How would I find the average rate of change over $8$ minutes, of a person that runs at a rate of $v(t)=x\sin(x^2-7x)$ ft/min? I missed when this was taught and I have no clue on how to do it. Help is greatly appreciated.
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$\begingroup$Hint: The average rate of change of a function $f(t)$ over the interval $[a,b]$ is $\dfrac{f(b)-f(a)}{b-a}$.
The numerator $f(b)-f(a)$ is the change in $f$, and $b-a$ measures how long it took.
In this problem, you have to calculate the change in displacement. That will require integration.
To integrate, we would need to know the interval, which has not been specified. We know "$b-a$" but do not know $b$ or $a$. Perhaps we are expected to assume that $a=0$ and $b=8$, but probably the problem as originally stated did not leave any ambiguity.
$\endgroup$ 9 $\begingroup$I am going to first rewrite the original question : "How would I find the average rate of change over the first 8 minutes of a person that runs at a rate of $v(t)=t\sin(t^2-7t)$ "
In general the average rate of change of a function $f(t)$ on an interval $[a,b]$ can be computed by the formula $$\dfrac{f(b)-f(a)}{b-a}$$
In the context of our problem let $f(t)$ measure the position or displacement of the runner at time $t$ with respect to a predefined origin. The difficulty here is that we are not given the position function $f(t)$, but rather we are given the (instantaneous) rate $v(t)$ at which the position function changes at time $t$. This rate of change of position at time $t$ is also known as velocity. Therefore we have $v(t)=f'(t)$.
Now by the fundamental theorem of calculus, the area under the $v(t)$ curve on the interval $[0,8]$ is equal to the change in position $f(8)-f(0)$. More formally, we have: $$f(8)-f(0) = \int_{0}^{8} f'(t)dt$$ and $$\int_{0}^{8} f'(t)dt =\int_{0}^{8} v(t)dt$$ so $$f(8)-f(0)=\int_{0}^{8} v(t)dt$$ Therefore by substitution we have average rate of change is equal to: $$\dfrac{f(8)-f(0)}{8-0} = \dfrac{\int_{0}^{8} f'(t)dt}{8-0}=\dfrac{\int_{0}^{8} v(t)dt}{8-0}=\dfrac{\int_{0}^{8}t\sin(t^2-7t)dt}{8-0}$$
Alternatively we can use the fact (derived from Riemann sums) that the average value of a continuous function $f(t)$ on an interval $[a,b]$ is equal to $\dfrac{\int_{a}^{b}f(t)}{b-a}$. We want the average value of the "instantaneous rate of change of position" or just "rate of change of position", where the instantaneous rate of change of position is defined as velocity $v(t)$. This gives us the result : $$\dfrac{\int_{0}^{8}v(t)dt}{8-0}=\dfrac{\int_{0}^{8}t\sin(t^2-7t)dt}{8-0}$$
So either approach results in the same solution. This is interesting because the term average in "average velocity" or "average rate of change" is not a coincidence. In the first approach we did not actually use an average but rather used the definition of average velocity as change in position over time. I have often wondered why it is called "average velocity" if you don't actually average the individual velocity values. But it turns out that you can average velocity values if you use Riemann sums and calculus. But it is easier for teachers to just give the definition that average velocity is change in position divided by change in time, or as the slope of two points on a position-time graph of the particle.
I have often wondered why it is called "average velocity" if you don't actually average the individual velocity values.
The conventional position function $s=s_0+v_0t + \dfrac{1}{2}at^2$ comes from a simple graphical analysis of the area under a velocity-time graph for constant acceleration. It could be written more simply as follows because $\Delta v = a\Delta t$.$$\Delta s = v_0\Delta t + \dfrac{1}{2}\Delta v\Delta t$$Given that every change in position requires a non-zero change in time, dividing both sides by $\Delta t$ it is an arithmetically valid and law-like maneuver.$$\dfrac{\Delta s}{\Delta t} = v_0 + \dfrac{1}{2}\Delta v$$A little basic algebra and you obtain the formula for average velocity that actually involves adding up some things and dividing by the number of things. $$\dfrac{\Delta s}{\Delta t} = v_0 + \dfrac{1}{2}v-\dfrac{1}{2}v_0$$$$\dfrac{\Delta s}{\Delta t} = \dfrac{1}{2}v+\dfrac{1}{2}v_0$$$$\dfrac{\Delta s}{\Delta t} = \dfrac{v+v_0}{2}=\bar{v}$$
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