Average rate of change of sin(x) on [0, pi/2]

We need the slope of the secant line joining $(0, f(0))$ and $(\pi/2, f(\pi/2))$. Since $f(x) = \sin(x)$, we have $f(0) = \sin(0) = 0$ and $f(\pi/2) = \sin(\pi/2) = 1$.

The average rate of change is $\dfrac{f(\pi/2)-f(0)}{\pi/2 - 0} = \dfrac{1 - 0}{\pi/2} = \dfrac{2}{\pi}$. This represents the slope of the secant line through the two endpoints on the graph of $\sin(x)$.

By the limit definition, the instantaneous rate of change at $x = \pi/4$ is $f'(\pi/4) = \lim_{x \to \pi/4} \frac{\sin(x) - \sin(\pi/4)}{x - \pi/4}.$ This limit represents the slope of the tangent line to the graph of $\sin(x)$ at the point where $x = \pi/4$.

Since the derivative of $\sin(x)$ is $\cos(x)$, this limit evaluates to $f'(\pi/4) = \cos(\pi/4) = \dfrac{\sqrt{2}}{2} \approx 0.707$.

From part (a), the average rate of change is $2/\pi \approx 0.637$, and from part (b), the instantaneous rate of change at $x=\pi/4$ is $\sqrt{2}/2 \approx 0.707$.

Since $\sqrt{2}/2 > 2/\pi$, the instantaneous rate of change (tangent slope) at the midpoint is larger than the average rate of change (secant slope) over the interval. This is consistent with $\sin(x)$ being concave up on $[0,\pi/2]$, so the tangent line at the midpoint is steeper than the chord connecting the endpoints.