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Thread: integrals of trigonometric functions

  1. #1
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    integrals of trigonometric functions

    a thief tries to enter a building by placing a ladder over a 9-ft-high fence so it rests against the building,which is 2 ft back from the fence.

    what is the length f the shortest ladder and the ground? (let Q be the angle between the ladder and the ground) express the length of the ladder in terms of Q,

    and then find the value of Q that minimizes the length of the ladder.
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  2. #2
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    let $\displaystyle x$ = distance from foot of the ladder to the fence

    $\displaystyle f\cos{Q} = x+2$

    $\displaystyle \cot{Q} = \frac{x}{9}$


    $\displaystyle f\cos{Q} = 9\cot{Q} + 2$

    $\displaystyle f = 9\csc{Q} + 2\sec{Q}$

    find $\displaystyle \frac{df}{dQ}$ and minimize $\displaystyle f$
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  3. #3
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    after I find out the derivetive, how can I minimize the f? let that equals to 0,but I don't know how to solve the value for Q
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  4. #4
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    $\displaystyle
    f = 9\csc{Q} + 2\sec{Q}
    $

    $\displaystyle
    \frac{df}{dQ} = -9\csc{Q}\cot{Q} + 2\sec{Q}\tan{Q} = 0
    $

    $\displaystyle \frac{-9\cos{Q}}{\sin^2{Q}} + \frac{2\sin{Q}}{\cos^2{Q}} = 0$

    $\displaystyle
    \frac{-9\cos^3{Q}}{\sin^2{Q}\cos^2{Q}} + \frac{2\sin^3{Q}}{\sin^2{Q}\cos^2{Q}} = 0
    $

    $\displaystyle
    -9\cos^3{Q} + 2\sin^3{Q} = 0
    $

    $\displaystyle
    9\cos^3{Q} = 2\sin^3{Q}
    $

    $\displaystyle
    \frac{9}{2} = \frac{sin^3{Q}}{\cos^3{Q}}
    $

    $\displaystyle \frac{9}{2} = \tan^3{Q}$

    $\displaystyle \sqrt[3]{\frac{9}{2}} = \tan{Q}$

    $\displaystyle Q = \arctan\left(\sqrt[3]{\frac{9}{2}}\right)$

    $\displaystyle
    Q \approx 58.8^{\circ}
    $

    now go back and determine the length of the ladder.
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