Todays Math Problem (RELATED RATES)

Boyles’ Law states that when a sample of gas is compressed at a constant temperature, the pressure P and the volume V satisfy the equation PV = C, where C is a constant. Suppose that at a certain instant the volume is 600 cm^3, the pressure is 150 kPa, and the pressure is increasing at a rate of 20 kPa/min. At what rate is the volume decreasing at this instant?

Hmm… we are given that PV = C, where C is a constant, So we can plug our values in for P and V where P = 150 kPa and V = 600 cm^3 therefore we get that 150*600 = C
But we are not asked to find what C equals we are asked to find out how fast the rate of V is decreasing over time. When dP/dt = 20 kPa/min
To figure this one out we need to take the derivatives of our equations using the rules for related rates. Which say that dV/dP = dV/dt*dt/dP so dV/dt = dV/dP*dP/dt and dP/dt = dP/dV*dV/dt so we get dV/dt*dP/dt = dV/dP*dP/dt*dP/dV*dV/dt = dV/dt * dP/dt so since everything else cancels out we will us dV/dt which reads the derivative of Volume in respect to time, and dP/dt which reads the derivative of Pressure in respect to time.

So lets see what we can find out.

Differentiating both sides of the equation PV = C we get P*dV/dt + V*dP/dt = 0 (by the product rule)
Subtracting V*dP/dt from both sides we get P*dV/dt = -V*dP/dt
Dividing P from both sides we get dV/dt = -V/P*dP/dt

Now we can plug in our known values, V = 600 P = 150 and dP/dt = 20
To get dV/dt = -600/150*(20) = -80

Remembering that we are asked to find out what the instantaneous rate of change is which is (dV/dt) when the instantaneous rate of change of pressure is 20 kPa/min which is (Pv/dt) and knowing that the Volume is cm^3 we get a final answer that tells us that as the pressure approaches 20 kPa/min the volume be decreasing at 80 cm^3/min.