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Parametric equations for lines and curves.

Logic will get you from A to B. Imagination will take you everywhere, Albert Einstein.

Recall

Definition. A vector $\vec{AB}$ is a geometric object or quantity that has both magnitude (or length) and direction. Vectors in an n-dimensional Euclidean space can be represented as coordinates vectors in a Cartesian coordinate system.

Definition. The magnitude of the vector $\vec{A}$, also known as length or norm, is given by the square root of the sum of its components squared, $|\vec{A}|~ or~ ||\vec{A}|| = \sqrt{a_1^2+a_2^2+a_3^2}$, e.g., $||< 3, 2, 1 >|| = \sqrt{3^2+2^2+1^2}=\sqrt{14}$, $||< 3, -4, 5 >|| = \sqrt{3^2+(-4)^2+5^2}=\sqrt{50}=5\sqrt{2}$, or $||< 1, 0, 0 >|| = \sqrt{1^2+0^2+0^2}=\sqrt{1}=1$.

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Equations of Lines

In geometry, understanding the relationship between points , lines, and planes is fundamental. A plane can be thought of as the intersection of three-dimensional figures.

A point is the intersection of two lines; a line is the intersection of two planes; and a plane can be thought as the intersection of three-dimensional figures.

Unlike in ℝ3 where the equation y = mx + b describe a line, in ℝ3, this equation describes a plane.

A line in three-dimensional space can be visualized as the path or trajectory of a moving point. Imagine a point Q(t) that moves along the line. At t = 0, it is at Q0, at t = 1 it is at Q1, and at any time t, it is at Qt. Besides, $\vec{Q_0Q(t)}=t\vec{Q_0Q_1}$ because it moves at a constant speed.

Let’s break it down with an example:

  1. Initial point. Suppose Q0 is the starting point on the line, given by the coordinates Q0 = (-1, 2, 2).
  2. Another point. Suppose Q1 is another point on the line, given by the coordinates Q1 = (1, 3, -1).
  3. Direction vector. To find the direction in which the line extends, we calculate the vector $\vec{Q_0Q_1}$ then $\vec{Q_0Q_1} = (1−(−1), 3−2, −1−2) = (2, 1, -3)$ (Figure i).
  4. Parametric Form of the Line. The line can be described parametrically as follows: $Q(t) = Q_0 + t\vec{Q_0Q_1}$.For our example, the coordinates Q(t) at any time t are give by:

Q(t) = $\bigl(\begin{smallmatrix}-1\\ 2\\ 2\end{smallmatrix}\bigr)+t\bigl(\begin{smallmatrix}2\\ -1\\ 3\end{smallmatrix}\bigr)$

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This results in the parametric equations:

$\begin{cases} x(t) = - 1 + 2t \\ y(t) = 2 + t \\ z(t) = 2 -3t \end{cases}$

First, Q0 $(x+2y+4z)\bigg|_{(-1, 2, 2)} = −1 + 2⋅2 + 4⋅2 = -1 + 4 + 8 = 11$ > 7 and Q1 = 1 + 2·3 + 4·(-1) =3 < 7 are not in the plane, so Q0 and Q1 are on opposite sides of the plane. Therefore, we can conclude that the line intersect the plane between these points.

To know the exact point where the line intersect the plane, we just need to plug the line’s equation (substitute the parametric equations of the line) in the plane’s equation, x(t) + 2y(t) + 4z(t) = 7 ↭ (-1 + 2t) + 2(2 + t) + 4(2 -3t) = 7 ↭[Simplifying] -8t + 11 = 7 ↭ -8t = -4 ↭ 2t = 1. Q(t) is in the plane ↭ t = 12, i.e., halfway Q0 and Q1, hence the line intersects the plane x + 2y + 4z = 7 at t = $Q(\frac{1}{2}) = (0, \frac{5}{2},\frac{1}{2})$.

Verification: $(x+2y+4z)\bigg|_{(0, \frac{5}{2},\frac{1}{2})} = 5 + 2 = 7.$

Checking Intersection of a Line and a Plane

  1. Line Intersects the Plane. If the parametric equations of the line, when substituted into the plane equation, yield a specific value of the parameter t, then the line intersects the plane at that value of t , e.g., Q(t) = $\bigl(\begin{smallmatrix}-1\\ 2\\ 2\end{smallmatrix}\bigr)+t\bigl(\begin{smallmatrix}2\\ -1\\ 3\end{smallmatrix}\bigr)$ intersects the plane x + 2y + 4z = 7 at t = $Q(\frac{1}{2}) = (0, \frac{5}{2},\frac{1}{2})$.

  2. Line lies in the plane: If, after substituting the parametric equations into the plane equation, the result simplifies to a true statement (like 7=7) for all values of t, the line lies entirely within the plane.

    Suppose we choose a point P0 on the plane x + 2y + 4z = 7, say P0 = (1, 1, 1), and a direction vector $\vec{d}$ that is orthogonal to the normal vector of the plane (hence, it is parallel to the plane) $\hat{\mathbf{n}} = ⟨1, 2, 4⟩$, e.g., $\vec{d} = ⟨0, -2, 1⟩$: ⟨0, -2, 1⟩·⟨1, 2, 4⟩ = 0 -4 + 4 = 0.

    The line Q(t) = $\bigl(\begin{smallmatrix}1\\ 1\\ 1\end{smallmatrix}\bigr)+t\bigl(\begin{smallmatrix}0\\ -2\\ 1\end{smallmatrix}\bigr)$ lies in the plane x + 2y + 4z = 7. Verification: 1 +2(1 -2t) + 4(1 +t) = 7 ↭ 1 +2 -4t +4 + 4t = 7 ↭ 7 = 7. Since the equation holds true for all t, the line lies entirely in the plane.

  3. Line is Parallel to the Plane. If, after substitution, the result is a contradiction (like a non-zero constant equal to zero), the line is parallel to the plane and does not intersect it.

    Suppose we choose a point P0 that is not the plane x + 2y + 4z = 7, say P0 = (-1, 2, 2), and a direction vector $\vec{d}$ that is orthogonal to the normal vector of the plane (hence, it is parallel to the plane) $\hat{\mathbf{n}} = ⟨1, 2, 4⟩$, e.g., $\vec{d} = ⟨0, -2, 1⟩$: ⟨0, -2, 1⟩·⟨1, 2, 4⟩ = 0 -4 + 4 = 0.

    The line Q(t) = $\bigl(\begin{smallmatrix}-1\\ 2\\ 2\end{smallmatrix}\bigr)+t\bigl(\begin{smallmatrix}0\\ -2\\ 1\end{smallmatrix}\bigr)$ is parallel to the plane x + 2y + 4z = 7. Verification: -1 + 2(2 -2t) + 4(2 + t) = 7 ↭ - 1 + 4 -4t + 8 + 4t = 7 ↭ 11 = 7 ⊥ Since the equation results in a contradiction, this confirms that the line is parallel to the plane and does not intersect it.

Parametric equations

A parametric equation defines a group of quantities as functions of one or more independent variables called parameters. In simpler terms, it is a set of equations where each equation describes one of the coordinates (or other dependent variables) in terms of an independent variable, known as the parameter, e.g., x = x(t), y = y(t), that defines or specifies the (dependent) variables x and y as functions of a parameter or independent variable t.

Parametric equations are often used to describe motions and paths in the plane or in space, where the x, y, and z coordinates vary with some other parameter, such as t (time) or even θ (for example, a particular angle).

Graphing a Parametrically Defined Curve

Sketch the curves described by the following parametric equations:

Graphing a Parametrically Defined Curve

Eliminating the Parameter

Eliminating the parameter involves rewriting parametric equations as a single equation in terms of x and y. This process can help you understand the graph of a curve more clearly by relating the variables directly without the intermediary parameter.

Let’s work through the examples provided and clarify each step.

Solve one of the equation for t. From (i), $t = \frac{x}{2}$, then substitute into the other equation (ii), $y = t^2 = (\frac{x}{2})^2 = \frac{x^2}{4}$, and this is the Cartesian form, $y = \frac{x^2}{4}$. This Cartesian form represents a parabola opening upwards.

Bibliography

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License and is based on MIT OpenCourseWare.
  1. NPTEL-NOC IITM, Introduction to Galois Theory.
  2. Algebra, Second Edition, by Michael Artin.
  3. LibreTexts, Calculus. Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
  4. Field and Galois Theory, by Patrick Morandi. Springer.
  5. Michael Penn, Andrew Misseldine, blackpenredpen, and MathMajor, YouTube’s channels.
  6. Contemporary Abstract Algebra, Joseph, A. Gallian.
  7. MIT OpenCourseWare, 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007, YouTube.
  8. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.
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