# Thread: Rates and Related Rates!!

1. ## Rates and Related Rates!!

A voltage V across a resistance R generates a current C = V/R. Suppose a constant voltage of 9 volts is put across a resistance that is decreasing at a rate of 0.3 ohms per second when the resistance is 4 ohms.

(a) At what rate is the current changing?
dC/dt =

(b) Is the current increasing or decreasing?

ok so if someone could like do explain how to do this problem that would help alot, i think you have to use the derivative of t somewhere in there...please help thanks

mathlete

2. first of all using the letter C for current is a really bad idea.

we know that the resistance is time dependent and thus the current will be also time dependent (Ohm's Law)

$\displaystyle \begin{gathered} C(t) = \frac{V} {{R(t)}} \hfill \\ \Leftrightarrow R(t)C(t) = V \hfill \\ C(t)\frac{{dR(t)}} {{dt}} + R(t)\frac{{dC(t)}} {{dt}} = 0 \hfill \\ \end{gathered}$

now all we have to do is use the given info:

$\displaystyle \begin{gathered} - \frac{9} {4} \cdot 0.3 + 4\frac{{dC}} {{dt}} = 0 \hfill \\ \Leftrightarrow \frac{{dC}} {{dt}} = \frac{{27}} {{160}} \hfill \\ \end{gathered}$