Lesson 3 Part 3
Remember that
r t fti  gtj  htk
represnts a curve in the space.
Read the pages 849-851
and note that if ft, gt, ht are twice differentiable functions of t
then
v 
dr
dt
and a 
d2 r
dt 2
are the velocity at acceleration vectors at time t
The speed is given by
v
Let us look at
exercise #16 on the page 854
To find the velocity, speed and the acceleration of the object that travels along
r 〈e t cos t, e t sin t, e t 
First note that
x  e t cos t
1
y  e t sin t
z  et
gives
x 2 y 2  e 2t cos 2 t  e 2t sin 2 t
or
x 2 y 2  e 2t cos 2 t  sin 2 t
or
x 2 y 2  z 2
which means that the curve is contained on the cone
2
3
The velocity vector
v  ddtr  dtd e t cos ti  e t sin tj  e t ke t cos t − e t sin ti e t sin t  e t cos tj  e t k
v e t cos t − e t sin ti e t sin t  e t cos tj  e t k
v
 e t cos t − e t sin t 2  e t sin t  e t cos t 2  e t  2
 e 2t cos 2 t  e 2t sin 2 t − 2e 2t cos t sin t  e 2t cos 2 t  e 2t sin 2 t  2e 2t cos t sin t  e 2t
 e 2t cos 2 t  e 2t sin 2 t  e 2t cos 2 t  e 2t sin 2 t  e 2t
 e 2t cos 2 t  sin 2 t  e 2t cos 2 t  sin 2 t  e 2t
4
 e 2t  e 2t  e 2t
 3e 2t
e t 3
The acceleration vector is
a ddtv e t cos t − e t sin t − e t sin t − e t cos ti e t sin t  e t cos t  e t cos t − e t sin tj  e t k
a −e t sin ti e t cos tj  e t k
APPLICATION:
Let us consider the following motion in a plane with the path described by the position vector
r  5 cos
1
4
t i  5 sin
1
4
t j
Note that
x  5 cos
1
4
y  5 sin
t
x 2 y 2  25 cos 2
x 2 y 2  25 cos 2
1
4
t 25 sin 2
1
4
t  sin 2
1
4
1
4
1
4
t
t
t
x 2 y 2  25
The path is slong the circle
5
5
P
4
3
rG
r
2
1
-5
-4
-3
-2
θ
-1
1
2
3
4
5
-1
-2
-3
-4
-5
Where  
Note that
1
4
d
dt
t

1
4
rad/per unit time
Note that
v  ddtr  dtd 5 cos
1
4
t i  5 sin
1
4
t j  −
5
4
sin
1
4
t
i  5 cos
1
4
t
j
You can easily note that
rv 0
6
5
v
4
3
r
2
1
-5
-4
-3
-2
-1
1
2
3
4
5
6
-1
-2
-3
-4
5
7
The acceleration vector is a 
a − 165 cos
Note that
1
4
a
5
i − 16
sin
t

1
4

dv
dt
d
dt
−
5
4
sin
1
4
t
i  5 cos
1
4
t
j
j
t
5
16
and also
a − 161 5 cos
1
4
t
i  5 sin
1
4
t
j
or
a − 161 r
8
5
v
r
4
a
3
2
1
-5
-4
-3
-2
-1
1
2
3
4
5
6
-1
-2
-3
4
9
In general if a particle moves along a circle given by r  b coswti  b sinwtj
then
v  ddtr  −wbsinwti  wbcoswtj
v  bw −sinwti  coswtj
v
|bw|
a ddtv
a −bw 2 coswti  sinwtj
a −w 2 r
.................
On the other hand we know the acceleration vector,
d2 r
dt 2
we can integrate twice and find the position vector function r t
Let us take up
#22 on the page 854
Given that
10
a  −cos ti − sin tj GIVEN THAT v 0  j  k
dv
dt
r 0  i
 −cos ti −sin tj
Integrate with respect to t
v  −sin ti cos tj C
Given that v 0  j  k
Therefore
−sin 0i cos 0j C j  k
j C j  k
Therefore C k
v  −sin ti cos tj C
→
dr
 −sin ti cos tj  k
dt
integrate with respect to t
r cos ti sin tj  tk D
Given that
11
r 0 i
cos 0i sin 0j 0k D i
i D i
→
D 0
r cos ti sin tj  2tj  tk
r cos ti sin tj  tk
12
13
A nice example of obtaining the position vector function r by integrating a twice
is the example where
a −gj
where g is the acceleration due to gravity, is described on the pages 852-853 of the text
I am going to show you the steps using an exercise in the text book:
#34 on the page 855
Even though you can just use the summary formulas, let us still work on it from scratch.
14
35 degrees
7 feet
4 feet
Considering this as a motion in the i-j plane
Note that the acceleration vector is
d2 r
dt 2
 ddtv  −g j
15
where g  32 ft/ sec 2
dv
dt
 −g j
→
v  −gt j C
If the speed at the time of release is v ∘ feet per second
the inital velocity v ∘ as a vector is
v ∘  v ∘ cos 35 ∘ i  v ∘ sin 35 ∘ j
that is, when t  0, v  v ∘  cos 35 ∘ i  sin 35 ∘ j
v ∘ cos 35 ∘ i  v ∘ sin 35 ∘ j  −g0 j C
therefore
C v ∘ cos 35 ∘ i  v ∘ sin 35 ∘ j
v  −gt j  v ∘ cos 35 ∘ i  v ∘ sin 35 ∘ j
v  v ∘ cos 35 ∘ i −gt  v ∘ sin 35 ∘ j
dr
dt
 v ∘ cos 35 ∘ i −gt  v ∘ sin 35 ∘ j
16
Integrate
r  v ∘ tcos 35 ∘ i  −g t2  v ∘ t sin 35 ∘ j D
2
Since, the ball was project initially from a height of 7 feet, taking the position of the Quarter back as the origin,
the position of the ball at time t  0 sec is
r ∘  7j
When t  0, we have r  r ∘  7j
7j  v ∘ 0cos 35 ∘ i  −g 02  v ∘ 0 sin 35 ∘ j D
2
D 7j
r  v ∘ tcos 35 ∘ i  −g t2  v ∘ t sin 35 ∘ j  7j
2
r  v ∘ tcos 35 ∘ i  7 − g t2  v ∘ t sin 35 ∘ j
2
Compare this with the equation
2
r t  v ∘ tcos i  h − g t2  v ∘ t sin  j on the page 853 of the text book
let us work on the problems now,
We have x  v ∘ tcos 35 ∘  and y  7 − g t2  v ∘ t sin 35 ∘
2
a)
17
35 degrees
R
7 feet
4 feet
Origin
30 yards=90feet
Note that (300,4) is a point on the curve that is given by
x  v ∘ tcos 35 ∘  and y  7 − g t2  v ∘ t sin 35 ∘
2
At the point, where the ball is received
v ∘ tcos 35 ∘   90 .......... (1)
18
7 − g t2  v ∘ t sin 35 ∘
2
 4 .......... (2)
Equation (1) gives
t
90
v ∘ cos 35 ∘ 
at R
Substitute the value is (2)
7 − 32 
90
v ∘cos 35 ∘ 
2
 v∘
2
90
v ∘ cos 35 ∘ 
sin 35 ∘
4
→
7 − 32 
90
v ∘cos 35 ∘ 
2

2
90
cos 35 ∘ 
sin 35 ∘
4
→
7 − 16
90
v ∘ cos 35 ∘ 
2

90
cos 35 ∘ 
sin 35 ∘
4
→
−16
90
v ∘ cos 35 ∘ 
2
 4 − 7 − cos9035 ∘  sin 35 ∘
→
19
90
v ∘ cos 35 ∘ 
2

−3−
90
cos 35 ∘ 
sin 35 ∘
−16
→
v ∘ cos 35 ∘ 
90
2

16
3
90
cos 35 ∘ 
sin 35 ∘
→
v 2∘ 
16
390 tan 35 ∘
900
cos 35 ∘
2
→
v∘ 
16
390 tan 35 ∘

54. 088 feet/sec
90
cos 35 ∘
b) The height is given by
y  7 − g t2  v ∘ t sin 35 ∘  7 − 16t 2  tsin 35 ∘ 
2
16
390 tan 35 ∘
90
cos 35 ∘
The max occurs at
t
sin 35 ∘ 
16
390 tan 35 ∘
32
90
cos 35 ∘

0. 969 5 sec
20
Therefore, the max height is
y
7 − 16
sin 35 ∘ 
16
390 tan 35 ∘
32
90
cos 35 ∘
2

16
390 tan 35 ∘
90
cos 35 ∘
sin 35 ∘ 
16
390 tan 35 ∘
32
90
cos 35 ∘
sin 35 ∘ 
22. 038 750 873
585 077 962 feet
c)
The time that it takes the football to reach the position R is
(as calculated in part a)
90
t  v ∘ cos
35 ∘ 
or
t
90
16
390 tan 35 ∘
90
cos 35 ∘
cos
35 ∘


2. 031 296 975 439 489 377 1 sec
Section 12.3: 3,5,7,11,13,15,17,19,21,27,29,31,35,41,45,49
21
22
23