Activity 3.2 Tangent Lines

In previous activities we saw that if $p$ is a position function, then the difference quotient for $p$ can be used to calculate average velocities and the expression $\lim\limits_{h \to 0}\frac{\fe{p}{t_0+h} - \fe{p}{t_0}}{h}$ calculates the instantaneous velocity at time $t_0\text{.}$

Graphically, the difference quotient of a function $f$ can be used to calculate the slope of secant lines to $f\text{.}$ What happens when we take the run of the secant line to zero? Basically, we are connecting two points on the curve that are really, really, (really), close to one another. As mentioned above, sending $h$ to zero turns an average velocity into an instantaneous velocity. Graphically, sending $h$ to zero turns a secant line into a tangent line.

Example 3.2.1.

The tangent line at $x = 4$ to the function $f$ defined by $\fe{f}{x} = \sqrt{x}$ is shown in Figure 3.2.2. Here is a calculation of the slope of this line.

\begin{align*} m_{\text{tan}} \amp= \lim_{h \to 0}\frac{\fe{f}{4+h} - \fe{f}{4}}{h}\\ \amp= \lim_{h \to 0}\frac{\sqrt{4+h} - \sqrt{4}}{h}\\ \amp= \lim_{h \to 0}\left(\frac{\sqrt{4+h} - 2}{h}\cdot\frac{\sqrt{4+h} + 2}{\sqrt{4+h} + 2}\right)\\ \amp= \lim_{h \to 0}\frac{4+h-4}{h\left(\sqrt{4+h} + 2\right)}\\ \amp= \lim_{h \to 0}\frac{h}{h\left(\sqrt{4+h} + 2\right)}\\ \amp= \lim_{h \to 0}\frac{1}{1\left(\sqrt{4+h} + 2\right)}\\ \amp= \frac{1}{\sqrt{4+0} + 2}\\ \amp= \frac{1}{4} \end{align*}

the curve y=sqrt(x) is plotted; beginning with a solid dot at (0,0), this curve briefly moves more upward than rightward, but quickly moves more and more rightward; it exits the graph on its right edge at about (11,3.3); the point (4,2) is on the curve, and has a solid dot to emphasize this; a straight line has been superimposed onto the graph that passes through (4,2), and is perfectly tangent to the curve; all of the curve is below the line, except for at the point (4,2) where they touch — Figure 3.2.2. $y = \fe{f}{x}$ and its tangent line at $x = 4$

You should verify that the slope of the tangent line shown in Figure 3.2.2 is indeed $\frac{1}{4}\text{.}$ You should also verify that the equation of the tangent line is $y = \frac{1}{4}x + 1\text{.}$

Exercises Exercises

Exercise Group.

Consider the function $g$ given by $\fe{g}{x} = 5 - \sqrt{4-x}$ graphed in the figure.

$a curve is plotted; this description will describe it from right to left; beginning with a solid dot at (4,5), this curve briefly moves more downward than leftward, but quickly moves more and more leftward; it exits the graph on its left edge at about (-7,1.7); the point (3,4) is on the curve, and has a solid dot to emphasize this; a straight line has been superimposed onto the graph that passes through (3,4), and is perfectly tangent to the curve; all of the curve is above the line, except for at the point (3,4) where they touch$

png svg pdf tex

$y = \fe{g}{x}$ and its tangent line at $x = 3$

1.

Find the slope of the tangent line shown in the figure using

\begin{equation*} m_{\text{tan}} = \lim_{h \to 0}\frac{\fe{g}{3+h} - \fe{g}{3}}{h}\text{.} \end{equation*}

Show work consistent with that illustrated in Example 3.2.1.

2.

Use the line in the figure to verify your answer to Exercise 3.2.1.

3.

State the equation of the tangent line to $g$ at $x = 3\text{.}$

Prev Top Next