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Thursday, May 25, 2006 

Scribe: Periodic Functions Problems

Today we had 2 periods and we were supposed to have a pre-test and we had but not the actual pre-test instead we answered 2 problems as our pre-test. So whoever missed the class here are your pre-test.

We did two problems and here is the first one:

LIFT OFF!!!

When a spaceship is fired into orbit from a site such as Cape Kennedy (which is not on the EQUATOR), it goes into an orbit that takes it alternately NORTH and SOUTH of the equator. Its DISTANCE FROM THE EQUATOR is approximately a SINUSOIDAL FUNCTION of time. Suppose that a spaceship is fired into orbit from Cape Kennedy. 10 min. after it leaves the Cape, it reaches its farthest distance NORTH of the equtor, 4000 km. HALF A CYCLE later it reaches its farthest distance SOUTH of the equator (on the other side of the Earth) also 4000 km. The spaceship COMPLETES an orbit once EVERY 90 min. (completes a cirlce around the world).

Let 'y' Be the number of kilometres the spaceship is north of the equator (consider distance south of the equator to be negative). Let 't' be the number of minutes that have elapsed since liftoff.

Tip: When trying to answer this kind of question try to write what you are given.

- 4000 km. (max and min) NORTH and SOUTH (this is the "A" )
- 10 min., from CAPE, it reached the farthest NORTH
- 90 min. to COMPLETE A CYCLE (this is the "B")
- HALF A CYCLE (90/2 = 45) it reached the farthest SOUTH (it took 45 min. to reach SOUTH from NORTH)
- EQUATOR as the SINUSOIDAL AXIS (this is the "D")


A.) Sketch a complete cycle of the graph of y versus t, and write the sinusoidal equation expressing y in terms of t.

This is how I got the C: (refer to the graph)
In 10 min. the spaceship reach the farthest NORTH and 45 min. to reach the farthest SOUTH from NORTH.
10 + 45 = 55 (half circle)
the middle number between 55 and 10 is 32.5
10 + 55 = 65 then 65/2 = 32.5 (quarter Circle)
now since I know the quarter circle
32.5 - 10 = 22.5
10 - 22.5 = -12.5 (C)

A = 4000 (max and min range)
B = 2(3.14)/90 (Period) (3.14 = pie)
C = x - -12.5 (Starting point)
D = 0 (Sinusoidal axis)

y = 4000sin((2*3.14/90)(x - -12.5))+0

window: Xmin=-20 Xmax=100 Xscl=10 Ymin=-5000 Ymax=5000 Yscl=1000















B.)Use your equation to predict the distance of the spaceship from the equator when:

Since we didnt talked about this in class here is what I came up.

Solution:(use calculator Ti-83)
Put in the equation "y = 4000sin((2*3.14/90)(x - -12.5))+0)" on "Y="; then "GRAPH"; then "2nd FUNCTION" "TRACE"(CALC); then choose "VALUE"; then plug in whatever the number is asked.
window: Xmin=-20 Xmax=100 Xscl=10 Ymin=-5000 Ymax=5000 Yscl=1000

i.) 25 min
ans.= 2000 km.

ii.) 41 min.
ans.= -2236.77 km. ( - = going south)

iii.) 24 hours. (adjust Xmax window setting = 1500)
24 x 60(min.) = 1440 min.
ans. 3064.17 km.

C.) Calculate the distance Cape Kennedy is from the Equator by solving for Y when t = 0

Solution: same as what I did in B.
Plug in 8 as your x value
ans. 3064.17km


Here is the second problem now:

EAST COAST HARBOUR

On May 3rd, the DEPTH of the water in an east coast harbour will vary over time as described by the equation:
"y = 2.3sin0.506(x+3.1)+2.8"
where x represents the time (hours), and y represents the depth (metres) of the water.
The time at x = 0 is midnight of May 2nd, and x = 12 is noon of May 3rd.

Since we are given the equation already:
A.) What are the minimum and maximum depths of water in the harbor?
D + A = MAX
2.8 + 2.3 = 5.1

D - A = MIN
2.8 - 2.3 = .5

B.)What is the average depth of the water in the harbour?

- you can refer to the SINUSODAL AXIS = 2.8
- or use this way (max + min)/2 (5.1 + .5)/2 = 2.8

C.)How much time is there between two high tides

Write the equation "y = 2.3sin0.506(x+3.1)+2.8" in "Y="; then "GRAPH"; then "2nd FUNCTION" "TRACE"(CALC); then choose "MAXIMUM";
Then "left bound and right bound and guess"; The X coordinate is the time
x = 12.4217 since there are no decimal hour: .42 X 60 = 25

window settings: Xmin=-2 Xmax=(30 or 50) Xscl=6 Ymin=-1 Ymax=6 Yscl=6

ans. 12:25 p.m. 12 hours and 25 minutes (from 0 hours)
















D.) What is the depth of the water at 8:00 AM?

With the same equation; press "GRAPH"; then "2nd FUNCTION" "TRACE"(CALC); then choose "VALUE"; then plug in 8 for x
ans. 1.377















E.) At what time in the afternoon is the water at it's lowest? How deep is the water at this time? (This is the time that little Ziggi likes to go kayaking in the harbour because there are no large boats moving at this time.)

With the same equation; put the .5 (lowest) in Y2(or in any Y); then "GRAPH"; then "2nd FUNCTION" "TRACE"(CALC); then choose "INTERSECTION"; it'll will give you the x coordinate which is
ans. 6.21 ; .21 X 60 = 6:12 am, 6:12 pm. etc...

D.)A commercial boat requires at least three metres of water to more around the harbour. Describe how you would determine when it is safe for the boat to operate in the harbor.

With the same equation; put a the 3 (required level) in Y2(or in any Y); then "GRAPH"; then "2nd FUNCTION" "TRACE"(CALC); then choose "INTERSECTION"; it'll give you Y coordinates. In able for the boat to operate he should ONLY operate in hours in the blue line.

















So that all folks. Sorry my scribe is too long. Anyways gotta sleep and Mohamed you are the scribe for tommorow.




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Sorry the numbers on the graph didnt seem to show up good as i would like it to be. (the second problem) the number after 6 was supposed to 18 but it look like 10.

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