Problem 45
Question
Find the parametric equation of the line in the \(x-y\) plane that goes through the indicated point in the direction of the indicated vector. $$ (-1,-2),\left[\begin{array}{r} 1 \\ -3 \end{array}\right] $$
Step-by-Step Solution
Verified Answer
Parametric equations: \\(x(t) = -1 + t\\), \\(y(t) = -2 - 3t\\)
1Step 1: Point and Direction Vector
Identify the given point, which is \((-1, -2)\), and the direction vector, which is \([1, -3]\). These will be used to write the parametric equations of the line.
2Step 2: Parametric Equation Setup
Each component of the point \((-1, -2)\) and the vector \[\begin{array}{r} 1 \ -3 \ \end{array}\] will be used to form the parametric equations of the line. This gives us two separate equations, one for \((x)\) and one for \((y)\).
3Step 3: X Equation Formulation
The equation for \((x)\) is formed from the point's x-value of -1 and the x-component of the direction vector, which is 1. Therefore, the parametric equation for \((x)\) is: \( x(t) = -1 + 1t = -1 + t.\)
4Step 4: Y Equation Formulation
Similarly, the equation for \((y)\) is formed from the point's y-value of -2 and the y-component of the direction vector, which is -3. Therefore, the parametric equation for \((y)\) is: \( y(t) = -2 - 3t.\)
5Step 5: Combine Parametric Equations
Combine the two equations to have the complete set of parametric equations for the line:- \(x(t) = -1 + t\)- \(y(t) = -2 - 3t\)
Key Concepts
Direction Vector and Its RoleUnderstanding Line Equations through ParametricsConnecting Parametric Equations and Calculus Education
Direction Vector and Its Role
In the realm of parametric equations, a direction vector is fundamental. This vector determines the direction in which the line extends. Think of it as an arrow pointing the way the line stretches across the plane. In our exercise, the direction vector is \([1, -3]\). Each component of this vector affects how the corresponding coordinates change.
A direction vector, by definition, shows how much the line rises or falls, and runs left or right.
A direction vector, by definition, shows how much the line rises or falls, and runs left or right.
- The first element, here being "1", indicates movement along the x-axis.
- The second one, "-3", describes movement along the y-axis.
Understanding Line Equations through Parametrics
Line equations are an essential part of understanding parametric equations. Unlike traditional line equations, which are often presented in the slope-intercept form \(y = mx + c\), parametric equations separate \(x\) and \(y\) into individual expressions that depend on a common parameter \(t\).
The parametric form of a line is given by:
\(x(t) = x_0 + at\)
\(y(t) = y_0 + bt\)
Where \((x_0, y_0)\) is a point on the line, and \((a, b)\) is the direction vector.
The parametric form of a line is given by:
\(x(t) = x_0 + at\)
\(y(t) = y_0 + bt\)
Where \((x_0, y_0)\) is a point on the line, and \((a, b)\) is the direction vector.
- This approach offers flexibility and clarity when plotting lines that aren't necessarily functions.
- In our specific problem, the parametric equations are \(x(t) = -1 + t\) and \(y(t) = -2 - 3t\).
Connecting Parametric Equations and Calculus Education
In calculus education, understanding parametric equations is a leap towards more advanced calculus topics. Parametric equations provide a solid foundation for curve sketching, a vital skill in calculus.
Unlike functions in the form \(y=f(x)\), parametric equations can easily deal with situations where a single x-value corresponds to multiple y-values, or where the function has no simple algebraic relationship.
This is particularly useful when describing motion or flow of multidimensional objects. For instance, consider a particle moving through space. Its path can be represented with parametric equations.
Unlike functions in the form \(y=f(x)\), parametric equations can easily deal with situations where a single x-value corresponds to multiple y-values, or where the function has no simple algebraic relationship.
This is particularly useful when describing motion or flow of multidimensional objects. For instance, consider a particle moving through space. Its path can be represented with parametric equations.
- Understanding the derivative of these equations helps find the velocity or acceleration of the particle.
- Grasping the integral of these equations can help determine the path length.
Other exercises in this chapter
Problem 44
Use a rotation matrix to rotate the vector \(\left[\begin{array}{l}-2 \\\ -3\end{array}\right]\) counterclockwise by the angle \(\pi / 9\).
View solution Problem 45
$$ A=\left[\begin{array}{rr} -1 & 1 \\ 2 & 3 \end{array}\right], \quad B=\left[\begin{array}{rr} 2 & -2 \\ 3 & 2 \end{array}\right] $$ Find the inverse (if it e
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$$ A=\left[\begin{array}{rr} -1 & 1 \\ 2 & 3 \end{array}\right], \quad B=\left[\begin{array}{rr} 2 & -2 \\ 3 & 2 \end{array}\right] $$ Find the inverse (if it e
View solution Problem 46
Find the parametric equation of the line in the \(x-y\) plane that goes through the indicated point in the direction of the indicated vector. $$ (-1,4),\left[\b
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