Answer :
We are given the temperature function for the lasagna as
[tex]$$
T(t) = e^{(4.86753-t)} + 70,\quad t\ge0,
$$[/tex]
where [tex]$t$[/tex] is in hours and [tex]$T(t)$[/tex] is in degrees Fahrenheit. The rate of cooling is determined by the derivative of [tex]$T(t)$[/tex] with respect to [tex]$t$[/tex]. We can compute the derivative as
[tex]$$
T'(t) = \frac{d}{dt} \left(e^{(4.86753-t)} + 70\right).
$$[/tex]
Since the derivative of a constant is zero and using the chain rule for the exponential part, we have
[tex]$$
T'(t) = -e^{(4.86753-t)}.
$$[/tex]
The quantity [tex]$T'(t)$[/tex] is negative because the temperature is decreasing, and the magnitude of this derivative, [tex]$\left| T'(t) \right| = e^{(4.86753-t)}$[/tex], represents the cooling rate.
To determine when the lasagna cools slowest, we need to find the time (from the provided options) at which [tex]$\left| T'(t) \right|$[/tex] is smallest. This corresponds to the value of [tex]$T'(t)$[/tex] that is closest to zero.
Let’s evaluate the derivative at the given times:
- At [tex]$t=3$[/tex] hours:
[tex]$$
T'(3) = -e^{(4.86753-3)} \approx -6.47229,
$$[/tex]
- At [tex]$t=5$[/tex] hours:
[tex]$$
T'(5) = -e^{(4.86753-5)} \approx -0.87593,
$$[/tex]
- At [tex]$t=6$[/tex] hours:
[tex]$$
T'(6) = -e^{(4.86753-6)} \approx -0.32224,
$$[/tex]
- At [tex]$t=7$[/tex] hours:
[tex]$$
T'(7) = -e^{(4.86753-7)} \approx -0.11854,
$$[/tex]
- At [tex]$t=9$[/tex] hours:
[tex]$$
T'(9) = -e^{(4.86753-9)} \approx -0.01604.
$$[/tex]
Since the absolute value of the derivative represents the rate of cooling, we compare the magnitudes:
[tex]$$
6.47229,\quad 0.87593,\quad 0.32224,\quad 0.11854,\quad 0.01604.
$$[/tex]
The cooling is slowest when the magnitude is the smallest, which occurs when [tex]$t = 9$[/tex] hours.
Thus, the lasagna cools slowest [tex]$9$[/tex] hours after being taken out of the oven.
[tex]$$
T(t) = e^{(4.86753-t)} + 70,\quad t\ge0,
$$[/tex]
where [tex]$t$[/tex] is in hours and [tex]$T(t)$[/tex] is in degrees Fahrenheit. The rate of cooling is determined by the derivative of [tex]$T(t)$[/tex] with respect to [tex]$t$[/tex]. We can compute the derivative as
[tex]$$
T'(t) = \frac{d}{dt} \left(e^{(4.86753-t)} + 70\right).
$$[/tex]
Since the derivative of a constant is zero and using the chain rule for the exponential part, we have
[tex]$$
T'(t) = -e^{(4.86753-t)}.
$$[/tex]
The quantity [tex]$T'(t)$[/tex] is negative because the temperature is decreasing, and the magnitude of this derivative, [tex]$\left| T'(t) \right| = e^{(4.86753-t)}$[/tex], represents the cooling rate.
To determine when the lasagna cools slowest, we need to find the time (from the provided options) at which [tex]$\left| T'(t) \right|$[/tex] is smallest. This corresponds to the value of [tex]$T'(t)$[/tex] that is closest to zero.
Let’s evaluate the derivative at the given times:
- At [tex]$t=3$[/tex] hours:
[tex]$$
T'(3) = -e^{(4.86753-3)} \approx -6.47229,
$$[/tex]
- At [tex]$t=5$[/tex] hours:
[tex]$$
T'(5) = -e^{(4.86753-5)} \approx -0.87593,
$$[/tex]
- At [tex]$t=6$[/tex] hours:
[tex]$$
T'(6) = -e^{(4.86753-6)} \approx -0.32224,
$$[/tex]
- At [tex]$t=7$[/tex] hours:
[tex]$$
T'(7) = -e^{(4.86753-7)} \approx -0.11854,
$$[/tex]
- At [tex]$t=9$[/tex] hours:
[tex]$$
T'(9) = -e^{(4.86753-9)} \approx -0.01604.
$$[/tex]
Since the absolute value of the derivative represents the rate of cooling, we compare the magnitudes:
[tex]$$
6.47229,\quad 0.87593,\quad 0.32224,\quad 0.11854,\quad 0.01604.
$$[/tex]
The cooling is slowest when the magnitude is the smallest, which occurs when [tex]$t = 9$[/tex] hours.
Thus, the lasagna cools slowest [tex]$9$[/tex] hours after being taken out of the oven.
