Exercise 1. Read Example 3 (p. 773 - 774). Summarize how we can obtain the position function of an object based on its acceleration and initial velocity and position. Be sure to specify the order in which the computations are done. We have a nice physical example of this by considering what happens with projectile motion, when an object with some initial motion moves under only the force of gravity.

Chapter 13 Vector-Valued Functions and Motion in Space 13.2 Integrals of Vector Functions; Projectile Motion 773 As in the integration of scalar functions, we recommend that you skip the steps in Equa- tions (1) and (2) and go directly to the final form. Find an antiderivative for each compo- nent and add a constant vector at the end. Definite integrals of vector functions are best defined in terms of components. The definition is consistent with how we compute limits and derivatives of vector functions. DEFINITION If the components of r() = f(1)i + g(1)j + h(0)k are integrable over [ a, b ], then so is r, and the definite integral of r from a to b is EXAMPLE 2 As in Example I, we integrate each component. ((cos ni + 1 - anko) ds = ("costar )it (far) - ( 2sai) k sint it ti - 2 k = [0 - 0]i + [* - 0]j - [#2 - 0'] k = wj - wk The Fundamental Theorem of Calculus for continuous vector functions says that rode = RID = R(b) - Ria) where K is any antiderivative of r, so that K'(f) = r(1) (Exercise 46). Notice that an anti- derivative of a vector function is also a vector function, whereas a definite integral of a vector function is a single constant vector,EXAMPLE 3 Suppose we do not know the path of a hang glider, but only its accel- eration vector a() = -(3 cos ni - (3sin nj + 2k. We also know that initially (at time 1 = 0) the glider departed from the point (4, 0, 0) with velocity v(0) = 3j. Find the glid- er's position as a function of 1. Solution Our goal is to find r(() knowing The differential equation: d'r = -(3 cos ni - (3 sin nj + 2k The initial conditions: v(0) = 3j and r(0) = 41 + 0j + 0k. Integrating both sides of the differential equation with respect to f gives v() = -(3 sin ni + (3cos nj + 2/k + C,. We use v(0) - 3j to find C: 3j = -(3 sin Oji + (3 cos O)j + (0)k + C, 3j = 31 + C1 C1 = 0. The glider's velocity as a function of time is " - vin) = -(3 sin ni + (3cos nj + 2rk.Chapter 13 Vector-Valued Functions and Motion in Space 774 Chapter 13 Vector-Valued Functions and Motion in Space Integrating both sides of this last differential equation gives r(?) = (3cos n)i + (3sinn)j + / k + Cz. We then use the initial condition r(0) = 4i to find Cz: 41 = (3 cos O)i + (3 sin O)j + (0')k + Cz 4i = 31 + (0)j + (O)k + C, C = i. The glider's position as a function of f is (4, 0, 0) r(0 = (1 + 3cos /)i + (3sinnj + /k. This is the path of the glider shown in Figure 13.9. Although the path resembles that of a FIGURE 13.9 The path of the hang helix due to its spiraling nature around the z-axis, it is not a helix because of the way it is glider in Example 3. Although the path rising. (We say more about this in Section 13.5.) spirals around the z-axis, it is not a helix.The Vector and Parametric Equations for Ideal Projectile Motion A classic example of integrating vector functions is the derivation of the equations for the motion of a projectile. In physics, projectile motion describes how an object fired at some angle from an initial position, and acted upon by only the force of gravity, moves in a ver- tical coordinate plane. In the classic example, we ignore the effects of any frictional drag on the object, which may vary with its speed and altitude, and also the fact that the force of gravity changes slightly with the projectile's changing height. In addition, we ignore the long-distance effects of Earth turning beneath the projectile, such as in a rocket launch or the firing of a projectile from a cannon. Ignoring these effects gives us a reasonable approximation of the motion in most cases. To derive equations for projectile motion, we assume that the projectile behaves like a particle moving in a vertical coordinate plane and that the only force acting on the project n sin a j tile during its flight is the constant force of gravity, which always points straight down. We assume that the projectile is launched from the origin at time / = 0 into the first quadrant with an initial velocity vo (Figure 13.10). If vo makes an angle a with the horizontal, then r = 0 at Ya cas ari timer = 0 Wo = (| vol cosa )i + (| volsin a )j. If we use the simpler notation up for the initial speed | vol, then Vo = (up cos a)i + (un sin a)j. (3) The projectile's initial position is To = 0i + 0j = 0. (4) Newton's second law of motion says that the force acting on the projectile is equal to the projectile's mass in times its acceleration, or m(d'r/dr ) if r is the projectile's position vector and r is time. If the force is solely the gravitational force -maj, then raxi+yi = -mgj and d'T= -gj. where g is the acceleration due to gravity. We find r as a function of / by solving the Horizontal range following initial value problem. FIGURE 13.10 (a) Position, velocity, Differential equation: d'r = -gj acceleration, and launch angle at / = 0. (b) Position, velocity, and acceleration Initial conditions: r= no and dI = Vo when / = 0 at a later time r