Suppose that during the first hour and 15 minutes of a physics experiment, the surface temperature of a small iron block is modeled by the exponential function f(t) = 15e^t , where f(t) is the Celsius temperature t hours after the experiment begins. Compute the average rate of change of temperature over the second half hour of the experiment (i.e., over the interval $0.5 \leq t \leq 1$ ). Round the answer to one decimal place. A) $32.1 ^ { \circ } \mathrm { C } \text { hour }$ B) $31.8 ^ { \circ } \mathrm { C } / \text { hour }$ C) $41 ^ { \circ } \mathrm { C } / \text { hour }$ D) $32.2 ^ { \circ } \mathrm { C } / \text { hour }$ E) $33.1 ^ { \circ } \mathrm { C } / \text { hour }$

Question

Quizplus · Accepted Answer

The average rate of change of a function over an interval $[a, b]$ is given by $\frac{f(b) - f(a)}{b - a}$. Here, $a = 0.5$ hours and $b = 1$ hour. Plugging these into the given function $f(t) = 15e^{t}$, we get $f(0.5) = 15e^{0.5}$ and $f(1) = 15e^{1}$. The average rate of change is $\frac{15e^{1} - 15e^{0.5}}{1 - 0.5} = \frac{15e - 15e^{0.5}}{0.5}$. Calculating this value gives approximately $32.1 ^ { \circ } \mathrm { C } / \text { hour }$, rounding to one decimal place.

Suppose That During the First Hour and 15 Minutes of a Physics