T.4 P= 60 (b/in? dp 1016/in? increasing per second de C= 5 16 unknown = do dt

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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The adiabatic law (no gain or loss of heat) for the expansion of air is ??^1.4=?, where P is the number of pounds per square unit of pressure, V is the number of cubic units of volume, and C is a constant, at a specific instant, the pressure is 60 lb/in2 and is increasing at the rate of 10 lb/in2 each second. If ?=516 . What is the rate of change of volume at this instant? 

(I'm thankful if you already answered the problem, but please don't do it twice. I'm trying to find different ways and perspective of solving it.)

T.4
PV=C
P= 60 6/in?
dp
=10(6/in2 increasing per second
de
C= 5
unknown = 0do
dt
16
solve for V
(60 16/in?)(v).4
V= 5
16
-0,02339 un 3
76
60
derive
1.4
vo.4
1a co.02339)4
6OCO. 3117023guul du = -oca.0 2339)
10(0.02334)7
आ२र० २३६५५)
0.05007656264
18.70214306
- -0.0028 in'/s
de
Transcribed Image Text:T.4 PV=C P= 60 6/in? dp =10(6/in2 increasing per second de C= 5 unknown = 0do dt 16 solve for V (60 16/in?)(v).4 V= 5 16 -0,02339 un 3 76 60 derive 1.4 vo.4 1a co.02339)4 6OCO. 3117023guul du = -oca.0 2339) 10(0.02334)7 आ२र० २३६५५) 0.05007656264 18.70214306 - -0.0028 in'/s de
Given
T.4
PV=C
P= 60 (6/in?
dp
de
10(6/in?
increasing per second
C= 5
16
unknown=
dt
solve fo r V
C60 16/in?)(v )'.4
5
V= 5
16
=0,02339in3
60
60
to
derive
1.4
vo.4/du
60C14(0.0233a dol t 10 co.02339)4
GO(O.3117023Qu4l do I -10(0.02339)
स्प
dt
1.4
10(0.02339)
t001023844)
-0,05207656264
18.70214306
dリ=-0.02784
-0.0028 in'/s.
Transcribed Image Text:Given T.4 PV=C P= 60 (6/in? dp de 10(6/in? increasing per second C= 5 16 unknown= dt solve fo r V C60 16/in?)(v )'.4 5 V= 5 16 =0,02339in3 60 60 to derive 1.4 vo.4/du 60C14(0.0233a dol t 10 co.02339)4 GO(O.3117023Qu4l do I -10(0.02339) स्प dt 1.4 10(0.02339) t001023844) -0,05207656264 18.70214306 dリ=-0.02784 -0.0028 in'/s.
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