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13 Nov 2019
14. In a murder investigation, the temperature of the corpse was 32.5â at 1:30 PM and 30.3°C an hour later. Normal body tem- perature is 37.0°C and the temperature of the surroundings was 200°C. When did the murder take place?
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Reid Wolff
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10 Feb 2019
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Related questions
18. 15 points BerrApCalc7 9.1.055. My Notes Ask Your Teacher A medical examiner called to the scene of a murder wil usually take the temperature of the body. A corpse cools at a rate proportional to the difference between its temperature and the temperature of the room. If y(t) is the temperature (in degrees Fahrenheit) of the body t hours after the murder, and if the room temperature is 70 then y satisfies y' =-0.32(y-70) y(0)-98.6 (body temperature initially 98.6) (a) Solve this differential equation and initial condition. y= (b) Use your answer to part (a) to estimate how long ago the murder took place if the temperature of the body when it was discovered was 90°. [Hint: Find the value of t that makes your solution equal 90°] (Round your answer to two decimal places.) t- hr Need Help? Read It 19. -15 points BerrApCalc7 9.1.056. My Notes Ask Your Teacher Hospital patients are often given glucose (blood sugar) through a tube connected to a bottle suspended over their beds. Suppose that this "drip" supplies glucose at the rate of 15 mg per minute, and each minute 10% of the accumulated glucose is consumed by the body. Then the amount y(t) of glucose (in excess of the normal level) in the body after t minutes satisfies the following -15 y(0) = 0 0.y Do you see why?) (zero excess glucose at t = 0) Solve this differential equation and initial condition y(t)- Need Help?Read It
Assignment 11: Problem 1 Problem List Next (1 point) Newton's Law of Cooling states that the rate of change of the temperature of an object is proportional to the difference between the temperature of the object, T, and the ambient (environmental) t differential equation = k E T) where k > 0 is a constant that represents the material properties and, E s the ambient temperkture. (We will assume that E is also constant.) The function emperature, E. This leads to the represents the temperature at time t that satisfies the differential equation. The time of death of a murder victim can be estimated from the temperature of the body if it is discovered early enough after the crime has occurred. Suppose that in a room whose ambient temperature is E = 19 degrees C, the temperature of the body upon discovery is T = 33 degrees, and that a second measurement, one hour later is T = 26 degrees. How many minutes before the body was discovered did the murder occur? (You should use the fact that just prior to death, the temperature of the victim was 37 degrees Celcius.) Time of deaths min You have attempted this problem 0 times. You have 10 attempts remaining.
scene of . murder wil umoly take the temperature ofthe body. A cor⢠cools at ante proportona. to A medical examiner of the reom. !tf yit is the temperature (in degrees tahrenheit) of the body t hours after the mundes, d er between its temperature and the temperature and if the room temperature is 60*, then y -0.32y-60) 0)-986(body temperature initialily 98.6) (a) Solve this differential equation and initial condition (b) Use your answer part (a) to te how long ago the murder took â¦"be temperature ofthe solution equal 9 J (Round your anower to two decimal places) t discovered was son tot, md vain of t that makes your Hospital petients are often given glucose (biood sugar) through a tube connected to a bottle suspended over their beds. Suppose that this "drip" supplies glucese at the rate of 15 mg per minute, and s censumed by the body. Then the amoune yt of glucose On excess of the normal level) in the body after t minuees satisfies the following. r""15-0.25y (Do you see why?) Solve this differential equation and initial condition Need Help?aes My Notes Ask Your you now have 57000, you expect to save an asdtional $2000 during each year, and aitl of this is deposited in a bank paying 10s interest Let yi) be your bank
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