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Math Help - Help finding x

  1. #1
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    Help finding x

    Hi can anybody show me how to find x?

    -180= -90 - tan^-1 (x/16) - tan^-1 (x/200)
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  2. #2
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    -180=-90-tan^{-1}(\frac{x}{16})-tan^{-1}(\frac{x}{200})

    Get rid of the negatives by adding the 90 to both sides and then multiplying by -1:

    90=tan^{-1}(\frac{x}{16})+tan^{-1}(\frac{x}{200})

    Take the derivatives:

    \frac{16}{x^{2}+256}+\frac{200}{x^{2}+40,000}=0

    Now, when we solve this we will get a complex solution, use the real part.

    Solving we get x=40\sqrt{2}i, \;\ x = -40\sqrt{2}i

    The 40\sqrt{2} is a solution.

    -90-tan^{-1}(\frac{40\sqrt{2}}{16})-tan^{-1}(\frac{40\sqrt{2}}{200})=-180
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  3. #3
    Moo
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    Hello,

    Why would the extremum of the function be the solution of the equation ? (just asking)
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  4. #4
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    Is

    Quote Originally Posted by galactus View Post
    -180=-90-tan^{-1}(\frac{x}{16})-tan^{-1}(\frac{x}{200})

    Get rid of the negatives by adding the 90 to both sides and then multiplying by -1:

    90=tan^{-1}(\frac{x}{16})+tan^{-1}(\frac{x}{200})

    Take the derivatives:

    \frac{16}{x^{2}+256}+\frac{200}{x^{2}+40,000}=0

    Now, when we solve this we will get a complex solution, use the real part.

    Solving we get x=40\sqrt{2}i, \;\ x = -40\sqrt{2}i

    The 40\sqrt{2} is a solution.

    -90-tan^{-1}(\frac{40\sqrt{2}}{16})-tan^{-1}(\frac{40\sqrt{2}}{200})=-180
    Is this actually a method of solving an equation?...if so could you possibly provide a site where I may read up on how to utilize it
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  5. #5
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    Hello, al2308!

    Solve for x\!:\;\;-180\;= \;-90 - \tan^{\text{-1}}\!\left(\frac{x}{16}\right) - \tan^{\text{-1}}\!\left(\frac{x}{200}\right)

    We have: . \tan^{\text{-1}}\!\left(\frac{x}{16}\right) + \tan^{\text{-1}}\!\left(\frac{x}{200}\right) \;=\;90


    Take the tangent of both sides: . \tan\left[\tan^{\text{-1}}\left(\frac{x}{16}\right) + \tan^{\text{-1}}\left(\frac{x}{200}\right)\right] \;=\;\tan90

    . . \frac{\tan\left[\tan^{-1}\left(\dfrac{x}{16}\right)\right] + \tan\left[\tan^{-1}\left(\dfrac{x}{200}\right)\right]} {1 - \tan\left[\tan^{-1}\left(\dfrac{x}{16}\right)\right]\left[\tan^{-1}\left(\dfrac{x}{200}\right)\right]} . =\;\tan90 \quad\Rightarrow\quad \frac{\dfrac{x}{16} + \dfrac{x}{200}}{1 - \left(\dfrac{x}{16}\right)\left(\dfrac{x}{200}\rig  ht)} \;=\;\tan90


    And we have: . \frac{\dfrac{x}{16} + \dfrac{x}{200}}{1 - \dfrac{x^2}{3200}} \;=\;\tan90


    Note: the right side is undefined.
    Hence, on the left side, the denominator is equal to zero.

    . . 1-\frac{x^2}{3200} \:=\:0\quad\Rightarrow\quad x^2 \:=\:3200 \quad\Rightarrow\quad x \:=\:\pm40\sqrt{2}

    [Check to see if any roots are extraneous.]

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  6. #6
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    Think of it geometrically. If you have a right-angled triangle in which the angles (apart from the right angle ) are α and β, then α + β = 90. But tan(β) = 1/tan(α) (because if one of them is "opposite over adjacent" then the other one is "adjacent over opposite"). In fact, this formula holds not just for acute angles, but for any values of α and β with α + β = 90.

    So if \tan^{-1}\Bigl(\frac x{16}\Bigr) + \tan^{-1}\Bigl(\frac x{200}\Bigr) = 90^{\circ} then  \frac x{200} = 1\left/\frac x{16}\right. = \frac{16}x. Thus x^2 = 16\times200 = 3200, and x = \pm40\sqrt2, as Soroban says.
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  7. #7
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    Quote Originally Posted by Moo View Post
    Hello,

    Why would the extremum of the function be the solution of the equation ? (just asking)
    Quote Originally Posted by Mathstud28 View Post
    Is this actually a method of solving an equation?...if so could you possibly provide a site where I may read up on how to utilize it
    With all due respect to galactus, no, his method does not work.

    -Dan
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  8. #8
    Eater of Worlds
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    Quote Originally Posted by topsquark View Post
    With all due respect to galactus, no, his method does not work.

    -Dan
    Yes, it do. The solution is 40\sqrt{2}. Try it with other problems of hat type.
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  9. #9
    Moo
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    But how can you explain it mathematically ? o.O
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  10. #10
    Eater of Worlds
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    That's a good question, Moo. I never bothered to look at why it worked. I will have to do that and get back. Okey-doke.

    But then again, perhaps it isn't viable. But I know that it happens to work with problems of this type.

    I even made up various other problems of this type only using values other than 16 and 200.

    If we do the same thing with, say, tan^{-1}(\frac{x}{22})+tan^{-1}(\frac{x}{168})=90

    It works also. Upon taking the derivatives and doing what we done previously, we get the solutions \pm{4\sqrt{231}}i.

    The real parts satisfy the equation. tan^{-1}(\frac{4\sqrt{231}}{22})+tan^{-1}(\frac{4\sqrt{231}}{168})=90

    I tried it with many others and it always seems to pan out. I stumbled upon this method myself, but never bothered trying to figure out why it works. May be something rather obvious. Maybe someone else has an idea. I will look at it over the weekend.

    Try another.

    tan^{-1}(x/4)+tan^{-1}(x/5)=90

    Derivative of tan^{-1}(x/4)=\frac{4}{x^{2}+16}

    Derivative of tan^{-1}(x/5)=\frac{5}{x^{2}+25}

    \frac{4}{x^{2}+16}+\frac{5}{x^{2}+25}=0

    x=\pm{2\sqrt{5}}i

    tan^{-1}(\frac{2\sqrt{5}}{4})+tan^{-1}(\frac{2\sqrt{5}}{5})=90
    Last edited by galactus; April 11th 2008 at 12:46 PM.
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  11. #11
    Eater of Worlds
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    When I tried this I was thinking along the lines of the derivatives being equal at the same point, then this point subbed into the arctans and summing to 90, though I did not try to prove it in a general case.

    If we just set the derivatives equal and solved for x:

    \frac{16}{x^{2}+256}=\frac{200}{x^{2}+40,000}

    x=\pm{40\sqrt{2}}

    The derivatives have equal slope at the points \pm{40\sqrt{2}}

    You really do not need to mess with the complex.

    These values will result in the the two angles summing to 90 degrees.
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