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Chapter 3: Problem 5

Write, in parametric form, the equation of the \(y\) axis.

Short Answer

Expert verified
The parametric form of the equation for the y-axis is \( x(t) = 0 \) and \( y(t) = t \).

Step by step solution

01

Understand the Concept of Parametric Equations

In parametric equations, points on a curve are represented as functions of a parameter, usually denoted by 't'. For a curve, this means expressing the coordinates (x, y) as (x(t), y(t)).
02

Identify the Characteristics of the y-axis

The y-axis consists of all points where the x-coordinate is always 0, regardless of the y-coordinate. Therefore, x is always 0.
03

Define x(t) and y(t)

Since the x-coordinate is always 0, we can write: x(t) = 0. The y-coordinate can be any real number, so we can let y(t) = t, where t is any real number.
04

Write the Parametric Equations

Combine the expressions from the previous step to write the parametric form of the y-axis: \( x(t) = 0 \)\( y(t) = t \)where t is a parameter that can take any real value.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

y-axis
The y-axis is an essential concept in the coordinate system. It is the vertical line that runs through the origin, where the x-coordinate is always zero, no matter what the y-coordinate is. This means every point on the y-axis has coordinates of the form (0, y). Understanding the y-axis is crucial because it helps you visualize how vertical lines behave in graphs.
For example, the point (0, 5) lies on the y-axis, as well as the point (0, -3). These points stretch infinitely in the positive and negative directions along the y-axis.
parametric form
In mathematics, parametric equations are a way to express coordinates using a parameter, denoted as 't'. This is different from the standard form where you have equations directly connecting x and y. Instead, parametric form allows for (x, y) to be expressed as functions of 't'.
For the y-axis example, the equations are:
\[x(t) = 0\]
\[y(t) = t\]
where t is any real number. This setup allows us to generate all the points on the y-axis through different values of 't'. By varying 't', we cover every possible point where x is zero and y can take any value.
coordinate system
The coordinate system, specifically the Cartesian coordinate system, consists of two perpendicular axes: the x-axis (horizontal) and the y-axis (vertical). These two axes intersect at a point called the origin, defined as (0, 0).
This system allows us to define any point in a plane using an ordered pair of numbers (x, y). For instance:
  • (2, 3) means moving 2 units along the x-axis and 3 units along the y-axis.
  • (-1, -4) indicates moving 1 unit left on the x-axis and 4 units down the y-axis.

The coordinate system is a powerful tool for graphing equations and visualizing more complex mathematical concepts.
real numbers
Real numbers are the set of all numbers that can be found on the number line. This includes both rational numbers (like 2, 1/2, and -3) and irrational numbers (like 鈭�2 and 蟺).
When working with parametric equations, especially for the y-axis, the parameter 't' can take on any real number value. This means:
  • For t = 0, the point is (0, 0).
  • For t = 5, the point is (0, 5).
  • For t = -3, the point is (0, -3).

By allowing 't' to be any real number, the parametric equations cover every possible point along the y-axis.

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Most popular questions from this chapter

Find \(A B, B A, A+B, 5 A, 3 B, 5 A-3 B\). Observe that \(A B \neq B A\). Show that \(\operatorname{det}(A B)=\operatorname{det}(B A)=(\operatorname{det} A)(\operatorname{det} B)\), but that \(\operatorname{det}(A+B) \neq \operatorname{det} A+\operatorname{det} B, \operatorname{det} 5 A \neq\) 5 det \(A\), and det \(3 B \neq 3\) det \(B\). (In Problem 2, show that det \(3 B=9\) det \(B\), and in Problem 3 . det \(3 B=27 \operatorname{det} B .)\) $$ A=\left(\begin{array}{rrr} 1 & 0 & 2 \\ 3 & -1 & 0 \\ 0 & 5 & 1 \end{array}\right), \quad B=\left(\begin{array}{rrr} 1 & 1 & 0 \\ 0 & 2 & 1 \\ 3 & -1 & 0 \end{array}\right) $$

Solve each set of equations by the method of finding the inverse of the coefficient matrix. $$ \left\\{\begin{array}{c} x-2 y=5 \\ 3 x+y=15 \end{array}\right. $$

The following matrix product is used in discussing a thick lens in air: $$ A=\left(\begin{array}{cc} 1 & (n-1) / R_{2} \\ 0 & 1 \end{array}\right)\left(\begin{array}{cc} 1 & 0 \\ d / n & 1 \end{array}\right)\left(\begin{array}{cc} 1 & -(n-1) / R_{1} \\ 0 & 1 \end{array}\right) $$ where \(d\) is the thickness of the lens, \(n\) is its index of refraction, and \(R_{1}\) and \(R_{2}\) are the radii of curvature of the lens surfaces. It can be shown that element \(A_{12}\) of \(A\) is \(-1 / f\) where \(f\) is the focal length of the lens. Evaluate \(A\) and det \(A\) (which should equal 1) and find \(1 / f\). [See the American Journal of Physics, \(48,396(1980) .]\)

Solve the following sets of simultaneous equations by reducing the matrix to row echelon form. $$ \begin{array}{r} \mid x-2 y+3 z=0 \\ x+4 y-6 z=0 \\ 2 x+2 y-3 z=0 \end{array} $$

Write a set of linear equations for each of the following problems and solve them using either determinants or row reduction. An object composed of \(x \mathrm{gm}\) of potassium (specific gravity \(0.8\) ) and \(y \mathrm{gm}\) of cesium (specific gravity \(2.0\) ) cannot be weighed in air or water because it would react with either. However, it weighs \(86 \mathrm{gm}\) in oil of specific gravity \(0.6\) and \(124 \mathrm{gm}\) in oil of specific gravity \(0.4\). Find \(x\) and \(y\).
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