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

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

Mathematics
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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
@jim_thompson5910 could you help me with this ? :/

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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
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
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
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?
The graph is where I got confused . I looked over it multiple times and I tried graphing but I just didn't understand
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)
they want you to plot these mins and maxes for 2 full cycles
I always forget how to graph lol . Just one second
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.
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
what do they mean by "use the diameter of the Ferris wheel to determine the correct vertical shift for your modeling function?"
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
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
D
yes, D = 110
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
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
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
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 ?
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
|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
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 :/
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 ?
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
I played around with the graph a bit, and I am just realizing that A could be negative, but that changes the phase shift.
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
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 ?
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
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 ?
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
1 Attachment
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
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
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
Then apply the arccosine function to both sides \[\Large \Large \cos\left(\frac{\pi}{45}(t-45)\right) = -0.1\] \[\Large \Large \arccos\left(\cos\left(\frac{\pi}{45}(t-45)\right)\right) = \arccos(-0.1)\] \[\Large \Large \frac{\pi}{45}(t-45) = ??\]
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
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)
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

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