Math  /  Calculus

QuestionMATH 1314 - College Algebra Lab 10
3. On a cold December day in Dallas, Detective Daniels went to an apartment complex to investigate a murder. When he arrived at noon, the sergeant informed the detective that they were having trouble determining the time of death. Detective Daniels measured the temperature of the body, finding it to be 77.9F77.9^{\circ} \mathrm{F}. He also noted that the thermostat in the room was set at 72F72^{\circ} \mathrm{F}. He then left for lunch, announcing that when he returned, he would tell them when the murder was committed. Upon his return at 1:00PM, he found the body temperature to be 75.6F75.6^{\circ} \mathrm{F}. At first it looks like Detective Daniels does not have enough information to find the time of death. However, Detective Daniels knows Newton's Law of Cooling, which can be used to predict the time for an object to cool to a given temperature. T(t)=Ts+(T0Ts)ektT(t)=T_{s}+\left(T_{0}-T_{s}\right) e^{-k t} T(t)T(t) is the temperature of the object at time tt in hours, TsT_{s} is the temperature of the surrounding environment (room temperature), T0T_{0} is the initial temperature of the object, and kk is the cooling rate. a. Using the temperatures of the body observed over one hour, along with the temperature of the room, find the cooling rate, kk. Round to five decimal places. b. Write the cooling function, T(t)T(t), using the fact that T0T_{0} was 98.6F98.6^{\circ} \mathrm{F} when the person was murdered. c. Let T(t)=77.9FT(t)=77.9^{\circ} \mathrm{F}. Use the cooling function from part b,T(t)b, T(t), to solve for tt, the number of hours since the body was murdered. Around what time was the murder committed? The tt value will be negative so count back the tt value in hours from noon to find the time. When answering the number of hours since the body was murdered, give the answer as a positive number.

Studdy Solution

STEP 1

1. The body follows Newton's Law of Cooling.
2. The surrounding temperature Ts T_s is constant at 72F 72^\circ \mathrm{F} .
3. The initial body temperature T0 T_0 at the time of death was 98.6F 98.6^\circ \mathrm{F} .
4. The body temperature at noon was 77.9F 77.9^\circ \mathrm{F} .
5. The body temperature at 1:00 PM was 75.6F 75.6^\circ \mathrm{F} .

STEP 2

1. Determine the cooling rate k k using the given temperatures.
2. Write the cooling function T(t) T(t) .
3. Solve for t t to find the time since the murder.

STEP 3

Use the formula T(t)=Ts+(T0Ts)ekt T(t) = T_s + (T_0 - T_s) e^{-kt} to set up equations for the two given temperatures.
At noon (t=0 t = 0 ): 77.9=72+(T072)ek0 77.9 = 72 + (T_0 - 72) e^{-k \cdot 0} Since e0=1 e^0 = 1 , this simplifies to: 77.9=72+(T072) 77.9 = 72 + (T_0 - 72) T0=77.9 T_0 = 77.9
At 1:00 PM (t=1 t = 1 ): 75.6=72+(77.972)ek1 75.6 = 72 + (77.9 - 72) e^{-k \cdot 1} 75.6=72+5.9ek 75.6 = 72 + 5.9 e^{-k}

STEP 4

Solve for k k using the equation from 1:00 PM: 75.672=5.9ek 75.6 - 72 = 5.9 e^{-k} 3.6=5.9ek 3.6 = 5.9 e^{-k} ek=3.65.9 e^{-k} = \frac{3.6}{5.9} k=ln(3.65.9) -k = \ln\left(\frac{3.6}{5.9}\right) k=ln(3.65.9) k = -\ln\left(\frac{3.6}{5.9}\right)
Calculate k k to five decimal places: kln(3.65.9)0.51382 k \approx -\ln\left(\frac{3.6}{5.9}\right) \approx 0.51382

STEP 5

Write the cooling function T(t) T(t) using T0=98.6 T_0 = 98.6 , Ts=72 T_s = 72 , and the calculated k k : T(t)=72+(98.672)e0.51382t T(t) = 72 + (98.6 - 72) e^{-0.51382 t} T(t)=72+26.6e0.51382t T(t) = 72 + 26.6 e^{-0.51382 t}

STEP 6

Set T(t)=77.9 T(t) = 77.9 to find t t : 77.9=72+26.6e0.51382t 77.9 = 72 + 26.6 e^{-0.51382 t} 5.9=26.6e0.51382t 5.9 = 26.6 e^{-0.51382 t} e0.51382t=5.926.6 e^{-0.51382 t} = \frac{5.9}{26.6} 0.51382t=ln(5.926.6) -0.51382 t = \ln\left(\frac{5.9}{26.6}\right) t=ln(5.926.6)0.51382 t = \frac{\ln\left(\frac{5.9}{26.6}\right)}{-0.51382}
Calculate t t : tln(5.926.6)0.513822.000 t \approx \frac{\ln\left(\frac{5.9}{26.6}\right)}{-0.51382} \approx -2.000
Since t t is negative, the murder occurred approximately 2 hours before noon.
The murder was committed around 10:00 AM.

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