1 pt) Coroners estimate time of death using the rule of thumb that a body cools about 2 degrees F during the first hour after death and about 1 degree F for each additional hour. Assuming an air temperature of 68 degrees F and a living body temperature of 98.6 degrees F, the temperature T(t) in degrees F of a body at a time t hours since death is given by

T(t)=68+30.6e^(−kt.)
For what value of k will the body cool by 2 degrees F in the first hour?
Using the value of k found above, after how many hours will the temperature of the body be decreasing at a rate of 1 degree F per hour?
Using the value of k found above, show by calculating both values that, 24 hours after death, the coroner's rule of thumb gives approximately the same temperature as the formula.

To answer the 1st question, just plug in t=2 and solve for k:

68+30.6e^(-2k) = 96.6

Now just use the amended formula for the rest. If you get stuck, come on back and show us whatcha got so far.

T(t)=68+30.6e^(−kt.)

After one hour the temp has dropped 2 degrees so T(t)=96.6, solve for K
k=0.06759

you just substitute into this equation using your k value for the rest of the equation except for part D which is just 98.6-25=73.6

K=.0337966456

Well, well, well, it seems we have a case of a chilling question here. Let's see if we can put the "fun" in "funeral" and solve this mystery!

To find the value of k that will make the body cool by 2 degrees F in the first hour, we can use the formula T(t) = 68 + 30.6e^(-kt). We know that the initial body temperature is 98.6 degrees F, and after one hour it should be 96.6 degrees F. So let's plug in those values and solve for k:

96.6 = 68 + 30.6e^(-k * 1)

Now, let's subtract 68 from both sides:

28.6 = 30.6e^(-k)

Dividing both sides by 30.6:

0.935 = e^(-k)

To get rid of that pesky exponential, we'll take the natural logarithm of both sides:

ln(0.935) = -k

Alright, let's calculate that value:

k ≈ -0.066

Now that we have our value of k, we can move on to the next part of the mystery!

To find the time at which the body temperature is decreasing at a rate of 1 degree F per hour, we need to find when the derivative of T(t) with respect to t is -1. The derivative of T(t) is given by:

T'(t) = -30.6ke^(-kt)

Setting that equal to -1:

-30.6ke^(-kt) = -1

Dividing both sides by -30.6k:

e^(-kt) = 1/30.6k

Now, let's take the natural logarithm of both sides again:

ln(e^(-kt)) = ln(1/30.6k)

Simplifying:

-kt = ln(1/30.6k)

Solving for t:

t = ln(1/30.6k) / -k

Now, let's plug in our value of k ≈ -0.066 and do some calculations:

t ≈ ln(1/(30.6*(-0.066))) / -(-0.066)

t ≈ 2.07 hours

Looks like our chilling detective work is paying off!

Finally, let's see how well the coroner's rule of thumb holds up against the formula 24 hours after death. We can calculate the temperature using both methods and compare.

Using the coroner's rule of thumb, the temperature would have cooled by 2 degrees F in the first hour, followed by 23 additional hours, each cooling the body by 1 degree F. Therefore, the temperature would be:

98.6 - 2 - 1(23) = 98.6 - 2 - 23 = 73.6 degrees F

Using the formula T(t) = 68 + 30.6e^(-kt), we can plug in t = 24 and k ≈ -0.066:

T(24) = 68 + 30.6e^(-0.066*24)

Calculating:

T(24) ≈ 68 + 30.6e^(-1.584)

T(24) ≈ 68 + 30.6(0.204)

T(24) ≈ 68 + 6.26

T(24) ≈ 74.26 degrees F

Well, well, what do we have here? The coroner's rule of thumb gives around 73.6 degrees F, while the formula gives approximately 74.26 degrees F. Close enough, I'd say! Looks like both methods have their merits.

And there you have it, my dear investigator. The body of knowledge has been revealed, and the mystery of temperatures has been solved!

To find the value of k that will make the body cool by 2 degrees F in the first hour, we can use the given equation for temperature T(t) and substitute t = 1 and T(t) = 2 into the formula:

T(t) = 68 + 30.6e^(-kt)

2 = 68 + 30.6e^(-k)

Subtracting 68 from both sides:

-66 = 30.6e^(-k)

Now, divide both sides by 30.6:

e^(-k) = -66/30.6

To isolate e^(-k), we take the natural logarithm (ln) of both sides:

ln(e^(-k)) = ln(-66/30.6)

Using the property of logarithms that ln(e^x) = x, we simplify:

-k = ln(-66/30.6)

Finally, multiply both sides by -1 to solve for k:

k = -ln(-66/30.6)

To determine the time, in hours, at which the temperature of the body is decreasing at a rate of 1 degree F per hour, we need to find when the derivative of T(t) is equal to -1. Taking the derivative of T(t) with respect to t:

T'(t) = -30.6k * e^(-kt)

Setting T'(t) equal to -1 and solving for t:

-1 = -30.6k * e^(-kt)

Dividing both sides by -30.6k:

1/(30.6k) = e^(-kt)

Taking the natural logarithm of both sides:

ln(1/(30.6k)) = -kt

To isolate t, divide both sides by -k:

t = ln(1/(30.6k)) / -k

Now that we have the value of k, we can calculate the temperature 24 hours after death using both the coroner's rule of thumb and the given formula.

For the coroner's rule of thumb, we use the rate of cooling stated in the beginning: 2 degrees F in the first hour and 1 degree F for each additional hour. Thus, after 24 hours, the temperature would have cooled by a total of 2 + (23 * 1) = 25 degrees F from the initial temperature of 98.6 degrees F.

Using the given formula for T(t):

T(t) = 68 + 30.6e^(-kt)

Plugging in t = 24 and the previously calculated value of k:

T(24) = 68 + 30.6e^(-k * 24)

We substitute the value of k found earlier:

T(24) = 68 + 30.6e^(-(-ln(-66/30.6)) * 24)

Calculating this expression will give the estimated temperature using both methods.