When an electric current passes through two resistors with a resistance r1 and r2. connected in parallel, the combined resistance, R, can be calculated from the equation: frac{1}{R}=frac{1}{2r1}+frac{1}{3r2} where R, r1 & r2 are greater than 0. Assume that r2 is constant. Show that RR is an increasing function of r1.

Question
Functions
asked 2021-01-16
When an electric current passes through two resistors with a resistance r1 and r2. connected in parallel, the combined resistance, R, can be calculated from the equation:
\(\displaystyle{\frac{{{1}}}{{{R}}}}={\frac{{{1}}}{{{2}{r}{1}}}}+{\frac{{{1}}}{{{3}{r}{2}}}}\)
where R, r1 & r2 are greater than 0. Assume that r2 is constant. Show that RR is an increasing function of r1.

Answers (1)

2021-01-17
Firstly, let's invert this equation so we don't have to do implicit differentiation.
\(\displaystyle{R}={\frac{{{1}}}{{{\frac{{{2}}}{{{r}{1}}}}+{\frac{{{1}}}{{{3}{r}{2}}}}}}}={\frac{{{1}}}{{{\frac{{{6}{r}{2}+{r}{1}}}{{{3}{r}{1}{r}{2}}}}}}}={\frac{{{3}{r}{1}{r}{2}}}{{{6}{r}{2}+{r}{1}}}}\)
To see if this function is increasing in r1, we need to take its derivative with respect to r1 and check that it is nonnegative for every value of r1.
So let's do that. Remember to use the quotient rule.
\(\displaystyle{R}'{\left({r}{1}\right)}=\frac{{{\left({3}{r}{2}\right)}{\left({6}{r}{2}+{r}{1}\right)}-{3}{r}{1}{r}{2}{\left({1}\right)}}}{{\left({1}\right)}^{{2}}}={\left({18}{r}{2}^{{2}}\right)}+{3}{r}{2}{r}{1}-{3}{r}{2}{r}{1}={18}{r}{2}^{{2}}\)
Because r2>0, we see that R'(r1) > 0 everywhere. Hence R is an increasing function of r1 Q.E.D.
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