Related Rates Question (Urgent)

Question

Experimentation shows that when air expands adiabatically (without losing or gaining heat), the volume V in cubic centimeters and the pressure P in kilopascals (kPa) are related by the eqation

PV^k = C

where k and C are constants. When the volume is 10 cm^3 and increasing by 4 cm^3/min, the pressure is 25 kPa and is decreasing by 14 kPa/min. What is the value of k?

Answer 1

P = C/V^k = C*V^-k

Differentiating this equation with respect to time give us an equation that relates the rate of change of volume to the rate of change of pressure. Using the chain rule and the power rule gives us,

P' = dP/dt = (d/dt)(CV^-k) = (dV/dt)*(d/dV)(CV^-k) = [d(C*V^-k)/dV]*(dV/dt) = -k*C*V^-k-1*V'.

Now dividing the first equation P by the second equation P' we see that the constant C and V^-k cancels out,

P/P' = (C*V^-k)/(-k*C*V^-k-1*V') = (V^-k)/(-k*V^-k*V^-1*V') = 1/(-k*V^-1*V') = -V/(k*V').

Now multiplying both sides by k and by P'/P gives us k in terms of V, V', P, and P', and we can then substitute in our known values V = 10, V' = 4, P = 25, P' = -14 to solve for k,

k = -(V*P')/(V'*P) = -(10*-14)/(4*25) = 7/5.

Answer 2

P V ^k = c

[d /d t] { P V ^k } = [d /d t] {c}

V ^k d P /d t +k P V ^k-1 d V /d t =0

V d P /d t +k P d V /d t =0

10 · (- 14 ) + k ·25 ·4 = 0

k = 1.4

Answer 3

PV^k = C

P = 25, V = 10, dV/dt = 4 and dP/dt = -14

(dP/dt)V^k + P(kV^k-1)(dV/dt) = 0

-14(10)^k + 25k(10^k-1)(4) = 0

100k(10^k-1) = 14(10^k)

10²k(10^k-1) = 14(10^k)

10^k+1k = 14(10^k)

10k = 14

k = 1.4

Answer 4

Try using the product rule to differentiate implicitly with respect to time, where P and V are variables and C and k are constant.

Then substitute the given information (units not displayed):

V = 10

dV/dt = 4

P=25

dP/dt = -14

Of course the derivative of C is 0, but k will remain in the relationship you end up with.

Then solve for k.