formula help.

[kricker]

L=2C+3.14(X+D)/2+(X -D)squared/4C


Please help, need to switch around formula to solve for C.
C=blah,blah,blah
Thankyou for any help!

 

[stapel]

You have:

. . . . .\(\displaystyle \large{L\mbox{ }=\mbox{ }2C\mbox{ }+\mbox{ }\frac{\pi}{2}(X\mbox{ }+\mbox{ }D)\mbox{ }+\mbox{ }\frac{(X\mbox{ }-\mbox{ }D)^2}{4C}}\)

To solve for C, a good first step would be to multiply through by C to clear that out of the denominator.

. . . . .\(\displaystyle \large{LC\mbox{ }=\mbox{ }2C^2\mbox{ }+\mbox{ }\frac{\pi}{2}(CX\mbox{ }+\mbox{ }CD)\mbox{ }+\mbox{ }\frac{1}{4}(X\mbox{ }-\mbox{ }D)^2}\)

Moving everything together and combining in terms of C:

. . . . .\(\displaystyle \large{0\mbox{ }=\mbox{ }2C^2\mbox{ }+[\frac{\pi}{2}X\mbox{ }+\mbox{ }\frac{\pi}{2}D\mbox{ }-\mbox{ }L]C\mbox{ }+\mbox{ }\frac{1}{4}(X\mbox{ }-\mbox{ }D)^2}\)

At this point, I think you'll have to apply the Quadratic Formula to find C in terms of X, L, and D. It won't be pretty. :shock:

Eliz.

 

[soroban]

Hello, kricker!

\(\displaystyle \text{Solve for }C:\;\;L\;=\;2C\,+\,\frac{\pi}{2}(X+D)\,+\,\frac{(X-D)^2}{4C}\)

Eliminate all denominators by multiplying through by \(\displaystyle 4C:\)

. . \(\displaystyle 4LC\;=\;8C^2\,+\,2\pi C(X+D)\,+\,(X-D)^2\)


And we have a quadratic in \(\displaystyle C:\;\;8C^2\:+\:[2\pi(X+D)\,-\,4L]C\:+\X-D)^2\;=\;0\)


The Quadratic Formula gives us:

. . \(\displaystyle \L C\;=\;\frac{-[2\pi(X+D)-4L]\,\pm\,\sqrt{[2\pi(X+D)-4L]^2\,-\,4\cdot8\cdot(X-D)^2}}{16}\) . . . . ugh!

 

[stapel]

...ugh!

Tol' ya it wouldn't be pretty! :wink:

Eliz.