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 August 28th, 2010, 04:26 AM #1 Newbie   Joined: Aug 2010 Posts: 6 Thanks: 0 ANOTHER MULTIVARIABLE PROBLEM! If R is the total resistance in three resistors, connected in parallel, with resistances x, y, and z, then 1/R = (1/x)+(1/y)+(1/z). if x = 25 ohms, y=40 ohms, and z=50 ohms, with possible errors of .5% in each case, estimate the maximum error of R?
 August 28th, 2010, 08:57 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 522 Math Focus: Calculus/ODEs Re: ANOTHER MULTIVARIABLE PROBLEM! First, we need to find R(x,y,z): 1/R = 1/x + 1/y + 1/z ? R = (xyz)/(xy + xz + yz) The total resistance in ohms will then be given by: R = (25?40?50)/(25?40 + 25?50 + 40?50) = 200/17 The differentials of the independent variables are: dx = ?x = ±0.005x, dy = ?y = ±0.005y, dz = ?z = ±0.005z The total differential of R is given by dR = (?R/?x)dx + (?R/?y)dy + (?R/?z)dx Next we compute the partial derivatives: ?R/?x = [(xy + xz + yz)yz - (xyz)(y + z)]/(xy + xz + yz)² = (xy²z + xyz² + y²z² - xy²z - xyz²)/(xy + xz + yz)² = [(yz)/(xy + xz + yz)]² Using the interchangeability of variables in R, we may state: ?R/?x = [(yz)/(xy + xz + yz)]² ?R/?y = [(xz)/(xy + xz + yz)]² ?R/?z = [(xy)/(xy + xz + yz)]² ?R ? dR = [(yz)/(xy + xz + yz)]²(±0.005x) + [(xz)/(xy + xz + yz)]²(±0.005y) + [(xy)/(xy + xz + yz)]²(±0.005z) ?R ? (±0.005R)[(yz + xz + xy)/(xy + xz + yz)] = ±0.005R = ±(1/17) ohms Thus, the maximum error of R is approximately 1/17 ohms. Note: The same result can be obtained by computing the increment of the dependent variable R: ?R = R[x(1 ± 0.005),y(1 ± 0.005),z(1 ± 0.005)] - R(x,y,z) = ±0.005R = dR Normally, the total differential is only an approximation to the increment, but in this case they are exactly equal. Each variable had the same relative error, so that contributed to the equality, along with the fact R was defined in such a way that the order of this relative error was one greater in the numerator than the denominator. Perhaps someone with more experience with this matter can explain it in more general terms. It began coming back to me, as I recalled something called exact differentials. Because of the interchangeability of the variables in R, it was a simple matter to show that this is an exact differential. You will probably be studying this topic soon, so I will leave it to your professor to explain the test for exactness.
August 29th, 2010, 11:19 AM   #3
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Re: ANOTHER MULTIVARIABLE PROBLEM!

Hello, person1200!

I worked the problem head-on . . .

Quote:
 $\text{If }R\text{ is the total resistance in three resistors, connected in parallel, with resistances }x,\, y,\, z,$ [color=beige]. . . [/color]$\text{then: }\;\frac{1}{R} \;=\;\frac{1}{x}\,+\,\frac{1}{y}\,+\,\frac{1}{z}$ $\text{if }x= 25,\;y=40,\;z=50\text{ with possible errors of 0.5\% in each case, estimate the maximum error of }R.$

$\text{Since } x=25,\:y=40,\:z=50\text{, then: }\;\frac{1}{R} \:=\:\frac{1}{25}\,+\,\frac{1}{40}\,+\,\frac{1}{50 } \:=\:\frac{17}{200}\;\;\;\Rightarrow\;\;\;R \:=\:\frac{200}{17}\;\;[1]$

$\text{We also have: }\;\begin{Bmatrix} dx=&(0.005)(25)=&0.125 \\ dy=&(0.005)(40)=&0.2\;\; \\ dz=&(0.005)(50)=&0.25\; \end{Bmatrix}\;\;[2]=$

$\text{The function is: }\;R^{-1} \;=\;x^{-1}\,+\,y^{-1}\,+\,z^{-1}$

$\text{The differential is: }\;-R^{^{-2}}d\!R \;=\;-x^{^{-2}}dx \,-\, y^{^{-2}}dy \,-\, z^{-2}dz$

[color=beige]. . . . . . . . . . . . . . . . . . . .[/color]$d\!R \;=\;R^2\left(\frac{dx}{x^2}\,+\,\frac{dy}{y^2}\,+ \,\frac{dz}{z^2}\right)$

$\text{Substitute [1] and [2]: }\;d\!R \;=\;\left(\frac{200}{17}\right)^2\left(\frac{0.12 5}{25^2}\,+\,\frac{0.2}{40^2}\,+\,\frac{0.25}{50^2 }\right)$

$\text{Therefore, the maximum error of }R\text{ is: }\;d\!R \;\approx\;0.058823529$

[color=beige]. . [/color] which agrees with MarkFL's solution.

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