Math 181 Miniproject 3: Texting Lesson.md
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Estimating the Average & Instantneous Rate of Change
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<div><img class="left"/><div class="alert gray">
Hi Isela,
So... what exactly is this assignment? it looks tricky , this assignment is estimating the average and instanteneous rate of change in a function and interpreting the values of a first derivative and a second derivative.
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What does that exactly mean? Can your perhaps provide me with an example?

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To calculates a car's average velocity we divide the number of miles traveled by the time elapsed, which gives the velocity in miles per hour.It's important to understand that the average rate of change of a function of f on the interval {a.b} is calculated by :

$AV[a,b]=f(b)-f(a)$/$(b-a)$

Which is the average rate of change, is simply the slope between two different points.

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What needs to be included?

In Contrast, The instantaneous rate of change at a single given point or single instant. The Instantaneous rate of change of a function at a given point is denoted by the derivative f'(a) at that point and its calculated by:

$f'(a)= lim(h to 0)$ $f(a+h)-f(a)/ h$
Instantaneous rate of change could measure the number of cells added to a bacteria culture per day, the number of additional gallons of gasoline consumed by increasing a car's velocity one mile per hour, or the number of dollars added to a mortgage payment for each percentage point increase in interest rate.

Hence, these both are slopes its just that the Average rate of change is the slope between two points and Instantaneous rate of change is the slope at one point.

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The derivative itself represents the slope of a particularly important line. Think of the first derivative as a rate of change, then the second derivative is the rate of change of the first derivative. If the second derivative is positive, then the rate of change is increasing. If the second Derivative is negative then the rate of change is decreasing.
For example,

A water balloon is tossed vertically in the air from a window. The balloon's height in feet at time t in seconds after being launched is given by
$s(t)=-16t^2+16t+32$

To Calculate the average rate of change of s on the time interval[1,2]

We use the formula
$AV[a,b]$= $s(b)-s(a)/$($b-a$)

Where we substitute the interval [1,2] into [a,b]:

$AV[1,2]$=$f(2)$-$f(1)$/($2-1$)
Then we substitute, f(2) and f(1) into our given equation:

$AV[1,2]$=$[-16(2)^2+16(2)+32$]-[$-16(1)^2+16(1)+32$]/($2-1$) =-32ft/s

We can interpret that after the ballon was tossed up, up being a positive direction from a window the balloon from 2 to 1 seconds started to decrease or starting falling at -32ft per second.

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More specifically we can use

$s'(a)= lim(h to 0)$ $s(a+h)-s(a)/ h$

to calculate the instantaneous rate of change of s with respect to time,t at the instant a=1

We Substitute 1 for a in the formula

$s'(1)= lim ($h to 0$) $s(1+h)$-$s(1)$/$h$

Then we substitute for s(1+h)- s(1) with our given equation.

=lim(h to 0) [$-16(1+h)^2+16(1+h)+32$]-[$-16(1)^2+16(1)+32$]/$h$

=lim ($h to 0$) -16($1+2h+h^2$)+16+16h+16+32/$h$
=lim ($h to 0$) $-16h-16h^2$/$h$
=lim ($h to 0$) h(-16-16h)/h

h cancel out and we plug in 0 into each because as h is approaching 0 it gets closer to
$-16-16(0)=-16

We can say that the instantaneouse rate of change at 1second is decreasing at a rate of 16ft per second per second.

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A study tip I highly suggest is too know the difference conceptually and mathmatically between Instantanouse rate of change and Average Rate of change.

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