Use Newton’s Law of Cooling, T = C + ( T 0 − C ) e k t , to solve this exercise. At 9:00 A.M., a coroner arrived at the home of a person who had died. The temperature of the room was 70°F, and at the time of death the person had a body temperature of 98.6°F. The coroner took the body’s temperature at 9:30 A.M., at which time it was 85.6°F, and again at 10:00 A.M., when it was 82.7°F. At what time did the person die?
Use Newton’s Law of Cooling, T = C + ( T 0 − C ) e k t , to solve this exercise. At 9:00 A.M., a coroner arrived at the home of a person who had died. The temperature of the room was 70°F, and at the time of death the person had a body temperature of 98.6°F. The coroner took the body’s temperature at 9:30 A.M., at which time it was 85.6°F, and again at 10:00 A.M., when it was 82.7°F. At what time did the person die?
Solution Summary: The author explains Newton's law of Cooling: the temperature of a heated object at time t is given by: T=C+(T_0-C)
Use Newton’s Law of Cooling,
T
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C
+
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T
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, to solve this exercise. At 9:00 A.M., a coroner arrived at the home of a person who had died. The temperature of the room was 70°F, and at the time of death the person had a body temperature of 98.6°F. The coroner took the body’s temperature at 9:30 A.M., at which time it was 85.6°F, and again at 10:00 A.M., when it was 82.7°F. At what time did the person die?
A 20 foot ladder rests on level ground; its head (top) is against a vertical wall. The bottom of the ladder begins by being 12 feet from the wall but begins moving away at the rate of 0.1 feet per second. At what rate is the top of the ladder slipping down the wall? You may use a calculator.
Explain the focus and reasons for establishment of 12.4.1(root test) and 12.4.2(ratio test)
use Integration by Parts to derive 12.6.1
Chapter 3 Solutions
MyLab Math with Pearson eText -- Standalone Access Card -- for Precalculus (6th Edition)
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