Question

in exercises 27–32 find a parametrization for the curve. 28. the line through the points (-3, -3) and (5, 1)

Explanation:

Step1: Find the direction vector

The direction vector \(\vec{v}\) between points \((-3, -3)\) and \((5, 1)\) is \(\langle 5 - (-3), 1 - (-3)
angle=\langle 8, 4
angle\). We can simplify it by dividing by 4, getting \(\langle 2, 1
angle\).

Step2: Use parametric equations formula

The parametric equations for a line through a point \((x_0,y_0)\) with direction vector \(\langle a,b
angle\) are \(x = x_0+at\) and \(y = y_0 + bt\), where \(t\in\mathbb{R}\). Let's use the point \((-3,-3)\) as \((x_0,y_0)\) and \(\langle 2,1
angle\) as \(\langle a,b
angle\). So we have:
\(x=-3 + 8t\) (or \(x=-3+2t\), both are correct, we can use the non - simplified direction vector as well) and \(y=-3 + 4t\) (or \(y=-3 + t\)). Another common way is to use the parameter \(t\) such that when \(t = 0\) we are at \((-3,-3)\) and \(t = 1\) we are at \((5,1)\). The parametric equations can be written as:
\(x=-3+(5 - (-3))t=-3 + 8t\)
\(y=-3+(1 - (-3))t=-3 + 4t\), where \(t\in\mathbb{R}\)

Answer:

A possible parametrization is \(x=-3 + 8t\), \(y=-3 + 4t\) for \(t\in\mathbb{R}\) (or equivalent forms like \(x=-3+2t\), \(y=-3 + t\) for \(t\in\mathbb{R}\))