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Parametric equations for lines and curves II

The excitement of learning separates youth from old age. As long as you’re learning, you’re not old, Rosalyn S.Yalow.

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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More parametric equations

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Solve for t in the first equation (i) $2t = x + 4 ↭ t = \frac{x+4}{2}$ and then substitute t into the second equation, $y = 4(\frac{x+4}{2})^2 =[\text{Simplify the expression}] = [\text{The 4’s cancel out:}]4·\frac{(x+4)^2}{4} = (x+4)^2$. Resulting Cartesian equation: y = (x +4)2.

Parametric curves have a direction of motion. The direction of motion is given by increasing t. So, when plotting parametric curves, we also include arrows that show the direction of motion.

This is the equation of a parabola that opens upwards, with its vertex at (−4, 0). As x increases, t also increases, indicating that the direction of the parabola is upwards.

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Solve each equation for the trigonometric function. From (i): $2cos(t) = x -3 ↭ cos(t) = \frac{x-3}{2}$ and (ii) $2sin(t) = y + 1↭ sin(t) = \frac{y+1}{2}$.

Substitute cos(t) and sin(t) from the above equations and use the Pythagorean identity: $sin^2(θ)+cos^2(θ) = 1⇒ \frac{(x-3)^2}{2^2}+\frac{(y+1)^2}{2^2} = 1 ⇒ (x-3)^2 + (y+1)^2 = 4.$ This is the equation of a circle with center at (3, -1) and radius 2. This sequence of points shows that as t increases from 0 to 2π, the curve traces a circle in a counterclockwise direction.

t x y
0 5 -1
π/2 3 1
π 1 -1
3π/2 3 -3
5 -1

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Create a Table of Values:

t x y
-3 -8 -7
-2 -5 -2
-1 -2 1
0 1 2
1 4 1
2 7 -2
3 10 -7

Solve for t in the first equation (i), $t = \frac{x-1}{3}$, and substitute into the second equation (ii) $y = 2 -t^2 = 2 -(\frac{(x-1)}{3})^2 ⇒y = \frac{-1}{9}(x-1)^2+2$. As t increases, x increases and y first increases to a maximum 2 (t = 0, (x, y) = (1, 2)) and then decreases.

The Cartesian equation is a parabola: y = a(x -h)2 + k, where the vertex is (h, k). In this example, the vertex is (1, 2). The coefficient $\frac{-1}{9}$ indicates the parabola opens downwards and is vertically compressed.

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Solve each equation for the trigonometric function. From (i): $cos(θ) = \frac{x}{4}$ and (ii) $sin(θ) = \frac{y}{3}$.

Substitute cos(t) and sin(t) from the above equations and use the Pythagorean identity: $sin^2(θ)+cos^2(θ) = 1⇒ \frac{x}{4}^2+\frac{y}{3}^2 = 1.$ This is the equation of an ellipse with center at the origin (0, 0). Since $\frac{-π}{2}≤θ≤\frac{π}{2}$, we are only considering the right half of the ellipse.

The semi-major axis is along the x-axis with a length of 4. The semi-minor axis is along the y-axis with a length of 3. This movement is in the counter-clockwise direction, starting from the bottom point (0, −3), moving through (4, 0), and ending at (0, 3).

t x y
-π/2 0 -3
0 4 0
π/2 0 3

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First, we solve for t in the first equation (i): $e^x = e^{ln(t+2)}↭ e^x = t + 2 ↭ t = e^x -2.$ Substitute this result into the second equation: $y = \frac{1}{e^x-2+1} = \frac{1}{e^x -1}$

There are three sections:

  1. x = 0, it is not defined because division by zero is undefined.
  2. x > 0 ⇒ ex > e0 = 1 ⇒ ex-1 > 0, and it is strictly increasing, so $\frac{1}{e^x -1}$ is positive and strictly decreasing. Futhermore, $\lim_{x \to 0+} \frac{1}{e^x -1} = ∞, \lim_{x \to ∞} \frac{1}{e^x -1} = 0$.
  3. x < 0 ⇒ ex < e0 = 1, ex -1 < 0, and it is strictly increasing, so $\frac{1}{e^x -1}$ is negative and strictly decreasing. Futhermore, $\lim_{x \to 0−} \frac{1}{e^x -1} = -∞, \lim_{x \to ∞} \frac{1}{e^x -1} = 0$.

Vertical asymptote at x = 0. Horizontal asymptote at y = 0. Direction of motion: From x→−∞, y→0, through the point t = 0, (ln(2), 1), to x→∞, y→0.

A working example: A cycloid

A cycloid is the path traced by a point P on the rim of a wheel with radius a, as it rolls along a straight line (the x axis) without slipping. P is a point on the rim of a wheel, and it starts at the origin (0, 0).

To describe the cycloid, we need to express the coordinates x and y of point P as functions of the angle θ by which the wheel has rotated. Refer to Figures ii and iii in the diagram for a visual aid.

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As the wheel rolls, the center O of the wheel moves horizontally. $\vec{OP}$ represents the distance traveled by the wheel’s center O. $\vec{AB}$ represents the vertical distance from the center of the wheel to the point A, that is, ⟨0, a⟩. $\vec{BP}$ represents the position of point P relative to the center of the wheel after it has rotated by an angle θ.

$\vec{OP} = \vec{OA} + \vec{AB} + \vec{BP}$ where $\vec{OA} = ⟨θ·a, 0⟩$ is (provided the angle θ is expressed in radians and the wheel is not slipping) the amount by which the wheel has rolled or moved horizontally. $\vec{OA}$ is equal to the Arc length from A to P, $\vec{AB} = ⟨0, a⟩, \vec{BP} = ⟨-a·sin(θ), -a·cos(θ)⟩$

Therefore, $\vec{OP} = \vec{OA} + \vec{AB} + \vec{BP} = ⟨aθ -asin(θ), a - acos(θ)⟩.$

Thus, the parametric equations for the cycloid are:

$\begin{cases} x(θ) = aθ -asin(θ)\\ y(θ) = a - acos(θ) \end{cases}$

Behaviour near the bottom point (θ ≈ 0)

When analyzing the cycloid near the origin, we simplify by setting a = 1 (Figure iv) This helps us understand how the cycloid behaves as the angle θ approaches zero.

Parametric Equations Simplified:

$\begin{cases} x(θ) = θ -sin(θ)\\ y(θ) = 1 - cos(θ) \end{cases}$

Using Taylor series approximations for small t, $f(t) ≈ f(0) + tf’(0) + \frac{t^2f’’(0)}{2} + \frac{t^3f’’’(0)}{6}$ is a pretty good approximation. In our particular case, $cos(θ) ≈ 1 -\frac{θ^2}{2}, sin(θ)≈θ -\frac{θ^3}{6}$

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Now, substitute these approximations into the parametric equations: $x(θ) = θ - sin(θ) ≈ θ -(θ -\frac{θ^3}{6}) = \frac{θ^3}{6}, y(θ) = 1 - cos(θ) ≈ 1 -(1 -\frac{θ^2}{2}) = \frac{θ^2}{2}$.

To find the slope near the origin, we compute the radio $\frac{y(θ)}{x(θ)}=\frac{\frac{θ^2}{2}}{\frac{θ^3}{6}} = \frac{6θ^2}{2θ^3} =[\text{Simplify the expression:}]\frac{3}{θ}$.

As θ approaches zero, the slope $\frac{3}{θ}$ approaches infinity. This indicates that the slope at the origin is vertical, confirming that the correct sketch of the cycloid near θ ≈ 0 is the one with a vertical tangent, as shown in Figure iv, sketch 4.

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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