A Function F is Continuous on 1 5 and Some of the Values of F Are Shown Below X1 4 5

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Before we embark on setting the groundwork for the derivative of a function, let's review some terminology and concepts. Remember that the slope of a line is defined as the quotient of the difference in y-values and the difference in x-values. Recall from Section 1.2 that a difference between two quantities is often denoted by the Greek symbol $\Delta$ - read "delta" as shown next, where delta notation is being used when calculating and interpreting the slope of a line.

Calculating and Interpreting the Slope of a Line.

Suppose we are given two points $\left(x_{1},y_{1}\right)$ and $\left(x_{2},y_{2}\right)$ on the line of a linear function $y = f(x)\text{.}$ Then the slope of the line is calculated by

\begin{equation*} m = \dfrac{\Delta y}{\Delta x} = \dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}\text{.} \end{equation*}

We can interpret this equation by saying that the slope $m$ measures the change in $y$ per unit change in $x\text{.}$ In other words, the slope $m$ provides a measure of sensitivity .

For example, if $y = 100x + 5\text{,}$ a small change in $x$ corresponds to a change one hundred times as large in $y\text{,}$ so $y$ is quite sensitive to changes in $x\text{.}$

Next, we introduce the properties of two special lines, the tangent line and the secant line, which are pertinent for the understanding of a derivative.

Secant Line.

Secant is a Latin word meaning to cut, and in mathematics a secant line cuts an arbitrary curve described by $y = f(x)$ through two points $P$ and $Q\text{.}$ The figure shows two such secant lines of the curve $f$ to the right and to the left of the point $P\text{,}$ respectively.

Since by necessity the secant line goes through two points on the curve of $y = f(x)\text{,}$ we can readily calculate the slope of this secant line.

Definition 4.1. Slope of Secant Line — Average Rate of Change.

Suppose we are given two points $\left(x_{1},y_{1}\right)$ and $\left(x_{2},y_{2}\right)$ on the secant line of the curve described by the function $y = f(x)$ as shown. Then the slope of the secant line is calculated by

\begin{equation*} m = \dfrac{\Delta y}{\Delta x} = \dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}\text{.} \end{equation*}

Note that we may also be given the change in $x$ directly as $\Delta x\text{,}$ i.e the two points are given as $\left(x,f(x)\right)$ and $\left(x+\Delta x, f(x + \Delta x)\right)\text{,}$ and so

\begin{equation*} m = \dfrac{\Delta y}{\Delta x} = \dfrac{f(x + \Delta x)-f(x)}{\Delta x}\text{.} \end{equation*}

Note:

In the above figure, the value of $\Delta x$ must be negative since it is on the left side of $x\text{.}$
The slope of the secant line is also referred to as the average rate of change of $f$ over the interval $\left[x,x+\Delta x\right]\text{.}$
The expression $\dfrac{f(x + \Delta x) - f(x)}{\Delta x}$ is referred to as the difference quotient.

Tangent Line.

Tangent is a Latin word meaning to touch, and in mathematics a tangent line touches an arbitrary curve described by $y = f(x)$ at a point $P$ but not any other points nearby as shown.

Since by definition the tangent line only touches one point on the curve of $y = f(x)\text{,}$ we cannot calculate the slope of this tangent line with our slope formula for a line. In fact, up to now, we do not have any way of calculating this slope unless we are able to use some geometry.

Suppose that $y$ is a function of $x\text{,}$ say $y = f(x)\text{.}$ Since it is often useful to know how sensitive the value of $y$ is to small changes in $x\text{,}$ let's explore this concept via an example, and see how this will inform us about the calculation of the slope of the tangent line.

Example 4.2. Small Changes in $x$.

Consider $\ds y=f(x)=\sqrt{625-x^2}$ (the upper semicircle of radius 25 centered at the origin), and let's compute the changes of $y$ resulting from small changes of $x$ around $x=7\text{.}$

Solution

When $x=7\text{,}$ we find that $\ds y=\sqrt{625-49}=24\text{.}$ Suppose we want to know how much $y$ changes when $x$ increases a little, say to 7.1 or 7.01.

Let us look at the ratio $\Delta y/\Delta x$ for our function $\ds y=f(x)=\sqrt{625-x^2}$ when $x$ changes from 7 to $7.1\text{.}$ Here $\Delta x=7.1-7=0.1$ is the change in $x\text{,}$ and

\begin{align*} \Delta y =f(x+\Delta x)-f(x)\amp =f(7.1)-f(7)\\ \amp =\sqrt{625-7.1^2}-\sqrt{625-7^2}\\ \amp \approx 23.9706-24=-0.0294\text{.} \end{align*}

Thus, $\Delta y/\Delta x\approx -0.0294/0.1=-0.294\text{.}$ This means that $y$ changes by less than one third the change in $x\text{,}$ so apparently $y$ is not very sensitive to changes in $x$ at $x=7\text{.}$ We say "apparently" here because we don't really know what happens between 7 and value at $7\text{.}$ This is not in fact the case for this particular function, but we don't yet know why.

The quantity $\Delta y/\Delta x\approx -0.294$ may be interpreted as the slope of the secant line through $(7,24)$ and $(7.1,23.9706)\text{.}$ In general, if we draw the secant line from the point $(7,24)$ to a nearby point on the semicircle $(7+\Delta x,\,f(7+\Delta x))\text{,}$ the slope of this secant line is the so-called difference quotient

\begin{equation*} \frac{f(7+\Delta x)-f(7)}{\Delta x}= \frac{\sqrt{625-(7+\Delta x)^2}-24}{\Delta x}\text{.} \end{equation*}

For example, if $x$ changes only from 7 to 7.01, then the difference quotient (slope of the secant line) is approximately equal to $(23.997081-24)/0.01=-0.2919\text{.}$ This is slightly different than for the secant line from $(7,24)$ to $(7.1,23.9706)\text{.}$

As $\Delta x$ is made smaller (closer to 0), $7+\Delta x$ gets closer to 7 and the secant line joining $(7,f(7))$ to $(7+\Delta x,f(7+\Delta x))$ shifts slightly, as shown in Figure 4.1. The secant line gets closer and closer to the tangent line to the circle at the point $(7,24)\text{.}$ (The tangent line is the line that just grazes the circle at that point, i.e., it doesn't meet the circle at any second point.) Thus, as $\Delta x$ gets smaller and smaller, the slope $\Delta y/\Delta x$ of the secant line gets closer and closer to the slope of the tangent line. This is actually quite difficult to see when $\Delta x$ is small, because of the scale of the graph. The values of $\Delta x$ used for the figure are $1\text{,}$ $5\text{,}$ $10$ and $15\text{,}$ not really very small values. The tangent line is the one that is uppermost at the right hand endpoint.

Figure 4.1. Secant lines approximating the tangent line.

So far we have found the slopes of two secant lines that should be close to the slope of the tangent line, but what is the slope of the tangent line exactly? Since the tangent line touches the circle at just one point, we will never be able to calculate its slope directly, using two "known" points on the line. What we need is a way to capture what happens to the slopes of the secant lines as they get "closer and closer" to the tangent line.

Instead of looking at more particular values of $\Delta x\text{,}$ let's see what happens if we do some algebra with the difference quotient using just $\Delta x\text{.}$ The slope of a secant line from $(7,24)$ to a nearby point $\left(7+\Delta x,f(7+\Delta x)\right)$ is given by

\begin{align*} \frac{f(7+\Delta x)-f(7)}{\Delta x}\amp =\frac{\sqrt{625-(7+\Delta x)^2} - 24}{\Delta x}\\ \amp = \left( \ds\frac{\sqrt{625-(7+\Delta x)^2} - 24}{\Delta x} \right) \left(\frac{\sqrt{625-(7+\Delta x)^2}+24}{\sqrt{625-(7+\Delta x)^2}+24} \right)\\ \amp =\frac{625-(7+\Delta x)^2-24^2}{\Delta x(\sqrt{625-(7+\Delta x)^2}+24)}\\ \amp =\frac{49-49-14\Delta x-\Delta x^2}{\Delta x(\sqrt{625-(7+\Delta x)^2}+24)}\\ \amp =\frac{625-(7+\Delta x)^2-24^2}{\Delta x(\sqrt{625-(7+\Delta x)^2}+24)}\\ \amp =\frac{49-49-14\Delta x-\Delta x^2}{\Delta x(\sqrt{625-(7+\Delta x)^2}+24)}\\ \amp =\frac{\Delta x(-14-\Delta x)}{\Delta x(\sqrt{625-(7+\Delta x)^2}+24)}\\ \amp =\frac{-14-\Delta x}{\sqrt{625-(7+\Delta x)^2}+24} \end{align*}

Now, can we tell by looking at this last formula what happens when $\Delta x$ gets very close to zero? The numerator clearly gets very close to $-14$ while the denominator gets very close to $\ds \sqrt{625-7^2}+24=48\text{.}$ The fraction is therefore very close to $-14/48 = -7/24 \cong -0.29167\text{.}$ In fact, the slope of the tangent line is exactly $-7/24\text{.}$

What about the slope of the tangent line at $x=12\text{?}$ Well, 12 can't be all that different from 7; we just have to redo the calculation with 12 instead of 7. This won't be hard, but it will be a bit tedious. What if we try to do all the algebra without using a specific value for $x\text{?}$ Let's copy from above, replacing 7 by $x\text{.}$

\begin{align*} \frac{f(x+\Delta x)-f(x)}{\Delta x}\amp =\frac{\sqrt{625-(x+\Delta x)^2} - \sqrt{625-x^2}}{\Delta x}\\ \amp =\frac{\sqrt{625-(x+\Delta x)^2} - \sqrt{625-x^2}}{\Delta x} \frac{\sqrt{625-(x+\Delta x)^2}+\sqrt{625-x^2}}{\sqrt{625-(x+\Delta x)^2}+\sqrt{625-x^2}}\\ \amp =\frac{625-(x+\Delta x)^2-625+x^2}{\Delta x(\sqrt{625-(x+\Delta x)^2}+\sqrt{625-x^2})}\\ \amp =\frac{625-x^2-2x\Delta x-\Delta x^2-625+x^2}{\Delta x(\sqrt{625-(x+\Delta x)^2}+\sqrt{625-x^2})}\\ \amp =\frac{\Delta x(-2x-\Delta x)}{\Delta x(\sqrt{625-(x+\Delta x)^2}+\sqrt{625-x^2})}\\ \amp =\frac{-2x-\Delta x}{\sqrt{625-(x+\Delta x)^2}+\sqrt{625-x^2}} \end{align*}

Now what happens when $\Delta x$ is very close to zero? Again it seems apparent that the quotient will be very close to

\begin{equation*} \frac{-2x}{\sqrt{625-x^2}+\sqrt{625-x^2}} =\frac{-2x}{2\sqrt{625-x^2}}=\frac{-x}{\sqrt{625-x^2}}\text{.} \end{equation*}

Replacing $x$ by 7 gives $-7/24\text{,}$ as before, and now we can easily do the computation for 12 or any other value of $x$ between $-25$ and 25.

So now we have a single expression, $\ds \dfrac{-x}{\sqrt{625-x^2}}\text{,}$ that tells us the slope of the tangent line for any value of $x\text{.}$ This slope, in turn, tells us how sensitive the value of $y$ is to small changes in the value of $x\text{.}$

To summarize, we computed the slope of the tangent line at a point $P = \left(x, f(x)\right)$ on the curve of a function $y = f(x)$ by forming the difference quotient and figuring out what happens when $\Delta x$ gets very close to $0\text{.}$ At this point, we should note that the idea of letting get closer and closer to $0$ is precisely the idea of a limit that we discussed in the last chapter. This leads us to the following definition.

Definition 4.3. Slope of Tangent Line—Instantaneous Rate of Change.

The slope of the tangent line to the graph of a function $y = f(x)$ at the point $P = \left(x,f(x)\right)$ is given by

\begin{equation*} m = \lim_{\Delta x \to 0} \dfrac{f(x + \Delta x)-f(x)}{\Delta x}\text{,} \end{equation*}

provided this limit exists.

Note: The slope of the tangent line is also referred to as the insantaneous rate of change of $f$ at $x\text{.}$

The expression$\ds \dfrac{-x}{\sqrt{625-x^2}}$ defines a new function called the derivative (see Section 4.2) of the original function (since it is derived from the original function). If the original is referred to as $f$ or $y$ then the derivative is often written as $f'$ or $y'\text{,}$ read "f prime" or "y prime". We also write

\begin{equation*} f'(x)=\dfrac{-x}{\sqrt{625-x^2}} \text{ or } y'=\ds \dfrac{-x}{\sqrt{625-x^2}}\text{.} \end{equation*}

At a particular point, say $x=7\text{,}$ we write $f'(7)=-7/24$ and we say that "$f$ prime of 7 is $-7/24$" or "the derivative of $f$ at 7 is $-7/24\text{.}$"

In the particular case of a circle, there's a simple way to find the derivative. Since the tangent to a circle at a point is perpendicular to the radius drawn to the point of contact, its slope is the negative reciprocal of the slope of the radius. The radius joining $(0,0)$ to $(7,24)$ has slope 24/7. Hence, the tangent line has slope $-7/24\text{.}$ In general, a radius to the point $\ds (x,\sqrt{625-x^2})$ has slope $\ds \sqrt{625-x^2}/x\text{,}$ so the slope of the tangent line is $\ds {-x/ \sqrt{625-x^2}}\text{,}$ as before. It is NOT always true that a tangent line is perpendicular to a line from the origin—don't use this shortcut in any other circumstance.

We now summarize our findings.

From Tangent Line Slope to Derivative.

Given a function $f$ and a point $x$ we can compute the derivative of $f(x)$ at $x$ as follows:

Form the difference quotient $\dfrac{f(x+\Delta x)-f(x)}{\Delta x}\text{,}$ which is the slope of a general secant line of the curve $f$ throught the points $P = \left(x,f(x)\right)$ and $Q = \left(x+\Delta x, f(x+\Delta x)\right)\text{.}$
Take the limits as $\Delta x$ goes to zero: $\lim\limits_{\Delta x \to 0} \dfrac{f(x+\Delta x)-f(x)}{\Delta x}\text{,}$ which is the slope of the tangent line of the curve $f$ at the point $P = \left(x,f(x)\right)\text{.}$
If this limit exists, then the derivative exists and is equal to this limit.

In other words,

\begin{equation*} f'(x)=\lim\limits_{\Delta x \to 0} \dfrac{f(x+\Delta x)-f(x)}{\Delta x}\text{,} \end{equation*}

provided the limit exists.

Interactive Demonstration. Use the sliders below to investigate the limit of secant lines (blue and green) as they approach the tangent line (red) to any point $P $ on the graph:

Subsection 4.1.1 Applications

We started this section by saying, "It is often useful to know how sensitive the value of $y$ is to small changes in $x\text{.}$" We have seen one purely mathematical example of this, involving the function $f(x)=\sqrt{625-x^2}\text{.}$ Here are some more applied examples.

With careful measurement it might be possible to discover that the height of a dropped ball $t$ seconds after it is released is $\ds h(t)=h_0-kt^2\text{.}$ (Here $h_0$ is the initial height of the ball, when $t=0\text{,}$ and $k$ is some number determined by the experiment.) A natural question is then, "How fast is the ball going at time $t\text{?}$" We can certainly get a pretty good idea with a little simple arithmetic.

Example 4.4. Analyzing Velocity.

Suppose that the height $h$ in metres of a dropped ball $t$ seconds after it is released is given by

\begin{equation*} h(t) = h_{0} - kt^{2} \end{equation*}

with $h_{0} = 100$ m and $k = 4.9\text{.}$ We will answer the question "How fast is the ball travelling at time $t = 2\text{?}$" by exploring average speed near time $t = 2$ and discussing the difference between speed and velocity.

Solution

We know that when $t=2$ the height is $100-4\cdot 4.9=80.4$ metres. A second later, at $t=3\text{,}$ the height is $100-9\cdot 4.9=55.9$ metres. The change in height during that second is $55.9-80.4=-24.5$ metres. The negative sign means the height has decreased, as we expect for a falling ball, and the number 24.5 is the average speed of the ball during the time interval, in metres per second.

We might guess that 24.5 metres per second is not a terrible estimate of the speed at $t=2\text{,}$ but certainly we can do better. At $t=2.5$ the height is $\ds 100-4.9(2.5)^2=69.375$ metres. During the half second from $t=2$ to $t=2.5\text{,}$ the change in height is $69.375-80.4=-11.025$ metres giving an average speed of $11.025/(1/2)=22.05$ metres per second. This should be a better estimate of the speed at $t=2\text{.}$ So it's clear now how to get better and better approximations: compute average speeds over shorter and shorter time intervals. Between $t=2$ and $t=2.01\text{,}$ for example, the ball drops 0.19649 metres in one hundredth of a second, at an average speed of 19.649 metres per second.

We still might reasonably ask for the precise speed at $t=2$ (the instantaneous speed) rather than just an approximation to it. For this, once again, we need a limit. Let's calculate the average speed during the time interval from $t=2$ to $t=2+\Delta t$ without specifying a particular value for $\Delta t\text{.}$ The change in height during the time interval from $t=2$ to $t=2+\Delta t$ is

\begin{align*} h(2+\Delta t)-h(2) \amp =(100-4.9(2+\Delta t)^2)-80.4\\ \amp =100-4.9(4+4\Delta t+\Delta t^2)-80.4\\ \amp =100-19.6-19.6\Delta t-4.9\Delta t^2-80.4\\ \amp =-19.6\Delta t-4.9\Delta t^2\\ \amp =-\Delta t(19.6+4.9\Delta t) \end{align*}

The average speed during this time interval is then

\begin{equation*} \frac{\Delta t(19.6+4.9\Delta t)}{\Delta t}=19.6+4.9\Delta t\text{.} \end{equation*}

When $\Delta t$ is very small, this is very close to 19.6. Indeed, $\lim\limits_{\Delta t\to 0}\left(19.6+4.9\Delta t\right)=19.6\text{.}$ So the exact speed at $t=2$ is 19.6 metres per second.

At this stage we need to make a distinction between speed and velocity. Velocity is signed speed, that is, speed with a direction indicated by a sign (positive or negative). Our algebra above actually told us that the instantaneous velocity of the ball at $t=2$ is $-19.6$ metres per second. The number 19.6 is the speed and the negative sign indicates that the motion is directed downwards (the direction of decreasing height).

In the language of the previous section, we might have started with $\ds f(x)=100-4.9x^2$ and asked for the slope of the tangent line at $x=2\text{.}$ We would have answered that question by computing

\begin{equation*} \begin{split} \lim_{\Delta x\to 0}\frac{f(2+\Delta x) - f(2)}{\Delta x} \amp=\lim_{\Delta x\to 0}\frac{-19.6\Delta x-4.9\Delta x^2}{\Delta x}\\ \amp=\lim_{\Delta x\to 0} \left(-19.6-4.9\Delta x\right)=-19.6 \end{split} \end{equation*}

The algebra is the same. Thus, the velocity of the ball is the value of the derivative of a certain function, namely, of the function that gives the position of the ball.

The upshot is that this problem, finding the velocity of the ball, is exactly the same problem mathematically as finding the slope of a curve. This may already be enough evidence to convince you that whenever some quantity is changing (the height of a curve or the height of a ball or the size of the economy or the distance of a space probe from earth or the population of the world) the rate at which the quantity is changing can, in principle, be computed in exactly the same way, by finding a derivative.

Example 4.5. Demand for Sweaters.

A clothing manufacturer has determined that the weekly demand function of their sweaters is given by

\begin{equation*} p = f(q) = 144-q^{2} \end{equation*}

where $p$ is measured in dollars and $q$ is measured in units of a thousand. Find the average rate of change in the unit price of a sweater if the quantity demanded is between $5000$ and $6000$ sweaters, between $5000$ and $5100$ sweaters, and between $5000$ and $5010$ sweaters. What is the instantaneous rate of change of the unit price when the quantity demanded is $5000$ units?

Solution

The average rate of change of the unit price of a sweater if the quantity demanded is between $q$ and $q + \Delta q$ is

\begin{equation*} \begin{split} \dfrac{f(q+\Delta q)-f(q)}{\Delta q} \amp = \dfrac{\left(144 - (q+\Delta q)^{2}\right) - \left(144 - q^{2}\right)}{\Delta q} \\ \amp = \dfrac{144 - q^{2} - 2q\Delta q - \Delta q^{2} - 144 + q^{2}}{\Delta q} \\ \amp = -2q - \Delta q \end{split} \end{equation*}

To find the average rate of change of the unit price of a sweater when the quantity demanded is between $5000$ and $6000$ sweaters (that is, over the interval $\left[5,6\right]$), we take $q = 5$ and $\Delta q = 1\text{,}$ obtaining

\begin{equation*} -2(5)-1 = -11 \end{equation*}

or -$$11$ per $1000$ sweaters. Similarly, taking $\Delta q = 0.1$ and $\Delta q = 0.01$ with $q=5\text{,}$ we find that the average rates of change of the unit price when the quantities demanded are between $5000$ and $5100$ and between $5000$ and $5010$ are -$$10.10$ and -$$10.01$ per $1000$ sweaters, respectively.

The instantaneous rate of change of the unit price of a sweater when the quantity demanded is $q$ units is given by

\begin{equation*} \begin{split} \lim_{\Delta q\to 0} \dfrac{f(q+\Delta q)-f(q)}{\Delta q} \amp = \lim_{\Delta q \to 0} \dfrac{\left(144 - (q+\Delta q)^{2}\right) - \left(144 - q^{2}\right)}{\Delta q} \\ \amp = \lim_{\Delta q \to 0} \left(-2q - \Delta q\right) \\ \amp = -2q \end{split} \end{equation*}

In particular, the instantaneous rate of change when the quantity demanded is $5000$ sweaters is

\begin{equation*} -2(5) = -10 \end{equation*}

or -$$10$ per $1000$ sweaters.

Exercises for Section 4.1.

Exercise 4.1.1.

Draw the graph of the function $\ds y=f(x)=\sqrt{169-x^2}$ between $x=0$ and $x=13\text{.}$ Find the slope $\Delta y/\Delta x$ of the secant line between the points of the circle lying over (a) $x=12$ and $x=13\text{,}$ (b) $x=12$ and $x=12.1\text{,}$ (c) $x=12$ and $x=12.01\text{,}$ (d) $x=12$ and $x=12.001\text{.}$ Now use the geometry of tangent lines on a circle to find (e) the exact value of the derivative $f'(12)\text{.}$ Your answers to (a)–(d) should be getting closer and closer to your answer to (e).

Answer

$-5\text{,}$ $-2.47106145\text{,}$ $-2.4067927\text{,}$ $-2.400676\text{,}$ $-2.4$

Solution

The graph of $y=f(x)=\sqrt{169-x^2}$ is a semi-circle of radius 13. We plot $f(x)$ below on the interval $[0,13]\text{:}$

We wish to estimate $f'(12)$ first using secant lines. Therefore, we set up the difference quotient at $x=12\text{:}$

\begin{equation*} \frac{\Delta y}{\Delta x} = \frac{f(12+\Delta x) - f(12)}{\Delta x} = \frac{\sqrt{162-(12+\Delta x)^2} - 5}{\Delta x} \end{equation*}

The slope of the secant line between $x=12$ and $x=13$ ($\Delta x = 1$) is given by

\begin{equation*} \frac{\Delta y}{\Delta x} = \frac{0-5}{1} = -5\text{.} \end{equation*}
The slope of the secant line between $x=12$ and $x=12.1$ ($\Delta x = 0.1$) is given by

\begin{equation*} \frac{\Delta y}{\Delta x} \approx \frac{4.75289 - 5}{0.1} = -2.471\text{.} \end{equation*}
The slope of the secant line between $x=12$ and $x=12.01$ ($\Delta x = 0.01$) is given by

\begin{equation*} \frac{\Delta y}{\Delta x} \approx \frac{4.97593 - 5}{0.01} = -2.407\text{.} \end{equation*}
A property of circles is that the tangent line at any point on the circle is perpendicular the the line connecting that point to the centre of the circle (in this case, the origin):

The slope of the line connecting the point $(12,5)$ to the origin is
\begin{equation*} \frac{\Delta y}{\Delta x} = \frac{5-0}{12 - 0} = \frac{5}{12}\text{.} \end{equation*}

Hence, the slope of the tangent line at $x=12$ must be

\begin{equation*} -\frac{1}{\frac{5}{12}} = -\frac{12}{5} = 2.4\text{.} \end{equation*}

We further notice that our answers to parts (a)-(d) are getting closer and closer to this value. Hence, $f'(12) = 2.4\text{.}$

Exercise 4.1.2.

Use geometry to find the derivative $f'(x)$ of the function $\ds f(x)=\sqrt{625-x^2}$ in the text for each of the following $x\text{:}$ (a) 20, (b) 24, (c) $-7\text{,}$ (d) $-15\text{.}$ Draw a graph of the upper semicircle, and draw the tangent line at each of these four points.

Answer

$-4/3\text{,}$ $-24/7\text{,}$ $7/24\text{,}$ $3/4$

Solution

The graph of $y=f(x)=\sqrt{625-x^2}$ is a semi-circle of radius 25. We plot $f(x)$ below on the interval $[-25,25]\text{:}$

We will use the fact that the tangent line at any point on the circle is perpendicular the the line connecting that point to the centre of the circle. For any point $(x, y) = (x, \sqrt{625-x^2})\text{,}$ the slope of the line connecting the origin to the point is given by

\begin{equation*} \frac{\Delta y}{\Delta x} = \frac{\sqrt{625-x^2}}{x}\text{,} \end{equation*}

and so the slope of the tangent line at this point must be the reciprocal of this value:

\begin{equation*} f'(x) = - \frac{x}{\sqrt{625-x^2}} \end{equation*}

At $x=20\text{,}$ we have

\begin{equation*} f'(20) = - \frac{20}{\sqrt{625-20^2}} = -\frac{20}{15} = -\frac{4}{3}\text{.} \end{equation*}
At $x=24\text{,}$ we have

\begin{equation*} f'(24) = - \frac{24}{\sqrt{625-24^2}} = -\frac{24}{7}\text{.} \end{equation*}
At $x=-7\text{,}$ we have

\begin{equation*} f'(-7) = -\frac{-7}{\sqrt{625-(-7)^2}} = \frac{7}{24}\text{.} \end{equation*}
At $x=-15\text{,}$ we have

\begin{equation*} f'(-15) = -\frac{-15}{\sqrt{625-(-15)^2}} = \frac{15}{20} = \frac{3}{4}\text{.} \end{equation*}

We illustrate with the following graph.

Exercise 4.1.3.

Draw the graph of the function $y=f(x)=1/x$ between $x=1/2$ and $x=4\text{.}$ Find the slope of the secant line between (a) $x=3$ and $x=3.1\text{,}$ (b) $x=3$ and $x=3.01\text{,}$ (c) $x=3$ and $x=3.001\text{.}$ Now use algebra to find a simple formula for the slope of the secant line between $(3,f(3))$ and $(3+\Delta x,f(3+\Delta x))\text{.}$ Determine what happens when $\Delta x$ approaches 0. In your graph of $y=1/x\text{,}$ draw the straight line through the point $(3,1/3)$ whose slope is this limiting value of the difference quotient as $\Delta x$ approaches 0.

Answer

$-0.107526881\text{,}$ $-0.11074197\text{,}$ $-0.1110741\text{,}$ $\ds{-1\over3(3+\Delta x)}\rightarrow {-1\over9}$

Solution

We plot the graph of $y = f(x) = 1/x$ over the interval $[0.5,4]$ below.

We first construct the difference quotient at $x=3\text{:}$

\begin{equation*} \begin{split} \frac{\Delta y}{\Delta x} \amp = \frac{f(3 + \Delta x) - f(3)}{\Delta x} \\ \amp = \frac{\frac{1}{3+ \Delta x} - \frac{1}{3}}{\Delta x} \\ \amp = \frac{\frac{3 - (3+\Delta x)}{3(3+\Delta x)}}{\Delta x} \\ \amp = -\frac{1}{3(3+\Delta x)} \end{split} \end{equation*}

We use this difference quotient to compute the following.

The slope of the secant line between $x=3$ and $x=3.1$ ($\Delta x = 0.1$) is

\begin{equation*} \frac{\Delta y}{\Delta x} = -\frac{1}{3(3+0.1)} \approx 0.1075\text{.} \end{equation*}
The slope of the secant line between $x=3$ and $x=3.01$ ($\Delta x = 0.01$) is

\begin{equation*} \frac{\Delta y}{\Delta x} = -\frac{1}{3(3+0.01)} \approx 0.1107\text{.} \end{equation*}
The slope of the secant line between $x=3$ and $x=3.001$ ($\Delta x = 0.01$) is

\begin{equation*} \frac{\Delta y}{\Delta x} = -\frac{1}{3(3+0.001)} \approx 0.1111\text{.} \end{equation*}

As $\Delta x \to 0\text{,}$ we see that the slope of the secant line between $(3,f(3))$ and $(3+\Delta x, f(3+\Delta x)$ becomes

\begin{equation*} \lim_{\Delta x \to 0} \frac{-1}{3(3+\Delta x)} = -\frac{1}{3(3+0)} = -\frac{1}{9}\text{.} \end{equation*}

The equation of the line which passes through the point $(3,f(3)) = (3, 1/3)$ and has slope $-1/9$ is

\begin{equation*} y - \frac{1}{3} = -\frac{1}{9} (x-3)\text{.} \end{equation*}

This is the equation of the tangent line at $x=3\text{,}$ shown below.

Exercise 4.1.4.

Find an algebraic expression for the difference quotient $\ds \bigl(f(1+\Delta x)-f(1)\bigr)/\Delta x$ when $\ds f(x)=x^2-(1/x)\text{.}$ Simplify the expression as much as possible. Then determine what happens as $\Delta x$ approaches 0. That value is $f'(1)\text{.}$

Answer

$\ds{3+3\Delta x+\Delta x^2\over1+\Delta x}\rightarrow3$

Solution

The difference quotient for $f(x) = x^2 - \frac{1}{x}$ is:

\begin{equation*} \begin{split} \frac{f(x+\Delta x) - f(x)}{\Delta x} \amp = \frac{1}{\Delta x} \left((x+\Delta x)^2 - \frac{1}{x+\Delta x} - \left(x^2 - \frac{1}{x}\right)\right) \\ \amp = \frac{1}{\Delta x} \left( x^2 + 2x\Delta x + \Delta x^2 - \frac{1}{x+\Delta x} - x^2 + \frac{1}{x} \right)\\ \amp = \frac{1}{\Delta x} \left(2x\Delta x + \Delta x^2 + \frac{- x + (x + \Delta x)}{x(x+\Delta x)}\right) \\ \amp = 2x + \Delta x + \frac{1}{x(x+\Delta x)} \end{split} \end{equation*}

Therefore, at $x=1\text{,}$ we have

\begin{equation*} \frac{f(1+\Delta x) - f(1)}{\Delta x} = 2 + \Delta x + \frac{1}{1+\Delta x} = \frac{3 + 3\Delta x + \Delta x^2}{1+\Delta x}\text{.} \end{equation*}

Taking the limit as $\Delta x \to 0\text{,}$ we find that

\begin{equation*} f'(1) = \lim_{\Delta x \to 0} \frac{3 + 3\Delta x + \Delta x^2}{1+\Delta x} = \frac{3+0+0}{1+0} = 3\text{.} \end{equation*}

Exercise 4.1.5.

Draw the graph of $\ds y=f(x)=x^3$ between $x=0$ and $x=1.5\text{.}$ Find the slope of the secant line between (a) $x=1$ and $x=1.1\text{,}$ (b) $x=1$ and $x=1.001\text{,}$ (c) $x=1$ and $x=1.00001\text{.}$ Then use algebra to find a simple formula for the slope of the secant line between $1$ and $1+\Delta x\text{.}$ (Use the expansion $\ds (A+B)^3=A^3+3A^2B+3AB^2+B^3\text{.}$) Determine what happens as $\Delta x$ approaches 0, and in your graph of $\ds y=x^3$ draw the straight line through the point $(1,1)$ whose slope is equal to the value you just found.

Answer

$3.31\text{,}$ $3.003001\text{,}$ $3.0000\text{,}$ $3+3\Delta x+\Delta x^2\rightarrow3$

Solution

We plot the graph of $y=f(x) = x^3$ on the interval $[0,1.5]$ below.

The slope of the secant line between $x=1$ and $x=1.1$ ($\Delta x = 0.1$)is

\begin{equation*} \frac{\Delta y}{\Delta x} = \frac{f(1.1)-f(1)}{0.1} = 3.31\text{.} \end{equation*}
The slope of the secant line between $x=1$ and $x=1.001$ ($\Delta x = 0.001$)is

\begin{equation*} \frac{\Delta y}{\Delta x} = \frac{f(1.001)-f(1)}{0.1} = 3.003001\text{.} \end{equation*}
The slope of the secant line between $x=1$ and $x=1.00001$ ($\Delta x = 0.00001$)is

\begin{equation*} \frac{\Delta y}{\Delta x} = \frac{f(1.00001)-f(1)}{0.1} = 3.0000300001\text{.} \end{equation*}

We now simplify the difference quotient at $x=1\text{:}$

\begin{equation*} \begin{split} \frac{f(1+\Delta x) - f(1)}{\Delta x} \amp = \frac{(1+\Delta x)^3 - 1}{\Delta x} \\ \amp = \frac{1 + 3\Delta x^2 +3\Delta x + \Delta x^3 - 1}{\Delta x}\\ \amp = 3\Delta x + 3 + \Delta x^2 \end{split} \end{equation*}

Therefore, as $\Delta x \to 0\text{,}$

\begin{equation*} f'(1) = \lim_{\Delta x \to 0} \left( 3\Delta x + 3 + \Delta x^2\right) = 3\text{.} \end{equation*}

The line with slope $3$ which passes through the point $(1,f(1)) = (1,1)$ is

\begin{equation*} y - 1 = 3(x-1)\text{.} \end{equation*}

This is the tangent line to $f$ at $x=1\text{:}$

Exercise 4.1.6.

Find an algebraic expression for the difference quotient $(f(x+\Delta x)-f(x))/\Delta x$ when $f(x)=mx+b\text{.}$ Simplify the expression as much as possible. Then determine what happens as $\Delta x$ approaches 0. That value is $f'(x)\text{.}$

Answer Solution

Let $f(x) = mx+b\text{.}$ The difference quotient is then

\begin{equation*} \begin{split} \dfrac{f(x+\Delta x) - f(x)}{\Delta x} \amp = \dfrac{m(x+\Delta x) + b - mx - b}{\Delta x} \amp = \dfrac{m\Delta x}{\Delta x} \amp = m \end{split} \end{equation*}

Therefore,

\begin{equation*} f'(x) = \lim\limits_{\Delta x \to 0} m = m\text{.} \end{equation*}

Exercise 4.1.7.

Sketch the unit circle. Discuss the behavior of the slope of the tangent line at various angles around the circle. Which trigonometric function gives the slope of the tangent line at an angle $\theta\text{?}$ Why? Hint

Think in terms of ratios of sides of triangles.

Solution

We consider the following diagram of the unit circle and some $\theta \in [0,\pi]\text{:}$

The point $(a,b)$ is then given by $(\cos(\theta), \sin(\theta))\text{.}$ And so the slope of the line connecting the origin to this point is

\begin{equation*} \frac{\Delta y}{\Delta x} = \frac{\cos(\theta)}{\sin(\theta)}\text{.} \end{equation*}

Now, since the slope of the tangent line $m$ at the point $(a,b)$ is perpendicular to the slope of the line connecting $(a,b)$ to the origin, we must have

\begin{equation*} m = -\frac{\sin(\theta)}{\cos(\theta)} = - \tan(\theta)\text{.} \end{equation*}

This holds for any $\theta \in [0, 2\pi]\text{.}$

Exercise 4.1.8.

Sketch the parabola $\ds y=x^2\text{.}$ For what values of $x$ on the parabola is the slope of the tangent line positive? Negative? What do you notice about the graph at the point(s) where the sign of the slope changes from positive to negative and vice versa?

Solution

We see that tangents lines which are tangent to the curve $y = x^{2}$ at points $x > 0$ have positive slope, and that tangent lines which are tangent to the curve at points $x \lt 0$ have negative slope. We also see that the line tangent to $x = 0$ has zero slope, and note that this corresponds to the vertex of the parabola: here, the minimum point on the graph.

Exercise 4.1.9.

An object is traveling in a straight line so that its position (that is, distance from some fixed point) is given by this table:

time (s)	0	1	2	3

distance (m)	0	10	25	60

Find the average speed of the object during the following time intervals: $[0,1]\text{,}$ $[0,2]\text{,}$ $[0,3]\text{,}$ $[1,2]\text{,}$ $[1,3]\text{,}$ $[2,3]\text{.}$ If you had to guess the speed at $t=2$ just on the basis of these, what would you guess?

Answer

$10\text{,}$ $25/2\text{,}$ $20\text{,}$ $15\text{,}$ $25\text{,}$ $35\text{.}$

Solution

Let $x$ be the distance travelled in metres and let $t$ be the time in seconds. Then the average speed of the object over some time interval is $\dfrac{\Delta x}{\Delta t}\text{.}$ Therefore:

For $t \in [0,1]\text{,}$ the average speed is $\dfrac{\Delta x}{\Delta t} = \dfrac{10-0}{1-0} = 10$ m/s.
For $t \in [0,2]\text{,}$ the average speed is $\dfrac{\Delta x}{\Delta t} = \dfrac{25-0}{2-0} = 12.5$ m/s.
For $t \in [0,3]\text{,}$ the average speed is $\dfrac{\Delta x}{\Delta t} = \dfrac{60-0}{3-0} = 20$ m/s.
For $t \in [1,2]\text{,}$ the average speed is $\dfrac{\Delta x}{\Delta t} = \dfrac{25-10}{2-1} = 15$ m/s.
For $t \in [1,3]\text{,}$ the average speed is $\dfrac{\Delta x}{\Delta t} = \dfrac{60-10}{3-1} = 25$ m/s.
For $t \in [2,3]\text{,}$ the average speed is $\dfrac{\Delta x}{\Delta t} = \dfrac{60-25}{3-2} = 35$ m/s.

We guess that the actual speed at $t=2$ is between 15 and 35 m/s.

Exercise 4.1.10.

Let $\ds y=f(t)=t^2\text{,}$ where $t$ is the time in seconds and $y$ is the distance in metres that an object falls on a certain airless planet. Draw a graph of this function between $t=0$ and $t=3\text{.}$ Make a table of the average speed of the falling object between (a) 2 sec and 3 sec, (b) 2 sec and 2.1 sec, (c) 2 sec and 2.01 sec, (d) 2 sec and 2.001 sec. Then use algebra to find a simple formula for the average speed between time $2$ and time $2+ \Delta t\text{.}$ (If you substitute $\Delta t=1, 0.1, 0.01,0.001$ in this formula you should again get the answers to parts (a)–(d).) Next, in your formula for average speed (which should be in simplified form) determine what happens as $\Delta t$ approaches zero. This is the instantaneous speed. Finally, in your graph of $\ds y=t^2$ draw the straight line through the point $(2,4)$ whose slope is the instantaneous velocity you just computed; it should of course be the tangent line.

Answer

$5\text{,}$ $4.1\text{,}$ $4.01\text{,}$ $4.001\text{,}$ $4+\Delta t\rightarrow 4$

Solution

The graph of $y=f(t)=t^2$ on the interval $[0,3]$ is shown below, where $y$ is the distance in metres of a falling object and $t$ is the time in seconds.

The average speed of the falling object is given by $\dfrac{\Delta y}{\Delta t}\text{.}$ Therefore:

Between $t=2$ and $t=3\text{,}$ the average speed is $\dfrac{\Delta y}{\Delta t} = \dfrac{3^2-2^2}{3-2} = 5$ m/s.
Between $t=2$ and $t=2.1\text{,}$ the average speed is $\dfrac{\Delta y}{\Delta t} = \dfrac{2.1^2-2^2}{2.1-2} = 4.1$ m/s.
Between $t=2$ and $t=2.01\text{,}$ the average speed is $\dfrac{\Delta y}{\Delta t} = \dfrac{2.01^2-2^2}{3-2} = 4.01$ m/s.
Between $t=2$ and $t=2.001\text{,}$ the average speed is $\dfrac{\Delta y}{\Delta t} = \dfrac{2.001^2-2^2}{3-2} = 4.001$ m/s.

Our formula for average speed between $t=2$ and $t=2+\Delta t$ can be simplified:

\begin{equation*} \begin{split} \frac{\Delta y}{\Delta t} \amp = \frac{f(2+\Delta t) - f(2)}{\Delta t}\\ \amp = \frac{(2+\Delta t)^2 - 4}{\Delta t}\\ \amp = \frac{(4+4\Delta t + \Delta t^2) - 4}{\Delta t}\\ \amp = \frac{4\Delta t + \Delta t^2}{\Delta t}\\ \amp = 2t +\Delta t \end{split} \end{equation*}

Therefore, the instantaneous speed is given by

\begin{equation*} \lim_{\Delta t \to 0} \left(4 + \Delta t\right) = 4\text{.} \end{equation*}

So at $t=2\text{,}$ the speed is 4 m/s (notice that our answers in (a)-(d) were approaching this value). The line which passes through the point $(2,4)$ and has slope $4$ is given by

\begin{equation*} y - 4 = 4(x-2)\text{,} \end{equation*}

and is plotted below. This is the tangent line to $y$ at $t=2\text{.}$

Exercise 4.1.11.

If an object is dropped from an 80-metre high window, its height $y$ above the ground at time $t$ seconds is given by the formula $\ds y=f(t)=80-4.9t^2\text{.}$ (Here we are neglecting air resistance.) Find the average velocity of the falling object between (a) 1 sec and 1.1 sec, (b) 1 sec and 1.01 sec, (c) 1 sec and 1.001 sec. Now use algebra to find a simple formula for the average velocity of the falling object between 1 sec and $1+\Delta t$ sec. Determine what happens to this average velocity as $\Delta t$ approaches 0. That is the instantaneous velocity at time $t=1$ second (it will be negative, because the object is falling).

Answer

$-10.29\text{,}$ $-9.849\text{,}$ $-9.8049\text{,}$ $-9.8-4.9\Delta t\rightarrow -9.8$

Solution

Let $y=f(t)=80-4.9t^2$ be the height in metres of a falling object above the ground at time $t$ seconds.

The average velocity of the falling object is given by $\dfrac{\Delta y}{\Delta t}\text{.}$ Notice that $y(1) = 80-4.9 = 75.1$ m. Therefore:

Between $t=1$ and $t=1.1\text{,}$ the average speed is $\dfrac{\Delta y}{\Delta t} = \dfrac{74.071-75.1}{0.1} = -10.29$ m/s.
Between $t=1$ and $t=1.01\text{,}$ the average speed is $\dfrac{\Delta y}{\Delta t} = \dfrac{75.00151-75.1}{0.01} = -9.849$ m/s.
Between $t=1$ and $t=1.001\text{,}$ the average speed is $\dfrac{\Delta y}{\Delta t} = \dfrac{75.0901951-75.1}{0.001} = -9.8049$ m/s.

The formula for average velocity between $t=1$ and $t=1+\Delta t$ can be simplified:

\begin{equation*} \begin{split} \frac{\Delta y}{\Delta t} \amp = \frac{f(1+\Delta t) - f(1)}{\Delta t}\\ \amp = \frac{(80-4.9(1+\Delta t)^2) - (80-4.9)}{\Delta t}\\ \amp = \frac{80- 4.9(1+2\Delta t + \Delta t^2) - 80 +4.9}{\Delta t}\\ \amp = \frac{-4.9-9.8\Delta t - 4.9\Delta t^2 + 4.9}{\Delta t}\\ \amp = -9.8 - 4.9\Delta t \end{split} \end{equation*}

Therefore, the instantaneous velocity is given by

\begin{equation*} \lim_{\Delta t \to 0} \left(-9.8 + 4.9\Delta t\right) = -9.8\text{.} \end{equation*}

So at $t=1\text{,}$ the velocity is -9.8 m/s (notice that our answers in (a)-(c) were approaching this value).

Exercise 4.1.12.

The following figure shows the devastating effect the opening of a new chain coffee shop had on a 3-generation rural café in a small town. The revenue of the chain coffee shop at time $t$ (in months) is given by $f(t)$ million dollars, whereas the revenue of the rural café at time $t$ is given by $g(t)$ million dollars. Answer the following questions by giving the value of $t$ at which the specifying event took place.

The revenue of the rural café is decreasing at the slowest rate.
Answer
The revenue of the rural café is decreasing at the fastest rate.
Answer
The revenue of the chain coffe shop first overtakes that of the rural café.
Answer
The revenue of the chain coffee shop is increasing at the fastest rate.
Answer

Exercise 4.1.13.

The demand function for tires is given by

\begin{equation*} p = f(q) = -0.1q^{2} - q + 125 \end{equation*}

where $p$ is measured in dollars and $q$ is measured in units of a thousand.

Find the average rate of change in the unit price of a tire if the quantity demanded is between $5000$ and $5050$ tires; between $5000$ and $5010$ tires.
Answer

-$$2.00$ per $1000$ tires.

Solution

Let's first form and simplify the difference quotient.

\begin{equation*} \begin{split} \dfrac{f(q+\Delta q) - f(q)}{\Delta q} \amp = \dfrac{\left(-0.1(q+\Delta q)^{2} - (q + \Delta q) + 125 \right) - \left(0.1q^{2} - q + 125\right)}{\Delta q} \\ \amp = \dfrac{-0.2q\Delta q-0.1\Delta q^{2} - \Delta q}{\Delta q}\\ \amp = -0.2q - 0.1\Delta q-1 \end{split} \end{equation*}

Therefore, when $q = 5$ for $\Delta q = 0.05$ and $0.01\text{,}$ we find that the difference quotient is equal to $-2.005$ and $-2.001\text{,}$ respectively. Therefore, when the quantity of tires demanded is between $5000$ and $5050\text{,}$ the unit price is decreasing by approximately $$2.00$ per $1000$ tires. And when the quantitity of tires demanded is between $5000$ and $5010\text{,}$ we again get that the unity price is decreasing by approximately $$2.00$ per $1000$ tires.
What is the rate of change of the unit price if the quantity demanded is $5000\text{?}$
Answer

-$$2.00$ per $1000$ tires.

Solution

We find $f'(x) = \lim\limits_{\Delta q \to 0}\left( -0.2q-0.1q\Delta q -1\right) = -0.2q - 1\text{.}$ Therefore, when the quantity demanded is $5000\text{,}$ the unit price of the tires is changing by $f'(5) = -2\text{.}$ That is, it is decreasing by $2.00 per $1000$ tires.

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Source: https://www.sfu.ca/math-coursenotes/Math%20157%20Course%20Notes/sec_Slope.html

Gatliff Upprall

A Function F is Continuous on 1 5 and Some of the Values of F Are Shown Below X1 4 5

Calculating and Interpreting the Slope of a Line.

Secant Line.

Definition 4.1. Slope of Secant Line — Average Rate of Change.

Tangent Line.

Example 4.2. Small Changes in \(x\).

Definition 4.3. Slope of Tangent Line—Instantaneous Rate of Change.

From Tangent Line Slope to Derivative.

Subsection 4.1.1 Applications

Example 4.4. Analyzing Velocity.

Example 4.5. Demand for Sweaters.

Exercises for Section 4.1.

Exercise 4.1.1.

Exercise 4.1.2.

Exercise 4.1.3.

Exercise 4.1.4.

Exercise 4.1.5.

Exercise 4.1.6.

Exercise 4.1.7.

Exercise 4.1.8.

Exercise 4.1.9.

Exercise 4.1.10.

Exercise 4.1.11.

Exercise 4.1.12.

Exercise 4.1.13.

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