[xofunkyfresh]

Has anyone every done the Sunrise over New York SL Type 2 portfolio? If so, great, PM me. LOL.
If not, can someone please help me determine the cosine equation of the following graph?
y=acos(b-c)+d
and so far, I figured that
a= 1.48
d= 7.20
I don't know how to solve the horizontal displacement and translation, can someone help >_<


Okay, I thought it'd be dumb to create another board for homework help for me, but I'm having problems solving a calculus question:

A ladder is to be carried horizontally around a corner from a corridor 'a' meters wide into a corridor 'b' meters wide.

What is the maximum length that the ladder can be?

[angelqian]

ok, so, I'll be general about this.

for one, I think your formula needs to be y = a * cos ( k * x - phi ) + d

I'm using standard mathematical notation for a wave function btw, so change your variables accordingly ( b = k etc ).

a describes the amplitude, so it is essentially [max(y) - min(y)] / 2, I divide by 2 because cosines are generally centered around 0.

k describes the wavenumber (2PI / wavelength) , or how "long" the sinusoid is. If we take the wavelength to be the whole 52 weeks, then k would be 2PI / 52.

phi represents the phase shift, which is how off center the cosine is, this is a poor definition though, but in your case, phi is about 0 (Since your cosine started at the max, if it were at the minimum, phi would be 180 degrees or PI radians). Note that a shift in the positive x direction is represented by a negative phi value.

d is the vertical displacement, and since cosines should average to 0, should equal the average of all the points for one period (you have one period here).

I gave you a lot of data but you should be able to figure everything out from here.

[xofunkyfresh]

Ah, so then 'k' would be 6.92.

Are the other variables correct?

Thanks so much (:

[angelqian]

k should be 2 PI / wavelength, your wavelength is 52, so k should be around 0.12 or something. (The bigger k is, the narrower the wave).

and no, d is not correct.

I think you assumed that your graph shows only the negative lobe of the cosine, but I believe it is an entire period (starts from max, goes to 0, goes to min, goes to 0, ends up at max) instead of (starts from 0, goes to min, goes back to 0).

for this case, you can simply average all the points to find d. (Around 5.5 - 6 ish)

[xofunkyfresh]

Oh I see where you're getting at, so D has to be 5.87 since its max+min/2 ??

Thanks a lot (:

[ChunJin]

Second problem is mildly easy once you start drawing it out and setting it up =p (similar triangles -> total length -> differentiating that -> setting to 0, then solving for x -> plug/chug & reduce). The only ugly part is reducing the formula to something beautiful; however, the unreduced form is decent. Just a few algebra things here and there.

It would have been alright to make another thread for the second question, but, it's okay.

Before I help you with problem #2. I really need to see what you've done in terms of work-wise. Whether it be a drawing or something written down. I'd love to help you cheat but that's not cool.

If you show a little work or how you'd go about the problem. I'll gladly post my solution with the work written out and guide you through it =).

[angelqian]

#2 can be solved by inspection with geometry, just draw it out. Interestingly though, this question probably involves a bit more calculus than you have taken to solve rigorously.