Construct an exponential function from given information and then use the function to solve a contextual problem. Instructions

Question

Lab Manual Lesson

Purpose

Construct an exponential function from given information and then use the function to solve a contextual problem.

Instructions

CSI Wake Tech - A Case for Homicide

The evening cleaning crew at a large firm discovers a dead body in the early morning hours. Police are immediately dispatched to the scene. Upon arrival at 5:00 am, police measure the temperature of the body to be 88.2°F. An hour later, the coroner arrives and again measures the temperature of the body. This time, the temperature is 86.7°F. The environmental system of the building maintains a constant temperature of 67°F.

Complete the following, including detailed explanations of your work:

1. Use the exponential topics discussed in the course to determine an exponential formula that models the temperature, B, of the body as a function of the time, t, in hours since the police arrived (t = 0). Round values to 2 decimal places as needed. Show how you arrived at this function, showing all necessary work.
2. Graph your function on a reasonable domain. Label and scale your graph. You can use tools such as desmos.com to graph your function.
3. Use your function to determine the temperature of the body if the coroner would have checked the body temperature again at 8:00 am.
4. Use your function to estimate the time of death, to the nearest half-hour. Show how you arrived at this answer, showing all necessary work.

Report

The report should be typed and include accurately drawn graphs of the function or electronically produced graphs.

Answer 1

(0, 88.2) <--- cops measured the body temp at 5 am, which shall be time zero.

(1, 86.7) <--- coroner measured body temp 1 hours later

y = k * b^(Ct)

(0,88.2) ---> k = 88.2

so

y = 88.2 * b^(ct)

for t=1

86.7 = 88.2 * b^c

86.7/88.2 = b^c

takes natural log of both sides,

-0.0171530792+ = c * ln(b) = c/logb(e) <--- change of base

-0.0171530792+ * logB(e) = c

plugs into the function

y = 88.2 * B^(-0.0171530792+ * logB(e) * t)

= 88.2 * exp( -0.0171530972+ * t) <--- properties of exponentials and logs

assuming normal body temp at time of death,

98.6 = 88.2 * exp(-0.0171530972+ * t)

1.1179138322- = exp(-0.0171530972 * t) <--- solves for t

Taking natural log of both sides:

0.11146429859584411443161068950363+ = -0.0171530972+ * t

t = -6.5

So the time of death is almost 6.5 hours BEFORE 5 am

time of death is around 10:30 pm

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