This section is dedicated to show important applications of vector valued functi

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This section is dedicated to show important applications of vector valued functions. We use these functions to represent the components of motion in space, specifically: the position of an object, velocity, and acceleration. Definition 1 Assume an object is moving on a curve T(t) in the space The position function of the object is given by T(t) (x(t),y(t),z(t The velocity function of the object is given by V(t) 7 (x (t),y(t),z (t)) The acceleration function of the object is given by a(t (t) T'(t) "(t),y'(t),z"(t)) The speed function of the object is given by lv(t) V x (t)2+ y +z (t) Example i Assume the object is moving along T(t) 2 ,3 The velocity function is vt T (t) (1,2,3t2). For example, at t 1, the velocity vector is VC1) (1,2,3). The acceleration function is given by t) V(t) (0,2,6t) Example 2 Let the position function of an object in the plane be r(t) (3cost,3sint). Note that x(t)2+y(t)2 9cos2 t +9 sin t 9. This shows that the trajectory of the moving object is a circle of radius 3 and centered at the origin. If we take r(t) (2 +t,-1-t, 3-2t) as the position function of an object, then the trajectory of motion is a line in the space passing through the point (2,-1,3) and in the direction of (1,-1,-2) Note that in the first definition, we used derivatives to find velocity from position and acceleration from velocity. However, we can reverse this process. If we are given the acceleration function, we can recover the velocity function using integration. Similarly, we can recover the position function from the velocity function using integration. Definition 2 Given the acceleration function a(t) f(t),g(t),h(t), the velocity function is i (t) Ta(t) dt f(t) dt, Tg (t) dt, h(t) dt). We can similarly find the position function r(t) i (t) dt. Example 3 Let a(t) cost, sint) with the initial velocity V(0) (0,1) and initial position T(0) (1,0) ro substituting 0 gives (0,1) (0+C cost 2) 0, C The position function can be found by integrating ii(t), thus r(t) cost C1, sint +2t Co). But substituting t 0 gives cost 2, sintt 2t 2, C r(t

Explanation / Answer

object is passing through (1,2,3) and moving vertically.

it means x and y co-ordinate is not changing.

so x co-ordinste will be 1

y co-ordinste will be 2

z will only change.

position function- (1,2,z) where z is passing through 3

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