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Thread: Angle optimization problem

  1. #1
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    Angle optimization problem

    anyone can you help me? please..
    ho to simplify this,
    [tanθ + (d/x)]/( 1+ (d/x)Ětanθ)= (d + 5.6)/x

    can you show me step by step?
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  2. #2
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    re: Angle optimization problem

    Simplify? One simplifies an expression ... you've posted an equation. Equations are usually solved.

    What variable are you solving for?
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    re: Angle optimization problem

    simplify to make the tanθ as the subject

    i have tried it. but just to make sure that I get the correct answer
    Last edited by nurininsyirah97; Apr 10th 2017 at 04:43 AM.
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  4. #4
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    re: Angle optimization problem

    Quote Originally Posted by nurininsyirah97 View Post
    simplify to make the tanθ as the subject

    i have tried it. but just to make sure that I get the correct answer
    Post your work, then ...
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    re: Angle optimization problem

    (tanθ+(d/X))/(1+tanθ(d/X) )=(d+5.6)/X
    Simplify the equation,
    tanθ+(d/X)=(d+5.6)(1+tanθ(d/X))/X
    tanθ=d/X+d^2/X^2 tanθ+5.6/x+5.6d/X^2 tanθ-d/X
    Factorize tanθ and get,
    tanθ(1-5.6d/X^2 -d^2/X^2 )=5.6/X
    tanθ=-5.6X/(-X^2+d^2+5.6d)

    is it correct?
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  6. #6
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    re: Angle optimization problem

    $\dfrac{x\tan{t}+d}{x+d \tan{t}}=\dfrac{d+5.6}{x}$

    $x^2\tan{t}+dx = dx + 5.6x + d^2\tan{t} + 5.6d \tan{t}$

    $x^2\tan{t} - d^2\tan{t} - 5.6d \tan{t} = 5.6x$

    $\tan{t}\bigg(x^2 -d^2 - 5.6d \bigg) = 5.6x$

    $\tan{t} = \dfrac{5.6x}{x^2-d^2-5.6d}$
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    re: Angle optimization problem

    ok cool. thanks for the answer. but then I want to derive it and find the critical point.
    sec^2 θ dθ/dX=5.6((-d^2-5.6d-X^2)/(X^2-d^2-5.6d)^2 )
    -d^2-5.6d-X^2=0
    X^2=-d^2-5.6d
    X=√(-d^2-5.6d)
    I don't think this is correct. can you help me?
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  8. #8
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    re: Angle optimization problem

    Quote Originally Posted by nurininsyirah97 View Post
    ok cool. thanks for the answer. but then I want to derive it and find the critical point.
    sec^2 θ dθ/dX=5.6((-d^2-5.6d-X^2)/(X^2-d^2-5.6d)^2 )
    -d^2-5.6d-X^2=0
    X^2=-d^2-5.6d
    X=√(-d^2-5.6d)
    I don't think this is correct. can you help me?
    what are you trying to accomplish here?

    it would help if you posted the original problem statement in its entirety.

    thread moved to calculus.
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    re: Angle optimization problem

    im sorry i didn't mean to trouble anybody its just that i confused.. im really sorry
    this is the real problem by daewoo_lowrider

    Hi, can someone please help me with this problem? It states:

    Suppose a try is scored at T, then the kicker can kick to the posts from anywhere along the line TF perpendicular to the try line PQ. (Refer to Figure 1) Where is the best place to kick from?
    Many kickers seem to think that the further they go back (within their kicking limit) the better. But this is not necessarily the case. The best position for the kick to be taken is clearly where angle ACB subtended by the posts along TF is a maximum.

    Here is the diagram:
    http://img.photobucket.com/albums/v3...ce/figure1.jpg

    Basically, I think I have to use trig, and a circle subtending the goal posts, but I'm stuck at that, I have hardly any measurements and angles (the distance between the goal posts is 5.6m.

    Thanks for any help,
    Daniel

    i found it in mhf but i dont understand the solution
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  10. #10
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    re: Angle optimization problem



    Suppose a try is scored at T, then the kicker can kick to the posts from anywhere along the line TF perpendicular to the try line PQ. (Refer to Figure 1) Where is the best place to kick from? ... the distance between the goal posts is 5.6m.
    let $\theta = \angle{ACB}$, $\phi = \angle{TCA}$, $|TA| = d$ and $|TC| = x$. Note $d$ is a constant and $x$ is variable.

    let the distance between the posts, $|AB| = 5.6 = r$, also a constant.

    $\phi = \text{arccot}\left(\dfrac{x}{d}\right)$

    $\phi + \theta = \text{arccot}\left(\dfrac{x}{d+r}\right)$

    $\theta = \text{arccot}\left(\dfrac{x}{d+r}\right) - \text{arccot}\left(\dfrac{x}{d}\right)$

    note the derivative of $\text{arccot}(u) = -\dfrac{1}{1+u^2} \cdot \dfrac{du}{dx}$ ...

    $\dfrac{d}{dx} \bigg[\theta = \text{arccot}\left(\dfrac{x}{d+r}\right) - \text{arccot}\left(\dfrac{x}{d}\right) \bigg]$

    $\dfrac{d \theta}{dx} = - \dfrac{1}{1+\left(\frac{x}{d+r}\right)^2} \cdot \dfrac{1}{d+r} + \dfrac{1}{1+\left(\frac{x}{d} \right)^2} \cdot \dfrac{1}{d} = \dfrac{d}{d^2+x^2} - \dfrac{d+r}{(d+r)^2+x^2}$

    $\dfrac{d}{d^2+x^2} - \dfrac{d+r}{(d+r)^2+x^2} = 0$

    $\dfrac{d}{d^2+x^2} = \dfrac{d+r}{(d+r)^2+x^2}$

    solving for x yields ...

    $x = \sqrt{d(d+r)}$
    Thanks from nurininsyirah97
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  11. #11
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    Re: Angle optimization problem

    may I ask? why you choose arccot? not arctan? wasn't it supposed to be tan(θ+β)=(d+5.6)/X from the beginning?
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  12. #12
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    Re: Angle optimization problem

    Quote Originally Posted by nurininsyirah97 View Post
    why you choose arccot? not arctan?
    Easier to calculate the derivative.
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  13. #13
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    Re: Angle optimization problem

    wowwwww!!!! you are genius. Thank you for everything. thank you very much. you just saved my life hahahaha. have a nice day skeeter..
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