Answer
a) $-144.7213595; 0$
b) $0$
Work Step by Step
We are given the function:
$h(t)=8\cos (12\pi t)$
a) The average rate of change on $[t_0,t_1]$ is:
$\dfrac{Δh}{Δt}=\dfrac{h(t_1)−h(t_0)}{t_1−t_0}$
Compute the average rate of change on $[0,0.1]$:
$\dfrac{Δh}{Δt}=\dfrac{h(0.1)−h(0)}{0.1-0}=\dfrac{8\cos(12\pi\cdot 0.1)-8\cos(12\pi\cdot 0)}{0.1}$
$\approx -144.7213595$
Compute the average rate of change on $[3,3.5]$:
$\dfrac{Δh}{Δt}=\dfrac{h(3.5)−h(3)}{3.5-3}=\dfrac{8\cos(12\pi\cdot 3.5)-8\cos(12\pi\cdot 3)}{0.5}$
$=0$
b) The instantaneous rate of change is the limit of the average rate of change.
In order to estimate the instantaneous rate of change at $t=0$, consider intervals $[t_1,t_0],[t_0,t_1]$ for $t_1$ close to $t_0$, where $t_0=3$ and build the table:
From the table we find that the instantaneous rate of of change at $t=3$ is approximately $0$.
