Operation Manual

14-14 Applications
8214APPS.DOC TI-82, Chapter 14, English Bob Fedorisko Revised: 02/09/01 9:27 AM Printed:
02/09/01 12:43 PM Page 14 of 20
Reservoir Problem
On the TI
.
82, parametric graphing can be used to animate a process, providing
valuable insight into dynamic problems such as water flow out of a reservoir.
Problem
A new park has a series of waterfalls, fountains, and pools (reservoirs). The
height of one of the reservoirs is 2 meters. Several holes of relatively small
diameter will be drilled in the side to make streams of water that fall into
the next pool.
At what height on the reservoir should a hole be placed to get the
maximum distance for the water jet? (Assume that the hole is at x=0, there
is no acceleration in the x-direction, and there is no initial velocity in the
y-direction.)
Integrating the definition of acceleration in both the x and y directions
twice yields the equations x=v
0
t and y=h
0
–(g
à
2)t
2
. Solving Bernoulli’s
equation for v
0
and substituting into v
0
t, we get the parametric equations
xt = t
‡
(2 g (2 – h
0
))
yt = h
0
– (g
à
2) t
2
where t is the time in seconds, h
0
is the height of the hole in the reservoir in
meters, and g is the acceleration of gravity (9.8 meters/sec
2
).
Procedure
1. Press
z
. Select
Par
,
Simul
, and the defaults.
2. Press
o
and
‘
all functions. Enter the equations to plot the water
jet for a hole at height
0.5
meters.
X
1T
=T
‡
(2
…
9.8(2–0.50))
Y
1T
=0.50–(9.8
à
2)T
2
3. Press
Í
to move to
X
2T
. Press
y
[
RCL
]
y
[
Y-VARS
]
2
(to display
the
Parametric...
menu)
1
(to select
X
1T
)
Í
. This recalls the
contents of
X
1T
into
X
2T
. Change the height from
0.50
to
0.75
meters.
Repeat the process to recall
Y
1T
into
Y
2T
and edit it.