12.8 Algebraic Vectors & Parametric Equations

1. 12.8 Algebraic Vectors & Parametric Equations

2. In 12-7, we focused on the geometric aspect of vectors. 12-8 focuses on the algebraic properties. (x, y) Note: the book will use ( , ) for a point and a vector! Be careful! We will use  ,  y x is the norm/ magnitude Unitvector: vector with magnitude 1 is the unit vector w/ same direction as (horizontal) (vertical) Component Form: Polar Form:

3. We can determine a vector if we know its initial (x1, y1) and terminal points (x2, y2). Ex 1) Given v with initial point (2, 3) & terminal point (7, 9), determine: component form polar form unit vector in same direction as

4. Vector operations: vector sum: vector difference: scalar multiplication: Ex 2)

5. In this picture, Q(x, y) is a point and P(a, b) is a point. P(a, b) If you wanted to get to Q(x, y) from P(a, b), we could add a vector (x, y) = (a, b) + vector Since we don’t know the size of the vector, we can multiply by a scalar to get to the point. Q(x, y) , the direction vector will be given to you, or you can find it by subtracting the 2nd point – 1st point that they give you. so, (x, y) = (a, b) + tc, d (x, y) = (a, b) + tc, td (x, y) = (a + tc, b + td) this leads to x = a + tc and y = b + td these two eqtns are called parametricequations with parameter t of the line

6. Ex 3) Determine a direction vector of the line containing the two points P(5, 8) and Q(11, 2). Then find the equation of the line & a pair of parametric equations of the line. direction vector = Q – P = (11 – 5, 2 – 8) = 6, –6 vector equation of line: (x, y) = (5, 8) + t6, –6 parametric: x = 5 + 6t and y = 8 – 6t We can also represent other graphs (not just lines) in parametric. Ex 4) Graph the curve with parametric equations x = 3cost and y = 5sint. Find an equation of the curve that contains no other variables but x & y. square Ellipse put together 1

7. Homework #1208 Pg 651 #1, 3, 6, 9, 11, 15–18, 20, 21, 24, 29, 31, 33, 35