Calc 3 Lecture Notes
Section 11.3
Page 1 of 6
Section 11.3: Motion in Space
Big idea: Vector-valued functions can be used to efficiently describe the position, velocity, and
acceleration of an object moving through space.
Big skill: You should be able to compute the velocity, acceleration, and force vectors for an
object given its position vector function, or derive its position vector function given the force on
it.
Foundational Concept: If the terminal point of a vector-valued function r(t) represents the
position of an object at time t, then r(t) represents the velocity vector of the object (and thus,
v(t) = r(t) represents the acceleration vector).
Reasoning:
For an object’s position vector at time t given by the vector-valued function
r  t   f  t  , g  t  , h  t  , we saw in section 11.2 that the derivative r  t   f   t  , g   t  , h  t 
is a vector that is tangent to the curve and that points in the direction of motion at time t, which is
consistent with what the direction of the velocity vector should be. So, it remains to be seen if
the magnitude of r(t) is really the speed of the object.
To see that this is true, start by computing the magnitude of r(t):
r '(t ) 
Also recall that the arc length s of the curve from t 0  t  t1 is given by
s
t1
t0
 f ' (t )2  g ' (t )2  h' (t )2
dt
The function that represents the arc length from t 0 to some arbitrary upper limit t is given by:
s (t )  
t
t0
 f ' (u )2  g ' (u )2  h' (u )2
du
Use Part II of the Fundamental Theorem of Calculus to calculate s(t):
s ' (t ) 
Thus we see that the vector r(t):
(a) points in the instantaneous direction of motion of the object
(b) has a magnitude equal to the rate of change of arc length with respect to time.
r(t) = v(t), the velocity vector.
Also, v(t) = r(t) = a(t), the acceleration vector.
Calc 3 Lecture Notes
Section 11.3
Page 2 of 6
You will be expected to do one of two calculations based on this interpretation:
1. Given r(t), find v(t) and a(t), by differentiation.
2. Given a(t), find v(t) and r(t), by integration.
Additional Fact: Newton’s Second Law states that F = ma, which thus relates force on an object
to its acceleration, which will be a starting point many times for #2.
Practice:
1. Find the force acting on a 2 kg ball moving in a circular path of radius 3 m, with the
period of rotation being 3.7 sec.
2. Find an expression for the centripetal acceleration of any object following a circular path
of radius R at a constant speed v. Relate this expression to the angular speed .
Calc 3 Lecture Notes
Section 11.3
Page 3 of 6
3. Find the velocity and position vectors given
a  t   10k , and initial conditions v  0  i  j- k and r  0   2i  3j
Calc 3 Lecture Notes
Section 11.3
Page 4 of 6
4. Suppose a projectile is fired from ground level with an initial speed of 500 feet per
second at an angle of 30  to the horizontal. Ignoring wind resistance and assuming that
gravity is the only force acting on the object, find: (a) the maximum height of the object;
(b) the time to impact; (c) the speed of the object at impact; and (d) the horizontal range.
Calc 3 Lecture Notes
Section 11.3
Page 5 of 6
5. Show that the general equations for a projectile launched with an initial velocity v0 at an
angle of inclination  from a location (x0, y0) are:
a  t   0,  g
v  t   v0 cos   , v0 sin    gt
1
r  t   x0  v0 cos   t , y0  v0 sin   t  gt 2
2
v sin  
Time to reach maximum altitude: tmax  0
g
1 v sin  
Maximum altitude: ymax  y0  0
2
g
Range: Find the time to reach the ground, then use that time to calculate horizontal
distance.
2
2
Calc 3 Lecture Notes
Section 11.3
Page 6 of 6
6. Suppose a 1 kg projectile is launched in the y-z plane with an initial speed of 300 m/s at
an angle of 52 above the horizontal (i.e., the y-axis) into a cross-breeze so that the force
it feels is given by F  1.0N,0, mg . Find (a) the maximum height of the projectile; (b)
the speed of the object at impact; and (c) the location of the impact.