Texas Instruments TI-82 User Manual - Page 206

Ferris Wheel Problem

Page 206 highlights

Ferris Wheel Problem Use two pairs of parametric equations to describe two objects in motion, a person on a ferris wheel and a ball thrown to that person. Determine when the two objects are closest. Problem The ferris wheel has a diameter of 20 meters (d) and is rotating counterclockwise at a rate of one revolution every 12 seconds (s). The following parametric equation describes the location of the person on the ferris wheel at time T, where a is the angle of rotation, the bottom center of the ferris wheel is (0,0), and the passenger is at the rightmost point (10,10) when T = 0. X(T) = r cos a Y(T) = r + r sin a where a = 2p T à s and r = d à 2 The ball is thrown from a height even with the bottom of the ferris wheel, but 25 meters (b) to the right of the bottom center of the ferris wheel (25,0), with velocity (v0) of 22 meters per second at an angle (q) of 66¡ from the horizontal. The following parametric equation describes the location of the ball at time T. X(T) = b - T v0 cos q Y(T) = T v0 sin q - (gà2) T2 (g = 9.8 m/sec2) Solution 1. Press z. Select Par, Connected, and Simul. Simultaneous MODE simulates what is happening with the two objects in motion over time. 2. Press o and turn off all functions. Press y [STAT PLOT] and turn off all stat plots. 3. Press p. Set the viewing WINDOW. Tmin = 0 Tmax = 12 Tstep = .1 Xmin = M13 Xmax = 34 Xscl = 10 Ymin = 0 Ymax = 31 Yscl = 10 4. Press o. Enter the expressions to define the path of the ferris wheel and the path of the ball. X1T = 10cos (pTà6) Y1T = 10+10sin (pTà6) X2T = 25-22Tcos 66¡ Y2T = 22Tsin 66¡-(9.8à2)T2 14-12 Applications

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14-12
Applications
Ferris Wheel Problem
Use two pairs of parametric equations to describe two objects in motion, a
person on a ferris wheel and a ball thrown to that person. Determine when the
two objects are closest.
Problem
The ferris wheel has a diameter of 20 meters (d) and is rotating
counterclockwise at a rate of one revolution every 12 seconds (s). The
following parametric equation describes the location of the person on the
ferris wheel at time T, where
a
is the angle of rotation, the bottom center of
the ferris wheel is (0,0), and the passenger is at the rightmost point (10,10)
when T = 0.
X(T) = r cos
a
where
a
= 2
p
T
à
s and r = d
à
2
Y(T) = r + r sin
a
The ball is thrown from a height even with the bottom of the ferris wheel,
but 25 meters (b) to the right of the bottom center of the ferris wheel (25,0),
with velocity (v
0
) of 22 meters per second at an angle (
q
) of 66
¡
from the
horizontal. The following parametric equation describes the location of the
ball at time T.
X(T) = b – T v
0
cos
q
Y(T) = T v
0
sin
q
– (g
à
2) T
2
(g = 9.8 m/sec
2
)
Solution
1.
Press
z
. Select
Par
,
Connected
, and
Simul
. Simultaneous
MODE
simulates what is happening with the two objects in motion over time.
2.
Press
o
and turn off all functions. Press
y
[
STAT PLOT
] and turn off
all stat plots.
3.
Press
p
. Set the viewing
WINDOW
.
Tmin = 0
Xmin =
M
13
Ymin = 0
Tmax = 12
Xmax = 34
Ymax = 31
Tstep = .1
Xscl = 10
Yscl = 10
4.
Press
o
. Enter the expressions to define the path of the ferris wheel
and the path of the ball.
X
1T
= 10cos (
p
T
à
6)
Y
1T
= 10+10sin (
p
T
à
6)
X
2T
= 25–22Tcos 66
¡
Y
2T
= 22Tsin 66
¡
–(9.8
à
2)T
2