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Math Help - Proportions & Integrals

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    Proportions & Integrals

    A searchlight is located at point A, 40 feet from a wall. The searchlight revolves counterclockwise at a rate of pi/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 frm 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 thetak the angle formed by AP and AB
    C) how fast(in lumens per second) is the strength of the light changing when theta=pi/4? Is it increasing or decreasing? Justify your answer.
    D) Find the value of theta between theta=0 and theta=pi/2 after which L is less than 1000 lumens
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    Quote Originally Posted by rawkstar View Post
    A searchlight is located at point A, 40 feet from a wall. The searchlight revolves counterclockwise at a rate of pi/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 frm 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.

    \textcolor{red}{10000 = \frac{k}{40^2}} ... solve for k

    B) express L as a function of thetak the angle formed by AP and AB

    \textcolor{red}{\cos{\theta} = \frac{40}{d}}

    solve for d in terms of \textcolor{red}{\theta}, then determine L as a function of \textcolor{red}{\theta}

    C) how fast(in lumens per second) is the strength of the light changing when theta=pi/4? Is it increasing or decreasing? Justify your answer.

    determine the value of \textcolor{red}{\frac{dL}{dt}} at the stated angle.

    D) Find the value of theta between theta=0 and theta=pi/2 after which L is less than 1000 lumens

    set the expression for L < 1000 and solve for \textcolor{red}{\theta}
    ...
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    Quote Originally Posted by rawkstar View Post
    A searchlight is located at point A, 40 feet from a wall. The searchlight revolves counterclockwise at a rate of pi/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 frm 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 thetak the angle formed by AP and AB
    C) how fast(in lumens per second) is the strength of the light changing when theta=pi/4? Is it increasing or decreasing? Justify your answer.
    D) Find the value of theta between theta=0 and theta=pi/2 after which L is less than 1000 lumens
    A) Find the constant of proportionality k.

    At the closest point P, L=10,000 Lumens
    A searchlight is located at point A, 40 feet from a wall. So d = 40
    Use L=k/d^2 to solve for k

    B) express L as a function of the angle formed by AP and AB

    I you connect A, P and B you will have a right triangle with AP=40, AB=d and measure of angle A= \theta
    Last edited by ione; March 2nd 2010 at 02:48 PM.
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    Quote Originally Posted by skeeter View Post
    ...

    im not sure how you would solve cos theta=40/d other than d=40/cos theta
    and then i have no idea how to determine L as a function of theta
    Last edited by rawkstar; March 2nd 2010 at 06:58 PM.
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  5. #5
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    Quote Originally Posted by rawkstar View Post
    im not sure how you would solve cos theta=40/d other than d=40/cos theta
    and then i have no idea how to determine L as a function of theta
    you're halfway there ... substitute \frac{40}{\cos{\theta}} in for d in the equation for L and simplify as much as possible.
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