Transcript No Slide Title

Recall:
z
A path in R3 can be represented as
c(t) = (x(t),y(t),z(t))= x(t)i + y(t)j + z(t)k
(the position vector)
y
x
The tangent or velocity vector to c(t) is
c  (t) = (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)).
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
d
—[p(t)c(t)] = —(p(t)c1(t) , p(t)c2(t) , p(t)c3(t)) =
dt
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) .
d
d
—[b(t) • c(t)] = —[b1(t) c1(t) + b2(t) c2(t) + b3(t) c3(t)] =
dt
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
( [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) .
d
—[c(q(t))] =
dt
d
—( c1(q(t)) , c2(q(t)) , c3(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.
Suppose the position vector c(t) describes a path.
(a)
Find
d
d
—||c(t)||2 = — [c(t) • c(t)] .
dt
dt
d
—[c(t) • c(t)] = c  (t) • c(t) + c(t) • c  (t) = 2 c(t) • c  (t) .
dt
(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 = 0 
dt
c(t) • c  (t) = 0. Thus, c(t) and c  (t) are orthogonal for all t.
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 < 0) when t > 2.
Newton’s 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
Note that t = 2r0/s
st
st
mass M, described by
r(t) = ( r0cos— , r0sin— ) . is the time after
r0
r0
one revolution
According to Newton’s 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
Consequently,
GmM
mr (t) = – ——— r(t) .
||r(t)||3

(Note that the acceleration vector is in the opposite direction of the
position vector.)
st
st
The planet’s circular path r(t) = ( r0cos— , r0sin— ) has radius ||r(t)|| =
r0
r0
r0 .
st
st

The velocity of the planet is r (t) = (– s sin— , s cos— ) .
r0
r0
The speed of the particle is ||r  (t)|| = s .
2
2
s
st
s
st


The acceleration of the particle is r (t) = (– — cos— , – — sin— ) =
r0
r0 r0
r0
s2
st
st
s2
– — ( r0cos— , r0sin— ) = – — r(t) .
r0 2
r0
r0
r0 2
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
r0 2
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.
We can then derive Kepler’s 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10–11
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 Kepler’s Law, we find that
–11)(5.981024)
GM
(6.6710
r03 = T2 —— = (86,400)2 —————————— =
(2)2
(2)2
7.541022 meters3
r0 = 4.23107 meters  26,200 miles