Given PV=4T (P is a function of V and T)
1) V=100 unit and is increase at a rate of 0.2 unit/s
2) T=300 unit and is increase at a rate of 0.1 unit/s
Determine dP/dt
partial derivation
2013-12-31 11:59 pm
回答 (4)
2014-01-02 5:38 am
✔ 最佳答案
PV=4TP=4T/V=4‧300/100=12
d(PV)/dt = d(4T)/dt
V‧dP/dt + P‧dV/dt = 4‧dT/dt
100‧dP/dt + 12‧0.2 = 4‧0.1
dP/dt= (0.4 - 2.4) / 100 = -0.02 unit/s
2014-01-01 6:51 pm
PV = 4T, P = 4T/V
Applying the chain rule
dP/dt = ((P/T) (dT/dt) + (P/V) (dV/dt )
= (4/V) (dT/dt) – (4T/V^2) (dV/dt )
= (4/100)(0.1) – (4(300)/100^2) (0.2)
= 0.004 – 0.024
= - 0.2
Note:
P/T = partial derivative of P with respect to T
P/V = partial derivative of P with respect to V
Applying the chain rule
dP/dt = ((P/T) (dT/dt) + (P/V) (dV/dt )
= (4/V) (dT/dt) – (4T/V^2) (dV/dt )
= (4/100)(0.1) – (4(300)/100^2) (0.2)
= 0.004 – 0.024
= - 0.2
Note:
P/T = partial derivative of P with respect to T
P/V = partial derivative of P with respect to V
2014-01-01 5:44 am
對了~~~~~~~~~~~~~~
2014-01-01 4:32 am
dP/dt = -0.02 ?
2014-01-01 11:47:32 補充：
我認同 Godfrey 的作答，應用 partial derivative 比較貼題。
2014-01-01 11:47:32 補充：
我認同 Godfrey 的作答，應用 partial derivative 比較貼題。
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