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Need Math WIZARDS help.. emergency

Discussion in 'The Lounge' started by 78Suburban, May 10, 2006.

  1. 78Suburban

    78Suburban 1/2 ton status

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    Here's the deal, I've got a really good buddy who is a ck5er on my messenger. He has night classes and won't be home untill 10 p.m. mountain time tonight. He has a trig test tomorrow. Here's one of the problems he told me he was having a hard time understanding how to do:

    The angle of elevation of the sun is decreasing at 1/4 rad/h. How fast is the shadow cast by a building of height 50 m lengthening, when the angle of evelvation of the sun is pi/4?





    So Does anyone know how to do this problem? Come on math geniuses, give use a little lesson in trig derivatives. :bow:
    You'll be helping a ck5er out!!!!! I don't know how many times this guy has helped me figure out things about how to work on trucks, lets help him figure out how to work on calculus. Math is not my thing, so I don't know how.

    Thanks,
    James
     
  2. BranndonC

    BranndonC 3/4 ton status

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  3. BigBen

    BigBen 1/2 ton status Premium Member

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    It's been a LONG while, but here we go.

    First we much find the relation between the main variables.
    This is drawn as a right triangle.
    (l) is the shadow length along the ground.
    (h) is the height of the building
    Theta is the angle we are talking about between the ground and the hypotenuse that reaches to the tip of the building.

    Relate l and theta with:
    Tangent = Opposite/Adjacent for right triangles.

    Tan(theta) = h/l
    solve for l
    l=h/tan(theta)
    or l = h*(Tan(theta))^-1
    Now we take the derivative with respect to time:

    (d/dt)l=(d/dt)(tan(theta))^-1*h

    Use the chain rule to solve and find dl/dt in terms of d(theta)/dt

    I belive it is: (I AM VERY RUSTY HERE)

    dl/dt=-1*(Tan(theta))*(Sec(theta))^2*(d(theta)/dt)*h

    we are looking for dl/dt
    we know that theta=pi/4 radians
    we know that d(theta)/dt=1/4 rad/hr
    we know that h = 50m

    I think you can "plug and chug" after this point.

    Once again, it's been a while.... but I think this is the direction to take on this one.

    If not..... someone will chime in shortly.

    -Ben
     
  4. BigBen

    BigBen 1/2 ton status Premium Member

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    That took a while to work out so......


    -pad-


    :D
     
  5. Russell

    Russell LB7 Tahoe Status Premium Member GMOTM Winner

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    Well, I am the buddy -- Things were looking very grim this morning when this was posted. I just got back from the Calc class, where unbelievably enough, only 3 people showed up (think there was a field trip or something) However, that also gave me the full class with the teacher to figure stuff out, and it was a hugely big time help :)

    Ben, you're right in how to do the problem, I just wasn't seeing all of the various relationships like I should have been.

    Now, off to studying for that Physics test tonight, lol
     
  6. BigBen

    BigBen 1/2 ton status Premium Member

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    Glad to hear you got it worked out.

    Also glad to know I haven't lost it all yet.

    -Ben
     

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