Related Rates Word Problem with Derivation

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October 7th 2009, 07:34 PM
Maziana

Related Rates Word Problem with Derivation

I keep getting the wrong answer to this related rate problem, so I would greatly appreciate it if someone could go over my work and tell me where I messed up and why. I'm not sure if I made a mistake early on in the problem, or if it's a little mistake somewhere at the end. Thanks a ton:

When air expands adiabatically (without gaining or losing heat), its pressure P and volume V are related by the equation PV1.4 = C, where C is a constant. Suppose that at a certain instant the volume is 750 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?

So I am given V = 750, P = 80, and dP/dt = -10.

PV^1.4 = C

1.4 lnPV = lnC

1.4 (1/PV) ( P(dP/dt) + V(dV/dt)) = 1/C (dC/dt)
[but since it is a constant, dC/dt goes to 0, right? Or is this wrong?]

1.4 (1/60000) ( 80(dV/dt) + 750(-10)) = 0

(2.333e-5) (80(dV/dt) - 7500) = 0

.0018667(dV/dt) - .175 = 0

.0018667(dV/dt) = .175

dV/dt = 93.75
October 7th 2009, 08:21 PM
PatrickFoster

I'm not sure about your method, honestly I didn't really review it (once you started taking the logs I zoned out).

But here's what I would do. Take the derivative of your equation with respect to time = t;

$\frac{d}{dt} [ PV^{1.4} = C ]$

Using product rule gets you:

$P'V^{1.4} + 1.4PV' = 0$

Solve for V' to get the rate the volume is increasing.

$<br /> V' = \frac{ -P'V^{1.4}}{1.4P}<br />$

The problem gives you P, P' and V, so you are all set to substitute in the values and get the answer. I got $945.949 \frac{cm^3}{min}$ .

Patrick
October 8th 2009, 04:25 AM
HallsofIvy

Quote:

Originally Posted by Maziana

I keep getting the wrong answer to this related rate problem, so I would greatly appreciate it if someone could go over my work and tell me where I messed up and why. I'm not sure if I made a mistake early on in the problem, or if it's a little mistake somewhere at the end. Thanks a ton:

When air expands adiabatically (without gaining or losing heat), its pressure P and volume V are related by the equation PV1.4 = C, where C is a constant. Suppose that at a certain instant the volume is 750 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?

So I am given V = 750, P = 80, and dP/dt = -10.

PV^1.4 = C

1.4 lnPV = lnC

This is incorrect. It is ln P+ 1.4ln V= ln C. Since the "1.4" powere is only on V, only ln(V) is multiplied by 1.4. But I can see no reason to take logarithms anyway. By the product rule $\frac{dP}{dt}V^{1.4}+ 1.4PV^{.4}dV/dt= 0$ . Now set P= 80, dP/dt= -10, and V= 750 and solve for dV/dt.

Quote:

1.4 (1/PV) ( P(dP/dt) + V(dV/dt)) = 1/C (dC/dt)
[but since it is a constant, dC/dt goes to 0, right? Or is this wrong?]

1.4 (1/60000) ( 80(dV/dt) + 750(-10)) = 0

(2.333e-5) (80(dV/dt) - 7500) = 0

.0018667(dV/dt) - .175 = 0

.0018667(dV/dt) = .175

dV/dt = 93.75

No, this is not correct.
October 8th 2009, 04:27 AM
HallsofIvy

Quote:

Originally Posted by PatrickFoster

I'm not sure about your method, honestly I didn't really review it (once you started taking the logs I zoned out).

But here's what I would do. Take the derivative of your equation with respect to time = t;

$\frac{d}{dt} [ PV^{1.4} = C ]$

Using product rule gets you:

$P'V^{1.4} + 1.4PV' = 0$

No, you left out the " $V^{0.4}$ "! It should be
$P'V^{1.4}+ 1.4PV^{0.4}V'= 0$

Quote:

Solve for V' to get the rate the volume is increasing.

$<br /> V' = \frac{ -P'V^{1.4}}{1.4P}<br />$

The problem gives you P, P' and V, so you are all set to substitute in the values and get the answer. I got $945.949 \frac{cm^3}{min}$ .

Patrick
October 8th 2009, 11:37 AM
PatrickFoster

Quote:

No, you left out the "http://www.mathhelpforum.com/math-he...0e3b1d87-1.gif"! It should be
http://www.mathhelpforum.com/math-he...3c1e13dc-1.gif

Agreed. Thanks for the correction.

The new answer is 66.9643.