmetry can reduce the amount of calculation needed and reveal how points relate. The following result gives a precise test for reflection across the \(x\)-axis.
To verify the result, evaluate both coordinate functions at \(-t\). The assumptions give the point \[ (x(-t),y(-t))=(x(t),-y(t)). \] The original point at \(t\) is \((x(t),y(t))\). Reflection across the \(x\)-axis keeps the first coordinate and changes the sign of the second, so the two points have exactly the required relationship. This proves the claim.
In the previous example, \(x(t)=t^2\) satisfies \(x(-t)=x(t)\), while \(y(t)=t\) satisfies \(y(-t)=-y(t)\). Thus calculating the points for positive parameter values also predicts the reflected points for negative values, provided those values belong to the parameter interval.
Worked Example: A Circle from Standard Angles
Worked Example: Sketching \(x=2\cos t,\ y=2\sin t\)
Use \(0\leq t\leq 2\pi\). Standard angles give exact sine and cosine values, so they produce convenient landmarks for a sketch.
For \(t=\frac{\pi}{4}\), \(\cos t=\sin t=\frac{\sqrt{2}}{2}\), so \(x=2\cdot\frac{\sqrt{2}}{2}=\sqrt{2}\) and \(y=2\cdot\frac{\sqrt{2}}{2}=\sqrt{2}\). For \(t=\frac{\pi}{2}\), the coordinates are \(2\cdot 0=0\) and \(2\cdot 1=2\). At \(t=\pi\), they are \(2(-1)=-2\) and \(2(0)=0\). The remaining coordinates follow from the corresponding standard-angle values.
The derivatives are \(x'(t)=-2\sin t\) and \(y'(t)=2\cos t\). For example, at \(t=\frac{\pi}{2}\), they equal \(-2\) and \(0\), respectively. These values are included as additional easy-to-check information, not as a method for finding the slope of the curve.
Plot the points at the top, right, bottom, and left of the coordinate plane, together with the additional point \((\sqrt{2},\sqrt{2})\). The parametrization describes a circle centered at the origin with radius \(2\); these points help place it accurately. At the two endpoints of the specified interval, direct substitution gives \((2,0)\) in both cases.
Check the coordinate calculations by substitution. At \(t=\pi\), \(x=2\cos\pi=-2\) and \(y=2\sin\pi=0\), matching \((-2,0)\). At \(t=2\pi\), \(x=2\cos(2\pi)=2\) and \(y=2\sin(2\pi)=0\), matching \((2,0)\). Substituting the other listed angles into \(2\cos t\) and \(2\sin t\) gives their table entries.
Using Coordinate Derivatives to Find a Landmark
For a differentiable coordinate function, its derivative can help identify a parameter value worth checking. In particular, an interior value where \(x'(t)=0\) or \(y'(t)=0\) is often useful to include in a point table. A zero derivative is a candidate for a local extreme value of that coordinate, but it does not by itself prove that an extreme occurs. The coordinate values and the surrounding behavior still need to be checked.
Worked Example: A Derivative-Selected Point on a Parabola
Worked Example: Sketching \(x=t,\ y=t^2-1\)
Consider \(-2\leq t\leq 2\). The derivative of the second coordinate is \(y'(t)=2t\), which is zero at \(t=0\). Since \(0\) is in the parameter interval, evaluate both coordinates there and also choose simple values on either side.
At \(t=0\), \(x(0)=0\) and \(y(0)=0^2-1=-1\), so the selected point is \((0,-1)\). At \(t=-2\), \(x=-2\) and \(y=(-2)^2-1=4-1=3\). At \(t=1\), \(x=1\) and \(y=1^2-1=0\). These substitutions illustrate how every pair in the table is obtained.
Because \(y=t^2-1\geq -1\) for every real \(t\), the point at \(t=0\) is the lowest point in this parametrized portion. The values \(t=-1\) and \(t=1\) give points at height zero, and the interval endpoints give \((-2,3)\) and \((2,3)\). Plotting these points produces a clear sketch of the relevant part of the parabola.
Check the key point by substituting \(t=0\) into the original coordinate functions: \((x(0),y(0))=(0,0^2-1)=(0,-1)\). The endpoint checks also agree: \(t=-2\) gives \((-2,3)\), and \(t=2\) gives \((2,3)\).
Common Pitfall: A Few Points Do Not Determine Every Detail
A point table is a sketching aid, not a substitute for reading the formulas and the parameter interval. For example, the circle example includes the point \((2,0)\) at both \(t=0\) and \(t=2\pi\). If you record only distinct points, you may miss that both parameter endpoints produce the same coordinate pair. The interval and parametrization are part of the description of the curve.
Another mistake is to treat a zero coordinate derivative as proof of a particular shape. Consider \(y=t^2-1\): \(y'(0)=0\) makes \(t=0\) worth checking, but the derivative equation alone does not tell you the coordinate value. You must substitute into the original functions to obtain \((0,-1)\). In addition, a derivative-zero value might not be in the allowed interval, and a zero derivative does not necessarily imply a maximum or minimum.
Check Your Understanding
For each question, use the parameter values and coordinate functions to justify your answer.
- For \(x=t^2,\ y=t\) on \(-2\leq t\leq 2\), what point corresponds to \(t=-1\), and what point corresponds to \(t=1\)?
- For \(x=3\cos t,\ y=3\sin t\), which parameter values among \(0,\frac{\pi}{2},\pi,\frac{3\pi}{2}\) give points on the coordinate axes?
- For \(x=t,\ y=t^2+2\), find the parameter value where \(y'(t)=0\), then calculate the corresponding point.
- Why is a value satisfying \(y'(t)=0\) a useful candidate to check but not, by itself, a proof of a local minimum or maximum?
- If \(x(-t)=x(t)\) and \(y(-t)=-y(t)\), how are the points corresponding to \(t\) and \(-t\) related?
