Problem 26 - Differential Equations and Linear Algebra

Question

At \(2 \mathrm{p} . \mathrm{m}\). on a cool \(\left(34^{\circ} \mathrm{F}\right)\) afternoon in March, Sherlock Holmes measured the temperature of a dead body to be \(38^{\circ} \mathrm{F}\). One hour later, the temperature was \(36^{\circ} \mathrm{F}\). After a quick calculation using Newton's law of cooling, and taking the normal temperature of a living body to be \(98^{\circ} \mathrm{F},\) Holmes concluded that the time of death was 10 a.m. Was Holmes right?

Step-by-Step Solution

Answer

Holmes' conclusion about the time of death being 10 a.m. is not correct. By using Newton's law of cooling and the given temperature measurements, we found that the time of death was approximately 5.66 hours before 2 p.m., not the 4 hours Holmes concluded.

Step 1: Write down Newton's law of cooling equation

Newton's law of cooling is given by the differential equation: \[\frac{dT}{dt} = k(T - T_s)\] where \(T\) is the temperature of the body, \(T_s\) is the temperature of the surroundings, and \(k\) is the constant of proportionality.

Step 2: Separate the variables and integrate

To solve the differential equation, we need to separate the variables and integrate. First, we rewrite the equation as: \[\frac{dT}{T - T_s} = k dt\] Now, integrate on both sides: \[\int \frac{dT}{T - T_s} = \int k dt\] The solution is: \[\ln |T - T_s| = kt + C\]

Step 3: Determine the constant C using initial conditions

We can use the initial condition at 2 p.m., where \(T = 38 °F\) and \(T_s = 34 °F\), to determine the constant \(C\). Substituting the values into the equation: \[\ln |38 - 34| = k(2) + C\] \[\ln 4 = 2k + C\] Now, let's keep this expression as equation (1).

Step 4: Determine the constant k using given conditions

At 3 p.m., the body's temperature was \(36 °F\). We can substitute these values into the equation to find the constant \(k\): \[\ln |36 - 34| = k(3) + C\] \[\ln 2 = 3k + C\] Using equation (1), we can write C as: \[C = \ln 4 - 2k\] Now, substitute the expression for C into the second equation: \[\ln 2 = 3k + (\ln 4 - 2k)\] \[\ln 2 = k + \ln 4\] Now, we can find k: \[k = \ln 2 - \ln 4 = \ln \frac{1}{2}\]

Step 5: Calculate the time of death

We know the normal body temperature is \(98 °F\). Let's substitute it into the equation to obtain the time of death: \[\ln |98 - 34| = \ln \frac{1}{2}t + \ln 4 - 2 \ln \frac{1}{2}t\] \[\ln 64 = \ln 4 - \ln 2t + \ln 4t^2\] \[\ln 64 = \ln \frac{4t^2}{2t}\] \[64 = \frac{4t^2}{2t}\] \[128t = 4t^2\] \[32 = t^2\] \[t = 4\sqrt{2}\] Since 2 p.m. was 4 hours after the time Holmes concluded as the time of death (10 a.m.), let's compare it with the value we obtained: \[4\sqrt{2} \approx 5.66\] As we can see, there is a difference between our result and Holmes' conclusion. So, we can say that Holmes' conclusion about the time of death being 10 a.m. is not correct.

Key Concepts

Differential Equations

Differential equations are fundamental tools in mathematics used to describe various phenomena. A differential equation involves functions and their derivatives, providing a way to model how a particular system changes over time. In the context of Newton's Law of Cooling, we use differential equations to model how the temperature of an object approaches the ambient temperature. This is expressed with the equation \(\frac{dT}{dt} = k(T - T_s)\), which features a derivative representing the rate of temperature change \(\frac{dT}{dt}\). Here:

• \(T\) represents the temperature of the body.
• \(T_s\) is the constant surrounding temperature.
• \(k\) is a positive constant associated with the rate of cooling.

This equation tells us that the rate of temperature change is directly proportional to the difference between the body's temperature and the ambient temperature. By solving this differential equation, we can predict how quickly the body temperature will change over time.

Temperature Change

Temperature change can be effectively described using Newton's Law of Cooling. This law shows how the temperature of an object decreases in an environment that has a different temperature. When a body is hotter than its surroundings:

• The temperature decreases towards the ambient temperature.
• The rate of temperature change is faster when the temperature difference is greater.

In our scenario, here's how it unfolds: At 2 p.m., the measured temperature is \(38^{\circ} \mathrm{F}\) and it drops to \(36^{\circ} \mathrm{F}\) by 3 p.m. Meanwhile, the surrounding temperature remains constant at \(34^{\circ} \mathrm{F}\). Using the initial and subsequent temperatures, we can solve the differential equation derived from Newton's Law of Cooling. The solution involves integrating the equation, which helps to calculate the cooling constant \(k\) and predict future temperatures or even backtrack to determine past temperatures, such as estimating the time of death in forensic science.

Problem-solving in Mathematics

Problem-solving in mathematics is the process of finding solutions to complex problems by breaking them down into manageable steps. In this particular exercise, we are faced with a real-world application of mathematics to solve a mystery - predicting the time of death using temperature data. This problem involves:

• Writing the differential equation representing cooling.
• Separating variables and integrating to find the general solution, \(\ln |T - T_s| = kt + C\).
• Using initial conditions to find the constant \(C\), and subsequent conditions to find \(k\).

Upon calculating \(k\), further steps include using this value to work backwards from the known times to hypothesize approximate past conditions, like the initial time when the body temperature might have been at a normal \(98^{\circ} \mathrm{F}\). By following these methodical steps, we can check Sherlock Holmes' deduction. His estimation of the time of death at 10 a.m. differed from our calculation, illustrating how this mathematical approach clarifies assumptions, providing more accuracy in predictions.