$$
\text{Average velocity} = \frac{\text{Change in position}}{\text{Change in time}} = \frac{s(b) – s(a)}{b-a}.
$$
In words, the average velocity of an object over an interval is the net change in position during the interval divided by the change in time. The average velocity also corresponds to the slope of the line connecting the points corresponding to \(t = a\) and \(t = b\).
$$
\begin{array}{c}\text{Instantaneous velocity at } \\ t = a\end{array} = \lim_{h \rightarrow 0} \frac{s(a+h) – s(a)}{h}.
$$
In words, the instantaneous velocity of an object at time \(t = a\) is given by the limit of the average velocity over an interval, as the interval shrinks around \(a\). The instantaneous velocity also corresponds to the slop of the curve at \(t = a\).
$$
y = \frac{x}{1+x}.
$$
If \(Q\) is the point \((x,x/(1+x))\), use your calculator to find the slope of the secant line \(PQ\) (correct to six decimal places) for the following values of \(x = 0.5, 0.9, 0.99, 0.999, 1.5, 1.1, 1.01, 1.001\) and use these values to guess the value of the slope of the tangent line to the curve at \(P(1,1/2)\). Also, find an equation for this tangent line.
The slope of the secant line connecting \(P\) to \(Q\) is given by
$$
m_{sec} = \frac{y_2-y_1}{x_2-x_1} = \frac{\frac{1}{2} – \frac{x}{1+x}}{1-x}
$$
The outputs from the calculator for the different values of \(x\) is summarized in the table below:
$$
\begin{array}{c|c}
x & m_{sec} \\ \hline
.5 & 0.333333\\
0.9 & 0.263158\\
0.99 & 0.251256\\
0.999 & 0.250125\\
\\
1.001 & 0.249875\\
1.01 & 0.248756\\
1.1 & 0.238095\\
1.5 & 0.2
\end{array}
$$
The slopes of the secant lines seem to suggest that the slope of the tangent line at \(P(1,1/2)\) is \(1/4\). An equation for the tangent line to the graph of \(y = x/(1+x)\) at \(P\) is given by
$$
y = f(a) + f'(a)(x-a) = 1/2 +1/4(x-1) \text{ or } y = 1/4 x + 1/4
$$
$$
\begin{array}{|l| c | c|c|c|c|c|}
\hline
t (\text{seconds}) & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline
m (\text{meters}) & 0 & 1.4 & 5.1 & 10.7 & 17.7 & 25.8\\ \hline
\end{array}
$$
Find the average velocity for the time period \([1,3]\).
The average velocity of the cyclist from 1 to 3 seconds is given by
$$
s_{avg} = \frac{s(b)-s(a)}{b-a}
$$
where \(b = 3\) and \(a = 1\). So,
$$
s_{avg} = \frac{s(3)-s(1) }{3-1} = \frac{10.7 – 1.4}{2} = 4.65
$$
So the average velocity of the cyclist for the time period \([1,3]\) is \(4.65 m/s\).
The instantaneous velocity when \(t = 1\) is given by
$$
s'(1) = \lim_{h \rightarrow 0} \frac{s(1+h)-s(1)}{h}
$$
Since we are just estimating the instantaneous velocity, then under suitable conditions (i.e. \(s\) is differentiable at \(t = 1\)) we can make use of the following approximation
$$
s'(1) \approx \frac{ s(1+h)-s(1)}{h}
$$
where \(h\) is sufficiently close. We choose \(h = .001\). So,
$$
s'(1) \approx \frac{ s(1+.001)-s(1)}{.001} \approx -6.26837
$$
So, \(-6.26837\) is an estimate for the instantaneous velocity of the particle when \(t = 1\).