A boat at anchor is bobbing up and down in the sea. The vertical distance, $y$, in meters, between the sea floor and the boat is given as a function of time, $t$, in hours, by
$$6 + \operatorname{sin} ((\frac{\pi t}{3}) + 1.5).$$
What is the average rate of change of $y$ over the 4 hour interval $1 \leq t \leq 4?$
So I think I know what to do. I found the derivative of the function which is $\operatorname{cos} ((\frac{\pi t}{3}) + 1.5) \cdot \frac{\pi}{3}.$ Do I plug in $1,2,3,4$ and then find and then add the answers and divide by four to find the mean? Thanks!
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$\begingroup$The average rate of change of $f(t)$ on the interval $a\le t\le b$ is just ${f(b)-f(a)\over b-a}$. So here---assuming I am interpreting your function correctly with the extra set of parentheses---you get:
$${f(4)-f(1)\over 4-1}\approx -0.373338.$$
To understand this geometrically, just visualize the secant line between the points $(1,f(1))$ and $(4,f(4))$ shown in black. The number above is the slope of this line. The original function is shown in blue. (Note how the axes are scaled though.)
Let $y(t) = 6 + \sin ((\frac{\pi t}{3}) + \frac{3}{2})$. The question is asking you to compute $\frac{1}{4-1} \int_1^4 y'(t) dt$. However, we have $\int_1^4 y'(t) dt = y(4)-y(1)$, so this reduces to computing $\frac{y(4)-y(1)}{4-1}$.
Hence the answer is $$\frac{y(4)-y(1)}{4-1} = \frac{1}{3}(\sin(\frac{4 \pi}{3}+\frac{3}{2})-\sin(\frac{\pi}{3}+\frac{3}{2})) = \frac{2}{3}\sin \frac{\pi}{2} \cos (\frac{5 \pi}{6}+\frac{3}{2}) = \frac{2}{3}\cos (\frac{5 \pi}{6}+\frac{3}{2})$$ (Thanks to Ross for catching my mistake.)
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