# 3.1 Defining the derivative  (Page 4/10)

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${v}_{\text{ave}}=\frac{s\left(t\right)-s\left(a\right)}{t-a}.$

As the values of $t$ approach $a,$ the values of ${v}_{\text{ave}}$ approach the value we call the instantaneous velocity    at $a.$ That is, instantaneous velocity at $a,$ denoted $v\left(a\right),$ is given by

$v\left(a\right)={s}^{\prime }\left(a\right)=\underset{t\to a}{\text{lim}}\frac{s\left(t\right)-s\left(a\right)}{t-a}.$

To better understand the relationship between average velocity and instantaneous velocity, see [link] . In this figure, the slope of the tangent line (shown in red) is the instantaneous velocity of the object at time $t=a$ whose position at time $t$ is given by the function $s\left(t\right).$ The slope of the secant line (shown in green) is the average velocity of the object over the time interval $\left[a,t\right].$

We can use [link] to calculate the instantaneous velocity, or we can estimate the velocity of a moving object by using a table of values. We can then confirm the estimate by using [link] .

## Estimating velocity

A lead weight on a spring is oscillating up and down. Its position at time $t$ with respect to a fixed horizontal line is given by $s\left(t\right)=\text{sin}\phantom{\rule{0.2em}{0ex}}t$ ( [link] ). Use a table of values to estimate $v\left(0\right).$ Check the estimate by using [link] .

We can estimate the instantaneous velocity at $t=0$ by computing a table of average velocities using values of $t$ approaching $0,$ as shown in [link] .

Average velocities using values of t Approaching 0
$t$ $\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}t-\text{sin}\phantom{\rule{0.1em}{0ex}}0}{t-0}=\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}t}{t}$
$-0.1$ $0.998334166$
$-0.01$ $0.9999833333$
$-0.001$ $0.999999833$
$0.001$ $0.999999833$
$0.01$ $0.9999833333$
$0.1$ $0.998334166$

From the table we see that the average velocity over the time interval $\left[-0.1,0\right]$ is $0.998334166,$ the average velocity over the time interval $\left[-0.01,0\right]$ is $0.9999833333,$ and so forth. Using this table of values, it appears that a good estimate is $v\left(0\right)=1.$

By using [link] , we can see that

$v\left(0\right)={s}^{\prime }\left(0\right)=\underset{t\to 0}{\text{lim}}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}t-\text{sin}\phantom{\rule{0.1em}{0ex}}0}{t-0}=\underset{t\to 0}{\text{lim}}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}t}{t}=1.$

Thus, in fact, $v\left(0\right)=1.$

A rock is dropped from a height of $64$ feet. Its height above ground at time $t$ seconds later is given by $s\left(t\right)=-16{t}^{2}+64,0\le t\le 2.$ Find its instantaneous velocity $1$ second after it is dropped, using [link] .

$-32$ ft/s

As we have seen throughout this section, the slope of a tangent line to a function and instantaneous velocity are related concepts. Each is calculated by computing a derivative and each measures the instantaneous rate of change of a function, or the rate of change of a function at any point along the function.

## Definition

The instantaneous rate of change    of a function $f\left(x\right)$ at a value $a$ is its derivative ${f}^{\prime }\left(a\right).$

## Chapter opener: estimating rate of change of velocity

Reaching a top speed of $270.49$ mph, the Hennessey Venom GT is one of the fastest cars in the world. In tests it went from $0$ to $60$ mph in $3.05$ seconds, from $0\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}100$ mph in $5.88$ seconds, from $0\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}200$ mph in $14.51$ seconds, and from $0\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}229.9$ mph in $19.96$ seconds. Use this data to draw a conclusion about the rate of change of velocity (that is, its acceleration ) as it approaches $229.9$ mph. Does the rate at which the car is accelerating appear to be increasing, decreasing, or constant?

First observe that $60$ mph = $88$ ft/s, $100$ mph $\approx 146.67$ ft/s, $200$ mph $\approx 293.33$ ft/s, and $229.9$ mph $\approx 337.19$ ft/s. We can summarize the information in a table.

$v\left(t\right)$ At different values of t
$t$ $v\left(t\right)$
$0$ $0$
$3.05$ $88$
$5.88$ $147.67$
$14.51$ $293.33$
$19.96$ $337.19$

Now compute the average acceleration of the car in feet per second on intervals of the form $\left[t,19.96\right]$ as $t$ approaches $19.96,$ as shown in the following table.

Average acceleration
$t$ $\frac{v\left(t\right)-v\left(19.96\right)}{t-19.96}=\frac{v\left(t\right)-337.19}{t-19.96}$
$0.0$ $16.89$
$3.05$ $14.74$
$5.88$ $13.46$
$14.51$ $8.05$

The rate at which the car is accelerating is decreasing as its velocity approaches $229.9$ mph $\text{(}337.19$ ft/s).

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fog(x)= f(g(x)) = x-2 = 3x+5-2 = 3x+3 f(x)/g(x)= x-2/3x+5
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