- 2.2 | The Derivative at a Point
- 2.3 | The Derivative of a Function
- 2.4 | Interpretations of the Derivative
- 2.5 | The Second Derivative
- 2.6 | Differentiability

If \(s(t)\) is the position of an object at time \(t\), then the

$$

\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 the object over the interval \(a \leq t \leq b\) is$$

\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 \(t = b\). Let \(s(t)\) be the position at time \(t\). Then the

$$

\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**at \(t= a\) is defined as$$

\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 \(t = a\).
The point \(P(1,1/2)\) lies on the curve

$$

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}

$$

$$

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

$$

The table shows the position of a cyclist.

$$

\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 \(4.65 m/s\).

$$

\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 displacement (in centimeters) of a particle moving back and forth along a straight line is given by the equation of motion \(s = 2 \sin \pi t + 3 \cos \pi t \), where \(t\) is measured in seconds. Estimate the instantaneous velocity of the particle when \(t = 1\).

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 \(t = 1\).

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 \(t = 1\).