Related Rates Word Problem Help? If two resistors with resistances $R_1$ and $R_2$ are connected in parallel then the total resistance $R$, measured in ohms, is given by $\frac{1}{R}=\frac{1}{R1}+\frac{1}{R2}$. If $R_1$ and $R_2$ are increasing at rates of o.3 ohms/second and 0.2 ohms/second, how fast is R changing when $R_1$=80 ohms and $R_2$=100 ohms?
 A: Hint: You know $\dfrac{dR_1}{dt}$ and $\dfrac{dR_1}{dt}$.
Differentiate your equation : $\frac{1}{R}=\frac{1}{R1}+\frac{1}{R2}$ with respect to $t$.
$\dfrac {dR}{dt}=d\frac{\dfrac{(R_1+R_2)}{R_1R_2}}{dt}$. You can get $\dfrac {dR}{dt}$
$d\dfrac{\frac{(R_1+R_2)}{R_1R_2}}{dt}= \dfrac{d}{dt}.\dfrac{(\dfrac{dR_1}{dt}+\dfrac{dR_2}{dt})-2(\dfrac{dR_1}{dt}.R_2+\dfrac{dR_2}{dt}.R_1)}{(R_1R_2)^2}$
Doesn't get more messier than this. ;).
A: simply calculate   $R$ for given  $R_1,R_2$;then increase $R_1,R_2$  by relevant  values and compare new result,generally they are constant right?if they are variables then express $R$ total as a  function of two variable and use  derivative
A: This problem is straight-forward if you take it a step at a time.  You have the equation that relates $R_1$ and $R_2$, namely $\frac{1}{R} = \frac{1}{R_1}+\frac{1}{R_2}$.  You know you want to differentiate that equation.  I recommend that you rewrite the fractions with negative exponents to make the differentiation more obvious.
$$\frac{1}{R} = \frac{1}{R_1}+\frac{1}{R_2}$$
$$R^{-1} = R_1^{-1}+R_2^{-1}$$
Now, differentiate with respect to time:
$$-1\cdot R^{-2}\frac{dR}{dt} = -1\cdot R_1^{-2}\frac{dR_1}{dt}+-1\cdot R_2^{-2}\frac{dR_2}{dt}$$
$$-R^{-2}\frac{dR}{dt} = -R_1^{-2}\frac{dR_1}{dt}+-R_2^{-2}\frac{dR_2}{dt}$$
Let's multiply through by -1 to get rid of all those negative terms:
$$R^{-2}\frac{dR}{dt} = R_1^{-2}\frac{dR_1}{dt}+R_2^{-2}\frac{dR_2}{dt}$$
It's probably easiest at this point to convert the negative exponents back to fractional form:
$$\frac{1}{R^2}\frac{dR}{dt} = \frac{1}{R_1^2}\frac{dR_1}{dt}+\frac{1}{R_2^2}\frac{dR_2}{dt}$$
$$\frac{\frac{dR}{dt}}{R^2} = \frac{\frac{dR_1}{dt}}{R_1^2}+\frac{\frac{dR_2}{dt}}{R_2^2}$$
Now, rearrange to solve for $\frac{dR}{dt}$ (multiply through by $R^2$):
$$\frac{dR}{dt} = R^{2}\left(\frac{\frac{dR_1}{dt}}{R_1^{2}}+\frac{\frac{dR_2}{dt}}{R_2^{2}}\right)$$
We were given the values of all of the variables except for $R$.  But we have the equation for $R$, viz., $\frac{1}{R} = \frac{1}{R_1}+\frac{1}{R_2}$, so we can solve for $R$ (we know $R_1$ and $R_2$).
Plug in all of the known values.
There you go.  One other thing to note.  Read your problem carefully and be sure to use negative values for any rate of change that is decreasing.
Alternative Notation
Sometimes the notation can get in the way.  Though I find the above $\frac{dR}{dt}$ notation more generally useful, here is an alternative using $R^{'}$ notation that differs only in the notation used - the equations are otherwise exactly the same:
Rewrite using negative exponents:
$$\frac{1}{R} = \frac{1}{R_1}+\frac{1}{R_2}$$
$$R^{-1} = R_1^{-1}+R_2^{-1}$$
Now, differentiate with respect to time (time does not appear in the notation).  Note:  Instead of using $R^{'}$ as I do here, you could (more properly) use $R(t)^{'}$, but, in my view, that tends to defeat the whole purpose of simplifying the notation.
$$-1\cdot R^{-2}\cdot R^{'} = -1\cdot R_1^{-2}\cdot R_1^{'}+-1\cdot R_2^{-2}\cdot R_2^{'}$$
$$-R^{-2}\cdot R^{'} = -R_1^{-2}\cdot R_1^{'}+-R_2^{-2}\cdot R_2^{'}$$
Let's multiply through by -1 to get rid of all those negative terms:
$$R^{-2}\cdot R^{'} = R_1^{-2}\cdot R_1^{'}+R_2^{-2}\cdot R_2^{'}$$
It's probably easiest at this point to convert the negative exponents back to fractional form:
$$\frac{1}{R^2}\cdot R^{'} = \frac{1}{R_1^2}\cdot R_1^{'}+\frac{1}{R_2^2}\cdot R_2^{'}$$
$$\frac{R^{'}}{R^2} = \frac{R_1^{'}}{R_1^2}+\frac{R_2^{'}}{R_2^2}$$
Now, rearrange to solve for $R^{'}$:
$$R^{'} = R^{2}\left(\frac{R_1^{'}}{R_1^{2}}+\frac{R_2^{'}}{R_2^{2}}\right)$$
Find the value of $R$, as above, and substitute known values to solve.
