A Veterinary doctor was examining a sick cat brought by a pet lover. When it was brought to the hospital, it was already dead. The pet lover wanted to find its time of death. He took the temperature of the cat at 11.30 pm which was `94.6^(@)F`. He took the temperature again after one hour, the temperature was lower than the first observation. It was `93.4^(@)F`. The room in which the cat was put is always at `70^(@)F`. The normal temperature of the cat is taken as `98.6^(@)F` when it was alive. The doctor estimated the time of death using Newton law of cooling which is governed by the differential equation: `(dT)/(dt) prop (T − 70),` where `70^(@)F` is the room temperature and T is the temperature of the object at time t.
Substituting the two different observations of T and t made, in the solution of the differential equation `(dT)/(dt)`=k(T-70) where k is a constant of proportion, time of death is calculated.
If the temperature was measured 2 hours after 11.30pm, will the time of death change? (Yes/No)

Answer

To find the time of death of the cat using Newton's Law of Cooling, we will follow these steps:

### Step 1: Set up the differential equation
According to Newton's Law of Cooling, the rate of change of temperature of an object is proportional to the difference between its temperature and the ambient temperature. The equation is given by:

\[ \frac{dT}{dt} = k(T - 70) \]

where \( T \) is the temperature of the object (the cat), \( k \) is a constant, and \( 70^\circ F \) is the room temperature.

### Step 2: Integrate the equation
We can rearrange and integrate both sides:

\[ \int \frac{1}{T - 70} dT = \int k \, dt \]

This gives us:

\[ \ln |T - 70| = kt + C \]

where \( C \) is the constant of integration.

### Step 3: Use the first observation
At \( t = 0 \) (11:30 PM), the temperature \( T = 94.6^\circ F \). Substituting these values into the equation:

\[ \ln |94.6 - 70| = k(0) + C \]

Calculating this gives:

\[ \ln 24.6 = C \]

### Step 4: Use the second observation
After 1 hour (at 12:30 AM), the temperature \( T = 93.4^\circ F \). Substituting these values into the equation:

\[ \ln |93.4 - 70| = k(1) + \ln 24.6 \]

This simplifies to:

\[ \ln 23.4 = k + \ln 24.6 \]

### Step 5: Solve for \( k \)
Rearranging gives:

\[ k = \ln 23.4 - \ln 24.6 \]

Using properties of logarithms, we can express this as:

\[ k = \ln\left(\frac{23.4}{24.6}\right) \]

### Step 6: Calculate the time of death
Now we know the normal temperature of the cat when alive is \( 98.6^\circ F \). We substitute this into our equation to find the time of death:

\[ \ln |98.6 - 70| = kT + \ln 24.6 \]

Calculating gives:

\[ \ln 28.6 = kT + \ln 24.6 \]

Solving for \( T \):

\[ kT = \ln 28.6 - \ln 24.6 \]

### Step 7: Substitute \( k \) back into the equation
Substituting \( k \) into the equation allows us to solve for \( T \):

\[ T = \frac{\ln\left(\frac{28.6}{24.6}\right)}{k} \]

### Step 8: Determine the time of death
After calculating \( T \), we find that the time of death is approximately \( -3.5 \) hours from 11:30 PM, which means the cat died around \( 8:00 PM \).

### Final Answer
If the temperature was measured 2 hours after 11:30 PM, the time of death would not change because the death occurred before 11:30 PM. Therefore, the answer is **No**.

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