A Preview of Calculus: Learn It 3

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A Preview of Calculus: Learn It 3

The Tangent Problem and Differential Calculus Cont.

Calculating Instantaneous Velocity

Let’s continue our investigation by exploring a related application, instantaneous velocity. Velocity may be thought of as the rate of change of position.

Suppose that we have a function, $s(t)$ , that gives the position of an object along a coordinate axis at any given time $t$ . Can we use these same ideas to create a reasonable definition of the instantaneous velocity at a given time $t=a$ ?

We start by approximating the instantaneous velocity with an average velocity.

Recall that the speed of an object traveling at a constant rate is the ratio of the distance traveled to the length of time it has traveled.

We define the average velocity of an object over a time period to be the change in its position divided by the length of the time period.

average velocity

Let $s(t)$ be the position of an object moving along a coordinate axis at time $t$ . The average velocity of the object over a time interval $[a,t]$ where $a < t$ (or $[t,a]$ if $t < a$ ) is

v_{avg} = \frac{s (t) - s (a)}{t - a}

$v_{\text{avg}}=\dfrac{s(t)-s(a)}{t-a}$

As $t$ is chosen closer to $a$ , the average velocity becomes closer to the instantaneous velocity.

Note that finding the average velocity of a position function over a time interval is essentially the same as finding the slope of a secant line to a function. Furthermore, to find the slope of a tangent line at a point $a$ , we let the $x$ -values approach $a$ in the slope of the secant line.

To find the instantaneous velocity at time $a$ , we let the $t$ -values approach $a$ in the average velocity. This process of letting $x$ or $t$ approach $a$ in an expression is called taking a limit.

instantaneous velocity

For a position function $s(t)$ , the instantaneous velocity at a time $t=a$ is the value that the average velocities approach on intervals of the form $[a,t]$ and $[t,a]$ as the values of $t$ become closer to $a$ , provided such a value exists.

A rock is dropped from a height of $64$ ft. It is determined that its height (in feet) above ground $t$ seconds later (for $0\le t\le 2$ ) is given by $s(t)=-16t^2+64$ .

Find the average velocity of the rock over each of the given time intervals. Use this information to guess the instantaneous velocity of the rock at time $t=0.5$ .

$[0.49,0.5]$
$[0.5,0.51]$

Substitute the data into the formula for the definition of average velocity.

\begin{array}{rcl} s (0.5) & = & - 16 (0.5)^{2} + 64 = 60 \\ s (0.49) & = & - 16 (0.49)^{2} + 64 = 60.1584 \\ s (0.51) & = & - 16 (0.51)^{2} + 64 = 59.8384 \\ v_{avg} for interval [0.49, 0.5] & = & \frac{s (0.5) - s (0.49)}{0.5 - 0.49} = \frac{60 - 60.1584}{0.01} = - 15.84 \\ v_{avg} for interval [0.5, 0.51] & = & \frac{s (0.51) - s (0.5)}{0.51 - 0.5} = \frac{59.8384 - 60}{0.01} = - 16.16 \end{array}

$\begin{array}{rcl} s(0.5) &=& -16(0.5)^2 + 64 = 60 \\ s(0.49) &=& -16(0.49)^2 + 64 = 60.1584 \\ s(0.51) &=& -16(0.51)^2 + 64 = 59.8384 \\ v_{\text{avg}} \text{ for interval [0.49, 0.5]} &=& \frac{s(0.5) - s(0.49)}{0.5 - 0.49} = \frac{60 - 60.1584}{0.01} = -15.84 \\ v_{\text{avg}} \text{ for interval [0.5, 0.51]} &=& \frac{s(0.51) - s(0.5)}{0.51 - 0.5} = \frac{59.8384 - 60}{0.01} = -16.16 \end{array}$

The instantaneous velocity is somewhere between $−15.84$ and $−16.16$ ft/sec. A good guess might be $−16$ ft/sec.

Watch the following video to see the worked solution to this example.

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