Report on Math

Introduction

According to the problem, there are two distinct data points. One is found through the reading of the second sentence where it says, at 5:00 a.m. when the police arrived at the scene they measured the corpses temp an ascertained it to be 88.20 F. Therefore this can be written as (5, 88.2) where 5 represent time and 88.2 represents temperature. Later on, in the question, it elaborates that a coroner makes to the scene and gets their measure of the body which is now 86.70 F and it clearly states that it was an hours later which leads us to conclude that (6, 86.7). And the concluding part of the problem now leads us to the final resolution of what the body temperature should be, therefore when the heat of the corpse had dissipated to the right environment the temperature was 670 F.

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Therefore,

The data that has been found so far can be fitted into an exponential form, which is given in a generic of y = c*a^x +k. Since most of the data in the question has been given in the form of time and temperature the variables will change, thus let time be the variable x(t) which can be seen as an independent variable also let us substitute the other variable which is y to represent the Temperature (T). Therefore, writing the equation using the exponential form it will give us

T = c *a^ t + k

However, using the function representation, we conclude that

T(t) = c *a^ t + k

Looking at the data, this can be seen as a declining function and leads us to the assumption that a would be any given number but is less than 1. By exploring the Behaviour of T(t) whereby t increases it is noticeable that in the end, the temperature will reach k asymptotically. Therefore, it is reasonable to conclude that k is a representation of a stable state temperature that will now take the form T_ss thus,

T(t) = c* a^ t +T_ss

From the above equation we substitute to the values in the question, it has been established that a stable state of the corpse is given by 670 F, and it has also been established that there are two forms of data curves that are (5, 88.2) and (6, 86.7) the two data points can fit in a format (t, T(t)). Therefore let's fit them in the already formulated equation. Taking the initial equation and rearrange it:

88.2 = c *a^ 5 + 67

Minus 67 on both ends

88.2 - 67 = (c * a^ 5) +67 - 67

Thus, 21.2 = c* a^5

Now applying the log properties to the equation would lead us to

Log 21.2 = log c* a^ 5

It can be rewritten as

Log 21.2 = Log c + log a^ 5

Get rid of the power

Log 21.2 = Log c + 5 * Log a

Therefore,

a = 19.7 divided by 21.2

a = 0.929

Conclusively, 21.2 divided by 0.929^5 is equals to 30.597 whereby T (t) equals to 30.597 multiplied by 0.929 raised to power t add 67

The time to which the body begun to cool can be done by the temperature of the body which is 98.6 0 F.

Thus,

98.6 = 30.597 * 0.929^ t + 67

Remove 67 through subtracting it on the left

31.6 = 30.597 * 0.929^ t

Take 30.597 to the left

31.6 divided by 30.597

1.033 = 0.929^ t

Application of Log we get t = -0.44 to 2 decimal places

It points us to the conclusion that the person in question must have died at 0.44 hours; therefore, we can conclude that the person must have died at 11:33 p.m.