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2nd Gen Help! Calculus Homework

Discussion in 'Off-Topic Discussion' started by xSpyderguyx, Oct 24, 2010.

  1. Oct 24, 2010 at 1:14 PM
    #1
    xSpyderguyx

    xSpyderguyx [OP] Well-Known Member

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    Hey guys, My son is in calculus and he has a rates of change word problem. It has been a long time since I have done these haha!

    I now realize how little I use certain maths when running a company.
    Please help me out! I tried once but was wrong!

    A street light is at the top of a 20 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of 6 ft/sec along a straight path. How fast is the tip of her shadow moving when she is 35 ft from the base of the pole?
     
  2. Oct 24, 2010 at 1:17 PM
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    xSpyderguyx

    xSpyderguyx [OP] Well-Known Member

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    He said a good tip was drawing a picture. I feel like the dumbest engineer ever :mad:
     
  3. Oct 24, 2010 at 1:30 PM
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    steve o 77

    steve o 77 braaap

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    If you had asked me this time last year I would have known lol. I have since forgotten.

    I know you need to make a function of some kind then take the derivative to get rate of change, then plug in 35'. I know that didn't help much, sorry.
     
  4. Oct 24, 2010 at 1:47 PM
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    RAT PRODUCTS

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    View the build. Too much to list.
    6x = 20x-20b
    -4x = -20b
    x = 5b/2
    we are given that her rate is 6 ft/s so db/dt=6. take the derivative of our function relating b to x with respect to time...
    = Interesting note here is that her position relative to the pole is not even in this equation, that means that the rate her shadow moves is constant through her entire walk...
    anyway...solve for dx/dt since we know db/dt= 6
    dx/dt= (5/2) * 6 = 15 ft/s <<<< rate of her shadow when she is 35 feet from the light
     
  5. Oct 24, 2010 at 1:48 PM
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    steve o 77

    steve o 77 braaap

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    :bowdown:
     
  6. Oct 24, 2010 at 1:54 PM
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    dalsmthme

    dalsmthme Well-Known Member

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    What he said......
     
  7. Oct 24, 2010 at 1:57 PM
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    theredofshaw

    theredofshaw Well-Known Member

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    we ought to start a section for education/school stuff...lol
     
  8. Oct 24, 2010 at 2:02 PM
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    xSpyderguyx

    xSpyderguyx [OP] Well-Known Member

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    He plugged in 15 ft/sec on the Webwork and it was incorrect?! That looked legit
     
  9. Oct 24, 2010 at 2:22 PM
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    Pugga

    Pugga Pasti-Dip Free 1983 - 2015... It was a good run

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    Don't feel bad, I haven't done problems like this in a looong time... My guess is 2.571 ft/s. I've got to say, I feel like my 4 years of engineering school has failed me big time...:eek:
     
  10. Oct 24, 2010 at 2:24 PM
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    Pugga

    Pugga Pasti-Dip Free 1983 - 2015... It was a good run

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    Haha, nice man, guess I was way the hell off! good job!
     
  11. Oct 24, 2010 at 2:30 PM
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    steve o 77

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    That was apparently wrong, or the stupid online thing is wrong.

    I seriously hate online homework lol.
     
  12. Oct 24, 2010 at 2:32 PM
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    113tac

    113tac Well-Known Member

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    i should know this! i am in calculus right now. what isu2014 looked about right. he has to find an equation that relates the info you know and the info you dont know.

    did it give any other information besides what you provided us?
     
  13. Oct 24, 2010 at 2:35 PM
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    xSpyderguyx

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    The 35 ft away from the pole does not matter. But 15 ft/sec was in correct.
     
  14. Oct 24, 2010 at 2:37 PM
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    steve o 77

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  15. Oct 24, 2010 at 2:38 PM
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    Pugga

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    Did the 2.571 ft/s work? I'm not sure if it hurts you to try to check it, if it does I wouldn't try it but that's what I came up with in a round about sort of way.
     
  16. Oct 24, 2010 at 2:39 PM
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    Krazie Sj

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    Depends on whether or not it's daylight savings or not.
     
  17. Oct 24, 2010 at 2:40 PM
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    Pugga

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    Good point, it never says the shadow is from the light pole, haha.
     
  18. Oct 24, 2010 at 2:43 PM
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    08pretaco

    08pretaco Almost there

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    i took online math try 15.0 ft/sec
     
  19. Oct 24, 2010 at 2:51 PM
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    113tac

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    20/6=y/(y-x)
    20y-20x=6y
    y=20/9x

    dx/dt=6

    dy/dt=20/9*dx/dt
    dy/dt=20/9*6=40/3 ft/sec


    try that. i found a similar problem in my calc book and that i just adhusted the problem to fit your numbers

    if you want me to try and explain it just let me know haha
     
  20. Oct 25, 2010 at 4:31 AM
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    Pugga

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    So OP, what was the correct answer?
     
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