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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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    $\begingroup$ No. It's an average rate of change over an interval, not an instantaneous rate of change at a point. Let $f(x)=6+\sin\bigl({\pi t\over3} +1.5\bigr)$. Simply compute $f(4)-f(1)\over 4-1$. $\endgroup$ – David Mitra Feb 5 '13 at 18:20
  • $\begingroup$ would I need to find the derivative? Or would it not matter? $\endgroup$ – user56852 Feb 5 '13 at 18:24
  • $\begingroup$ There is no need to do that. The average rate of change of a function $f$ over an interval $[a,b]$ is by definition ${f(b)-f(a)\over b-a}$. $\endgroup$ – David Mitra Feb 5 '13 at 18:25
  • $\begingroup$ Blech... I meant "$f(t)$" in my first comment. $\endgroup$ – David Mitra Feb 5 '13 at 18:28
  • $\begingroup$ Oh I understood what you meant :) so I got an answer of -.37, does this seem plausible? I think I did this right. $\endgroup$ – user56852 Feb 5 '13 at 18:29
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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.)

Mathematica graphics

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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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  • $\begingroup$ I'm having a bad compute day... $\endgroup$ – copper.hat Feb 5 '13 at 19:14
  • $\begingroup$ @RossMillikan: Thanks, I was missing a factor of $2$. Thanks for spotting. $\endgroup$ – copper.hat Feb 5 '13 at 19:16

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