Question

Find the average rate of change of the function over the given interval. Compare this average rate of change with the instantaneous rates of change at the endpoints of the interval. $$ f(x)=\cos x, \quad\left[0, \frac{\pi}{3}\right] $$

Short answer

The average rate of change of the function over the given interval is -\(\frac{3}{2\pi}\). The instantaneous rates of change at the endpoints of the interval are 0 and -\(\frac{\sqrt{3}}{2}\) respectively. Consequently, for this problem, the average rate of change does not equal the instantaneous rates of change at the endpoints of the interval.

Step-by-step solution

Calculation of the average rate

The average rate of change of the function is given by \((f(b)-f(a)) / (b-a)\). Substituting the given values, \(a=0\) and \(b=\frac{\pi}{3}\), the average rate is therefore \((\cos(\frac{\pi}{3}) - \cos(0)) / (\frac{\pi}{3}-0)\)

Simplification of the average rate

\(\cos(\frac{\pi}{3})\) is equal to \(\frac{1}{2}\) and \(\cos(0)\) is equal to 1. Substituting these values, the average rate will be \((\frac{1}{2} - 1) / (\frac{\pi}{3}-0) = -\frac{1}{2} / (\frac{\pi}{3}) = -\frac{3}{2\pi}\)

Calculation of the instantaneous rates

The derivative of \(\cos(x)\) is \(-\sin(x)\). Therefore, the instantaneous rates of change at the endpoints of the interval, \(x=0\) and \(x=\frac{\pi}{3}\), will be -\(\sin(0)\) and -\(\sin(\frac{\pi}{3})\) respectively.

Simplification of the instantaneous rates

\(\sin(0)\) is equal to 0 and \(\sin(\frac{\pi}{3})\) is equal to \(\frac{\sqrt{3}}{2}\). Therefore, the instantaneous rates of change at the endpoints of the interval will be -0 and -\(\frac{\sqrt{3}}{2}\) respectively.