Group Second to None

Okay, late I know, but unfortunately our group didn't post in time, So I'll try to do this quick fast.

Anyways, this is our solution.

SLIDE 17:

This equation gives the depth of the water, h meters, at an ocean port at any time, t hours during a certain day

h(t) = 2.5 sin[2pi(t - 1.5)/12.4] + 4.3

A) Explain the significance of each number in the equation

I) 2.5 - This is parameter A, which will determine the amplitude of the function


II) 12.4 is the Period of the function, which is obtained from parameter B - 2pi/12.4

III) 1.5 - This is parameter C, which is the phase shift, in which this function, the graph shifts to the right 1.5 units/hours.

IV)4.3 - This is parameter D, which is the sinusoidal axis, which shifts the sinusoidal axis up 4.3 metres..



B) What is the minimum depth of the water? When does it occur?




Now, We can go backwards and use that point, but since we can use whichever point, we decided to use the next one, which minimum depth, 1.8 metres, occurs at 10.8 hours.



C) Determine the depth of the water at 9:30 am.

So then we just plug it in.

h(9.5) = 2.5sin[2pi(9.5 - 1.5)/12.4] + 4.3
*t is 9.5 because time ~ 9:30 AM ~ is converted to 9.5 because :30 minutes is .5 hours.*

in which we get... 2.323 metres.


D) Determine one time when the water is 4.0 metres deep.

So then we go something like...

4 = 2.5sin[2pi(t-1.5)/12.4] + 4.3



First we subtracted 4.3 from both sides like so.

-0.3 = 2.5sin[2pi(t-1.5)/12.4]


Then we added the phase shift to both sides to get...

1.2 = 2.5 sin[2pi(t)/12.4]


Then we divided parameter A out.

0.48 = sin[2pi(t)/12.4)


We then divided 2pi/12.4 in which .48 would be multiplied by the reciprocal of 2pi/12.4, and moved sine to the other side to change it to ARCSine to isolate t.

ARCSine(0.9473) = t

Which would equal to --> 1.2447 hours.


To make it efficient, we'll multiply the .2247 by 60 so


Depth of water of 4 metres occurs at 1:13:48 am.



Again, sorry for posting so late. Original personnel that was to post, did not post. Feel free to comment, as we are supposed to.

Nothing Less Than The Best....Group

SLIDE 18

On a typical day at an ocean port, the water has a maximum depth of 20 m at 8:00 a.m. The minimum depth of 8 m occurs 6.2 hours later. Assume that the relation between the depth of the water and time is a sinusoidal function.

Let's draw a graph!


a) What is the period of the function?

From the information we have been given...
* We can set the 8am as t = 0 hours.
* A maximum value is when t = 0 hours and when d = 20.
* A minimum value is when t = 6.2 and d = 8.

We can see from the graph that the period is 12.4 hours.

b) Write an equation for the depth of the water at any time, t hours.

cosine equation's parameters...
A = 6
B = (2pi) / 12.4 = pi/6.2
C = 0
D = 14

To get A, amplitude, calculate the distance from the sinusoidal axis to a maximum value or minimum value.
sinusoidal axis = (20+8)/2 = 14
amplitude = 14-8 = 6

B = pi/period = pi/6.2

C, the phase shift, is 0.

D is the sinusoidal axis, 14.

D(t) = 6cos [(pi/6.2)t] + 14


c) Determine the depth of the water at 10:00 a.m.

10:00 am = 2 hrs from when t = 0 or 8:00 am. Plug in the 2 as t into the equation to get the answer.

D(2) = 6 cos [(pi/6.4)2] + 14 = 17.1738 metres

d) Determine one time when the water is 10 m deep.

The wave is 10 metres deep, so the qestion is asking for what the time is when D = 10. Plug in 10 as D, then solve for t.

10 = 6cos[(pi/6.2)t] +14
-4 = 6cos[(pi/6.2)t]
-4/6 = cos[(pi/6.2)t]
arc cos(-4/6) = (pi/6.2)t
2.3005 = (pi/6.2)t
2.3005/(pi/6.2) = t
2.3005 x 6.2/pi = t
14.2632/pi = t
t = 4.5401

Convert the 4.5401 into "actual time" because we use hours:minutes:seconds to show time, so...

4.5401 hrs + 8am = 12.5401 hrs

Obviously, its not efficient to say .5401 hrs so we convert that to minutes.

0.5401 x 60 = 32.406 min.

12:32:24pm

We can round that to 12:30pm.

Trig Assignment team "Jabawakeez"?

So these are the solutions for the question:

Tidal forces are greatest when Earth, the sun, and the moon are in line. When this occurs at the Annapolis Tidal Generating Station, the water has a maximum depth of 9.6 m at 4:30 am and a minimum depth of 0.4m 6.2 hours later.

Here's how our graph looks like. In the question we were given the max and min values of the wave. The max is at 4:30 and we changed the time to its number value by taking the minutes and dividing it by 60 so its new value was 4.5 on the graph. Now it is easier to add 6.2 hours to it, so the min will be at 10.7 on the x axis. With the max and min we can figure out 1/2 of the period. Since the min and max are 6.2 hours apart we use that. 6.2 * 2 = 12.4. But here's the tricky part. If you do 12.4*(1/4) it will not equal 4.5. So we can't do the 1,2,3,4 ticks and automatically write in the values. What we did was 4.5+12.4 to make it easier to graph. The value in the 4th tick on the x axis is now 16.9. The values on the x axis are now 1/4, 1/2, 3/4 of 16.9. Then you'll just put the points on at the max and min. Then we know the patterns of the curve (after the max) will go to the avg value, then min, avg value again, and then the max.

a)Here's an equation in cose. To find D we found the average of the min and max which was 5. We can now find parameter a by subtracting 5 from the max 9.6. B=2pi/period so in our equation it'll be 2pi/12. In our graph here it does not start at the max where cosine starts so it must've moved by 4.5 our parameter c.

b) First we changed 2:46pm to its number value. Since it has passed into night time we can add the 12 hours from the morning. Then all we do is add the 2 hours and 46 minutes. But we divide 46/60 first. So its number value should be 14.7667. We take this value and plug it into t. This can be done on the calculator...and will get an answer of h=7.1641m.

c)In this question we just plugged in 2 into the equation and solved for t.

WORD PROBLEMS...TEAM KA-BLAMO! Problems?? Word, man...

I think it's original to call our fantabulous team of three Team Ka-Blamo [members: Jamie, Kristina and Eleven]

Well, our problem of course, was on slide 15 and it asks:

Well, we're going to do things the old school modern way...actually write the stuff on paper...[because we can't afford a tablet] then scan it and upload...that's the way we do things. We spent like five hours getting everything well, thorough enough via MSN...

A FERRIS WHEEL HAS A RADIUS OF 20 METERS. IT ROTATES ONCE EVERY 40 SECONDS. PASSENGERS GET ON AT POINT S WHICH IS 1 METERS ABOVE GROUND LEVEL. SUPPOSE YOU GET ON AT S AND THE WHEEL STARTS TO ROTATE.

a.] Graph how your height above the ground varies during the first two cycles...
The graph we drew looks pointy but it is CURVED sorry.



That's just a graph based on the given info.... and we nearly got confused because of the minimum value being 1 m off the ground instead of touching the x-axis. But then we figured that ferris WHEELS [typo in your slide.] don't touch the ground or else they would scrape the concrete.

b.] Write an equation that expresses your height as a function of the elapsed time.

Basically, all we did was find each parameter for both the sine and cosine function even though we only needed one.



c.] Determine your height above the ground after 45 seconds.

All that needs to be done here is sub the t in the function H(t) with 45 seconds and the result will be the new height at this time according to the graph, assuming the revolving doesn't stop and is continuous.



d] Determine one time when your height is 35m above the ground.

Replace H(t) with 35 and isolate t to get time in seconds.



That wasn't much explanation in words but you know... we spent a lot of time getting these answers and we hope this is what you guys are looking for....I'm famished. G'night...I'll edit this later....haha