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Math Help - simple alegbra question:

  1. #1
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    simple alegbra question:

    Hi I have the following I need to solve for x

    .44m = Sqrt(((4*pi*6m + 2*Pi*6m) * x)^2 + (2*Pi*3.5m*x)^2)

    It should be a ridiculously low number, to the third decimal place is all im looking for.

    I think my problem is that im not taking the square root properly of the terms which involve x....it's been so long :S

    In case anyone is wondering what this formula is for, x is the standard deviation for a measurement for both the height and the radius. .44 is the basically the cumulative standard deviation allowed for the area of a cylindrical tank. This is like grade 10 or 11 math, and I completely forget how to reduce this to solve for x


    Alternatively if anyone wants to check my result for the partial derivatives for the surface area id appreciate it

    Surface area of a cylinder = 2*Pi*r^2 + 2*pi*r*h

    partial SA in respect to r= 4*pi*r + 2*pi*h

    partial Surface area in respect to h= 2* Pi * r
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  2. #2
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    Quote Originally Posted by Ronball View Post
    Hi I have the following I need to solve for x

    .44m = Sqrt(((4*pi*6m + 2*Pi*6m) * x)^2 + (2*Pi*3.5m*x)^2)

    It should be a ridiculously low number, to the third decimal place is all im looking for.

    I think my problem is that im not taking the square root properly of the terms which involve x....it's been so long :S

    In case anyone is wondering what this formula is for, x is the standard deviation for a measurement for both the height and the radius. .44 is the basically the cumulative standard deviation allowed for the area of a cylindrical tank. This is like grade 10 or 11 math, and I completely forget how to reduce this to solve for x


    Alternatively if anyone wants to check my result for the partial derivatives for the surface area id appreciate it

    Surface area of a cylinder = 2*Pi*r^2 + 2*pi*r*h

    partial SA in respect to r= 4*pi*r + 2*pi*h

    partial Surface area in respect to h= 2* Pi * r

    0.44m=\sqrt{[(4 \pi 6m+2 \pi 6m)x]^2+[2 \pi 3.5mx]^2}

    0.44m = \sqrt{[36 \pi m x]^2+[7 \pi m x]^2}

    squaring both sides,

    [0.44m]^2=[36 \pi m x]^2+[7 \pi m x]^2

    0.1936m^2=1296\pi^2m^2x^2+49\pi^2m^2x^2

    0.1936m^2=1345\pi^2m^2x^2

    x^2=\frac{0.1936m^2}{1345\pi^2m^2}

    x^2=\frac{0.1936}{1345\pi^2}

    x=\sqrt{\frac{0.1936}{1345\pi^2}}

    x=\frac{0.44}{\pi \sqrt{1345}}

    now, simplify it.
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  3. #3
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    You're a good man, thanks alot!

    Gotta bang the rust off hehe!
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