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4.7 Parametrization of implicitly defined functions

Lesson

Introduction

Learning objective

  • 4.7.A - Represent a curve in the plane parametrically.
  • 4.7.B - Represent conic sections parametrically.

Represent a curve in the plane parametrically

In mathematical terms, curves can be represented in a variety of ways. Two common methods include the explicit representation, where y is a function of x (y=f(x)), and the parametric representation, which provides more flexibility.

In parametric representation, we describe the x and y coordinates as functions of a third variable, usually denoted as t. This t parameter can be thought of as a sort of "time" variable; as t changes, the values of x and y change as well, tracing out the curve in the process. The pair of functions x(t) and y(t) that arise from this method are referred to as the parametric equations of the curve.

Validating this parametric representation involves substituting x(t) and y(t) back into the original equation in place of x and y. If the equation remains valid for all values of t in its domain, then we can say the parametric representation is correct.

Moving into more specific scenarios, consider a function f that is a function of x. In this situation, the equation y=f(x) can be parametrized. Instead of y being directly dependent on x, we express both y and x as functions of t, such that (x(t),\, y(t)) = (t,\, f(t)). In other words, we set x=t, and for each value of t, we find the corresponding y value by substituting t into function f.

Moreover, if function f is invertible, its inverse can also be parametrized. The parametric equations for the inverse function become (x(t),\, y(t)) = (f(t), t). Here, we set y=t and determine the corresponding x value for each t by substituting t into the inverse function.

In the context of curves in a plane, they can be parametrically represented by a pair of functions x(t) and y(t) that satisfy the original equation for all values of t in the domain. Functions and their inverses can be parametrized as (t,\, f(t)) and (f(t),\, t). For conic sections like parabolas, ellipses, and hyperbolas, specific formulas can be used for their parametrization. Ellipses and hyperbolas, in particular, make use of trigonometric functions in their parametrization.

Examples

Example 1

We have a curve defined by the equation y = x^{2}. The task is to represent this curve in parametric form.

Worked Solution
Create a strategy

In order to parametrize the curve, we want to express both x and y in terms of a new variable, often denoted by t. For simple functions like y = x^{2}, a common strategy is to let x = t, then substitute this into the equation to find the corresponding y in terms of t.

Apply the idea
\displaystyle y\displaystyle =\displaystyle t^{2}Substitute x=t

Therefore, the parametric equations for the curve are x(t) = t and y(t) = t^{2}.

Reflect and check

To check if our parametric equations are correct, we can substitute y(t) and x(t) back into the original equation and see if it holds. If y = x^{2}, then t^{2} should equal (t)^{2}, which is true. Therefore, our parametric representation is correct. As an alternative approach, we could have also chosen x = t^{2} and y = t^{4}, which would have resulted in a different, but valid parametrization of the curve.

Example 2

Given the function f(x) = \sqrt{x}, parametrize its inverse. The given function is f(x) = \sqrt{x}. The task is to represent the inverse of this function in parametric form.

Worked Solution
Create a strategy

The inverse function of f(x) = \sqrt{x} is f^{-1}(x) = x^{2}, because squaring is the operation that undoes the square root. To parametrize this inverse function, we can express both x andy as functions of a new variable, t.

Apply the idea

For the inverse function, we let y = t and then find the corresponding x by substituting into the equation of the inverse function.

So, if y = t, then x = t^{2}, because the inverse function is f^{-1}(x) = x^{2}.

Therefore, the parametric equations for the inverse function are x(t) = t^{2} and y(t) = t.

Reflect and check

To check our answer, we can substitute x(t) and y(t) into the equation of the inverse function and see if it holds. If x = y^{2}, then t^{2} should equal (t)^{2}, which is true. Therefore, our parametric representation of the inverse function is correct. An alternative approach could involve graphing the original function and its inverse to visually verify the parametrization.

Idea summary

Parametric equations for a curve are typically represented as x(t) and y(t).

Parametric equations for a curve are typically represented as x(t) and y(t). For a function f of x, the equation y=f(x) can be parametrized as (x(t),\, y(t)) = (t,\, f(t)).

If the function f is invertible, the inverse can also be parametrized. The parametric equations for an invertible function's inverse are (x(t),\, y(t)) = (f(t),\, t).

Representing conic sections parametrically

In the mathematical study of conic sections, which include parabolas, ellipses, and hyperbolas, we can utilize specific formulas to achieve parametric representations. This allows us to express the x and y coordinates in terms of a third variable, typically denoted as t. In the case of ellipses and hyperbolas, trigonometric functions play an integral role in this parametrization.

For a parabola, the method for parametrization is similar to any equation that can be solved for either x or y. The most common form of a parabolic equation is y = x^{2}. In this form, the parabola opens upward, and the vertex of the parabola is at the origin (0,\, 0). If we can rearrange the equation to solve for x, we can represent the curve parametrically as (x(t),\, y(t)) = (f(t),\, t). This means that we find a function f such that when we input t, it gives us the x-coordinate. And then, for every such t, the y-coordinate is simply t. In other words, we're making t play the role that yusually does, and x is calculated as a function of this t.

We can solve our parabolic equation for y, the parametrization would be (x(t),\, y(t)) = (t,\, f(t)). In this scenario, we're letting t play the role that x usually does, and y is calculated as a function of this t. That is, for every t, the x-coordinate is simply t, and the y-coordinate is found by plugging t into our function f.

In both cases, as t varies, the pair of functions (x(t),\, y(t)) define the points on the parabola, thus giving us a way to trace out the curve. The choice between these two approaches usually depends on the form of the parabolic equation we're given and which variable it's easier to solve for.

An ellipse, a closed curve in which the sum of the distances from two points (foci) to every point on the curve is constant, can be parametrized using the trigonometric functions \cos and \sin. The specific functions used are x(t) = h + a \cos t and y (t) = k + b \sin t, where a and b are the semi-major and semi-minor axes respectively, h and k are the center of the ellipse, and t ranges from 0 to 2\pi. As t varies, these equations trace out the ellipse in the plane. The semi-major axis is the longest diameter of the ellipse, while the semi-minor axis is the shortest. These lengths determine the overall size and shape of the ellipse.

A hyperbola, a curve where the difference of distances from any point on the curve to two fixed points (foci) is constant, can also be parametrized using trigonometric functions.

Parametrizing a hyperbola involves representing all points on the hyperbola in terms of a third variable, typically denoted as t. This is achieved by using trigonometric functions, specifically secant (\sec) and tangent (\tan), which have properties that align with the characteristics of a hyperbola.

For a hyperbola that opens left and right, we use the functions x(t) = h + a \sec t and y(t) = k + b \tan t. Here, h and k denote the coordinates of the center of the hyperbola, the midpoint between its two foci. The terms a and b represent distances from the center of the hyperbola to the vertices (the points where the hyperbola intersects its major axis) and to the conjugate axis respectively. This conjugate axis is a line segment perpendicular to the major axis that passes through the center of the hyperbola. For a hyperbola that opens up and down, the functions switch to x(t) = h + a \tan t and y(t) = k + b \sec t.

In both cases, the variable t varies from 0 to 2\pi. As t changes in this range, the secant and tangent functions generate the wide range of values needed to trace out the open, ever-expanding shape of the hyperbola in each direction.

Examples

Example 3

Given the parabola defined by the equation y = 2x^{2} + 3x - 1, derive the parametric equations for this parabola.

Worked Solution
Create a strategy

The equation is already solved for y, indicating the relationship between y and x. You should utilize this clarity by substituting x with t, which will give you a new equation expressed in terms of t. By doing this, you will be able to express both x and y as functions of t, resulting in the parametric representation that you're seeking.

Apply the idea

Let x = t. This gives you the first parametric equation: x(t) = t.

\displaystyle y\displaystyle =\displaystyle 2x^{2} + 3x - 1Write the original equation
\displaystyle y\displaystyle =\displaystyle 2t^{2} + 3t - 1Substitute x=t
\displaystyle y(t)\displaystyle =\displaystyle 2t^{2} + 3t - 1Second parametric equation

The parametric equations are, therefore, x(t) = t and y(t) = 2t^{2} + 3t - 1.

These equations represent the parabola parametrically.

Reflect and check

To verify your solution, you can substitute x(t) and y(t) back into the original equation. If y(t) = 2(x(t))^{2} + 3x(t) - 1 holds true for all t in the domain, then the parabola has been correctly parametrized.

Example 4

Given an ellipse with a semi-major axis of length 4, a semi-minor axis of length 2, and its center at the origin (0,\,0). Your task is to parametrize this ellipse.

Worked Solution
Create a strategy

Understand the properties of the ellipse. The parameters a and b denote the lengths of the semi-major and semi-minor axes respectively, while h and k are the coordinates of the center. You'll use these to write the standard parametric equations for an ellipse: x(t) = h + a \cos t and y(t) = k + b \sin t.

Apply the idea
  1. Identify h = k = 0 (center), a = 4 (semi-major axis), and b = 2 (semi-minor axis) from the problem statement.

  2. Substitute these values into the standard parametric equations to get x(t) = 4 \cos t and y(t) = 2 \sin t for 0 \leq t \leq 2\pi.

These are the parametric equations for the ellipse.

Reflect and check

Confirm the solution by substituting x(t) and y(t) into the standard equation for an ellipse centered at the origin: \dfrac{(x^{2})}{(a^{2})} + \dfrac{(y^{2})}{(b^{2})} = 1. If this equation is satisfied for all t in the given range, then the ellipse has been correctly parametrized.

Example 5

Let's examine a hyperbola that has vertices at (\pm 3,\,0), a center at the origin, and a conjugate axis of length 4. This hyperbola opens towards the left and right. The challenge here is to determine the parametric representation of this hyperbola.

Worked Solution
Create a strategy

In this case, you are tasked with finding the parametric representation of a hyperbola. This means you need to express both x and y in terms of a third variable, t. The relationships between x,\, y, and t in this scenario will be defined by the standard parametric equations for a hyperbola that opens left and right: x(t) = h + a \sec t and y(t) = k + b \tan t.

Apply the idea
  1. Identify the given parameters: h = k = 0 (center of the hyperbola), a = 3 (distance from center to a vertex), and b = 2 (half the length of the conjugate axis).

  2. Substitute these values into the standard parametric equations to derive the specific equations for this hyperbola: x(t) = 3 \sec t and y(t) = 2 \tan t for 0 \leq t \leq 2\pi.

These equations represent the hyperbola parametrically. They tell us the x and y coordinates for any point on the hyperbola, given a specific value of t.

Reflect and check

To confirm the accuracy of these parametric equations, substitute x(t) and y(t) back into the standard equation for a hyperbola that opens left and right: \dfrac{(x^{2})}{(a^{2})} - \dfrac{(y^{2})}{(b^{2})} = 1. This should hold true for all values of t in the range 0 \leq t \leq 2\pi.

Idea summary

The parametric form of a parabola, y = f(x), can be represented as x(t) = t and y(t) = f(t). The points (x(t),\, y(t)) thus trace out the curve of the parabola.

Parametrization of an ellipse uses the trigonometric functions sine and cosine. The typical parametric equations are given by x(t) = h + a \cos(t) and y(t) = k + b \sin(t), where (h,\, k) is the center of the ellipse, a and b are the lengths of the semi-major and semi-minor axes respectively, and t ranges from 0 to 2\pi.

Hyperbolas can be parametrized using secant and tangent functions. For a hyperbola that opens left and right, we use x(t) = h + a \sec(t) and y(t) = k + b \tan(t). For one that opens up and down, we switch to x(t) = h + a \tan(t) and y(t) = k + b \sec(t). In all cases, the parametric equations provide a convenient way to represent and study these curves, enabling us to calculate coordinates, plot points, and understand the behavior of the curves as t varies.

Outcomes

4.7.A

Represent a curve in the plane parametrically.

4.7.B

Represent conic sections parametrically.

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