**Period 2:**

Today we were doing a ‘workshop’ where we did some practice questions to review for our transformations test on **Friday, October 19, 2007**.

Mr. K started the class by putting everyone in little groups. He then put up questions for the groups to work on and discuss.***Note: The text in red are little notes to help you slove the equation***

Here’s the first question:**---------------------------------------------------------------------------------------------------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) = 2.5sin[(2π)(t - 1.5)/12.4] +4.3**

a) Explain the significance of each number in the equation:

(i) 2.5 (ii) 12.4 (iii) 1.5 (vi) 4.3

a) Explain the significance of each number in the equation:

(i) 2.5 (ii) 12.4 (iii) 1.5 (vi) 4.3

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

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

d) Determine one time when the water is 4.0 meters deep.

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

d) Determine one time when the water is 4.0 meters deep.

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Here are the solutions:

*Reminder: t = time in hours and h = height in meters.*

**a) Explain the significance of each number in the equation:**

**(i) 2.5 - Amplitude**

In the equation, 2.5 is where ‘a’ is and ‘a’ determines the amplitude. Therefore, 2.5 is the amplitude.

**(ii) 12.4 -Period**

In the square brackets it shows you (2π)(t – 1.5)/12.4, well, that’s the same thing as saying [(2π)/12.4] [(t – 1.5)/12.4]. (2π)/12.4 is where ‘b’ is and we all know that ‘b’

**is not**the period, it

**determines**the period. Therefore, 12.4 is the period.

**(iii) 1.5**

**- Phase Shift**

This is where parameter ‘c’ is and ‘c’ determines the phase shift. 1.5 is the phase shift.

**(iv) 4.3 -Vertical shift, Average, or sinusoidal axis**

This is where parameter ‘d’ is and ‘d’ determines the vertical shift. It is also called the ‘average’ and the ‘sinusoidal axis’. 4.3 is the vertical shift.

**b) What is the minimum depth of the water?**

Average = 4.3

Amplitude = 2.5

4.3 - 2.5 = 1.8

*Since you’re looking for the minimum, you must subtract from the average.*

The minimum depth is 1.8 meters.

**When does it occur?**

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

*Replace h(t) with 1.8*

1.8 = 2.5sin[(2π)(t - 1.5)/12.4] + 4.3

Let [(2π)(t - 1.5)/12.4] be θ

1.8 = 2.5sinθ + 4.3

-2.5 = 2.5sinθ

-1 = sinθ

*Remember from Circular functions, the sine of -1 is (3π)/2*

(3π)/2 = θ

(3π)/2 = (2π)(t - 1.5)/12.4

58.4336 = (2π)(t - 1.5)

9.3 = t - 1.5

t = 10.8 hours

*This is how to find out what time 10.8 hours is in time:

10 hours is 10:00 am.

0.8 in percent is 80%

Here's how to find 80% of 60 minutes:

(0.8)(60) = 48 minutes

Therefore, 10.8 hours = 10:48am*

The minimum depth of the water occurs in 10.8 hours or at 10:48am.

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

9:30 = 9.5 hours

t = 9.5

h(9.5) = 2.5sin[(2π)(t - 1.5)/12.4] + 4.3

*Substitute in 9.5 for ‘t’*

h(9.5) = 2.5sin[(2π)(9.5 - 1.5)/12.4] + 4.3

*Do what’s in the brackets first*

h(9.5) = 2.5sin[(2π)(8)/12.4] + 4.3

h(9.5) = 2.5sin(4.0538) + 4.3

*Insert into your calculator*

h(9.5) = 2.3231

h(9.5) = 2.3 m

**The depth of the water at 9:30 am is approximately 2.3 meters.**

**d) Determine one time when the water is 4.0 meters deep.**

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

*Substitute h(t) with 4.0*

4.0 = 2.5sin[(2π)(t - 1.5)/12.4] + 4.3

Let [(2π)(t - 1.5)/12.4] be θ

4.0 = 2.5sinθ+ 4.3

-0.3 = 2.5sinθ

-0.3/2.5 = sinθ

-0.1203 = θ

*Remember this: ‘Let [(2π)(t - 1.5)/12.4] be θ’. Now put in (2π)(t - 1.5)/12.4 where θ is *

-0.1203 = (2π)(t - 1.5)/12.4

*Earlier I said that (2π)(t - 1.5)/12.4 = [(2π)/12.4] [(t – 1.5)/12.4]. To get rid of the [(2π)/12.4], you have to divide both sides by [(2π)/12.4] since it’s being multiplied. But in grade 12, we don’t divide; instead we multiply by the reciprocal.*

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

-0.2374 = t – 1.5

1.2626 = t

t = 1.26 h

**One time when the water is 4.0 meters deep is in approximately 1.26 hours or at about 1:25 am.**

If you graphed this function, it would look something like this:

Image from www.fooplot.com

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

After that question we got this question:**---------------------------------------------------------------------------------------------------A Ferris wheel has a radius of 20 m. It rotates once every 40 seconds. Passengers get on at point s, which is 1 m above the 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.b) Write an equation that expresses your height as a function of the elapsed time.c) Determine your height above the ground after 45 seconds.d) Determine one time when your height is 35 m above the ground.---------------------------------------------------------------------------------------------------**

Here are the solutions:

*Reminder: t = time in seconds and h = height in meters.*

**a) Graph how your height above the ground varies during the first two cycles.**

Here’s what your graph should look like:

Image from http://www.fooplot.com/

In the words of Mr. K “Once you have one cycle, you have it all”... or something like that.

**- End of Period 2 –**

**Period 4**

We started this class by getting back into the groups we were in during second period and then Mr. K, discussed the Flickr assignment.

A student asked if we could go and use pictures off the Internet. The answer to that is ‘no’. You have to take your own pictures of things, not people. Then we got into the whole copyright thing.

Then we got rubrics to see what your photo should have on it to get full marks and bonus marks.

After the Flickr discussion, we continued where we left off from period 2.

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

*You can write the equation in both sine and cosine.*

Sine:

**20sin(π/20)(t - 10) + 21**

A = 20

B = p/20

C = 10

D = 21

Cosine:

**-20cos[(π/20)(t)] + 21**

A = -20

B = p/20

C = 0

D = 21

*Originally it was 2π/40, I just reduced it to make it π/20. You'll should end up with the same answer.*

**c) Determine your height above the ground after 45 seconds.**

*Take the equation you found in the previous question. I'll use cosine just because*

h(45) = -20cos[(π/20)(t)] + 21

*Replace ‘t’ with 45*

h(45) = -20cos[(π/20)(45)] + 21

h(45) = -20cos(45π/20) + 21

*45π/20 is reduced to 9π/4*

h(45) = -20cos(9π/4) + 21

*If you remember from the circualr functions, cos9π/4 is √2/2*

h(45) = -20(√2/2) + 21

*When you multiply -20 into √2/2, you get -20√2/2. Then it reduces to -10√2*

h(45) = -10√2 + 21

*That's what your answer would be if there was no calculators allowed. But if there was, this would be your answer*

h(45) = 6.8579

h(45) = 6.9 m

**Your height above the ground would be approximately 6.9 meters after 45 seconds.**

That's where the class ended... BUT... we had one more question to do, so I'm going to give it a shot. Here's what I got:

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

h(t) = -20cos[(π/20)(t)] + 21

*Replace h(t) with 35*

35 = -20cos[(π/20)(t)] + 21

Let [π/20)(t)] be θ

35 = -20cosθ + 21

14 = -20cosθ

-0.7 = cosθ

2.3462 = θ

2.3462 = (π/20)(t)

*Multiply both sides by 20/π [the reciprocal of π/20]*

14.9363 = t

t = 14.9

**One time when the your height is 35 m above the ground is at approximately 14.9**

**seconds.**

Alright, I'm finally done. This took me quite some time to do. Anyways... The next scribe is le joséph..

REMINDER: There's a pretest tomorrow morning and our Transformations test is on

**Friday, October 19, 2007, Period 2.**

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