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@jim_thompson5910 could you help me with this ? :/

did I write that ?

no it's printed on page 1

apex is confusing and doesn't do the right information

doesn't give *

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

it didn't give that . Idk

|dw:1440386808612:dw|

|dw:1440386828053:dw|

since the ride starts 10 ft off the ground

oh so it'd be 10 ? lol

yeah

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

I have the diameter as 200

oh nvm I misread that part

other than that, page 1 looks great

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

yeah, in this case, "feet"

same for the diameter

okay I just wanted to be sure

ready for #2 ?

yes

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

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

I always forget how to graph lol . Just one second

so don't plot all three of those points ?

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

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

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

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

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

do I connect the dots or leave them as is ?

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

Okay so that one is done

what did you get for #3

I'm reading it right now

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

yes

so it'd be 210 ?

they want halfway to the top though

|dw:1440388468748:dw|

would you divide 210 by 2 ?

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

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

so 110 ?

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

so their elevation would be 110 feet ?

at the midway point, yes

I like the midpoint method better

Midpoint takes out a step

so I'd graph it at 110 ?

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

on number 4

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

|dw:1440389236868:dw|
so like this ?

yeah

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

uh , how do you figure that out ?

look back on page 1

yes, D = 110

Yeah I just subtracted it and got 100

amplitude is distance ?

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

|dw:1440389822564:dw|

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

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

what do you mean?

or is all just the same ?

those are 2 different ways to get the same thing

so it doesn't matter which path you take

ooh okay , I just wanted to make sure

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

I'd leave it in terms of pi

reduce the fraction as much as possible though

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

yeah

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

sin

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

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

Okay , it makes sense

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

|dw:1440390964651:dw|

|dw:1440390975447:dw|

|dw:1440390992114:dw|

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

yeah it's 45 if we use cosine

Since with cosine x = 0

what do you mean?

idk . I'm probably confusing myself

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

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

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

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

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

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

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

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

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

exactly

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

okay , I get it

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

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

one moment

oh, that's A*f

f is the trig function
look back at page 1

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

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

where did you write that ?

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

oooh okay , I remember

and t would b 90 , correct ?

be *

t is just a variable that represents the time in seconds

wait so how exactly is the equation set up ?

what do you mean?

I plug everything into that equation right ?

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

They want the function or equation that models the curve

So no solving ?

nope, they just want you to write the equation

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

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

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

yes

what do I do from there ?

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

now divide both sides by 100

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

yeah or -0.1

I feel like decimal is better

what's arccos ?

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

so csc ?

no, that's a common misconception

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

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

oh okay , so I haven't learned arccos

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

I have cos^-1

ok use that

so I find arccos-0.1

yeah what is that numeric value?

95.74 ?

make sure you're in radian mode

1.67

yes or -1.67

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

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

I'd multiply by the reciprocal

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

-75.15/pi = -23.9209879 approximately

now just one more step to go

adding the 45 over

yes

so it would equal 21.08 ?

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

so roughly 21 seconds

I rounded the 21.07901205

that works

thank you so much !

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

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

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

you're welcome