y

1
Recall
z
A path in R3 can be represented as c(t) (the
position vector)
(x(t),y(t),z(t)) x(t)i y(t)j z(t)k
y
The tangent or velocity vector to c(t) is c ? (t)

x
(x ? (t),y ? (t),z ? (t)) x ? (t)i y ? (t)j
z ? (t)k
If c(t) represents the path of a moving particle,
then we often let
v(t) c ? (t) represent its velocity vector,
a(t) v ? (t) c ? ? (t) represent its
acceleration vector,
s v(t) c ? (t) represent its speed.
Let b(t) (b1(t) , b2(t) , b3(t)) and c(t)
(c1(t) , c2(t) , c3(t)).
2
Let b(t) (b1(t) , b2(t) , b3(t)) and c(t)
(c1(t) , c2(t) , c3(t)). d b(t) c(t) dt
d (b1(t) c1(t) , b2(t) c2(t) , b3(t)
c3(t)) dt
(b1 ? (t) c1 ? (t) , b2 ? (t) c2 ? (t) , b3 ?
(t) c3 ? (t))
(b1 ? (t) , b2 ? (t) , b3 ? (t)) (c1 ? (t) , c2
? (t) , c3 ? (t))
b ? (t) c ? (t) .
d p(t)c(t) dt
d (p(t)c1(t) , p(t)c2(t) , p(t)c3(t)) dt
(p ? (t)c1(t) p(t)c1 ? (t) , p ? (t)c2(t)
p(t)c2 ? (t) , p ? (t)c3(t) p(t)c3 ? (t))
(p ? (t)c1(t) , p ? (t)c2(t) , p ? (t)c3(t))
(p(t)c1 ? (t) , p(t)c2 ? (t) , p(t)c3 ? (t))
p ? (t)(c1(t) , c2(t) , c3(t)) p(t)(c1 ? (t) ,
c2 ? (t) , c3 ? (t))
p ? (t)c(t) p(t)c ? (t) .
3
d b1(t) c1(t) b2(t) c2(t) b3(t) c3(t) dt
d b(t) c(t) dt
b1 ? (t)c1(t) b1(t)c1 ? (t) b2 ?
(t)c2(t) b2(t)c2 ? (t) b3 ? (t)c3(t)
b3(t)c3 ? (t)
b1 ? (t)c1(t) b2 ? (t)c2(t) b3 ?
(t)c3(t) b1(t)c1 ? (t) b2(t)c2 ? (t)
b3(t)c3 ? (t)
b ? (t)c(t) b(t)c ? (t) .
d b(t) ? c(t) dt
d (b2(t)c3(t) b3(t)c2(t) , b3(t)c1(t)
b1(t)c3(t) , b1(t)c2(t) b2(t)c1(t)) dt
4
( b2 ? (t)c3(t) b2(t)c3 ? (t) b3 ?
(t)c2(t) b3(t)c2 ? (t) ,

,

)

( b2 ? (t)c3(t) b3 ? (t)c2(t) b2(t)c3 ?
(t) b3(t)c2 ? (t) ,

,

)

(b2 ? (t)c3(t) b3 ? (t)c2(t) , , )
(b2(t)c3 ? (t) b3(t)c2 ? (t) , ,
)
b ? (t) ? c(t) b(t) ? c ? (t) .
5
d ( c1(q(t)) , c2(q(t)) , c3(q(t)) ) dt
d c(q(t)) dt
( c1 ? (q(t)) q ? (t) , c2 ? (q(t)) q ? (t) , c3
? (q(t)) q ? (t) )
q ? (t) ( c1 ? (q(t)) , c2 ? (q(t)) , c3 ? (q(t))
)
q ? (t) c ? (q(t)) .
All of the preceding derivative formulas, except
the one concerning cross product, can be extended
to Rn for any n. These are the differentiation
rules displayed on page 262 of the text.
6
Suppose the position vector c(t) describes a path.
d d (a) Find c(t)2 dt dt
c(t) c(t) .
d c(t) c(t) dt
c ? (t) c(t) c(t) c ? (t) 2 c(t) c ?
(t) .
(b)
If c(t) has constant length, that is, c(t) is
a constant, what kind of path would this describe
in R2 or in R3?
a path which lies on either a circle or a sphere
centered at the origin
(c)
If c(t) has constant length, then why must c(t)
and c ? (t) be orthogonal for all t?
d If c(t) is a constant, then c(t)2
dt
0 ?
c(t) c ? (t) 0. Thus, c(t) and c ? (t) are
orthogonal for all t.
7
Suppose c(t) (x(t) , y(t) , z(t)) describes a
path of a particle where a(t) k. When t 0,
the position of the particle is (0,0,1), and the
velocity is i j. Find c(t), and find the
values of t ? 0 for which the particle is below
the xy plane.
a(t) k ? v(t)
p1i p2j ( t p3)k , for some constants p1
, p2 , p3 .
v(0) i j ? v(t)
i j tk
v(t) i j tk ? c(t)
(t q1) i (t q2) j ( t2/2 q3)k ,
where q1 , q2 , q3 are constants.
c(0) k ? c(t)
ti tj (1 t2/2)k .
The particle is below the xy plane (z lt 0) when
t gt ?2.
8
Newtons Second Law (page 264 of the text) states
that if F(t) is the force at time t acting on a
particle with mass m, then
F(t) ma(t) .
Let us consider a circular path of a planet of
mass m orbiting the sun with mass M, described by
Note that t 2?r0/s is the time after one
revolution
st st r(t) ( r0cos , r0sin )
. r0 r0
According to Newtons Law of Gravitation, the
force acting on the planet is described by
GmM F(t) r(t) , where G is the
gravitational constant. r(t)3
GmM Consequently, mr ? ? (t)
r(t) . r(t)3
(Note that the acceleration vector is in the
opposite direction of the position vector.)
9
st st The planets circular path
r(t) ( r0cos , r0sin ) has radius r(t)
r0 r0
r0 .
st st ( s sin , s cos ) .
r0 r0
The velocity of the planet is r ? (t)
s .
The speed of the particle is r ? (t)
s2 st s2 st ( cos ,
sin ) r0 r0 r0 r0
The acceleration of the particle is r ? ? (t)
s2 r(t) . r02
st st ( r0cos , r0sin )
r0 r0
s2 r02
mr ? ? (t) ma(t) is called the centripetal
force, and we now see that
GmM ms2 r(t) mr ? ?
(t) r(t) ? s2 GM / r0 .
r03 r02
Since the period for each of sin(kx) and cos(kx)
is 2?/k, then the period for the orbiting body is
T
2?r0/s.
10
We can then derive Keplers Law (page 266 of the
textbook), which says that the square of the
period is proportional to the cube of the radius
(2?)2 T2 r03 . GM
While we have considered a planet orbiting the
sun, all our derivations can be applied to a
satellite orbiting the earth. A satellite is said
to be in geosynchronous orbit around the earth,
if it is always over the same point above the
equator. Given G 6.67?1011 when working with
meters, kilograms, and seconds, and given the
mass of the earth is 5.98?1024 kilograms, find
the radius necessary for a geosynchronous orbit
above the equator.
We want the period of the orbit to be 1 day
60?60?24 seconds 86,400 seconds. From Keplers
Law, we find that r03
GM T2 (2?)2
(6.67?1011)(5.98?1024) (86,400)2
(2?)2
7.54?1022 meters3
4.23?107 meters ? 26,200 miles
r0