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.

## 19 comments:

Yeah just a tad late but hey you got it in and that is what counts :) Your 1st question seems the most different from the other groups questions which is interesting. I'm not sure what that question was asking for but I understand what you did to answer it and it makes sense.

The Questions in bold.

you handed your work in and that's what counts. What I really liked about your solutions is particularly in question d. I appreciated how you explained each step..that is crucial especially if someone **coughs, me** has trouble with solving this kind of problem...but I did have to catch you on one thing...proper math terms... now look whose talking..but it was two in the morning when our group finished, but still..the proper use of math terms is a beautiful thing..

Question isn't in bold. I'll fix it so that you can see A,B,C,D etc.

Jamie, I'll see if I can catch what your talking about and fit in proper terms.

Well I was already talking to you Lawrence when you decided to get your post up so you already know that I feel its good you got it up :p

Anyways, like Jamie said, I really appreciate how you answered part D.) step by step. Especially in those word problems its easy to get lost in your steps, and illustrating exactly how you did it helps both the people looking at your work, and yourself (as it makes it easier to catch mistakes).

I vote correct!

welll, it's not that bad... it's quite understandable.. it's just if you want perfection, you know. but then again i wouldn't know about much of that. haha its just the "get rid of" that bothers me.

In your part D, I don't understand why'd you 'get rid of' your phase shift before you divided the 2.5 on both sides.

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

Let X = 2pi(t-1.5)/12.4.

4 = 2.5 sin X + 4.3

-0.12 = sin X

-0.1203 = X

-0.1203 = 2pi(t-1.5)/12.4

[(-0.1203)(12.4)/(2pi)] + 1.5 = t

1.2626 = t

...which converts to 1:15:45 a.m.

Yeah, I just tried your part D as well. I was confused when I read the part where you said you added the phase shift to both sides. Since after I tried it, I ended up getting the same answer that Joseph has shown. Other than that, everything is good. Be careful of the order in which you solve those types of questions next time, okay? :P

yes yes part d is kinda confusing. I noticed you didn't make that the numbers in the brackets or paranthesis...whatever they're called into theta. After doing the question i got 1:15:45 am too joseph.

First of all good job on the detailed answer. I really like how you put a lot of space so your solution will be easy to follow.After doing the whole question myself, I therefore conclude that Zeph's answer is the right one. There is something wrong with the algebra. There are many different ways to check this problem. One of them is to use your calculator. First of all, you need to plug in the function. Then press the 2ND button then TRACE then press 1.After that plug in your answer as the x value then press ENTER. You should see that the y value should be very close to 4. Thats the quick way of doing it.

-m@rk

Better late than never right?! (: Anyways, i found this problem particulary easier than the others because the equation was already there. BUT on part D I got a different answer. I believe it should be 1:15am?!

i got 1.2626 hrs for d

which is like 1:15am

After careful examination between Group Second to None and joseph i have come to conclude that joseph's answer is correct

I loved how you explained every little detail. It gave me a great understanding, made it seem like I was doing the questions myself. The graph was also very neat. Unlike ours. I noticed that someone recognized a flaw. Close to flawless.

Two minutes off guys. Give me a break. Lol.

I like how you posted your answers. It was neat and organized. And I did not have to refer to the slides to see what is what. I think Zeph has got the correct answer for part D

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