# Rates and Related Rates!!

• April 2nd 2008, 10:46 AM
mathlete
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
• April 2nd 2008, 10:53 AM
Peritus
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)

$
\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:

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