Parametrization of Implicitly Defined Functions - AP PreCalculus Study Guide

The Core Idea: Parametrization of Implicitly Defined Functions

In mathematics, we often encounter relations that are defined "implicitly," where the $x$ and $y$ variables are intertwined in a single equation, such as the equation of a circle $x^2+y^2=9$. These relations are not always functions (a circle, for instance, fails the vertical line test). Parametrization provides a powerful method to describe the coordinates $(x,y)$ of a point on the curve by expressing both $x$ and $y$ as separate functions of a third, independent variable called a parameter, usually denoted by $t$.

This topic focuses on how to convert implicitly defined relations, specifically circles and ellipses, into a pair of parametric equations, $x(t)$ and $y(t)$. By doing so, we can trace the path of a point along the curve as the parameter $t$ changes. This technique also extends to standard functions of the form $y=f(x)$, allowing us to represent any function in a parametric form. The core of this process for circles and ellipses relies on the fundamental properties of trigonometric functions, sine and cosine, to satisfy the implicit equations.

Key Formulas: Standard Parametrizations

The following are the standard parametrization formulas derived directly from the essential knowledge for this topic.

Function Parametrization

For any function defined explicitly as $y=f(x)$, a straightforward parametrization is given by:

• $x(t)=t$

• $y(t)=f(t)$

In this construction, the parameter $t$ simply takes the place of the independent variable $x$.

Circle Parametrization

For a circle centered at the origin $(0,0)$ with radius $r$, defined by the implicit equation $x^2+y^2=r^2$, the standard parametrization is:

• $x(t)=r\cos (t)$

• $y(t)=r\sin (t)$

To trace the entire circle exactly once, the parameter $t$ typically ranges over the interval $0\le t<2\pi$.

Ellipse Parametrization

For an ellipse centered at $(h,k)$ with a horizontal semi-axis of length $a$ and a vertical semi-axis of length $b$, defined by the implicit equation $\frac{(x-h{)}^2}{a^2}+\frac{(y-k{)}^2}{b^2}=1$, the standard parametrization is:

• $x(t)=h+a\cos (t)$

• $y(t)=k+b\sin (t)$

Similar to the circle, the parameter interval $0\le t<2\pi$ is used to trace the entire ellipse once.

Understanding the Pythagorean Connection

The reason sine and cosine are central to parametrizing circles and ellipses is the Pythagorean identity: ${\cos }^2(t)+{\sin }^2(t)=1$. This identity is the algebraic key that allows us to satisfy the implicit equations for these shapes.

Consider the parametrization for a circle, $x(t)=r\cos (t)$ and $y(t)=r\sin (t)$. If we substitute these expressions back into the implicit equation $x^2+y^2=r^2$, we can verify that the equation holds true for any value of t:

$$(r\cos (t){)}^2+(r\sin (t){)}^2=r^2$$

$$r^2{\cos }^2(t)+r^2{\sin }^2(t)=r^2$$

$$r^2({\cos }^2(t)+{\sin }^2(t))=r^2$$

$$r^2(1)=r^2$$

$$r^2=r^2$$

This confirms that any point $(r\cos (t),r\sin (t))$ lies on the circle of radius r.

Similarly, for the ellipse parametrization $x(t)=h+a\cos (t)$ and $y(t)=k+b\sin (t)$, we can rearrange to isolate the trigonometric parts:

$$x-h=a\cos (t)⟹\frac{x-h}{a}=\cos (t)$$

$$y-k=b\sin (t)⟹\frac{y-k}{b}=\sin (t)$$

Substituting these into the Pythagorean identity ${\cos }^2(t)+{\sin }^2(t)=1$ gives us the original implicit equation for the ellipse:

$${\left(\frac{x-h}{a}\right)}^2+{\left(\frac{y-k}{b}\right)}^2=1$$

$$\frac{(x-h{)}^2}{a^2}+\frac{(y-k{)}^2}{b^2}=1$$

This demonstrates that the trigonometric parametrization is a valid and direct method for representing these specific implicitly defined relations.

Core Concepts & Rules

• Parametrization of Relations: Any implicitly defined relation can be expressed using a set of parametric equations, where both $x$ and $y$ are functions of a parameter t`. - **Simple Function Parametrization:** The most direct way to parametrize a function $y = f(x) is to set the parameter $t$ equal to $x$, resulting in the parametric equations $x(t)=t$ and $y(t)=f(t)$.

• Standard Circle Parametrization: A circle with the equation $x^2+y^2=r^2$ is parametrized by $x(t)=r\cos (t)$ and $y(t)=r\sin (t)$. The value $r$ is the radius of the circle.

• Standard Ellipse Parametrization: An ellipse with the equation $\frac{(x-h{)}^2}{a^2}+\frac{(y-k{)}^2}{b^2}=1$ is parametrized by $x(t)=h+a\cos (t)$ and $y(t)=k+b\sin (t)$. The point $(h,k)$ is the center, $a$ is the horizontal semi-axis length, and $b$ is the vertical semi-axis length.

• Parameter Interval for Closed Curves: For both the standard circle and ellipse parametrizations, the interval $0\le t<2\pi$ is used to generate the complete curve exactly one time.

Step-by-Step Example 1: Parametrizing a Basic Circle

Problem: Determine a parametrization for the circle defined by the equation $x^2+y^2=81$.

Step 1: Identify the type of relation and its key parameters.

The equation $x^2+y^2=81$ is in the form $x^2+y^2=r^2$, which represents a circle centered at the origin $(0,0)$.

We can identify $r^2=81$. Taking the square root gives the radius, $r=9$.

Step 2: Apply the standard parametrization formula for a circle.

The formula is given by:

$x(t)=r\cos (t)$

$y(t)=r\sin (t)$

Step 3: Substitute the specific parameter value.

Substitute the radius $r=9$ into the formulas:

$x(t)=9\cos (t)$

$y(t)=9\sin (t)$

Step 4: State the parameter interval for a full curve.

To trace the entire circle once, we use the standard interval for the parameter $t$:

$0\le t<2\pi$

Final Answer:

A parametrization for the circle $x^2+y^2=81$ is given by the equations $x(t)=9\cos (t)$ and $y(t)=9\sin (t)$ for $0\le t<2\pi$.

Step-by-Step Example 2: Parametrizing a Translated Ellipse

Problem: Find a set of parametric equations that describes the ellipse given by the relation $9(x-5{)}^2+4(y+1{)}^2=36$.

Step 1: Convert the implicit equation to standard form.

The standard form for an ellipse is $\frac{(x-h{)}^2}{a^2}+\frac{(y-k{)}^2}{b^2}=1$. To get the given equation into this form, we must divide the entire equation by 36.

$$\frac{9(x-5{)}^2}{36}+\frac{4(y+1{)}^2}{36}=\frac{36}{36}$$

$$\frac{(x-5{)}^2}{4}+\frac{(y+1{)}^2}{9}=1$$

Step 2: Identify the type of relation and its key parameters.

The equation is now in the standard form for an ellipse. We can extract the center `(h, k)and the semi-axis lengths $a and $b$.

• Comparing $(x-5{)}^2$ to $(x-h{)}^2$, we find $h=5$.

• Comparing $(y+1{)}^2$ to $(y-k{)}^2$, we find $k=-1$. The center is $(5,-1)$.

• The term under $(x-h{)}^2$ is $a^2$. So, $a^2=4⟹a=2$.

• The term under $(y-k{)}^2$ is $b^2$. So, $b^2=9⟹b=3$.

Step 3: Apply the standard parametrization formula for an ellipse.

The formula is given by:

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

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

Step 4: Substitute the extracted parameter values.

Substitute $h=5$, $k=-1$, $a=2$, and $b=3$ into the formulas:

$x(t)=5+2\cos (t)$

$y(t)=-1+3\sin (t)$

Step 5: State the parameter interval.

To trace the entire ellipse once, we use the standard interval:

$0\le t<2\pi$

Final Answer:

A parametrization for the ellipse $9(x-5{)}^2+4(y+1{)}^2=36$ is given by $x(t)=5+2\cos (t)$ and $y(t)=-1+3\sin (t)$ for $0\le t<2\pi$.

Using Your Calculator

While finding a parametrization is an analytical process, a graphing calculator is an excellent tool for verifying that your parametric equations correctly represent the original implicit relation.

To graph and verify the parametrization from Example 2 ($x(t)=5+2\cos (t)$, $y(t)=-1+3\sin (t)$):

1. Set the Mode: Press the [MODE] button. Navigate down to the function type line and select PARAMETRIC (or PAR). Also, ensure that the angle mode is set to RADIAN. Press [ENTER] on both.

2. Enter the Equations: Press the [Y=] button. You will now see input fields for $X_{1}T$ and $Y_{1}T$.

   • In $X_{1}T$, enter `5 + 2cos(T)$.The$[T,\theta,n,x]button is used to type the parameter $T.

   • In $Y_{1}T$, enter -1 + 3sin(T).

3. Set the Parameter Window: Press the [WINDOW] button. This is the most critical step for parametric graphing.

   • $Tmin=0$: This is the starting value for the parameter.

   • $Tmax=2\pi$: To enter 2π, you can type `2 * [2nd] [^]. This ensures the full curve is drawn. A decimal approximation like $6.283 also works.

   • $Tstep=0.1$: This value determines how often the calculator plots a point. A smaller value creates a smoother curve but takes longer to graph. $0.1$ or $\pi /24$ are good starting points.

4. Set the Viewing Window: Adjust $Xmin$, $Xmax$, $Ymin$, and $Ymax$ to fit the graph. For our ellipse centered at $(5,-1)$ with $a=2$ and $b=3$:

   • The x-values range from $h-a=5-2=3$ to $h+a=5+2=7$.

   • The y-values range from $k-b=-1-3=-4$ to $k+b=-1+3=2$.

   • A good window would be:

     • $Xmin=0$

     • $Xmax=10$

     • $Ymin=-5$

     • $Ymax = 3`

5. Graph: Press the [GRAPH] button. You should see the ellipse being drawn on the screen, centered at $(5, -1)`, confirming your parametrization is correct.

AP Exam Quick Hit

Common Question Types

• Direct Parametrization from Standard Form: You will be given an equation of a circle or ellipse in standard form and asked to select the correct parametrization from multiple-choice options.

  • Example: "Which of the following is a parametrization of the ellipse $\frac{x^2}{49} + \frac{(y-2)^2}{4} = 1$?"

  • Answer:$x(t)=7\cos (t),y(t)=2+2\sin (t)$

• Parametrization from General Form: You will be given the equation of a circle or ellipse that requires algebraic manipulation (like completing the square or dividing by a constant) before you can identify the parameters $h,k,a,b,$ or r`. - *Example:* "Find a set of parametric equations for the relation $x^2 + 6x + y^2 - 4y = 3." (This would first require completing the square to get $(x+3{)}^2+(y-2{)}^2=16$).

• Parametrizing a Basic Function: You may be asked to provide a simple parametrization for a standard function.

  • Example: "Determine a parametrization for the function $g(x)=\sqrt{x^2+1}$."

  • Answer:$x(t)=t,y(t)=\sqrt{t^2+1}$

Common Mistakes

• Using Squared Values for Radii: For an equation like $x^2+y^2=16$, a common error is to use $16$ in the parametrization instead of the radius $r=\sqrt{16}=4$. The correct form is $x(t)=4\cos (t)$, not $x(t)=16\cos (t)$. The same applies to $a^2$ and $b^2$ for ellipses.

• **Sign Errors on the Center `(h, k)Formula28^2$ and $(y-k{)}^2$. For an equation like $\frac{(x+3{)}^2}{16}+\frac{(y-1{)}^2}{9}=1$, students may incorrectly identify $h$ as $3$ instead of $-3$. Remember to take the opposite sign of the number shown in the parentheses.

• **Swapping $a$ and b`:** In the standard parametrization formula provided, $a is associated with the $cos(t)$ term and the horizontal shift, while $b$ is associated with the sin(t)` term and the vertical shift. Be careful not to swap them, even if the major axis of the ellipse is vertical. Stick to the formula: $a^2 is under the x-term, $b^2$ is under the y-term.

• **Forgetting the Center `(h, k):** When parametrizing a translated ellipse or circle, it is easy to find $a and $b$ but forget to add the horizontal shift $h$ and vertical shift $k$. The final equations must be $x(t)=h+a\cos (t)$ and $y(t)=k+b\sin (t)$.