Answer :
We are given the temperature function
[tex]$$
T(t) = e^{(4.86753 - t)} + 70,\quad t \ge 0,
$$[/tex]
and we wish to find the rate at which the temperature is changing at [tex]$t = 2$[/tex] hours.
Step 1. Compute the derivative of [tex]$T(t)$[/tex]
The derivative of the constant 70 is 0, so we only differentiate the term [tex]$e^{(4.86753-t)}$[/tex]. Recall that the derivative of [tex]$e^u$[/tex] with respect to [tex]$t$[/tex] is [tex]$e^u \cdot u'$[/tex], where [tex]$u = 4.86753-t$[/tex]. Since
[tex]$$
u' = \frac{d}{dt}(4.86753-t) = -1,
$$[/tex]
the derivative is:
[tex]$$
T'(t) = \frac{d}{dt}\left(e^{(4.86753-t)}\right) = e^{(4.86753-t)} \cdot (-1) = -e^{(4.86753-t)}.
$$[/tex]
Step 2. Evaluate the derivative at [tex]$t = 2$[/tex]
Substitute [tex]$t = 2$[/tex] into the derivative:
[tex]$$
T'(2) = -e^{(4.86753-2)}.
$$[/tex]
Compute the exponent:
[tex]$$
4.86753 - 2 = 2.86753.
$$[/tex]
Thus,
[tex]$$
T'(2) = -e^{2.86753}.
$$[/tex]
Step 3. Approximate the value
The calculation shows that
[tex]$$
e^{2.86753} \approx 17.59.
$$[/tex]
So, the rate of change of the temperature at 2 hours is:
[tex]$$
T'(2) \approx -17.59 \text{ degrees/hour}.
$$[/tex]
Conclusion
The temperature of the lasagna is decreasing at a rate of approximately [tex]$17.59^{\circ}F$[/tex] per hour at [tex]$t = 2$[/tex] hours.
The correct answer is:
C. [tex]\(-17.59\)[/tex] degrees/hour.
[tex]$$
T(t) = e^{(4.86753 - t)} + 70,\quad t \ge 0,
$$[/tex]
and we wish to find the rate at which the temperature is changing at [tex]$t = 2$[/tex] hours.
Step 1. Compute the derivative of [tex]$T(t)$[/tex]
The derivative of the constant 70 is 0, so we only differentiate the term [tex]$e^{(4.86753-t)}$[/tex]. Recall that the derivative of [tex]$e^u$[/tex] with respect to [tex]$t$[/tex] is [tex]$e^u \cdot u'$[/tex], where [tex]$u = 4.86753-t$[/tex]. Since
[tex]$$
u' = \frac{d}{dt}(4.86753-t) = -1,
$$[/tex]
the derivative is:
[tex]$$
T'(t) = \frac{d}{dt}\left(e^{(4.86753-t)}\right) = e^{(4.86753-t)} \cdot (-1) = -e^{(4.86753-t)}.
$$[/tex]
Step 2. Evaluate the derivative at [tex]$t = 2$[/tex]
Substitute [tex]$t = 2$[/tex] into the derivative:
[tex]$$
T'(2) = -e^{(4.86753-2)}.
$$[/tex]
Compute the exponent:
[tex]$$
4.86753 - 2 = 2.86753.
$$[/tex]
Thus,
[tex]$$
T'(2) = -e^{2.86753}.
$$[/tex]
Step 3. Approximate the value
The calculation shows that
[tex]$$
e^{2.86753} \approx 17.59.
$$[/tex]
So, the rate of change of the temperature at 2 hours is:
[tex]$$
T'(2) \approx -17.59 \text{ degrees/hour}.
$$[/tex]
Conclusion
The temperature of the lasagna is decreasing at a rate of approximately [tex]$17.59^{\circ}F$[/tex] per hour at [tex]$t = 2$[/tex] hours.
The correct answer is:
C. [tex]\(-17.59\)[/tex] degrees/hour.
