Vector-Valued Functions

Unit Vectors:

A unit vector is a vector with length 1.  It can be found by dividing a vector by its magnitude.

Find the unit vector for v = < 3,-1 >.
  Answer:  magnitude = 
                   unit vector = 

If u is a unit vector in standard position, then u = <cos q, sin q>.
 

Linear Combinations:

i = <1,0> and j = <0,1> are standard unit vectors

Let v = <3,-1>.  Write v as a linear combination of i and j.
  Answer:  3i - j

Find the direction angle for v = 3i - j.
  Answer:  arctan q = -1/3
                             q = -18.435o  or341.565o since the vector is in the fourth quadrant
 

Limits and Continuity:

Let r(t) = f(t)i + g(t)j
The limit as t approaches c of r(t) is the sum of the limits on f(t)i and g(t)j as t approaches c.
r(t) is continuous at a point t = c in its domain if the limit of r(t) as t approaches c is equal to r(c).  Both f and g must be continuous at t = c.

Check for continuity:
r(t) = (t cos t)i + (t sin t)j
  Answer:  Since t cos t and t sin t are both continuous functions, this vector is continuous.

r(t) = (1/t)i + (sin t)j
  Answer: This vector is continuous over its domain (nonzero real numbers).  However, it is not
                    continuous at t = 0 since 1/t is not continuous there.

Derivatives: dr/dt = (df/dt)i + (dg/dt)j
 

Differentiation rules:

1.  d(C)   =  0  (Constant Rule)
     dt

2.  d (cu)  =  c  du    (where u is a vector and c is a scalar)
      dt               dt

3.  d  (u + v)   =  du  +  dv  (where u and v are vectors)
        dt                dt       dt

4.  d  (u - v)   =   du  -  dv    (where u and v are vectors)
         dt               dt      dt

Dot Product  Rule:
5.  d ( u × v)  =  du  ×  v   +   u ×  dv   (where u and v are vectors)
          dt            dt                       dt

6.  dr / ds  =  (dr / dt  )(dt / ds )  (where r is a vector)
 

Position, Velocity, and Acceleration:

v(t) = dr/dt   is the particle’s velocity vector and is tangent to the curve.
||v(t)|| is the particle’s speed.  (magnitude of v)
a(t) = dv/dt = d2r/dt2 is the particle’s acceleration vector.
   v    is the direction of motion.  This is a unit vector.
||v||
Velocity = ||v||    v
                        ||v||

Problem:
r(t) = (cos 2t)i + (2 sin t)j at t = 0

1.  Draw the graph of the path of the particle.

2.  Find the velocity and acceleration vectors.
     v(t) = (-2 sin 2t)i + (2 cos t)j
     a(t) = (-4 cos 2t)i - (2 sin t)j

3.  Find the particle’s speed and direction of motion at the given value of t.
     speed = ||(-2 sin 2t)i + (2 cos t)j|| = 
     At t = 0, this speed = 

4.  Write the particle’s velocity at that time as the product of its speed and direction.
      2((-2 sin 2t)i + (2 cos t)j)
                    2
     At t = 0, this is  2j = j
                              2

Integrals:

To integrate a vector, integrate each component.

Problems:
1.  Solve for r when dr/dt = (t3 + 4t)i + tj when r(0) = i + j
     Answer:  r = (3t + 2t2 + C1 )i + (t2/2 + C2 )j
                      i + j = C1i + C2 j  (substituting r(0) = i + j  in)
                      Both C 's are 1.
                      So  r = (3t + 2t2)i + (t2/2)j
 
 
 


Problems (There are no questions here yet)

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