Solve the Problem $V$ (In Volts) Is Dropping as the Battery Wears Out

Multiple Choice

Solve the problem.
-A simple electrical circuit consists of a resistor connected between the terminals of a battery. The voltage $V$ (in volts) is dropping as the battery wears out. At the same time, the resistance $R$ (in ohms) is increasing as the resistor heats up. The power $P$ (in watts) dissipated by the circuit is given by $P = \frac { V ^ { 2 } } { R }$ . Use the equation $\frac { \mathrm { dP } } { \mathrm { dt } } = \frac { \partial \mathrm { P } } { \partial \mathrm { V } } \frac { \mathrm { dV } } { \mathrm { dt } } + \frac { \partial \mathrm { P } } { \partial \mathrm { R } } \frac { \mathrm { dR } } { \mathrm { dt } }$
to find how much the power is changing at the instant when $R = 5$ ohms, $V = 4$ volts, $d R / d t = 1$ ohms/sec and $d V$ , $= - 0.04$ volts/sec.

A) -0.58 watts
B) 0.58 watts
C) 0.7 watts
D) -0.7 watts