4.7 Parametrization of Implicitly Defined Functions

Parametrizing an implicitly defined function means rewriting a curve like $x^2 + y^2 = 1$ as a pair of equations $x(t)$ and $y(t)$ that both depend on a parameter $t$. A correct parametrization, when you substitute $x(t)$ and $y(t)$ back into the original equation, satisfies that equation for every $t$ in the domain.

Why This Matters for the AP Precalculus Exam

Unit 4 topics, including this one, are not assessed on the AP Precalculus Exam. The exam covers material from Units 1, 2, and 3. Whether you study Unit 4 depends on your school's course plan.

That said, parametrizing implicitly defined functions is still worth learning. It pulls together skills from earlier in Unit 4 (parametric functions, implicitly defined functions, and conic sections) and builds reasoning you will use in calculus, physics, and computer graphics. If your class covers this topic, you may use a graphing calculator to graph parametric curves, so practice setting viewing windows and parameter restrictions.

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Key Takeaways

• A parametrization writes a curve as $(x(t), y(t))$, where both coordinates depend on a parameter $t$.
• A parametrization is correct only if substituting $x(t)$ and $y(t)$ into the original equation makes it true for every $t$ in the domain.
• For $y = f(x)$, the simplest parametrization is $(x(t), y(t)) = (t, f(t))$. If $f$ is invertible, its inverse can be written as $(f(t), t)$.
• A parabola is parametrized by solving for one variable: solve for $y$ to get $(t, f(t))$, or solve for $x$ to get $(f(t), t)$.
• An ellipse centered at $(h, k)$ is parametrized by $x(t) = h + a\cos t$ and $y(t) = k + b\sin t$ for $0 \le t \le 2\pi$.
• A hyperbola uses secant and tangent: which one goes with $x$ versus $y$ depends on whether it opens left-right or up-down.

Representing Curves in the Plane Parametrically

A parametrization expresses a curve as a set of equations $x(t)$ and $y(t)$ that both depend on a parameter $t$. When you substitute these into the equation that defines the curve implicitly, they should satisfy the equation for every value of $t$ in the domain.

For example, the equation $x^2 + y^2 = 1$ describes the unit circle centered at the origin. A parametrization of this circle is $(x(t), y(t)) = (\cos t, \sin t)$ with domain $0 \le t \le 2\pi$. Substituting into the implicit equation gives $\cos^2 t + \sin^2 t = 1$, which is always true.

The key point is that a parametrization gives a specific way to write the curve, in terms of a set of equations, that satisfies the implicit equation for every value of the parameter in the domain.

Functions and Their Inverses

If $f$ is a function of $x$ and $y = f(x)$ is the graph of $f$, then the curve can be parametrized as $(x(t), y(t)) = (t, f(t))$, where $x(t) = t$ and $y(t) = f(t)$. You can picture this as moving along the x-axis: at each input $t$, you plot the point $(t, f(t))$ on the graph of $f$.

When $f$ is invertible, it has an inverse function $f^{-1}$. The inverse can be parametrized as $(x(t), y(t)) = (f(t), t)$ for an appropriate interval of $t$. Notice that the roles of the two coordinates are swapped.

One detail to watch: the domain of $t$ for the inverse parametrization comes from the range of the original function.

Representing Conic Sections Parametrically

In Topic 4.6, you saw conic sections, the shapes created by intersecting a plane and a cone, such as circles, ellipses, parabolas, and hyperbolas. Each type has its own standard equation and properties. Here is how to parametrize each one.

Parabolas

A parabola, like any equation involving two variables, can be written in parametric form. The approach is to solve for one variable and let the parameter stand in for the other.

If the equation can be solved for $x$, parametrize it as $(x(t), y(t)) = (f(t), t)$, where $x$ is expressed in terms of $t$ and $y$ is just $t$.

If the equation can be solved for $y$, parametrize it as $(x(t), y(t)) = (t, f(t))$, where $y$ is expressed in terms of $t$ and $x$ is just $t$.

Either way, the parabola is represented as a set of points $(x(t), y(t))$ that satisfy the original equation for different values of $t$ in the domain.

Ellipses

An ellipse centered at $(h, k)$ can be parametrized by:

$x(t) = h + a\cos t$

$y(t) = k + b\sin t$

where $a$ is the horizontal radius and $b$ is the vertical radius. The domain is $0 \le t \le 2\pi$, which traces a complete counterclockwise revolution. These trigonometric equations let you express the position of a point on the ellipse at any angle $t$.

Hyperbolas

A hyperbola can also be parametrized using trigonometric functions, but the setup depends on the direction it opens.

For a hyperbola centered at $(h, k)$ that opens left and right, use:

$x(t) = h + a\sec t$

$y(t) = k + b\tan t$

for $0 \le t \le 2\pi$.

For a hyperbola centered at $(h, k)$ that opens up and down, use:

$x(t) = h + a\tan t$

$y(t) = k + b\sec t$

for $0 \le t \le 2\pi$.

Remember that $\sec t = \frac{1}{\cos t}$ and that $\tan t = \frac{\sin t}{\cos t}$, so these parametrizations are undefined wherever $\cos t = 0$.

How to Use This on the AP Precalculus Exam

Unit 4 is not tested on the AP Precalculus Exam, so treat the steps below as practice for class assessments and as preparation for calculus.

Problem Solving

• To parametrize $y = f(x)$, set $x(t) = t$ and $y(t) = f(t)$. To parametrize an invertible function's inverse, swap to $(f(t), t)$.
• For a conic, first identify the center $(h, k)$ and the values of $a$ and $b$ from the standard equation, then plug into the right trig template.
• Match the hyperbola template to its orientation. Left-right opening pairs $\sec t$ with $x$ and $\tan t$ with $y$; up-down opening pairs $\tan t$ with $x$ and $\sec t$ with $y$.

How to Check Your Work

• Substitute $x(t)$ and $y(t)$ back into the original equation. A correct parametrization makes the equation true for every $t$ in the domain.
• For the ellipse, the identity $\cos^2 t + \sin^2 t = 1$ confirms the parametrization. For the hyperbola, $\sec^2 t - \tan^2 t = 1$ does the same job.
• Confirm the parameter interval covers the full curve and does not include values where the functions are undefined.

Technology

• When graphing a parametric curve, set an appropriate viewing window and a parameter range for $t$. Without the right interval, you may see only part of the curve or an unexpected direction of travel.

Common Misconceptions

• A parametrization is not unique. The same curve can be written with different equations and can even be traced in different directions, so your answer may look different from a classmate's and still be correct.
• Substituting back into the original equation is what proves a parametrization works. Just having two equations that depend on $t$ does not guarantee they trace the right curve.
• For an inverse function, the parameter domain comes from the range of the original function, not its domain. Reusing the original domain can produce the wrong piece of the curve.
• The hyperbola templates are easy to mix up. The trig pairing depends on whether the hyperbola opens left-right or up-down, so check the orientation before you write $\sec t$ or $\tan t$.
• Secant and tangent are undefined where $\cos t = 0$, so a hyperbola parametrization has gaps in $t$. Treating the interval as fully continuous will give you incorrect points.

Related AP Precalculus Guides

• 4.10 Matrices
• 4.12 Linear Transformations and Matrices
• 4.5 Implicitly Defined Functions
• 4.9 Vector-Valued Functions
• 4.13 Matrices as Functions
• 4.4 Parametrically Defined Circles and Lines