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Homework Help: Solving for a variable

  1. Nov 30, 2004 #1
    i have two equations r^2/wc
    divided by r^2 + (1/(wc)^2)
    equals 13026

    and r/wc
    divided by r^2 + (1/(wc)^2)
    equals 951

    where w is a known quantity. how do I solve for r and c.
  2. jcsd
  3. Nov 30, 2004 #2


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    Science Advisor

    You want to solve
    [tex]\frac{\frac{r^2}{wc}}{r^2+ \frac{1}{w^2c^2}}= 13026[/tex]
    [tex]\frac{\frac{r}{wc}}{r^2+ \frac{1}{w^2c^2}}= 13026[/tex]
    for r and c given that w is a known constant.

    The first thing I would do is notice that the left hand side of the first equation is just the left hand side of the second equation times r:
    [tex]\frac{\frac{r^2}{wc}}{r^2+ \frac{1}{w^2c^2}}= r\frac{\frac{r}{wc}}{r^2+ \frac{1}{w^2c^2}}= r(951)= 13026[/tex]
    so r= 13026/951= 13.7 (approximately). That was easy!

    If we let x= wc then the second equation can be written
    [tex]\frac{\frac{r}{x}}{r^2+\frac{1}{x^2}}= 951.[/tex]
    Multiply both numerator and denominator of the fraction by x2 to get
    [tex]\frac{rx}{r^2x^2+ 1}= 951[/tex]
    Now, I would be inclined to let y= rx so that
    [tex]\frac{y}{y^2+ 1}= 951[/tex]
    Multiplying on both sides by y2+1 gives the quadratic equation
    951y2- y- 951= 0. That can be solved by the quadratic formula:
    the two solutions are approximately 0.0329 and -0.0319. Since I don't know where this problem came from I don't know if both of these are valid or not.
    Now we have rx= 0.0329 and rx= -0.0319 and we know that r= 13.7 so
    x= 0.0329/13.7= 0.0024 and x= -0.0319/13.7= -0.0023.
    If x= wc= 0.0024, c= 0.0024/w.
    If x= wc= -0.0023, c= -0.0023/w.
  4. Nov 30, 2004 #3
    thanks for the help it is appreciated
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