# simple alegbra question:

• Nov 1st 2008, 08:59 AM
Ronball
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 :D

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
• Nov 1st 2008, 09:20 AM
Shyam
Quote:

Originally Posted by Ronball
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 :D

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.
• Nov 1st 2008, 09:32 AM
Ronball
You're a good man, thanks alot!

Gotta bang the rust off hehe!