2. Compute the orthogonal trajectories of the family of curves given by
where is an arbitrary constant.
Solution: Differentiating we get
At the point if any curve intersects orthogonally, then (if its slope is ) we must have
Solving this differential equation, we get
NEWTON’S LAW OF COOLING
2. At 9am , a thermometer reading 70ºF is taken outdoors, where the temperature is 15ºF. At 9:05am, the thermometer reading is 45ºF. At 9:10am, the thermometer is taken back indoors, where the temperature is fixed at 70ºF.
(a). Find the reading at 9:20am
(b). when the reading, to the nearest degree, will show the correct (70ºF) indoor temperature.
To = 70° F
Ta = 15° F
T(5) = 45° F
45 - 15 = (70 - 15)e^-5k
30 = 55e^-5k
ln (30/55) / -5 = k
k = 0.1213
T(10) - 15 = 55e^-1.213
T(10) = 55e^-1.213 + 15
T(10) = 31.4°
Apply new condtions
To = 31.4°
Ta = 70°
k = 0.1213
T(10) - 70 = (31.4 - 70)e^-
T(10) = -38.6e^-1.213 + 70
T(10) = 58.5°
-1 = -38.6e^-(0.1213)t
t = ln ( 1/38.6 ) / -0.1213
t ≈ 30.1 minutes
GROWTH AND DECAY
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