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All Genius' are Welcome!

  1. Nov 8, 2009 #1
    Simple yet challenging calc problem - please help!

    http://img252.imageshack.us/i/en8t.jpg/

    a searchlight is located at point A 40 feet from a wall. The searchlight revolves counterclockwise at a rate of π/30 radians per second. At any point B on the
    wall, the strength of the light L, is inversely proportional to the square of the distance d from A; that is, at any point on the wall L = k/d^2 . At the closest point P, L = 10,000 lumens.

    a) Find the constant of proportionality k.
    b) Express L as a function of θ , the angle formed by AP and AB.
    c) How fast (in lumens/second) is the strength of the light changing when θ =π/4? Is it
    increasing or decreasing? Justify your answer.
    d) Find the value of θ between θ =0 and θ =π/2 after which L<1000 lumens.

    I was only able to do part a) the other parts are rather confusing. If anyone could help that would be greatly appreciated.
    My Attempt:

    a) (40)^2 + (10,000)^2 = d^2
    d = 10,000

    L = k/d^2
    10,000 = k/(10,000)^2
    k = 1E12
     
    Last edited: Nov 8, 2009
  2. jcsd
  3. Nov 8, 2009 #2
    b)

    L = [.5(40 x 10,000)] x sin(theta)
    L = 200,000 x sin(theta)

    I'm sorry, I'm really confused and just need guidance to know if I'm doing this problem right, or if I'm completely wrong.
     
  4. Nov 8, 2009 #3
    d)

    999.999 =< 200,000 x sin(theta)
    theta =< 0.005

    I have a feeling I'm doing this terribly wrong =/
     
  5. Nov 8, 2009 #4
    For A, you are given three unknowns (one of which is a constant) and the values for the two variables (what is the distance between point A and point P?). I'm sure that you can figure A out from there.

    For B, whats the relationship between d and theta?

    For C, think instantaneous rate of change.

    For D, how what would you do if you were asked to find of x for y = 2x so that every value of x after that produced a y value thats greater than 10?

    My answers are below in a spoiler so that you can check yours.
    a) L(d) = k/d2
    L(40) = 10000
    10000 = k/402
    16000000 (1.6 x 107) = k

    b) cos(θ) = A/H = 40/d
    d = 40/cos(θ)

    L(θ) = k/(40/cos(θ))2 = 10000cos2(θ)

    c) L(θ) = 10000cos2(θ)
    dL/dθ = 10000•2•cos(θ)•-sin(θ) = -20000cos(θ)sin(θ)
    when θ = π/4 --> dL/dθ = -20000cos(π/4)sin(π/4) = -10000
    Since the derivative is negative, L is decreasing when θ = π/4

    d) L(θ) = 1000
    1000 = 10000cos2(θ)
    1/10 = cos2(θ)
    1/[tex]\sqrt{10}[/tex] = cos(θ)
    cos-1(1/[tex]\sqrt{10}[/tex]) = θ
    which is about .398π
     
    Last edited: Nov 8, 2009
  6. Nov 8, 2009 #5
    Thanks so much ƒ(x), i personally loved your "hints" in the beginning because they really helped me look at the problems differently rather than me just getting an answer.

    Thanks again.
     
  7. Nov 8, 2009 #6
    Welcome
     
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