2.1: Secant and Tangent Lines

$\newenvironment {prompt}{}{} \newcommand {\DeclareMathOperator }[2]{\@OldDeclareMathOperator {##1}{##2}\immediate \write \myfile {\unexpanded {\DeclareMathOperator }{\unexpanded {##1}}{\unexpanded {##2}}}} \newcommand {\reserved@a }[2]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument } \newcommand {\reserved@a }[1]{} \newcommand {\reserved@a }[2]{} \newcommand {\AppendGraphicsExtensions }[0]{\@ifundefined {Gin@extensions}{\let \Gin@extensions \@empty }{}\@ifstar {\grfext@Append \grfext@Check }{\grfext@Append \grfext@@Add }} \newcommand {\PrependGraphicsExtensions }[0]{\@ifundefined {Gin@extensions}{\let \Gin@extensions \@empty }{}\@ifstar {\grfext@Prepend \grfext@Check }{\grfext@Prepend \grfext@@Add }} \newcommand {\RemoveGraphicsExtensions }[1]{\@ifundefined {Gin@extensions}{\def \Gin@extensions {}}{\edef \grfext@tmp {\zap@space ##1 \@empty }\@for \grfext@ext :=\grfext@tmp \do {\def \grfext@next {\let \grfext@tmp \Gin@extensions \@expandtwoargs \@removeelement \grfext@ext \Gin@extensions \Gin@extensions \ifx \grfext@tmp \Gin@extensions \let \grfext@next \relax \fi \grfext@next }\grfext@next }}\grfext@Print \RemoveGraphicsExtensions } \newcommand {\epstopdfsetup }[0]{\setkeys {ETE}} \newcommand {\epstopdfcall }[1]{\ifETE@InsideSetfile \expandafter \@firstoftwo \else \expandafter \@secondoftwo \fi {‘##1}{\Gin@base \Gin@ext }} \newcommand {\epstopdfDeclareGraphicsRule }[4]{\ifx \\##4\\\@PackageError {epstopdf-base}{Conversion command is missing}\@ehc \else \begingroup \@ifundefined {Gin@rule@##1}{}{\@PackageInfo {epstopdf-base}{Redefining graphics rule for ‘##1’}}\endgroup \@namedef {Gin@rule@##1}####1{{##2}{##3}{\epstopdfcall {##4}}}\fi } \newcommand {\GetTitleStringSetup }[0]{\setkeys {gettitlestring}} \newcommand {\GetTitleString }[0]{\ifGTS@expand \expandafter \GetTitleStringExpand \else \expandafter \GetTitleStringNonExpand \fi } \newcommand {\GetTitleStringExpand }[1]{\def \GetTitleStringResult {##1}\begingroup \GTS@DisablePredefinedCmds \GTS@DisableHook \edef \x {\endgroup \noexpand \def \noexpand \GetTitleStringResult {\GetTitleStringResult }}\x } \newcommand {\GetTitleStringNonExpand }[1]{\def \GetTitleStringResult {##1}\global \let \GTS@GlobalString \GetTitleStringResult \begingroup \GTS@RemoveLeft \GTS@RemoveRight \endgroup \let \GetTitleStringResult \GTS@GlobalString } \newcommand {\GTS@DisableHook }[0]{} \newcommand {\label@hook }[0]{} \newcommand {\@currentlabelname }[0]{} \newcommand {\reserved@a }[0]{} \newcommand {\@safe@activestrue }[0]{} \newcommand {\@safe@activesfalse }[0]{} \newcommand {\nameref }[0]{\@ifstar \T@nameref \T@nameref } \newcommand {\Sectionformat }[2]{##1} \newcommand {\vnameref }[1]{\unskip ~\nameref {##1}\@vpageref [\unskip ]{##1}} \newcommand {\ref }[0]{\@ifstar \@refstar \T@ref } \newcommand {\pageref }[0]{\@ifstar \@pagerefstar \T@pageref } \newcommand {\nameref }[0]{\@ifstar \@namerefstar \T@nameref } \newcommand {\reserved@a }[2]{} \newcommand {\reserved@a }[0]{\AtBeginDocument } \newcommand {\reserved@a }[1]{} \newcommand {\reserved@a }[2]{}$

Take a look at the following video which recaps the ideas from the section.

Below is a video showing a worked example.

Given a function $f(x)$ and two points $P$ and $Q$ on $y=f(x)$ corresponding to $x=a$ and $x=b$ , respectively, the secant line is the line connecting $P$ and $Q$ on the graph, and its slope is the average rate of change of $f(x)$ on the interval $[a,b]$ .

Consider $f(x) = x^3-x$ . Find:

The average rate of change of $f(x)$ on the interval $[1,3]$ .
This is the slope of the line connecting the two points on the graph of $f(x) = x^3-x$ at $x=1$ and $x=3$ . We can first get the $y$ -coordinates of these points. Notice $f(1) = \answer {0}, \quad f(3) = \answer {24}.$ We now compute the slope of the line through $(1,\answer {0})$ and $(3,\answer {24})$ . Computing $\frac {\Delta y}{\Delta x}$ , we get an answer of $\answer {12}$ .
The equation of the secant line connecting the points on $f(x) = x^3-x$ with $x=1$ and $x=3$ .
We already know the $y$ -coordinates from the previous part, as well as the slope of the line. Using point-slope form of a line, we get an equation of $y - \answer {0} = \answer {12}(x-1),$ which we can simplify to $y = \answer {12x-12}$ .

Suppose a car starts moving at noon and moves along a straight road. The distance traveled by the car $t$ hours after noon is given by $f(t) = 5t^2+10t$ miles.

How far has the car moved after $3$ hours? $\answer {75}$ mileshoursmiles/hour .
What was the average speed of the car over the first two hours of the trip? $\answer {20}$ mileshoursmiles/hour .
To calculate the average speed of the car, we do the change in distance divided by the change in time (notice, also, that this is the slope of the secant line connecting the points $t=0$ and $t=2$ on the graph of $f(t)$ ). We compute: $f(0) = \answer {0}, \quad f(2) = \answer {40},$ so the average speed is $\frac {f(2)-f(0)}{2-0} = \answer {20}.$

Sam is busy gardening in the Shire and records the temperature at different points of the day. He has recorded the following data: $\begin {array}{|c|c|c|c|c|c|} \hline Time & 9AM & 10AM & 11AM & 12PM & 1PM \\ \hline Temperature (^\circ F) & 70 & 72 & 74 & 75 & 74 \\ \hline \end {array}$ Let $T(t)$ be the temperature at time $t$ , where $t$ is in hours after 9:00 AM.

Based on the table, what is $T(3)$ ? $\answer {75}$ hoursdegrees Fahrenheitdegrees Fahrenheit per hour .
Is the slope of the secant line through the points $(3,T(3))$ and $(4,T(4))$ positive, negative or zero? positivenegativezero
Is the slope of the secant line through $(2,T(2))$ and $(4,T(4))$ positive, negative, or zero? positivenegativezero
What is the average rate of change of the temperature between 9 AM and 11 AM? $\answer {2}$ hoursdegrees Fahrenheitdegrees Fahrenheit per hour .

Find the slope of the tangent line to the curve $y = x^3$ at $x=1$ . You can use a calculator to help you.
Let $f(x) = x^3$ . To calculate the slope of the tangent line, we will calculate the slope of the secant line through $(1,f(1))$ other points $Q$ and then let $Q$ get closer and closer to $x = 1$ . For example, if we take the second point to be $(2,f(2))$ , then the slope of the secant line is $\frac {f(2)-f(1)}{2-1} = \answer {7}.$ If we take the second point to be $(1.1,f(1.1))$ , then the slope of the secant line is (use a calculator) $\frac {f(1.1)-f(1)}{1.1-1} = \frac {\answer {.331}}{.1} = \answer {3.31}.$ Similarly, if the second point is $(1.01,f(1.01))$ , then the slope of the secant line is (round to four digits after the decimal point) $\frac {f(1.01)-f(1)}{1.01-1} = \answer {3.0301}.$ If the second point is $(1.001,f(1.001))$ , then the slope of the secant line is (round to four digits after the decimal point) $\frac {f(1.001)-f(1)}{1.001-1} = \answer {3.003},$ and if the second point is $(1.0001, f(1.0001))$ , then the slope of the secant line is (round to four digits after the decimal point) $\frac {f(1.0001)-f(1)}{1.0001-1} = \answer {3.0003}.$ Looking at all these slopes, it appears that as the second point gets closer to $(1,f(1))$ , the slopes are approaching $\answer {3}$ . This is the slope of the tangent line to $y=x^3$ at $x=1$ .
What is the equation of the tangent line to $y = x^3$ at $x=1$ ?
From the previous part, we know the slope is $3$ . We also have the point $(1,1)$ on the curve. Using point-slope form of a line, we get $y - 1 = \answer {3}(x-\answer {1}),$ and we can (if we want) simplify this to $y = \answer {3x-2}.$

Suppose $f(x)$ is a function with $f(2) = -5$ . Suppose the slopes of the secant lines connecting $(2,f(2))$ and $(2+h,f(2+h))$ for different values of $h$ are shown in the table below: $\begin {array}{c|c|c|c|c} h & .1 & .01 & .001 & .0001 \\ \hline slope & 4.1 & 4.01 & 4.001 & 4.0001 \\ \end {array}$

Based on this table, the slope of the tangent line to $y=f(x)$ at $x=2$ appears to be $\answer {4}$ .
The equation of the tangent line to $y=f(x)$ at $x=2$ is $y - \answer {-5} = \answer {4}(x-2),$ which simplifies to $y = \answer {4x-13}.$

Consider the function $f(x) = 3x+2$ , and let $P$ be the point $(1,5)$ .

If $Q$ is any other point on the graph of $y = f(x)$ , then the secant line through $P$ and $Q$ has what slope? $\answer {3}$
What is the slope of the tangent line to $y = f(x)$ at the point $P$ ? $\answer {3}$ .
What is the equation of the tangent line to $y = f(x)$ at the point $P$ ? $y = \answer {3x+2}$ .