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Mathematicians... I need help!

Discussion in 'BBS Hangout' started by Hayesfan, Jun 25, 2010.

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  1. Hayesfan

    Hayesfan Member

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    Okay trying to help a friend with his homework... and this stuff is beyond me.

    Here are the two problems:

    #1 -
    On a certain spring day with 12 hours of daylight, the light intensity I takes on its largest value of 530 calories/cm^2 at midday. If t = 0 corresponds to sunrise, find a formula I = a sin bt that fits this information

    #2 (he sent me a screen shot)
    [​IMG]

    let me know if the picture doesn't work.. and I will just write the question here
     
  2. Bandwagoner

    Bandwagoner Member

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    Link doesn't work.

    First answer should be

    A = 530
    B = 0.2618

    in radians
     
  3. Sweet Lou 4 2

    Sweet Lou 4 2 Member

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    Problem is that the earth is curved so I <> 0 at t=0. Fail of a problem.

    you solve these problems by setting up two equations. The first one is:

    530 = a sin (b * 6)

    but you also know that
    0 = a sin (b * 0)

    and

    0 = a sin (b *12)


    go have fun!
     
  4. Hayesfan

    Hayesfan Member

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    Thanks for the answer... (edit - thanks lou! That's what I thought, but I had no idea how to do it. I haven't done these problems since high school).

    Here's the second one.


    Scientists sometimes use the formula f(t)=a sin (bt+c) +d to simulate temperature variations during the day, with time t in hours, temperature f(t) in degrees C and t=0 corresponding to midnight. Assume that f(t) is decreasing at midnight. Determine the values of a, b, c, and d that fit the information.

    The temperature at midnight is 11 degrees C and the high and low temperatures are 21 deg C and 1 deg C.

    This is the answer he has (c is what it looks like he is filling in)
    A=10
    B = pi/12
    C= -5*Pi/6
    D= 11
     
  5. RoxSqaud

    RoxSqaud Member

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    I was told there would be no math.
     
  6. Sweet Lou 4 2

    Sweet Lou 4 2 Member

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    Same thing, set up the conditions

    you know that at midnight, the temp is 11 degrees so

    11 = a sin (b * 0 + c) + d

    or

    11 = a sin (b * 24 + c) + d

    so that gives you

    a sin (c ) + d = 11

    and sin (c)= sin (24b + c)

    since you know the max and min are 21 and 1

    you can say

    21 = a sin (b*t1 + c) + d
    1 = a sin (b * t2 + c) + d

    i think you'd take the derivatives of those as well to give yourself two more equations and from that can solve the problem.
     
  7. Bandwagoner

    Bandwagoner Member

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    No way man these are just function modifiers. No need to do all of that.
     
  8. B-Bob

    B-Bob "94-year-old self-described dreamer"
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    Intensity should be given in power per area, so more problem fail. Hayesfan, this friend needs to get to a better course, prof or textbook.
     
  9. Bandwagoner

    Bandwagoner Member

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    All of those A B C D do a specific change to the shape of the sin curve.

    My answer above is actually pi/12 but did not know if you were looking for that or a decimal.

    It looks like he has it but double check with a graphing calculator. The point of the exercise is to know:

    What does the value of A,B,C,D do to the shape of a Sin curve.

    Once you know that very well it is easy to fill in the numbers to change the curve to set position.
     
  10. Bandwagoner

    Bandwagoner Member

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  11. Hayesfan

    Hayesfan Member

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    thanks casey... I will tell him what you said and see if he has any questions :)
     
  12. Hayesfan

    Hayesfan Member

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    Need a bit more assistance :)

    Here's the next problems I'm dealing with.

    Can any of you explain this to me so that I understand what the heck to tell him??
     
  13. Qball

    Qball Member

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    So what's the answer to the above Casey?
     
  14. SwoLy-D

    SwoLy-D Member

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    Shouldn't HE be getting a ClutchFans account? :grin:

    Also, do you tell him you're getting the answers here?

    Please read: http://en.wikipedia.org/wiki/Trigonometric_functions - If you understand the Pythagorean Theorem, then you understand these functions. :eek: Start there.
     
  15. Hayesfan

    Hayesfan Member

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    LOL maybe I should tell him to. This trig class is killing me and I'm not even in it! I was doing good on some of the word problems... but when they get complicated like this I just want to cry. ;)
     
  16. Air Langhi

    Air Langhi Contributing Member

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    this one is really easy without substitution:

    cos(x)-cos(2x)=0

    cos(x) = cos(2x)

    0 would be the obvious solution if you think of the unit circle
     
  17. DonkeyMagic

    DonkeyMagic Member
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    the answer is....the answer is....4 :confused:

    [​IMG]
     
    1 person likes this.
  18. Hayesfan

    Hayesfan Member

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    The professor has said that you're not supposed to move any of these to the other side of the formula, so I'm pretty sure that wouldn't work.

    Thank you though.
     
  19. AustinBriggs

    AustinBriggs Member

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    1 person likes this.
  20. francis 4 prez

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    it's cos(x) - 2cos^2x + 1, not -1. so you get (2cos(x)+1)(cos(x)-1)=0.
     

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