Answer is: option3
50 lb/in² per secSolution:
We are given Boyle’s Law:
\[ P V = c \]
where \( P \) is pressure, \( V \) is volume, and \( c \) is a constant. We need to determine how fast the pressure \( P \) is increasing \(\frac{dP}{dt}\) given that the volume \( V \) is decreasing at a rate of \(\frac{dV}{dt} = -10\) in³/sec when \( P = 100 \) lb/in² and \( V = 20 \) in³.
Differentiating both sides of \( P V = c \) with respect to time \( t \):
\[ \frac{d}{dt} (PV) = \frac{d}{dt} (c) \]
Since \( c \) is a constant, its derivative is zero:
\[ P \frac{dV}{dt} + V \frac{dP}{dt} = 0 \]
Solving for \( \frac{dP}{dt} \):
\[ V \frac{dP}{dt} = - P \frac{dV}{dt} \]
\[ \frac{dP}{dt} = -\frac{P}{V} \frac{dV}{dt} \]
Substituting the given values:
\[ \frac{dP}{dt} = -\frac{100}{20} \times (-10) \]
\[ \frac{dP}{dt} = \frac{100}{20} \times 10 \]
\[ \frac{dP}{dt} = 5 \times 10 \]
\[ \frac{dP}{dt} = 50 \text{ lb/in²/sec} \]
Final Answer: 50 lb/in²/sec