Could anyone help with this activity I have to do? It's trigonometry

- anonymous

Could anyone help with this activity I have to do? It's trigonometry

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- anonymous

- anonymous

I'd really appreciate it if someone could explain so I can figure it out . I'm terrible at trig . I'm taking an algebra 2 class and apparently trig is apart of algebra 2

- anonymous

@jim_thompson5910 could you help me with this ? :/

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## More answers

- jim_thompson5910

Your worksheet is a bit misleading. It says that B = 2pi/(period) but that's only true for sine, cosine, cosecant, secant. It doesn't apply to tangent and cotangent. That rule would be B = pi/(period). Luckily we won't be dealing with tangent or cotangent

- anonymous

did I write that ?

- jim_thompson5910

no it's printed on page 1

- anonymous

apex is confusing and doesn't do the right information

- anonymous

doesn't give *

- jim_thompson5910

you wrote the correct max elevation
what is the min elevation?

- anonymous

it didn't give that . Idk

- jim_thompson5910

|dw:1440386808612:dw|

- jim_thompson5910

|dw:1440386828053:dw|

- jim_thompson5910

since the ride starts 10 ft off the ground

- anonymous

oh so it'd be 10 ? lol

- jim_thompson5910

yeah

- jim_thompson5910

and that would make the diameter 210 - 10 = 200 and not 210

- anonymous

I have the diameter as 200

- jim_thompson5910

oh nvm I misread that part

- jim_thompson5910

don't forget the units. When it asks for the time to complete 1 revolution. Don't just write "90". Write "90 seconds". Same applies for the other values too

- jim_thompson5910

other than that, page 1 looks great

- anonymous

do the max and min have units ? oh and the diameter ?

- jim_thompson5910

yeah, in this case, "feet"

- jim_thompson5910

same for the diameter

- anonymous

okay I just wanted to be sure

- jim_thompson5910

ready for #2 ?

- anonymous

yes

- jim_thompson5910

I'm guessing you haven't started at all? Or do you have an idea where to start?

- anonymous

The graph is where I got confused . I looked over it multiple times and I tried graphing but I just didn't understand

- jim_thompson5910

At time t = 0 seconds, the height is h = 10 ft
So one ordered pair is (0,10) and this is a min
--------------------------------------
45 seconds later at t = 45, the height moves to h = 210
one max point is (45,210)
--------------------------------------
another 45 seconds pass by and we're now at t = 90 seconds
we're back at the min height again and now at the point (90, 10)

- jim_thompson5910

they want you to plot these mins and maxes for 2 full cycles

- anonymous

I always forget how to graph lol . Just one second

- jim_thompson5910

For #2, they don't want the full graph of the sinusoidal curve (ie they don't want sine or cosine plotted). They just want those discrete points.

- anonymous

so don't plot all three of those points ?

- jim_thompson5910

You do plot those points and any others that I didn't list until you've done 2 full cycles

- jim_thompson5910

If you start at t = 0,
1 full cycle ends at t = 90
2 full cycles end at t = 180

- anonymous

|dw:1440387875856:dw|
so it'd look somewhat like that ?

- anonymous

that's a horrible drawing I know lol . My mousepad sucks

- jim_thompson5910

that's ok. You should get something like this (see attached). It looks like you did, so you are correct for #2

##### 1 Attachment

- anonymous

do I connect the dots or leave them as is ?

- jim_thompson5910

For #2, they just want the min/max points. That's all. So you leave them as is.

- anonymous

Okay so that one is done

- jim_thompson5910

what did you get for #3

- anonymous

I'm reading it right now

- anonymous

wouldn't the top of the ferris wheel be the max ?

- jim_thompson5910

yes

- anonymous

so it'd be 210 ?

- jim_thompson5910

they want halfway to the top though

- jim_thompson5910

|dw:1440388468748:dw|

- anonymous

would you divide 210 by 2 ?

- jim_thompson5910

you would find the midpoint of h = 10 and h = 210
add the values up and divide by 2:
(10+210)/2 = ??

- jim_thompson5910

`would you divide 210 by 2 ?`
only if the starting height was 0

- anonymous

so 110 ?

- jim_thompson5910

yes, that's your midline
|dw:1440388774233:dw|

- anonymous

so their elevation would be 110 feet ?

- jim_thompson5910

at the midway point, yes

- anonymous

what do they mean by "use the diameter of the Ferris wheel to determine the correct vertical shift for your modeling function?"

- jim_thompson5910

well I guess another way to do is to divide the diameter in half
200/2 = 100
the radius is 100. Add that radius to the starting height: 100+10 = 110

- jim_thompson5910

I like the midpoint method better

- anonymous

Midpoint takes out a step

- anonymous

so I'd graph it at 110 ?

- jim_thompson5910

are you still on #3 ? or #4 now?

- anonymous

on number 4

- jim_thompson5910

yeah you'll have a dashed/dotted line that is horizontal and going through y = 110

- anonymous

|dw:1440389236868:dw|
so like this ?

- jim_thompson5910

yeah

- jim_thompson5910

which letter represents the vertical shift? A, B, C, or D?

- anonymous

uh , how do you figure that out ?

- jim_thompson5910

look back on page 1

- anonymous

D

- jim_thompson5910

yes, D = 110

- jim_thompson5910

for # 5, you can find the distance using two methods
a) subtract the max and min height, then divide by 2
(max - min)/2 = (210-10)/2 = 200/2 = 100
OR
b) Find the distance from the center (110) to either the peak or valley
Center to peak: 210 - 110 = 100
Center to valley: 110 - 10 = 100
either way, the radius of this wheel is 100 ft. This is the amplitude of the sinusoidal curve
the amplitude is |A|, so A = 100

- anonymous

Yeah I just subtracted it and got 100

- anonymous

amplitude is distance ?

- jim_thompson5910

the distance from midline to the max
or
the distance from midline to the min

- jim_thompson5910

|dw:1440389822564:dw|

- jim_thompson5910

or half the distance from the very top to the very bottom
|dw:1440389869160:dw|

- anonymous

so it just depends on exactly what they're asking ?

- jim_thompson5910

what do you mean?

- anonymous

or is all just the same ?

- jim_thompson5910

those are 2 different ways to get the same thing

- jim_thompson5910

so it doesn't matter which path you take

- anonymous

ooh okay , I just wanted to make sure

- anonymous

for 6, do I keep it in terms of pi or simplify it completely ?

- jim_thompson5910

I'd leave it in terms of pi

- jim_thompson5910

reduce the fraction as much as possible though

- anonymous

so it'd be \[B=\frac{ 2\pi }{ 90 } = \frac{ \pi }{ 45 }\]

- jim_thompson5910

yeah

- jim_thompson5910

which trig function will you use for #7 ? sine or cosine?

- anonymous

sin

- jim_thompson5910

either works, but cosine is actually easier to work with
if you graph the parent function y = cos(x), you'll notice it has the peak max point at (0,1). We want the peak max point to happen when t = 45
so that's why C = 45 is the phase shift. We shift everything over 45 units to the right

- jim_thompson5910

A = 100
B = pi/45
C = 45
D = 110
\[\Large H(t) = A\cos\left(B(t-C)\right) + D\]
\[\Large H(t) = 100\cos\left(\frac{\pi}{45}(t-45)\right) + 110\]
attached is the graph

##### 1 Attachment

- jim_thompson5910

you could use sine but the phase shift would be different. Everything else would stay the same

- anonymous

Okay , it makes sense

- anonymous

in number 8 , what do they mean by horizontal shift ?

- jim_thompson5910

the parent function y = cos(x) has a max when x = 0
but we want to force the max to be when x = 45
so that's why we do a horizontal shift and why C = 45

- jim_thompson5910

|dw:1440390964651:dw|

- jim_thompson5910

|dw:1440390975447:dw|

- jim_thompson5910

|dw:1440390992114:dw|

- anonymous

So it is 45 . I'm used to shifts when it comes to 2 graphs

- jim_thompson5910

yeah it's 45 if we use cosine

- anonymous

Since with cosine x = 0

- jim_thompson5910

what do you mean?

- anonymous

idk . I'm probably confusing myself

- anonymous

so for number 9 , where are they getting at with the parent function being flipped ? I don't even know what the parent function is :/

- jim_thompson5910

in this case, we used the parent function y = cos(x)

- jim_thompson5910

it's just the base function that you apply the transformations to

- jim_thompson5910

A = 100 which is positive, we aren't flipping the function over

- anonymous

if it were to be flipped , how would I know that was the case ?

- jim_thompson5910

well I started with y = cos(x) which has A = 1 as a positive number
shifting everything over to the right 45 units doesn't flip things

- jim_thompson5910

I played around with the graph a bit, and I am just realizing that A could be negative, but that changes the phase shift.

- jim_thompson5910

So there are a few ways to get the h(t) function

- jim_thompson5910

if A = +100, then the phase shift is C = 45
if A = -100, then the phase shift is C = 0

- anonymous

But since it's dealing with a ferris wheel , it couldn't have a negative value since you can't go in circles negative times right ?

- jim_thompson5910

t and h can't be negative, but A could be

- jim_thompson5910

the amplitude is |A|
the absolute value ensures we have a positive amplitude

- jim_thompson5910

on your worksheet, draw vertical bars around the "A"
where it says `A is the amplitude` on page 1

- anonymous

oh okay , so it doesn't really matter if it was negative cause it regardless it's going to be positive since it's absolute value ?

- jim_thompson5910

exactly

- jim_thompson5910

eg:
y = -2*cos(x) has an amplitude of |-2| = 2

- anonymous

okay , I get it

- anonymous

in the equation on number 10, what's "Af" ?

- jim_thompson5910

I don't see #10 in any of the attachments

- anonymous

one moment

- anonymous

##### 1 Attachment

- jim_thompson5910

oh, that's A*f

- jim_thompson5910

f is the trig function
look back at page 1

- anonymous

so it's the amplitude times the trig function ?
100cosx ? or ...

- jim_thompson5910

yeah like how I wrote above when I wrote out H(t)

- anonymous

where did you write that ?

- jim_thompson5910

it's buried a bit
when I wrote
\[\Large H(t) = 100\cos\left(\frac{\pi}{45}(t-45)\right) + 110\]

- anonymous

oooh okay , I remember

- anonymous

and t would b 90 , correct ?

- anonymous

be *

- jim_thompson5910

t is just a variable that represents the time in seconds

- jim_thompson5910

you plug in values of t to get corresponding values of h
eg: plug in t = 0 to get h = 10
plug in t = 45 to get h = 210
etc etc

- anonymous

wait so how exactly is the equation set up ?

- jim_thompson5910

what do you mean?

- anonymous

I plug everything into that equation right ?

- jim_thompson5910

I'd just leave it as
\[\Large H(t) = 100\cos\left(\frac{\pi}{45}(t-45)\right) + 110\]

- jim_thompson5910

They want the function or equation that models the curve

- anonymous

So no solving ?

- jim_thompson5910

nope, they just want you to write the equation

- anonymous

and for number 11 , i use that same equation, I'm just solving for t ?

- jim_thompson5910

correct. replace h(t) with 100 and solve for t

- anonymous

To start I'd subtract 110 over to 100 right ?

- jim_thompson5910

yes

- anonymous

what do I do from there ?

- jim_thompson5910

you should have this
\[\Large 100\cos\left(\frac{\pi}{45}(t-45)\right) = -10\]

- jim_thompson5910

now divide both sides by 100

- anonymous

so it'd be equal to -1/10 ?

- jim_thompson5910

yeah or -0.1

- anonymous

I feel like decimal is better

- jim_thompson5910

Then apply the arccosine function to both sides
\[\Large \Large \cos\left(\frac{\pi}{45}(t-45)\right) = -0.1\]
\[\Large \Large \arccos\left(\cos\left(\frac{\pi}{45}(t-45)\right)\right) = \arccos(-0.1)\]
\[\Large \Large \frac{\pi}{45}(t-45) = ??\]

- anonymous

what's arccos ?

- jim_thompson5910

arccos is \(\Large \cos^{-1}\), ie inverse cosine

- anonymous

so csc ?

- jim_thompson5910

no, that's a common misconception

- jim_thompson5910

if you want to say 1/cos, you say secant

- jim_thompson5910

you don't say \(\Large \cos^{-1}\) to mean secant
\(\Large \cos^{-1} \ne \sec\)

- anonymous

oh okay , so I haven't learned arccos

- jim_thompson5910

your calculator should have a cos^(-1) or arccos button

- anonymous

I have cos^-1

- jim_thompson5910

ok use that

- anonymous

so I find arccos-0.1

- jim_thompson5910

yeah what is that numeric value?

- anonymous

95.74 ?

- jim_thompson5910

make sure you're in radian mode

- anonymous

1.67

- jim_thompson5910

yes or -1.67

- jim_thompson5910

we will use the negative form to figure out when the height reaches 100 ft for the first time (the +1.67 is used to figure out when the height reaches 100 ft for the second time)

- jim_thompson5910

so solve for t
\[\Large \Large \frac{\pi}{45}(t-45) = -1.67\]

- anonymous

do I distribute or multiply pi/45 by the recipricol on both sides ?

- jim_thompson5910

I'd multiply by the reciprocal

- anonymous

so it'd be \[t-45=\frac{ -75.15 }{ \pi }\]

- jim_thompson5910

-75.15/pi = -23.9209879 approximately

- jim_thompson5910

now just one more step to go

- anonymous

adding the 45 over

- jim_thompson5910

yes

- anonymous

so it would equal 21.08 ?

- jim_thompson5910

I'm getting 21.07901205 with a calculator
and geogebra is telling me 21.06521
so it's pretty close

- jim_thompson5910

so roughly 21 seconds

- anonymous

I rounded the 21.07901205

- jim_thompson5910

that works

- anonymous

thank you so much !

- jim_thompson5910

and desmos says the same thing more or less
https://www.desmos.com/calculator/swknd4tau1

- jim_thompson5910

you'll have to click on the first intersection point between the curve and the line

- anonymous

You really helped a lot lol . I can finally release some stress . Thank you

- jim_thompson5910

you're welcome

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