NCSSM Online AP Calculus (BC) Distance Education & Extended Programs Lesson2.3...IntroductiontoVectors INTRODUCTION In this lesson, we will discuss the basics of vectors and their relationship to parametrically defined curves. A vector in the 𝑥𝑦-plane can be described as a directed line segment from a point of origin to another point (𝑎, 𝑏). A vector labeled 𝑣 from the origin to the point (3, 5) can be denoted as 3, 5 . The x component, or horizontal component, of the vector 𝑣 is 3. The y component, or vertical component, is 5. Note that we use the angled brackets to denote a vector; however, some mathematicians express vectors in the context of parametric equations using parentheses. Also notice that the v has an arrow over it to help us distinguish v as a vector. Since parametric equations can represent the path of a moving object, and they are expressed as a pair of components in two-dimensional space (horizontal: 𝑥(𝑡), and vertical: 𝑦(𝑡)), vectors allow us to express this pair in a single form, namely 𝑥 𝑡 ,𝑦(𝑡) . In the context of parametric equations, a vector provides two important pieces of information—namely the direction of the object and its speed. As we learned earlier in this unit, we can also analyze the behavior of a moving object given the parametric equations that model that behavior by looking at the rates of change of the object’s position in the horizontal direction !"

!" and that in the vertical direction !"

!". We can then express this in vector

form as the velocity vector: dxdt

, dydt

.

Even further, we can describe the acceleration of the object in each of the horizontal and vertical

directions by finding the acceleration vector, which is defined as: d2xdt2

, d2ydt2

SPEED OF AN OBJECT Consider an object’s path defined parametrically by 𝑥 𝑡 ,𝑦(𝑡) . In one unit of time, the object moves !"

!"

units in the horizontal direction and !"!"

in the vertical direction. Using the Pythagorean Theorem, we know,

then, that the distance the object traveled over that one unit of time is !"!"

!+ !"

!"

! units.

So, the speed of the object is

NCSSM Online AP Calculus (BC) Distance Education & Extended Programs Example #1: A particle moves in the xy-plane so that the position of the particle at any time t is given by 𝑥 𝑡 = 2𝑒!! + 𝑒!!! and 𝑦 𝑡 = 3𝑒!! − 𝑒!!!. We will find the velocity and acceleration vectors for the particle in terms of t , evaluate them at 𝑡 = 0, and find the speed of the particle at time 0t = .

With position vector 2𝑒!! + 𝑒!!! , 3𝑒!! − 𝑒!!! , we find each of the following:

𝑑𝑥𝑑𝑡 = 6𝑒!! − 7𝑒!!!

𝑑𝑦𝑑𝑡 = 9𝑒!! + 2𝑒!!!

So, our velocity vector is 6𝑒!! − 7𝑒!!! , 9𝑒!! + 2𝑒!!! , and at 𝑡 = 0, we have −1, 11 . Since the x-component is negative at 𝑡 = 0 and the y-component is positive, we know that the object is moving to the left and up at that instant (faster in the vertical direction).

We also find the following:

𝑑!𝑥𝑑𝑡! = 18𝑒!! + 49𝑒!!!

𝑑!𝑦𝑑𝑡! = 27𝑒!! − 4𝑒!!!

So, our acceleration vector is 18𝑒!! + 49𝑒!!! , 27𝑒!! − 4𝑒!!! , and at 𝑡 = 0, we have 67, 23 . Since each component is positive at 𝑡 = 0, we know that the object is speeding up

both vertically and horizontally.

The speed of the object in terms of 𝑡 is 6𝑒!! − 7𝑒!!! ! + 9𝑒!! + 2𝑒!!! !

At 𝑡 = 0, we find that the object’s speed is 1! + 11! = 122 ≈ 11.05 position units per unit time.

Example #2: Consider a particle moving in the xy-plane whose path is given by the parametric equations 𝑥 = 𝑡! − 3𝑡! and 𝑦 = 2𝑡! − 3𝑡! − 12𝑡. When is the particle at rest?

Well,weknowthatthespeedoftheparticlecanbecalculatedbyfinding !"!"

!+ !"

!"

!.

So,thespeedoftheparticleis 3𝑡! − 6𝑡 ! + 6𝑡! − 6𝑡 − 12 !.Thisisequalto0when3𝑡! − 6𝑡 ! + 6𝑡! − 6𝑡 − 12 ! = 0.Thismeansthateachofthetermsinthesummustbe0,sincetheyarebothpositive.3𝑡! − 6𝑡 ! = 0when3𝑡! − 6𝑡 = 0,whichiswhen𝑡 = 0and2.Also, 6𝑡! − 6𝑡 − 12 ! = 0when6𝑡! − 6𝑡 − 12 = 0.Factoring,weget6 𝑡! − 𝑡 − 2 = 0.Thisistruewhen𝑡 = 2.Therefore,theparticleisatrestwhen𝑡 = 2.

NCSSM Online AP Calculus (BC) Distance Education & Extended Programs TOTAL DISTANCE TRAVELED BY AN OBJECT Remembering that the area under a velocity curve gives to us the total distance traveled by an object, we can use the formula for the speed of an object whose motion is described parametrically to determine the total distance that the object traveled. Total distance traveled can thus be given by the integral of speed over an interval.