Partial derivatives and percent error

  • Thread starter MeMoses
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  • #1
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Homework Statement


When two resistors R1 and R2 are connected in parallel, their effective resistance R = (R1R2)/(R1+R2). Show that is R1 and R2 are both increased by a small percentage c, then the percentage increase of R is also c.


Homework Equations





The Attempt at a Solution


I had dR1/dt=c and dR2/dt=c. Then dF/dt=∂F/∂R1*dR1/dt + ∂F/∂R2*dR2/dt which I then simplified to c(R1^2+R2^2)/(R1+R2)^2, which does not simplify to c so I clearly did something wrong. Thanks in advance for any help.
 

Answers and Replies

  • #2
Curious3141
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Why bring "t" into it at all? t (which usually represents time) has no bearing here. And what is F?

You're given R as a function of two variables R1 and R2.

Find the partial derivatives of R wrt R1, then R2 (hint: they are quite similar (symmetric) in form).

Let c/100 = p (this is the proportional change corresponding to a percentage change of c)

You're given δR1/R1 = δR2/R2 = p, meaning δR1 = pR1 and δR2 = pR2.

What you want to do then is to use this:

δR ≈ (∂R/∂R1)*δR1 + (∂R/∂R2)*δR2

Work through the algebra, simplify to find δR, then divide that by the expression for R to find δR/R, and prove that it's equal to p (meaning that its percentage change is also c).

(Back after a long hiatus for a quick visit, and loving the quick symbols pane. Because I've forgotten all my LaTex! LOL).
 

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