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Unformatted text preview: s us only the CHANGE in height. It does NOT tell us the height
at instant t i . We must know the initial height at time t = 0, which in this case is the
constant of integration C 0 . Then :
HEIGHT at instant t i = CHANGE in height + INITIAL height
= y(t i ) + C 0
Example 2 We know the relationship between temperature t measured
in F and C .
F (t) = 9 5 t C + 32
F ’(t) =
∫ oF ’(t) dt = 9 5 t oC + C 0
o o How do we determine the constant of integration C0 ? We must know F (t) at t = 0.
We may conduct an experiment. We immerse the F thermometer and the C
thermometer in a beaker of ice which we know is 0 C . Then we read the F
thermometer to get C 0 = 32. So:
F (t) = 9 5 t C + 32
Ex ample 3: A pump delivers with a rate of flow f ’ (t) = 60 litres/minute.
o What is the volume of water in the tank after 5 minutes ?
f(t) = ∫ f ’ (t) . dt = 60 t + C0
t=5 ∫ f ’ (t) . dt = t=0 t=5 5 t=0 0 ∫ 60 . dt =[60 t ] litres = 300 litres 165 However, this is NOT the volume of water in the tank. This is only the CHANGE in
volume. We must know the INITIAL volume, say C0 = 500 litres. Then the volume of
water in the tank at t = 5 minutes is :
VOLUME (at t = 5 minutes) = CHANGE in volume + INITIAL volume
= 300 litres + 500 litres
= 800 litres.
This example may be easily adapted to compute the charge on a capacitor and also
the time taken to charge or discharge the capacitor given f ’ (t) = the rate of flow of
charge = current i = dq/dt. Let us look at this very interesting in example in a more
general way .
STOP TIME ∫ ( RATE of FLOW). dt = CHANGE in VOLUME START TIME There are 4 entities involved: f ’(t) = RATE of FLOW, CHANGE in VOLUME, START TIME
and STOP TIME. Usually we know f ’ (t) . Now we have 3 possibilities.
1. If we know the START TIME and STOP TIME we may compute the CHANGE in
VOLUME for the given RATE of FLOW f ’ (t) . 2. If we know the START TIME and CHANGE in VOLUME we may determine the
STOP TIME. Then ( STOP TIME − START TIME) will tell us how much time it took
to pump a cer...
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- Fall '09