When air expands adiabatically (without gaining or losing heat), its pressure P and volume V are related by the equation PV^{1.4}=C where C is a constant. Suppose that at a certain instant the volume is 430 cubic centimeters and the pressure is 89 kPa and is decreasing at a rate of 13 kPa/minute. At what rate in cubic centimeters per minute is the volume increasing at this instant?
PV^1.4 = C
using the product and chain rules,
V^1.4 dP/dt + 1.4PV^.4 dV/dt = 0
now just plug in your numbers
To find the rate at which the volume is increasing at a certain instant, we need to use calculus to differentiate the given equation.
Step 1: Differentiate the given equation PV^1.4 = C with respect to time (t) using the product rule of differentiation.
d/dt(PV^1.4) = d/dt(C)
Step 2: Apply the power rule to differentiate V^1.4.
P * 1.4 * V^(1.4-1) * dV/dt = 0
Step 3: Simplify the equation, as we are given the values of P and dP/dt.
89 * 1.4 * (430)^(1.4-1) * dV/dt = -13
Step 4: Solve for dV/dt, which is the rate of change of volume at the specified instant.
dV/dt = -13 / (89 * 1.4 * (430)^(1.4-1))
By substituting the given values into this equation, we can find the rate at which the volume is increasing (dV/dt) at the specified instant.