💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here! # In a murder investigation, the temperature of the corpse was $32.5^o C$ at 1:30 PM and $30.3^o C$ an hour later. Normal body temperature is $37.0^o C$ and the temperature of the surroundings was $20.0^o C.$ When did the murder take place?

## The murder took place $11 : 55 \mathrm{AM}$

Derivatives

Differentiation

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##### Top Calculus 1 / AB Educators  ##### Heather Z.

Oregon State University ##### Kristen K.

University of Michigan - Ann Arbor Lectures

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### Video Transcript

in this problem, we get to be forensic scientists and figure out using science and math when someone was murdered. So this is a Newton's law of cooling problem. And if you read the section of the book, you notice that Newton's Law of cooling is based on the same kind of exponential growth and decay equation y equals. Why not eat to the K T derived from a differential equation? And we can replace why, for this particular kind of application with t minus T's of S, where T is the final temperature of the object, Tisa best is the temperature of the surroundings, and we can replace Why not with t not minus tisa Best t not is the initial temperature of the object and Tisa best is till this is still the temperature of the surroundings. So for this problem, we know the temperature of the surroundings is 20 degrees Celsius, and we're also told that the body temperature goes from 32.5 degrees to 30.3 degrees in one hour, so we can use that piece of information and our model in sulphur K. So let's plug those numbers into our model, so we have 30.3 minus 20 temperature. Final temperature minus surrounding temperature equals 32.5 minus 20. Initial temperature minus surrounding temperature times E to the K times 11 for one hour. Okay, let's go ahead and simplify this insult for Kit. So we end up with 12.5. Oops. Back up a second. Here we end up with 10.3 equals 12.5 times eat of the K. Then we divide both sides by 12.5 and we have 0.8 to 4. So 0.8 to 4 equals each of the K, and then we take the natural log of both sides, and that gives us K. All right, so we put that into our model, and we now have t minus 20 equals t not minus 20 times E to the natural log of 0.8 to 4 times the temperature. So we're going to use that model to help us figure out the time of death. Okay, so this is the point where we utilize the information that normal body temperature is 37 degrees. So we're going to assume that when the person was alive, their body temperature was 37 degrees. That's going to be our new T, not our new starting temperature. And we want to find out the time difference between that temperature and when the temperature of the body was 32.5 degrees at 1:30 p.m. So let's use 32.5 as our final temperature. And let's use 37 as our initial temperature, and we're going to substitute that into our equation and solve for T. Okay, so we've 12.5 equals 17 e to the natural log of point A to four t. We can divide both sides by 17 and we end up with 25. 34th equals e to the natural log of 0.8 to 4 times t. And then we're going to take the natural log of both sides. So we have the natural log of 25. 34th equals the natural log of point a to four times teeth. Now, to get t by itself, let's divide both sides by the natural log of point A to four, and then this whole expression goes in the calculator and we find the approximate value. Now this gives us approximately 1.59 hours. If you convert that into minutes, you get one hour and 35 minutes. So that tells us that one hour and 35 minutes before 1:30 p.m. Was when the person was murdered. So one hour and 35 minutes before 1:30 p.m. Is 11. 55 a. M. Oregon State University

#### Topics

Derivatives

Differentiation

##### Top Calculus 1 / AB Educators  ##### Heather Z.

Oregon State University ##### Kristen K.

University of Michigan - Ann Arbor Lectures

Join Bootcamp