Question

Sinusoidal equations

Answer 1

I'm very unsure if I am graphing this correctly, some assistance would be greatly appreciated.

Answer 2

We looking at Mark ....blah blah

Answer 3

The sine graph looks right to me assuming the point is on the tip of the blade.

Answer 4

that was a question
i just forgot the ?

Answer 5

yes mark :)

Answer 6

the point is on the colored in dot lol @aum

Answer 7

Is there a diagram of the paddle wheel accompanying this question?

Answer 8

no but i could draw one?

Answer 9

"... a point on the paddle blade ...."
You have to make the assumption the point was at the tip of the paddle blade in order to have 16' above water at max and 2 feet below water at min.

Answer 10

yup! got that! the midline would be 7

Answer 11

If the point was at a radius of 5' from the center of the paddle wheel then you will need additional information or a figure to indicate how much of the paddle wheel is submerged under water.

Answer 12

wouldn't that be 2 ft under the water..

Answer 13

If the point was at the tip of the paddle wheel (that is, at radius 9').

Answer 14

yuh,

Answer 15

diameter was 18 feet
so radius is 9 feet
\[\text{ y coordinate is } y=9 \sin(\theta) \]
1 revolution ever 10 secs
so we can write theta in terms of the time unit seconds
\[\theta=\frac{1 rev}{10 \sec} t \sec =\frac{ 2 \pi rad }{10 } t= \frac{\pi t }{5} rad \]

Answer 16

so what I'm saying is you can write your equation y up there in terms of t

Answer 17

I don't know if we are still looking at the graph or now

Answer 18

not*

Answer 19

this equation doesn't include anything about the water though

Answer 20

it is about the wheel and time

Answer 21

my equation that is

Answer 22

There should be a phase shift and a vertical shift too in the equation.

Answer 23

I guess we are suppose to be graphing the distance the blade is from the water per sec

Answer 24

so far we have
\[y=9\sin( \frac{\pi}{5} t)\]
and we know the highest we want this function is at the y value 16
we can can get there might letting the vertical shift be 7 if I'm not mistaken
because the highest sin(pi/5t) can be is 1
and 1(9)=9
and 9+7=16

Answer 25

\[y=9\sin(\frac{\pi}{5}t)+7\]

Answer 26

ok we also have to consider that happens when t=4

Answer 27

sorry i lost wifi

Answer 28

Is my graph correct..

Answer 29

y = Asin(Bt + C) + D
A = 9; B = 2pi/10 = pi/5
y = 9sin(pi/5 * t + C) + D
D = 7 because max of sine is 1 and 9 + D = 16 and so D = 7.
y = 9sin(pi/5 * t + C) + 7
put t = 4 and y = 16 and solve for C.

Answer 30

so I think we can get the equation if we find a such that
\[16=9 \sin(\frac{\pi}{5}(4)+a)+7\]

Answer 31

oh that is what aum is saying

Answer 32

4pi/5 + C = pi/2
C = -3pi/10

y = 9sin(pi/5 * t - 3pi/10) + 7

Answer 33

exactly

Answer 34

this is mine:
9cos(pi/5(t-4))+7

Answer 35

is that at all right?

Answer 36

either one of those work

Answer 37

looks dif from yall..
why did you get a 3pi/10?

Answer 38

gj @karatechopper

Answer 39

wait that means i did it right?! OMG MYIN
YASSSSS
THANK YOU MYIN AND AUM! GREATLY APPRECIATED

Answer 40

we used the sin function instead

Answer 41

oh.
Now thats a problem..

Answer 42

http://www.wolframalpha.com/input/?i=9+sin%28pi%2F5*t-3pi%2F10%29%2B7%3D9cos%28pi%2F5%28t-4%29%29%2B7

Answer 43

how to get the sin one?
i think i know how but i want to see if from yalls perspective..

Answer 44

it says true
but it says true for integers

Answer 45

sine and cosine differ by pi/2.

Answer 46

?

Answer 47

Estoy confuzzled.

Answer 48

Move a sine curve to the left by pi/2 you will get the cosine curve.

Answer 49

but doesn't that very by period?

Answer 50

I am talking about comparing sin(x) and cos(x).

Answer 51

I was taught that the things that change are
function
phase shift
and reflection

Answer 52

\[\sin(\frac{\pi}{5}t-\frac{3 \pi}{10})=\cos(\frac{\pi}{5}t-\frac{4\pi}{5}) \\ \text{ recall } \sin(\frac{\pi}{2}-x)=\cos(x) \\ \frac{\pi}{2}-(\frac{\pi}{5}t-\frac{3 \pi}{10})=-\frac{ \pi}{5}t+\frac{8 \pi}{10} =\frac{-\pi}{5}t+\frac{4 \pi}{5}=-(\frac{\pi}{5}t-\frac{4\pi}{5}) \\ \text{ so } \\ \cos(\frac{\pi}{5} t- \frac{4 \pi}{5}) \\ =\cos(-(\frac{\pi}{5}t-\frac{4\pi}{5}))=\cos(\frac{\pi}{2}-(\frac{\pi}{5}t-\frac{3\pi}{10}))=\sin(\frac{\pi}{5}t-\frac{3\pi}{10})\]
i had to verify it for myself lol

Answer 53

I don't think I have been taught that quite yet.

Answer 54

so to verify the equations were the same
I used the fact that sin and cos are co-functions cos(pi/2-x)=sin(x)
and that cos is even function that is cos(-x)=cos(x)

Answer 55

ahh ok

Answer 56

would yall mind assisting me on one more graph?

Answer 57

you can post it
but i will have to come back and look at it
i shouldn't be way too long

Answer 58

alrighty:) thanks!

Answer 59

it says the min is at t= 2.9 and we have 200 foxes there
it says the max is at t=5.1 and we have 800 foxes there
would should be able to find the amplitude from this pretty easily
800-200=600
600/2=300
so the amp is 300
so we know we have something like y=300*sin(theta)

Answer 60

ok but we also need to consider at what times those min and max occur

Answer 61

and i guess we do have a vertical shift as well
the lowest value was 200

Answer 62

\[y=300 \sin( b \theta+c)+500 \\ y(2.9)=200 \\ y(5.1)=800\]
this should give us enough info to find b and c

Answer 63

I put 500 as the vertical shift because
the lowest value sin can give is -1
and 300(-1)=-300
so I was thinking -300+what=200

Answer 64

the what being 500

Answer 65

oops that theta is suppose to be t in case

Answer 66

I think I should change the sign of 300 because the min occurs before the max
and on the regular sine graph the max occurs before the mine
so I think we have
\[y=-300\sin(bt+c)+500\]
I think the period can be found by doing 2*(5.1-2.9)
like the period can be found on the regular sine graph be doing 2*(where first max occurs-where first min occurs)
so I think the period is 4.4 here
that is \[\frac{2 \pi}{b}=4.4 \\ \frac{b}{2\pi}=\frac{1}{4.4} =\frac{10}{44}=\frac{5}{22} \\ b=\frac{10 \pi}{22}=\frac{5 \pi}{11}\]
\[y=-300\sin(\frac{5 \pi}{11}t+c)+500\]
now we can try to use one of those points to find c
\[800=-300\sin(\frac{5 \pi}{11}(5.1)+c)+500\]
\[-1=\sin(\frac{5 \pi }{11} \frac{51}{10} +c) \\ -1=\sin( \frac{51 \pi}{22}+c) \\ \frac{51 \pi}{22}+c=\frac{3\pi}{2} \\ c=\frac{3 \pi}{2}-\frac{51\pi}{22}=\frac{33-51}{22} \pi =\frac{-18}{22} \pi=\frac{-9}{11} \pi\]

Answer 67

\[y=-300\sin( \frac{5 \pi}{11} t+\frac{-9}{11})+500\]
I have to check to see if this right
I think it is
I might be making it also more complicated than it really is

Answer 68

oops I miss the pi there

Answer 69

\[y=-300\sin( \frac{5 \pi}{11} t+\frac{-9 \pi}{11 })+500\]

Answer 70

so we do have that min at 2.9 and we do have that max at 5.1

Answer 71

so that is good news

Answer 72

@aum using any formulas here is there a faster way to get that equation?

Answer 73

I feel like it took way to much effort for some reason

Answer 74

@zepdrix is there a faster way to get that equation I just got or an equivalent equation?

Answer 75

the only way i do these equations is compare that graph to the original graph y=sin(theta)

Answer 76

And arrange things so they fit the parent graph pattern

Answer 77

Hmm I was kind of doing it the way Chop was: Trying to write it in terms of Cosine, that seems easier to me at least.

The distance from the min to max is `half of the period`.
Ah ok looks like you figured all that out already :)

\[\Large\rm y=a \cos(stuff)-thing\]Our b term is given by: \(\Large\rm b=\frac{2\pi}{4.4}\)
Half of the distance from our min to max is 300, so our amplitude is 300.
And our midpoint is shifted up 500,
aaaand our max is shifted 5.1 to the right,\[\Large\rm y=300\cos\left[\frac{2\pi}{4.4}(x-5.1)\right]+500\]I think cosine makes it work out a tiny bit faster :O
Mmmm whatev :) Your approach seemed to work out well hehe!

Answer 78

how would the graph look?

Answer 79

im struggling to find b and the phase shift because i feel like my graph is completely off.

Answer 80

If I'm reading the instructions correctly, then it looks like your graph is very close to correct.
I would just fix that little leg that is reaching down to 0.