A tray of lasagna comes out of the oven at [tex]200^{\circ} F[/tex] and is placed on a table where the surrounding room temperature is [tex]70^{\circ} F[/tex]. The temperature [tex]T[/tex] (in [tex]{}^{\circ} F[/tex]) of the lasagna is given by the function [tex]T(t) = e^{(4.86753-t)} + 70, \, 0 \leq t[/tex], where [tex]t[/tex] is time (in hours) after taking the lasagna out of the oven.

What is the rate of change in the temperature of the lasagna exactly 2 hours after taking it out of the oven?

A. -19.22 degrees/hour
B. -15.36 degrees/hour
C. -17.59 degrees/hour
D. -22.37 degrees/hour
E. -20.21 degrees/hour

We start with the temperature function for the lasagna:

[tex]$$
T(t) = e^{(4.86753-t)} + 70.
$$[/tex]

Step 1. Differentiate the function

To find the rate of change of temperature, we differentiate [tex]$T(t)$[/tex] with respect to time [tex]$t$[/tex]. Since the derivative of a constant is zero, we only need to differentiate the exponential part. Applying the chain rule to [tex]$e^{(4.86753-t)}$[/tex], we have:

[tex]$$
\frac{d}{dt}\left(e^{(4.86753-t)}\right) = e^{(4.86753-t)} \cdot \frac{d}{dt}(4.86753-t).
$$[/tex]

Noting that

[tex]$$
\frac{d}{dt}(4.86753-t) = -1,
$$[/tex]

the derivative of the temperature function is:

[tex]$$
T'(t) = -e^{(4.86753-t)}.
$$[/tex]

Step 2. Evaluate at [tex]$t = 2$[/tex] hours

Substitute [tex]$t = 2$[/tex] into the derivative:

[tex]$$
T'(2) = -e^{(4.86753-2)} = -e^{2.86753}.
$$[/tex]

Step 3. Compute the value

Evaluating the exponential term,

[tex]$$
e^{2.86753} \approx 17.59,
$$[/tex]

we obtain:

[tex]$$
T'(2) \approx -17.59 \text{ degrees per hour}.
$$[/tex]

Conclusion

The lasagna is cooling at approximately [tex]$-17.59$[/tex] degrees per hour exactly 2 hours after it comes out of the oven. This corresponds to option C:

[tex]$$\boxed{-17.59 \text{ degrees/hour}}.$$[/tex]