Hello, this is my first post here. One of the things I like to do in my spare time is generalize mathematical formulas.
On the subreddit r/calculus , this question was posted.
The question essentially asks what is the change in resistance with respect to time.
The formula for two resistors in parallel is the following.
1 R Ξ£ 2 = 1 R 1 + 1 R 2
\frac{1}{R_{\Sigma 2}} = \frac{1}{R_1} + \frac{1}{R_2}
R Ξ£2 β 1 β = R 1 β 1 β + R 2 β 1 β
Note: I used the
R Ξ£ 2 R_{\Sigma 2} R Ξ£2 β
to indicate the Sum of Two Resistors. (
Ξ£ \Sigma Ξ£
often used for "Sum". I will use this notation in future posts.
What trips up a lot of students with this problem is that they feel overwhelmed by how they should calculate the problem.
To fix that, they should adjust the formula before plug in the numbers.
1 R Ξ£ 2 = 1 R 1 + 1 R 2 β‘ 1 R Ξ£ 2 = R 2 R 1 R 2 + R 1 R 1 R 2 1 R Ξ£ 2 = R 1 R 1 R 2 + R 2 R 1 R 2 1 R Ξ£ 2 = R 1 + R 2 R 1 R 2 β‘ R Ξ£ 2 = R 1 R 2 R 1 + R 2 β
\begin{aligned}
\frac{1}{R_{\Sigma 2}} &= \frac{1}{R_1} + \frac{1}{R_2} && \square \cr
\frac{1}{R_{\Sigma 2}} &= \frac{R_2}{R_1 R_2} + \frac{R_1}{R_1 R_2} \cr
\frac{1}{R_{\Sigma 2}} &= \frac{R_1}{R_1 R_2} + \frac{R_2}{R_1 R_2} \cr
\frac{1}{R_{\Sigma 2}} &= \frac{R_1 + R_2}{R_1 R_2} && \square \cr
R_{\Sigma 2} &= \frac{R_1 R_2}{R_1 + R_2} && \blacksquare
\end{aligned}
R Ξ£2 β 1 β R Ξ£2 β 1 β R Ξ£2 β 1 β R Ξ£2 β 1 β R Ξ£2 β β = R 1 β 1 β + R 2 β 1 β = R 1 β R 2 β R 2 β β + R 1 β R 2 β R 1 β β = R 1 β R 2 β R 1 β β + R 1 β R 2 β R 2 β β = R 1 β R 2 β R 1 β + R 2 β β = R 1 β + R 2 β R 1 β R 2 β β β β β‘ β‘ β β
Since this is a problem with respect to time, we need to think of this problem like this.
1 R Ξ£ 2 ( t ) = 1 R 1 ( t ) + 1 R 2 ( t )
\frac{1}{R_{\Sigma 2}(t)} = \frac{1}{R_1(t)} + \frac{1}{R_2(t)}
R Ξ£2 β ( t ) 1 β = R 1 β ( t ) 1 β + R 2 β ( t ) 1 β
Or, since we adjusted our formula, this.
R Ξ£ 2 ( t ) = R 1 ( t ) β
R 2 ( t ) R 1 ( t ) + R 2 ( t )
R_{\Sigma 2}(t) = \frac{R_1(t) \cdot R_2(t)}{R_1(t) + R_2(t)}
R Ξ£2 β ( t ) = R 1 β ( t ) + R 2 β ( t ) R 1 β ( t ) β
R 2 β ( t ) β
We need to find the derivative of this problem.
R Ξ£ 2 β² ( t ) = ( R 1 ( t ) β
R 2 ( t ) R 1 ( t ) + R 2 ( t ) ) β²
R'_{\Sigma 2}(t) = \left( \frac{R_1(t) \cdot R_2(t)}{R_1(t) + R_2(t)} \right)'
R Ξ£2 β² β ( t ) = ( R 1 β ( t ) + R 2 β ( t ) R 1 β ( t ) β
R 2 β ( t ) β ) β²
Let's make some substitutions. Let us set the numerator to
u ( t ) u(t) u ( t )
and the denominator to
v ( t ) v(t) v ( t )
.
R Ξ£ 2 β² ( t ) = ( u ( t ) v ( t ) ) β² = ( R 1 ( t ) β
R 2 ( t ) R 1 ( t ) + R 2 ( t ) ) β²
R'_{\Sigma 2}(t) = \left(\frac{u(t)}{v(t)}\right)' = \left( \frac{R_1(t) \cdot R_2(t)}{R_1(t) + R_2(t)} \right)'
R Ξ£2 β² β ( t ) = ( v ( t ) u ( t ) β ) β² = ( R 1 β ( t ) + R 2 β ( t ) R 1 β ( t ) β
R 2 β ( t ) β ) β²
We can apply the quotient rule
R Ξ£ 2 β² ( t ) = ( u ( t ) v ( t ) ) β² = u β² ( t ) β
v ( t ) β u ( t ) β
v β² ( t ) [ v ( t ) ] 2
R'_{\Sigma 2}(t) = \left(\frac{u(t)}{v(t)}\right)' = \frac{u'(t) \cdot v(t) - u(t) \cdot v'(t)}{[v(t)]^2}
R Ξ£2 β² β ( t ) = ( v ( t ) u ( t ) β ) β² = [ v ( t ) ] 2 u β² ( t ) β
v ( t ) β u ( t ) β
v β² ( t ) β
Let us find the derivatives of
u ( t ) u(t) u ( t )
and
v ( t ) v(t) v ( t )
. We will apply the product rule
u β² ( t ) u'(t) u β² ( t )
and the Sum Rule
v β² ( t ) v'(t) v β² ( t )
.
u ( t ) = R 1 ( t ) β
R 2 ( t ) β
β βΉ β
β u β² ( t ) = R 1 β² ( t ) β
R 2 ( t ) + R 1 ( t ) β
R 2 β² ( t ) β‘ v ( t ) = R 1 ( t ) + R 2 ( t ) β
β βΉ β
β v β² ( t ) = R 1 β² ( t ) + R 2 β² ( t ) β‘
\begin{aligned}
u(t) &= R_1(t) \cdot R_2(t) &\implies u'(t) &= R_1'(t) \cdot R_2(t) + R_1(t) \cdot R_2'(t) && \square \cr
v(t) &= R_1(t) + R_2(t) &\implies v'(t) &= R_1'(t) + R_2'(t) && \square
\end{aligned}
u ( t ) v ( t ) β = R 1 β ( t ) β
R 2 β ( t ) = R 1 β ( t ) + R 2 β ( t ) β βΉ u β² ( t ) βΉ v β² ( t ) β = R 1 β² β ( t ) β
R 2 β ( t ) + R 1 β ( t ) β
R 2 β² β ( t ) = R 1 β² β ( t ) + R 2 β² β ( t ) β β β‘ β‘ β
Since the denominator of the quotient rule is
[ v ( t ) ] 2 [v(t)]^2 [ v ( t ) ] 2
, we should find the square of
v ( t ) v(t) v ( t )
.
[ v ( t ) ] 2 = [ R 1 ( t ) + R 2 ( t ) ] 2 = [ R 1 ( t ) ] 2 + 2 β
R 1 ( t ) β
R 2 ( t ) + [ R 2 ( t ) ] 2 β‘
\begin{aligned}
[v(t)]^2 &= [R_1(t) + R_2(t)]^2 \cr
&= [R_1(t)]^2 + 2 \cdot R_1(t) \cdot R_2(t) + [R_2(t)]^2 && \square
\end{aligned}
[ v ( t ) ] 2 β = [ R 1 β ( t ) + R 2 β ( t ) ] 2 = [ R 1 β ( t ) ] 2 + 2 β
R 1 β ( t ) β
R 2 β ( t ) + [ R 2 β ( t ) ] 2 β β β‘ β
Let's return to the numerator,
u β² ( t ) β
v ( t ) β u ( t ) β
v β² ( t ) u'(t) \cdot v(t) - u(t) \cdot v'(t) u β² ( t ) β
v ( t ) β u ( t ) β
v β² ( t )
. We should find
u β² ( t ) β
v ( t ) u'(t) \cdot v(t) u β² ( t ) β
v ( t )
and
u ( t ) β
v β² ( t ) u(t) \cdot v'(t) u ( t ) β
v β² ( t )
. This is quite long, so let's wrap this around each term.
u β² ( t ) β
v ( t ) = ( R 1 β² ( t ) β
R 2 ( t ) + R 1 ( t ) β
R 2 β² ( t ) ) β
( R 1 ( t ) + R 2 ( t ) ) = R 1 β² ( t ) β
R 2 ( t ) β
R 1 ( t ) + R 1 β² ( t ) β
R 2 ( t ) β
R 2 ( t ) + R 1 ( t ) β
R 2 β² ( t ) β
R 1 ( t ) + R 1 ( t ) β
R 2 β² ( t ) β
R 2 ( t ) u ( t ) β
v β² ( t ) = ( R 1 ( t ) β
R 2 ( t ) ) β
( R 1 β² ( t ) + R 2 β² ( t ) ) = R 1 ( t ) β
R 2 ( t ) β
R 1 β² ( t ) + R 1 ( t ) β
R 2 ( t ) β
R 2 β² ( t )
\begin{aligned}
u'(t) \cdot v(t) &= (R_1'(t) \cdot R_2(t) + R_1(t) \cdot R_2'(t)) \cdot (R_1(t) + R_2(t)) \cr
&= R_1'(t) \cdot R_2(t) \cdot R_1(t) \cr
&+ R_1'(t) \cdot R_2(t) \cdot R_2(t) \cr
&+ R_1(t) \cdot R_2'(t) \cdot R_1(t) \cr
&+ R_1(t) \cdot R_2'(t) \cdot R_2(t) \cr
\cr
u(t) \cdot v'(t) &= (R_1(t) \cdot R_2(t)) \cdot (R_1'(t) + R_2'(t)) \cr
&= R_1(t) \cdot R_2(t) \cdot R_1'(t) \cr
&+ R_1(t) \cdot R_2(t) \cdot R_2'(t)
\end{aligned}
u β² ( t ) β
v ( t ) u ( t ) β
v β² ( t ) β = ( R 1 β² β ( t ) β
R 2 β ( t ) + R 1 β ( t ) β
R 2 β² β ( t )) β
( R 1 β ( t ) + R 2 β ( t )) = R 1 β² β ( t ) β
R 2 β ( t ) β
R 1 β ( t ) + R 1 β² β ( t ) β
R 2 β ( t ) β
R 2 β ( t ) + R 1 β ( t ) β
R 2 β² β ( t ) β
R 1 β ( t ) + R 1 β ( t ) β
R 2 β² β ( t ) β
R 2 β ( t ) = ( R 1 β ( t ) β
R 2 β ( t )) β
( R 1 β² β ( t ) + R 2 β² β ( t )) = R 1 β ( t ) β
R 2 β ( t ) β
R 1 β² β ( t ) + R 1 β ( t ) β
R 2 β ( t ) β
R 2 β² β ( t ) β
Recall that there is a minus sign in front of
u ( t ) β
v β² ( t ) u(t) \cdot v'(t) u ( t ) β
v β² ( t )
. We will need to negate the sign on the two terms distributed. We will also group similar terms in the
u β² ( t ) β
v ( t ) u'(t) \cdot v(t) u β² ( t ) β
v ( t )
equation. We will remove the terms that cancel each other out, and add what remains to get our numerator.
u β² ( t ) β
v ( t ) β u ( t ) β
v β² ( t ) = ( R 1 β² ( t ) β
R 2 ( t ) + R 1 ( t ) β
R 2 β² ( t ) ) β
( R 1 ( t ) + R 2 ( t ) ) = R 1 β² ( t ) β
R 2 ( t ) β
R 1 ( t ) + R 1 β² ( t ) β
[ R 2 ( t ) ] 2 + [ R 1 ( t ) ] 2 β
R 2 β² ( t ) + R 1 ( t ) β
R 2 β² ( t ) β
R 2 ( t ) β R 1 ( t ) β
R 2 ( t ) β
R 1 β² ( t ) β R 1 ( t ) β
R 2 ( t ) β
R 2 β² ( t ) = R 1 β² ( t ) β
[ R 2 ( t ) ] 2 + [ R 1 ( t ) ] 2 β
R 2 β² ( t ) β‘
\begin{aligned}
u'(t) \cdot v(t) - u(t) \cdot v'(t)
&= (R_1'(t) \cdot R_2(t) + R_1(t) \cdot R_2'(t))
\cdot (R_1(t) + R_2(t)) \cr
&= \cancel{R_1'(t) \cdot R_2(t) \cdot R_1(t)} \cr
&+ R_1'(t) \cdot [R_2(t)]^2 \cr
&+ [R_1(t)]^2 \cdot R_2'(t) \cr
&+ \bcancel{R_1(t) \cdot R_2'(t) \cdot R_2(t)} \cr
&- \cancel{R_1(t) \cdot R_2(t) \cdot R_1'(t)} \cr
&- \bcancel{R_1(t) \cdot R_2(t) \cdot R_2'(t)} \cr
&= R_1'(t) \cdot [R_2(t)]^2 + [R_1(t)]^2 \cdot R_2'(t) && \square
\end{aligned}
u β² ( t ) β
v ( t ) β u ( t ) β
v β² ( t ) β = ( R 1 β² β ( t ) β
R 2 β ( t ) + R 1 β ( t ) β
R 2 β² β ( t )) β
( R 1 β ( t ) + R 2 β ( t )) = R 1 β² β ( t ) β
R 2 β ( t ) β
R 1 β ( t ) β + R 1 β² β ( t ) β
[ R 2 β ( t ) ] 2 + [ R 1 β ( t ) ] 2 β
R 2 β² β ( t ) + R 1 β ( t ) β
R 2 β² β ( t ) β
R 2 β ( t ) β β R 1 β ( t ) β
R 2 β ( t ) β
R 1 β² β ( t ) β β R 1 β ( t ) β
R 2 β ( t ) β
R 2 β² β ( t ) β = R 1 β² β ( t ) β
[ R 2 β ( t ) ] 2 + [ R 1 β ( t ) ] 2 β
R 2 β² β ( t ) β β β‘ β
Let's put our equations together.
R Ξ£ 2 β² ( t ) = R 1 β² ( t ) β
[ R 2 ( t ) ] 2 + [ R 1 ( t ) ] 2 β
R 2 β² ( t ) [ R 1 ( t ) ] 2 + 2 β
R 1 ( t ) β
R 2 ( t ) + [ R 2 ( t ) ] 2 β‘
\begin{aligned}
R_{\Sigma 2}'(t) &= \frac{R_1'(t) \cdot [R_2(t)]^2 + [R_1(t)]^2 \cdot R_2'(t)}{[R_1(t)]^2 + 2 \cdot R_1(t) \cdot R_2(t) + [R_2(t)]^2} && \square
\end{aligned}
R Ξ£2 β² β ( t ) β = [ R 1 β ( t ) ] 2 + 2 β
R 1 β ( t ) β
R 2 β ( t ) + [ R 2 β ( t ) ] 2 R 1 β² β ( t ) β
[ R 2 β ( t ) ] 2 + [ R 1 β ( t ) ] 2 β
R 2 β² β ( t ) β β β β‘ β
You know what, let's factor the denominator. It seems pointless to expand it.
R Ξ£ 2 β² ( t ) = R 1 β² ( t ) β
[ R 2 ( t ) ] 2 + [ R 1 ( t ) ] 2 β
R 2 β² ( t ) [ R 1 ( t ) + R 2 ( t ) ] 2 β
\begin{aligned}
R_{\Sigma 2}'(t) &= \frac{R_1'(t) \cdot [R_2(t)]^2 + [R_1(t)]^2 \cdot R_2'(t)}{[R_1(t) + R_2(t)]^2} && \blacksquare
\end{aligned}
R Ξ£2 β² β ( t ) β = [ R 1 β ( t ) + R 2 β ( t ) ] 2 R 1 β² β ( t ) β
[ R 2 β ( t ) ] 2 + [ R 1 β ( t ) ] 2 β
R 2 β² β ( t ) β β β β β
Perfect!
Take note that in the numerator there is no negative sign like is typical in a quotient rule. Because we cancelled out some terms, we can simply add everything in the numerator.
We can now solve the problem with that equation.
Let's consider
R 1 β² ( t ) = 1 Ξ© / s R_1'(t) = 1\Omega/\text{s} R 1 β² β ( t ) = 1Ξ©/ s
and
R 2 β² ( t ) = 1.5 Ξ© / s R_2'(t) = 1.5\Omega/\text{s} R 2 β² β ( t ) = 1.5Ξ©/ s
. (These are our derived values because these values are in ohms per second, so resistance with respect to time.)
If
R 1 ( t ) = 50 Ξ© R_1(t) = 50\Omega R 1 β ( t ) = 50Ξ©
and
R 2 ( t ) = 75 Ξ© R_2(t) = 75\Omega R 2 β ( t ) = 75Ξ©
, what is the rate of change of
R ( t ) R(t) R ( t )
?
R Ξ£ 2 β² ( t ) = R 1 β² ( t ) β
[ R 2 ( t ) ] 2 + [ R 1 ( t ) ] 2 β
R 2 β² ( t ) [ R 1 ( t ) + R 2 ( t ) ] 2 R Ξ£ 2 β² ( t ) = 1 β
[ R 2 ( t ) ] 2 + [ R 1 ( t ) ] 2 β
1.5 [ R 1 ( t ) + R 2 ( t ) ] 2 R 1 β² ( t ) = 1 Ξ© / s , R 2 β² ( t ) = 1.5 Ξ© / s R Ξ£ 2 β² ( t ) = 1 β
[ 75 ] 2 + [ 50 ] 2 β
1.5 [ 50 + 75 ] 2 R 1 ( t ) = 50 Ξ© , R 2 ( t ) = 75 Ξ© R Ξ£ 2 β² ( t ) = 1 β
5625 + 2500 β
1.5 [ 125 ] 2 R Ξ£ 2 β² ( t ) = 5625 + 3750 15625 R Ξ£ 2 β² ( t ) = 9375 15625 R Ξ£ 2 β² ( t ) = 0.6 Ξ© / s β
\begin{aligned}
R_{\Sigma 2}'(t) &= \frac{R_1'(t) \cdot [R_2(t)]^2 + [R_1(t)]^2 \cdot R_2'(t)}{[R_1(t) + R_2(t)]^2} \cr
R_{\Sigma 2}'(t) &= \frac{1 \cdot [R_2(t)]^2 + [R_1(t)]^2 \cdot 1.5}{[R_1(t) + R_2(t)]^2} && R_1'(t) = 1\Omega/\text{s}, R_2'(t) = 1.5\Omega/\text{s} \cr
R_{\Sigma 2}'(t) &= \frac{1 \cdot [75]^2 + [50]^2 \cdot 1.5}{[50 + 75]^2} && R_1(t) = 50\Omega, R_2(t) = 75\Omega\cr
R_{\Sigma 2}'(t) &= \frac{1 \cdot 5625 + 2500 \cdot 1.5}{[125]^2}\cr
R_{\Sigma 2}'(t) &= \frac{5625 + 3750}{15625}\cr
R_{\Sigma 2}'(t) &= \frac{9375}{15625}\cr
R_{\Sigma 2}'(t) &= 0.6\Omega/\text{s} && \blacksquare
\end{aligned}
R Ξ£2 β² β ( t ) R Ξ£2 β² β ( t ) R Ξ£2 β² β ( t ) R Ξ£2 β² β ( t ) R Ξ£2 β² β ( t ) R Ξ£2 β² β ( t ) R Ξ£2 β² β ( t ) β = [ R 1 β ( t ) + R 2 β ( t ) ] 2 R 1 β² β ( t ) β
[ R 2 β ( t ) ] 2 + [ R 1 β ( t ) ] 2 β
R 2 β² β ( t ) β = [ R 1 β ( t ) + R 2 β ( t ) ] 2 1 β
[ R 2 β ( t ) ] 2 + [ R 1 β ( t ) ] 2 β
1.5 β = [ 50 + 75 ] 2 1 β
[ 75 ] 2 + [ 50 ] 2 β
1.5 β = [ 125 ] 2 1 β
5625 + 2500 β
1.5 β = 15625 5625 + 3750 β = 15625 9375 β = 0.6Ξ©/ s β β R 1 β² β ( t ) = 1Ξ©/ s , R 2 β² β ( t ) = 1.5Ξ©/ s R 1 β ( t ) = 50Ξ© , R 2 β ( t ) = 75Ξ© β β
Our solution is 0.6 ohms per second . π
This was a wonderful problem to solve.
In my next post, I want to show how we could generalize the formula we used. What if we had more than two resistors in parallel.
I look forward to using this blog to share math problems as well as code examples. Thanks for reading!
