# Rate of change, linear approximation?

For part a, how do I find the average rate of change between the two values? I know how to find the instantaneous rate of change, but not the average. For part b, I know the equation for a tangent line is $F'(x)=F(a)+F'(a)(x-a)$ but I'm not sure where to go from there. And all I have to do for c is find the derivative of the function given and then input the values 65mph and 24mph and subtract them to get the difference? Then I do the same thing, except with variables. For d, is $d,(v)$ a coordinate point? I don't quite understand this question. The same goes for the next. It might just be how they are phrasing it. Exercise a:

The average rate of change of $F$ between $a$ and $b$ is \begin{align*} &\frac{F(b)-F(a)}{b-a}\\ =&\frac{230\text{ ft}-172\text{ ft}}{60\text{ mph}-50\text{ mph}}\\ =&\frac{58\text{ ft}}{10\text{ mph}}\\ \approx&1.1\cdot 10^{-3}\text{ h} \end{align*}

Alright, I'll go through it all. If you or anybody else believes I am incorrect, please state.

Exercise a: Slo has pretty much answered this, so I will not. I'll just point out how he got his answer as $1.1 \cdot 10^{-3} h$, as he does not really explain it. Basically 58foot=0.0109848miles. Therefore we can say $$\frac{58ft}{10mph}=\frac{0.0109848miles}{10mph}=1.09848 \cdot 10^{-3} hours.$$ From there it is clear he just rounded.

Exercise b: Firstly, your equation for a tangent line approximation is wrong, and I will give you the right equation based on the function in this question. The equation is $$d(v)=d(a)+d'(a)(v-a).$$

We can use this to estimate $d(90)$ as follows. Firstly, we need to find a point "close" to do $90$, and since we're only given $80$, we'll use that. So we get \begin{align} d(90)&=d(80)+d'(80)(90-80) \\ &=d(80)+10d'(80). \end{align} Now the next logical question you might be asking is, how do we work out $d'(80)$? Well we can simply use the same equation, but this time for $a=70$, because if $90$ is close to $80$, then we can say $70$ is as well. This then gives \begin{align} d(70)&=d(80)+d'(80)(70-80) \\ &= d(80)-10d'(80) \\ &=395-10d'(80) \\ &=305. \end{align} I got the values in the equation above from the table provided in the question. Anyhow, now solve for $d'(80)$ to get $$d'(80)=9.$$ Plugging this back into our initial equation gives \begin{align} d(90)&=395+10 \cdot 9 \\ &=485. \end{align} Now, this is an underestimate. The reason for this is that $d'(v)>0$, so the function is increasing, and from sketching the graph with the values given, it is clear the tangent line will be beneath the graph at $v=90$.

Note: I didn't put $\approx$ anywhere above because I think it is obvious everything is being approximated. I should of really done that if we are being anal.

Exercise c: I think to do this, we just take $d(v)=0.05v^2+0.86v+2.5$ and differentiate to give $d'(v)=0.1v+0.86$. Then we calculate $d'(25)$ and $d'(65)$. From this, we get \begin{align} &d'(25)=3.36ft/s \\ &d'(65)=7.36ft/s. \end{align} The difference is then $4ft/s$.

Then do the same but with $a$ and $b$ to get the difference as $0.1|b-a| ft/s$ It is in absoloute value because it doesn't specify if $a$ or $b$ is the quicker speed.

Try exercise d and e yourself. Tell me what you get with your workings.