anonymous
  • anonymous
Could anyone help with this activity I have to do? It's trigonometry
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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
  • anonymous
@jim_thompson5910 could you help me with this ? :/

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jim_thompson5910
  • 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
  • anonymous
did I write that ?
jim_thompson5910
  • jim_thompson5910
no it's printed on page 1
anonymous
  • anonymous
apex is confusing and doesn't do the right information
anonymous
  • anonymous
doesn't give *
jim_thompson5910
  • jim_thompson5910
you wrote the correct max elevation what is the min elevation?
anonymous
  • anonymous
it didn't give that . Idk
jim_thompson5910
  • jim_thompson5910
|dw:1440386808612:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1440386828053:dw|
jim_thompson5910
  • jim_thompson5910
since the ride starts 10 ft off the ground
anonymous
  • anonymous
oh so it'd be 10 ? lol
jim_thompson5910
  • jim_thompson5910
yeah
jim_thompson5910
  • jim_thompson5910
and that would make the diameter 210 - 10 = 200 and not 210
anonymous
  • anonymous
I have the diameter as 200
jim_thompson5910
  • jim_thompson5910
oh nvm I misread that part
jim_thompson5910
  • 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
  • jim_thompson5910
other than that, page 1 looks great
anonymous
  • anonymous
do the max and min have units ? oh and the diameter ?
jim_thompson5910
  • jim_thompson5910
yeah, in this case, "feet"
jim_thompson5910
  • jim_thompson5910
same for the diameter
anonymous
  • anonymous
okay I just wanted to be sure
jim_thompson5910
  • jim_thompson5910
ready for #2 ?
anonymous
  • anonymous
yes
jim_thompson5910
  • jim_thompson5910
I'm guessing you haven't started at all? Or do you have an idea where to start?
anonymous
  • 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
  • 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
  • jim_thompson5910
they want you to plot these mins and maxes for 2 full cycles
anonymous
  • anonymous
I always forget how to graph lol . Just one second
jim_thompson5910
  • 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
  • anonymous
so don't plot all three of those points ?
jim_thompson5910
  • jim_thompson5910
You do plot those points and any others that I didn't list until you've done 2 full cycles
jim_thompson5910
  • jim_thompson5910
If you start at t = 0, 1 full cycle ends at t = 90 2 full cycles end at t = 180
anonymous
  • anonymous
|dw:1440387875856:dw| so it'd look somewhat like that ?
anonymous
  • anonymous
that's a horrible drawing I know lol . My mousepad sucks
jim_thompson5910
  • jim_thompson5910
that's ok. You should get something like this (see attached). It looks like you did, so you are correct for #2
anonymous
  • anonymous
do I connect the dots or leave them as is ?
jim_thompson5910
  • jim_thompson5910
For #2, they just want the min/max points. That's all. So you leave them as is.
anonymous
  • anonymous
Okay so that one is done
jim_thompson5910
  • jim_thompson5910
what did you get for #3
anonymous
  • anonymous
I'm reading it right now
anonymous
  • anonymous
wouldn't the top of the ferris wheel be the max ?
jim_thompson5910
  • jim_thompson5910
yes
anonymous
  • anonymous
so it'd be 210 ?
jim_thompson5910
  • jim_thompson5910
they want halfway to the top though
jim_thompson5910
  • jim_thompson5910
|dw:1440388468748:dw|
anonymous
  • anonymous
would you divide 210 by 2 ?
jim_thompson5910
  • 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
  • jim_thompson5910
`would you divide 210 by 2 ?` only if the starting height was 0
anonymous
  • anonymous
so 110 ?
jim_thompson5910
  • jim_thompson5910
yes, that's your midline |dw:1440388774233:dw|
anonymous
  • anonymous
so their elevation would be 110 feet ?
jim_thompson5910
  • jim_thompson5910
at the midway point, yes
anonymous
  • 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
  • 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
  • jim_thompson5910
I like the midpoint method better
anonymous
  • anonymous
Midpoint takes out a step
anonymous
  • anonymous
so I'd graph it at 110 ?
jim_thompson5910
  • jim_thompson5910
are you still on #3 ? or #4 now?
anonymous
  • anonymous
on number 4
jim_thompson5910
  • jim_thompson5910
yeah you'll have a dashed/dotted line that is horizontal and going through y = 110
anonymous
  • anonymous
|dw:1440389236868:dw| so like this ?
jim_thompson5910
  • jim_thompson5910
yeah
jim_thompson5910
  • jim_thompson5910
which letter represents the vertical shift? A, B, C, or D?
anonymous
  • anonymous
uh , how do you figure that out ?
jim_thompson5910
  • jim_thompson5910
look back on page 1
anonymous
  • anonymous
D
jim_thompson5910
  • jim_thompson5910
yes, D = 110
jim_thompson5910
  • 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
  • anonymous
Yeah I just subtracted it and got 100
anonymous
  • anonymous
amplitude is distance ?
jim_thompson5910
  • jim_thompson5910
the distance from midline to the max or the distance from midline to the min
jim_thompson5910
  • jim_thompson5910
|dw:1440389822564:dw|
jim_thompson5910
  • jim_thompson5910
or half the distance from the very top to the very bottom |dw:1440389869160:dw|
anonymous
  • anonymous
so it just depends on exactly what they're asking ?
jim_thompson5910
  • jim_thompson5910
what do you mean?
anonymous
  • anonymous
or is all just the same ?
jim_thompson5910
  • jim_thompson5910
those are 2 different ways to get the same thing
jim_thompson5910
  • jim_thompson5910
so it doesn't matter which path you take
anonymous
  • anonymous
ooh okay , I just wanted to make sure
anonymous
  • anonymous
for 6, do I keep it in terms of pi or simplify it completely ?
jim_thompson5910
  • jim_thompson5910
I'd leave it in terms of pi
jim_thompson5910
  • jim_thompson5910
reduce the fraction as much as possible though
anonymous
  • anonymous
so it'd be \[B=\frac{ 2\pi }{ 90 } = \frac{ \pi }{ 45 }\]
jim_thompson5910
  • jim_thompson5910
yeah
jim_thompson5910
  • jim_thompson5910
which trig function will you use for #7 ? sine or cosine?
anonymous
  • anonymous
sin
jim_thompson5910
  • 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
  • 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
jim_thompson5910
  • jim_thompson5910
you could use sine but the phase shift would be different. Everything else would stay the same
anonymous
  • anonymous
Okay , it makes sense
anonymous
  • anonymous
in number 8 , what do they mean by horizontal shift ?
jim_thompson5910
  • 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
  • jim_thompson5910
|dw:1440390964651:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1440390975447:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1440390992114:dw|
anonymous
  • anonymous
So it is 45 . I'm used to shifts when it comes to 2 graphs
jim_thompson5910
  • jim_thompson5910
yeah it's 45 if we use cosine
anonymous
  • anonymous
Since with cosine x = 0
jim_thompson5910
  • jim_thompson5910
what do you mean?
anonymous
  • anonymous
idk . I'm probably confusing myself
anonymous
  • 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
  • jim_thompson5910
in this case, we used the parent function y = cos(x)
jim_thompson5910
  • jim_thompson5910
it's just the base function that you apply the transformations to
jim_thompson5910
  • jim_thompson5910
A = 100 which is positive, we aren't flipping the function over
anonymous
  • anonymous
if it were to be flipped , how would I know that was the case ?
jim_thompson5910
  • 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
  • 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
  • jim_thompson5910
So there are a few ways to get the h(t) function
jim_thompson5910
  • jim_thompson5910
if A = +100, then the phase shift is C = 45 if A = -100, then the phase shift is C = 0
anonymous
  • 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
  • jim_thompson5910
t and h can't be negative, but A could be
jim_thompson5910
  • jim_thompson5910
the amplitude is |A| the absolute value ensures we have a positive amplitude
jim_thompson5910
  • jim_thompson5910
on your worksheet, draw vertical bars around the "A" where it says `A is the amplitude` on page 1
anonymous
  • 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
  • jim_thompson5910
exactly
jim_thompson5910
  • jim_thompson5910
eg: y = -2*cos(x) has an amplitude of |-2| = 2
anonymous
  • anonymous
okay , I get it
anonymous
  • anonymous
in the equation on number 10, what's "Af" ?
jim_thompson5910
  • jim_thompson5910
I don't see #10 in any of the attachments
anonymous
  • anonymous
one moment
anonymous
  • anonymous
1 Attachment
jim_thompson5910
  • jim_thompson5910
oh, that's A*f
jim_thompson5910
  • jim_thompson5910
f is the trig function look back at page 1
anonymous
  • anonymous
so it's the amplitude times the trig function ? 100cosx ? or ...
jim_thompson5910
  • jim_thompson5910
yeah like how I wrote above when I wrote out H(t)
anonymous
  • anonymous
where did you write that ?
jim_thompson5910
  • jim_thompson5910
it's buried a bit when I wrote \[\Large H(t) = 100\cos\left(\frac{\pi}{45}(t-45)\right) + 110\]
anonymous
  • anonymous
oooh okay , I remember
anonymous
  • anonymous
and t would b 90 , correct ?
anonymous
  • anonymous
be *
jim_thompson5910
  • jim_thompson5910
t is just a variable that represents the time in seconds
jim_thompson5910
  • 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
  • anonymous
wait so how exactly is the equation set up ?
jim_thompson5910
  • jim_thompson5910
what do you mean?
anonymous
  • anonymous
I plug everything into that equation right ?
jim_thompson5910
  • jim_thompson5910
I'd just leave it as \[\Large H(t) = 100\cos\left(\frac{\pi}{45}(t-45)\right) + 110\]
jim_thompson5910
  • jim_thompson5910
They want the function or equation that models the curve
anonymous
  • anonymous
So no solving ?
jim_thompson5910
  • jim_thompson5910
nope, they just want you to write the equation
anonymous
  • anonymous
and for number 11 , i use that same equation, I'm just solving for t ?
jim_thompson5910
  • jim_thompson5910
correct. replace h(t) with 100 and solve for t
anonymous
  • anonymous
To start I'd subtract 110 over to 100 right ?
jim_thompson5910
  • jim_thompson5910
yes
anonymous
  • anonymous
what do I do from there ?
jim_thompson5910
  • jim_thompson5910
you should have this \[\Large 100\cos\left(\frac{\pi}{45}(t-45)\right) = -10\]
jim_thompson5910
  • jim_thompson5910
now divide both sides by 100
anonymous
  • anonymous
so it'd be equal to -1/10 ?
jim_thompson5910
  • jim_thompson5910
yeah or -0.1
anonymous
  • anonymous
I feel like decimal is better
jim_thompson5910
  • 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
  • anonymous
what's arccos ?
jim_thompson5910
  • jim_thompson5910
arccos is \(\Large \cos^{-1}\), ie inverse cosine
anonymous
  • anonymous
so csc ?
jim_thompson5910
  • jim_thompson5910
no, that's a common misconception
jim_thompson5910
  • jim_thompson5910
if you want to say 1/cos, you say secant
jim_thompson5910
  • jim_thompson5910
you don't say \(\Large \cos^{-1}\) to mean secant \(\Large \cos^{-1} \ne \sec\)
anonymous
  • anonymous
oh okay , so I haven't learned arccos
jim_thompson5910
  • jim_thompson5910
your calculator should have a cos^(-1) or arccos button
anonymous
  • anonymous
I have cos^-1
jim_thompson5910
  • jim_thompson5910
ok use that
anonymous
  • anonymous
so I find arccos-0.1
jim_thompson5910
  • jim_thompson5910
yeah what is that numeric value?
anonymous
  • anonymous
95.74 ?
jim_thompson5910
  • jim_thompson5910
make sure you're in radian mode
anonymous
  • anonymous
1.67
jim_thompson5910
  • jim_thompson5910
yes or -1.67
jim_thompson5910
  • 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
  • jim_thompson5910
so solve for t \[\Large \Large \frac{\pi}{45}(t-45) = -1.67\]
anonymous
  • anonymous
do I distribute or multiply pi/45 by the recipricol on both sides ?
jim_thompson5910
  • jim_thompson5910
I'd multiply by the reciprocal
anonymous
  • anonymous
so it'd be \[t-45=\frac{ -75.15 }{ \pi }\]
jim_thompson5910
  • jim_thompson5910
-75.15/pi = -23.9209879 approximately
jim_thompson5910
  • jim_thompson5910
now just one more step to go
anonymous
  • anonymous
adding the 45 over
jim_thompson5910
  • jim_thompson5910
yes
anonymous
  • anonymous
so it would equal 21.08 ?
jim_thompson5910
  • jim_thompson5910
I'm getting 21.07901205 with a calculator and geogebra is telling me 21.06521 so it's pretty close
jim_thompson5910
  • jim_thompson5910
so roughly 21 seconds
anonymous
  • anonymous
I rounded the 21.07901205
jim_thompson5910
  • jim_thompson5910
that works
anonymous
  • anonymous
thank you so much !
jim_thompson5910
  • jim_thompson5910
and desmos says the same thing more or less https://www.desmos.com/calculator/swknd4tau1
jim_thompson5910
  • jim_thompson5910
you'll have to click on the first intersection point between the curve and the line
anonymous
  • anonymous
You really helped a lot lol . I can finally release some stress . Thank you
jim_thompson5910
  • jim_thompson5910
you're welcome

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