In the description

Question

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 500 cm³ and the pressure is 80 kPa and is decreasing at a rate of 10 kPa/min. At what rate is the volume increasing at this instant? (Round your answer to the nearest whole number.)

Answer 1

From the description, we want to find dV/dt when V=500 cm^3 and P = 80kPa and dP/dt = -10 kPa/min

For the equation PV^1.4 = C, take the derivative of both sides using the product and chain rules:

dP/dt * V^1.4 + P(1.4V^.4)dV/dt = 0

solving for dV/dt:

dV/dt = -(V^1.4)(dP/dt)/(P(1.4V^.4)(-10)) = -V(dP/dt)/(-14P) = (500)(10)/((14)(80)) = 5000/1120 = 4.464...

~=5 cm^3/min

Notice that dP/dt is negative because it is decreasing. Also note that V^1.4/V^.4 = V^1

Answer 2

This is a related rates problem. We are already given the relationship between the two variables, which is nice. To get an equation relating their rates, just implicitly differentiate both sides of the equation (you will be differentiating with respect to time t in this case). First you will have to do the product rule since on the left we have two functions multiplied together. Remember that both Pressure and Volume are going to be dependent on time so you need to take that into account. For example, d/dt V^1.4 = 1.4 V^0.4 (dV/dt) as given by the chain rule. The derivative of the constant C is, as always, zero. After that implicit differentiation step you will have an equation relating the two rates. Into that equation you will plug in the conditions given in the problem statement and solve for the desired unknown which in this problem is the rate volume is increasing (dV/dt).

Hope that helps!