Boyle's law states that if the temperature of a gas remains constant, then PV = c. Where P is the pressure, V is the volume, and c is a constant. Given a quantity of gas at constant temperature, if V is decreasing at a rate of 13 in.^3/s, at what rate is P increasing when
P = 90lb/in.^2 and V = 50 in.^3?
Well, given Boyle's law that says PV = c, we can differentiate both sides with respect to time t to get d(PV)/dt = 0.
By applying the product rule, we have dP/dt * V + P * dV/dt = 0.
Since we are given that V is decreasing at a rate of 13 in.^3/s, we know that dV/dt = -13 in.^3/s.
Simplifying the equation, we have dP/dt * V = -P * dV/dt.
Now let's plug in the specific values given: P = 90 lb/in.^2 and V = 50 in.^3.
We have dP/dt * 50 = -90 * -13.
Dividing both sides by 50, we get dP/dt = (-90 * -13) / 50.
So, dP/dt = 23.4 lb/(in.^2 * s).
So there you have it! The pressure is increasing at a rate of 23.4 lb/(in.^2 * s). Who knew Boyle's law could be so entertaining?
To find the rate at which P is increasing, we need to differentiate the equation PV = c with respect to time.
Differentiating both sides of the equation with respect to time, we get:
(dP/dt)V + P(dV/dt) = 0
Since the temperature is constant, (dP/dt)V = -P(dV/dt)
Now we can substitute the given values into the equation:
P = 90 lb/in^2
V = 50 in^3
(dV/dt) = -13 in^3/s
Substituting these values into the equation:
90(dV/dt) = -P(dV/dt)
90(-13) = -P(dV/dt)
-1170 = -50(dP/dt)
Now we can solve for (dP/dt), the rate at which P is increasing:
dP/dt = -1170 / -50
dP/dt = 23.4 lb/(in^2 * s)
Therefore, the rate at which P is increasing is 23.4 lb/(in^2 * s).
To find the rate at which P is increasing, we can use the relationship between P and V given by Boyle's law. Boyle's law states that if the temperature of a gas remains constant, then the product of the pressure and volume is constant.
Mathematically, this can be expressed as PV = c, where P is the pressure, V is the volume, and c is a constant.
To find the rate at which P is increasing, we need to differentiate both sides of the equation with respect to time (t). Since V is decreasing at a rate of 13 in.^3/s, we can express this as dV/dt = -13 in.^3/s (negative because V is decreasing).
Differentiating both sides of the equation PV = c with respect to time, we get:
d(PV)/dt = d(c)/dt
Using the product rule of differentiation, this becomes:
V * dP/dt + P * dV/dt = 0
Since we're interested in finding the rate at which P is increasing (dP/dt), we can rearrange the equation as follows:
dP/dt = -(P * dV/dt) / V
Substituting the given values, we have P = 90 lb/in.^2 and V = 50 in.^3, and dV/dt = -13 in.^3/s. Plugging these values into the equation, we can calculate the rate at which P is increasing:
dP/dt = -(90 * (-13)) / 50 lb/in.^2/s
Simplifying the expression, we find:
dP/dt = 234 / 50 lb/in.^2/s
Therefore, the rate at which P is increasing when P = 90 lb/in.^2 and V = 50 in.^3 is approximately 4.68 lb/in.^2/s.
PV = c
by product rule, implicitly,
dP/dt (V) + dV/dt (P) = 0
sub in the given values, solve for dP/dt