diff --git a/publication/publication.ptx b/publication/publication.ptx index 7280fdc..9724d71 100644 --- a/publication/publication.ptx +++ b/publication/publication.ptx @@ -21,8 +21,8 @@ - - + + @@ -121,7 +121,7 @@ - + /> - + @@ -171,13 +171,14 @@ - + /> + @@ -208,7 +209,7 @@ - + diff --git a/source/bookinfo.xml b/source/bookinfo.xml index e541491..07269b7 100644 --- a/source/bookinfo.xml +++ b/source/bookinfo.xml @@ -52,10 +52,11 @@ Preview Activity - Motivating Questions + Section Objectives + Foundations   - Question + diff --git a/source/chap-1.xml b/source/chap-1.xml index 3c04e0a..44245be 100644 --- a/source/chap-1.xml +++ b/source/chap-1.xml @@ -5,11 +5,11 @@ - + - + diff --git a/source/companion-template.xml b/source/companion-template.xml new file mode 100644 index 0000000..0c5ed3c --- /dev/null +++ b/source/companion-template.xml @@ -0,0 +1,61 @@ + + + +
+ Section Title + +
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    + Objective 1 +

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    + Objective 2 +

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    + Foundational knowledge or technique 1 +

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    + Foundational knowledge or technique 1 +

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+ + + + + + + + + Additional Section Exercises + + + + + + + + Foundational Exercises + + + + +
diff --git a/source/main.ptx b/source/main.ptx index cc365c1..d1bb2c4 100644 --- a/source/main.ptx +++ b/source/main.ptx @@ -8,14 +8,14 @@ - + + + diff --git a/source/sec-1-1-vel.xml b/source/sec-1-1-vel.xml index db64c78..8f233f5 100644 --- a/source/sec-1-1-vel.xml +++ b/source/sec-1-1-vel.xml @@ -3,289 +3,59 @@
How do we measure velocity? - +

  • - How is the average velocity of a moving object connected to the values of its position function? + Objective 1

  • - How do we interpret the average velocity of an object geometrically on the graph of its position function? -

    -
    • - -
  • -

    - How is the notion of instantaneous velocity connected to average velocity? + Objective 2

- - Introduction -

- Calculus can be viewed broadly as the study of change. - A natural and important question to ask about any changing quantity is - how fast is the quantity changing? -

- -

- We begin with a simple problem: - a ball is tossed straight up in the air. - How is the ball moving? - Questions like this one are central to our study of differential calculus. -

- - - - - - - Position and average velocity -

- Any moving object has a position - position - that can be considered a function of time. - When the motion is along a straight line, - the position is given by a single variable, - which we denote by s(t). - For example, - s(t) might give the mile marker of a car traveling on a straight highway at time t in hours. - Similarly, the function s described in Preview Activity is a position function, - where position is measured vertically relative to the ground. -

- -

- On any time interval, a moving object also has an - average velocity. - For example, - to compute a car's average velocity we divide the number of miles traveled by the time elapsed, - which gives the velocity in miles per hour. - Similarly, the value of AV_{[0.5,1]} in Preview Activity - gave the average velocity of the ball on the time interval [0.5,1], - measured in feet per second. -

- -

- In general, we make the following definition: -

- - - Average Velocity -

- For an object moving in a straight line with position function s(t), the - average velocity - average velocity of the object on the interval from t = a to t = b, - denoted AV_{[a,b]}, is given by the formula - - AV_{[a,b]} = \frac{s(b)-s(a)}{b-a} - . -

- - -

- Note well: the units on AV_{[a,b]} are - units of s per unit of t, - such as miles per hour or - feet per second. -

- - - - - - Instantaneous Velocity -

- Whether we are driving a car, - riding a bike, or throwing a ball, - we have an intuitive sense that a moving object has a velocity at any given moment a number that measures how fast the object is moving right now. - For instance, - a car's speedometer tells the driver the car's velocity at any given instant. - In fact, the velocity on a speedometer is really an average velocity that is computed over a very small time interval. - If we let the time interval over which average velocity is computed become shorter and shorter, - we can progress from average velocity to - instantaneous velocity. -

- -

- Informally, we define the instantaneous velocity - instantaneous velocity - of a moving object at time t = a to be the value that the average velocity approaches as we take smaller and smaller intervals of time containing t = a. - We will develop a more formal definition of instantaneous velocity soon, - and this definition will be the foundation of much of our work in calculus. - For now, it is fine to think of instantaneous velocity as follows: - take average velocities on smaller and smaller time intervals around a specific point. - If those average velocities approach a single number, - then that number will be the instantaneous velocity at that point. -

- - - -

- - At this point we have started to see a close connection between average velocity and instantaneous velocity. - Each is connected not only to the physical behavior of the moving object but also to the geometric behavior of the graph of the position function. - We are interested in computing average velocities on the interval [a,b] for smaller and smaller intervals. - In order to make the link between average and instantaneous velocity more formal, - think of the value b as b = a + h, - where h is a small (non-zero) number that is allowed to vary. - Then the average velocity of the object on the interval [a,a+h] is - - AV_{[a,a+h]} = \frac{s(a+h)-s(a)}{h} - , - with the denominator being simply h because (a+h) - a = h. - Note that when h \lt 0, - AV_{[a,a+h]} measures the average velocity on the interval [a+h,a]. -

- -

- To find the instantaneous velocity at t = a, - we investigate what happens as the value of h approaches zero. -

- - - Computing instantaneous velocity for a falling ball - -

- The position function for a falling ball is given by s(t) = 16 - 16t^2 - (where s is measured in feet and t in seconds). -

- -

-

    -
  1. - Find an expression for the average velocity of the ball on a time interval of the form [0.5, 0.5+h] where - -0.5 \lt h \lt 0.5 and h \ne 0. -
    2. - -
  2. - Use this expression to compute the average velocity on - [0.5,0.75] and [0.4,0.5]. -
    3. - -
  3. - Make a conjecture about the instantaneous velocity at t = 0.5. -
    4. -
-

- - -

-

    -
  1. - We make the assumptions that - -0.5 \lt h \lt 0.5 and h \ne 0 because h cannot be zero - (otherwise there is no interval on which to compute average velocity) - and because the function only makes sense on the time interval 0 \le t \le 1, - as this is the duration of time during which the ball is falling. - We want to compute and simplify - - AV_{[0.5, 0.5+h]} = \frac{s(0.5+h) - s(0.5)}{(0.5+h) - 0.5} - . - We start by finding s(0.5+h). - To do so, we follow the rule that defines the function s, and then expand and simplify. Doing so, - - s(0.5+h) \amp = 16 - 16(0.5 + h)^2 - \amp = 16 - 16(0.25 + h + h^2) - \amp = 16 - 4 - 16h - 16h^2 - \amp = 12 - 16h - 16h^2 - . - Now, returning to our computation of the average velocity, - we find that - - AV_{[0.5, 0.5+h]} \amp = \frac{s(0.5+h) - s(0.5)}{(0.5+h) - 0.5} - \amp = \frac{(12 - 16h - 16h^2) - (16 - 16(0.5)^2)}{0.5 + h - 0.5} - \amp = \frac{12 - 16h - 16h^2 - 12}{h} - \amp = \frac{-16h - 16h^2}{h} - \amp = \frac{h(-16 - 16h)}{h} - . - At this point, we note two things: - first, the expression for average velocity clearly depends on h, - which it must, - since as h changes the average velocity will change. - Further, we note that since h can never equal zero, - we may remove the common factor of h from the numerator and denominator. - It follows that - - AV_{[0.5, 0.5+h]} = -16 - 16h - . -
    2. - -
  2. - From this expression we can compute the average for any small positive or negative value of h. - For instance, to obtain the average velocity on - [0.5,0.75], we let h = 0.25, - and the average velocity is AV_{[0.5, 0.75]} = -16 - 16(0.25) = -20 ft/sec. - To get the average velocity on [0.4, 0.5], - we let h = -0.1, and compute the average velocity as - - AV_{[0.4, 0.5]} = -16 - 16(-0.1) = -14.4\ \text{ft/sec} - . -
    3. - -
  3. - We can even explore what happens to - AV_{[0.5, 0.5+h]} = -16 - 16h as h gets closer and closer to zero. - As h approaches zero, - -16h will also approach zero, - so it appears that the instantaneous velocity of the ball at t = 0.5 should be -16 ft/sec. -
    4. -
-

- - - - - - - - Summary - -

+

    -
  • - For an object moving in a straight line with position function s(t), the - average velocity - average velocity of the object on the interval from t = a to t = b, - denoted AV_{[a,b]}, is given by the formula - - AV_{[a,b]} = \frac{s(b)-s(a)}{b-a} - . -
    • -

  • - The average velocity on [a,b] can be viewed geometrically as the slope of the line between the points (a,s(a)) and - (b,s(b)) on the graph of y = s(t), - as shown in Figure. + Foundational knowledge or technique 1

    - -
    - The graph of position function s together with the line through (a,s(a)) and (b,s(b)) whose slope is m = \frac{s(b)-s(a)}{b-a}. The line's slope is the average rate of change of s on the interval [a,b]. - -
    -
    • -
  • - Given a moving object whose position at time t is given by a function s, - the average velocity of the object on the time interval [a,b] is given by AV_{[a,b]} = \frac{s(b) - s(a)}{b-a}. - Viewing the interval [a,b] as having the form [a,a+h], - we equivalently compute average velocity by the formula AV_{[a,a+h]} = \frac{s(a+h) - s(a)}{h}. -
    • - -
  • - The instantaneous velocity of a moving object at a fixed time is estimated by considering average velocities on shorter and shorter time intervals that contain the instant of interest. +

    + Foundational knowledge or technique 1 +

-

+ + + + + + + + + Additional Section Exercises + + + + + - + Foundational Exercises + + + +
diff --git a/source/sec-1-2-lim.xml b/source/sec-1-2-lim.xml index 8402ec9..3af3281 100644 --- a/source/sec-1-2-lim.xml +++ b/source/sec-1-2-lim.xml @@ -7,612 +7,94 @@

  • - What is the mathematical notion of limit - and what role do limits play in the study of functions? + Determine the limit of a function at a point using graphical, numerical, and algebraic methods.

  • - What is the meaning of the notation \lim_{x \to a} f(x) = L? -

    -
    • - -
  • -

    - How do we go about determining the value of the limit of a function at a point? -

    -
    • - -
  • -

    - How do we manipulate average velocity to compute instantaneous velocity? + Explain how average velocity and instantaneous velocity are connected using the notion of limit.

- - Introduction -

- In we used a function, s(t), - to model the location of a moving object at a given time. - Functions can model other interesting phenomena, - such as the rate at which an automobile consumes gasoline at a given velocity, - or the reaction of a patient to a given dosage of a drug. - We can use calculus to study how a function value changes in response to changes in the input variable. -

- -

- Think about the falling ball whose position function is given by s(t) = 64 - 16t^2. - Its average velocity on the interval [1,x] is given by - - AV_{[1,x]} = \frac{s(x) - s(1)}{x-1} = \frac{(64-16x^2) - (64-16)}{x-1} = \frac{16 - 16x^2}{x-1} - . -

- -

- Note that the average velocity is a function of x. - That is, the function g(x) = \frac{16 - 16x^2}{x-1} tells us the average velocity of the ball on the interval from t = 1 to t = x. - To find the instantaneous velocity of the ball when t = 1, - we need to know what happens to g(x) as x gets closer and closer to 1. - But also notice that g(1) is not defined, - because it leads to the quotient 0/0. -

- -

- This is where the notion of a limit comes in. - By using a limit, - we can investigate the behavior of g(x) as x gets arbitrarily close, - but not equal, to 1. - We first use the graph of a function to explore points where interesting behavior occurs. -

- - - - - - - The Notion of Limit -

- Limits give us a way to identify a trend in the values of a function as its input variable approaches a particular value of interest. - We need a precise understanding of what it means to say - a function f has limit L as x approaches a. To begin, - think about a recent example. -

- -

- In Preview Activity, - we saw that - as x gets closer and closer - (but not equal) - to 0, g(x) gets as close as we want to the value 4. - At first, this may feel counterintuitive, - because the value of g(0) is 1, not 4. - But limits describe the behavior of a function - arbitrarily close to a fixed input, - and the value of the function at - the fixed input does not matter. - More formally, - What follows here is not what mathematicians consider the formal definition of a limit. - To be completely precise, - it is necessary to quantify both what it means to say - as close to L as we like - and sufficiently close to a. - That can be accomplished through what is traditionally called the epsilon-delta definition of limits. - The definition presented here is sufficient for the purposes of this text. - we say the following. -

- - - -

- Given a function f, - a fixed input x = a, and a real number L, - we say that f has limit L as x approaches a, - limitdefinition - and write - - \lim_{x \to a} f(x) = L - - provided that we can make f(x) as close to L as we like by taking x sufficiently close - (but not equal) - to a. - If we cannot make f(x) as close to a single value as we would like as x approaches a, - then we say that f does not have a limit as x approaches a. -

- - - - - -

- For the function g pictured in Figure from Preview Activity, what can we say about the values of - - \lim_{x \to -1} g(x), \ \lim_{x \to 0} g(x), \ \lim_{x \to 1} g(x), \ \text{and} \ \lim_{x \to 2} g(x) - ? -

-

- For the function g pictured in Preview Activity, what can we say about the values of - - \lim_{x \to -1} g(x), \ \lim_{x \to 0} g(x), \ \lim_{x \to 1} g(x), \ \text{and} \ \lim_{x \to 2} g(x) - ? -

- - -

- When working from a graph, - it suffices to ask if the function approaches - a single value from each side of the fixed input. - The function value at the fixed input is irrelevant. - Based on the graph of g in Figure, we can conclude that - - \lim_{x \to -1} g(x) = 3, \ \lim_{x \to 0} g(x) = 4, \ \text{and} \ \lim_{x \to 2} g(x) = 1 - , - because at each of those x-values, the function approaches the same height from both sides of the point of interest. -

- -

- However, g does not have a limit as x \to 1 due to the jump in the graph at x = 1. - If we approach x = 1 from the left, - the function values tend to get close to 3, - but if we approach x = 1 from the right, - the function values get close to 2. - There is no single number that all of these function values approach. - This is why the limit of g does not exist at x = 1. -

- - - -

- For any function f, - there are typically three ways to answer the question - does f have a limit at x = a, and if so, - what is the limit? The first is to reason graphically as we have just done with the example from Preview Activity. - If we have a formula for f(x), - there are two additional possibilities: - -

    -
  1. - Evaluate the function at a sequence of inputs that approach a on either side - (typically using some sort of computing technology), - and ask if the sequence of outputs seems to approach a single value. -
    2. - -
  2. - Use the algebraic form of the function to understand the trend in its output values as the input values approach a. -
    3. -
- - The first approach produces only an approximation of the value of the limit, - while the latter can often be used to determine the limit exactly. -

- - - Limits of Two Functions - -

- For each of the following functions, - we'd like to know whether or not the function has a limit at the stated a-values. - Use both numerical and algebraic approaches to investigate and, - if possible, estimate or determine the value of the limit. - Compare the results with a careful graph of the function on an interval containing the points of interest. -

- -

-

    -
  1. - f(x) = \frac{4-x^2}{x+2}; - a = -1, a = -2 -
    2. - -
  2. - g(x) = \sin\left(\frac{\pi}{x}\right); - a = 3, a = 0 -
    3. -
-

- - -

- a. We first construct a graph of f along with tables of values near a = -1 and a = -2. - - - - - Values of <m>f</m> near <m>x=-1</m>. - - - x - f(x) - - - -0.9 - 2.9 - - - -0.99 - 2.99 - - - -0.999 - 2.999 - - - -0.9999 - 2.9999 - - - -1.1 - 3.1 - - - -1.01 - 3.01 - - - -1.001 - 3.001 - - - -1.0001 - 3.0001 - - - -
- - - Values of <m>f</m> near <m>x=-2</m>. - - - x - f(x) - - - -1.9 - 3.9 - - - -1.99 - 3.99 - - - -1.999 - 3.999 - - - -1.9999 - 3.9999 - - - -2.1 - 4.1 - - - -2.01 - 4.01 - - - -2.001 - 4.001 - - - -2.0001 - 4.0001 - - - -
- -

- Plot of f(x) on [-4,2]. - -
- -

-

- From Table, - it appears that we can make f as close as we want to 3 by taking x sufficiently close to -1, - which suggests that \lim_{x \to -1} f(x) = 3. - This is also consistent with the graph of f. - To see this a bit more rigorously and from an algebraic point of view, - consider the formula for f: - f(x) = \frac{4-x^2}{x+2}. - As x \to -1, - (4-x^2) \to (4 - (-1)^2) = 3, - and (x+2) \to (-1 + 2) = 1, so as x \to -1, - the numerator of f tends to 3 and the denominator tends to 1, hence \lim_{x \to -1} f(x) = \frac{3}{1} = 3. -

- -

- The situation is more complicated when x \to -2, - because f(-2) is not defined. - If we try to use a similar algebraic argument regarding the numerator and denominator, - we observe that as x \to -2, - (4-x^2) \to (4 - (-2)^2) = 0, - and (x+2) \to (-2 + 2) = 0, so as x \to -2, - the numerator and denominator of f both tend to 0. - We call 0/0 an indeterminate form. - indeterminate form - This tells us that there is more work to do to try to determine the limit's value. - From Table - and Figure, - it appears that f should have a limit of 4 at x = -2. -

- -

- To see algebraically why this is the case, observe that - - \lim_{x \to -2} f(x) = \amp \lim_{x \to -2} \frac{4-x^2}{x+2} - = \amp \lim_{x \to -2} \frac{(2-x)(2+x)}{x+2} - . -

- -

- It is important to observe that, - since we are taking the limit as x \to -2, - we are considering x values that are close, - but not equal, to -2. - Because we never actually allow x to equal -2, - the quotient \frac{2+x}{x+2} has value 1 for every possible value of x. - Thus, we can simplify the most recent expression above, - and find that - - \lim_{x \to -2} f(x) = \lim_{x \to -2} 2-x - . - This limit is now easy to determine, - and its value clearly is 4. - Thus, from several points of view we've seen that \lim_{x \to -2} f(x) = 4. -

- -

- b. Next we turn to the function g, - and construct two tables and a graph. -

- - - - - Values of <m>g</m> near <m>x=3</m>. - - - x - g(x) - - - 2.9 - 0.88351 - - - 2.99 - 0.86777 - - - 2.999 - 0.86620 - - - 2.9999 - 0.86604 - - - 3.1 - 0.84864 - - - 3.01 - 0.86428 - - - 3.001 - 0.86585 - - - 3.0001 - 0.86601 - - - -
- - - Values of <m>g</m> near <m>x=0</m>. - - - x - g(x) - - - -0.1 - 0 - - - -0.01 - 0 - - - -0.001 - 0 - - - -0.0001 - 0 - - - 0.1 - 0 - - - 0.01 - 0 - - - 0.001 - 0 - - - 0.0001 - 0 - - - -
- -
- Plot of g(x) on [-4,4]. - -
- - -

- First, as x \to 3, it appears from the table values - that the function is approaching a number between 0.86601 and 0.86604. - From the graph it appears that g(x) \to g(3) as x \to 3. - The exact value of g(3) = \sin(\frac{\pi}{3}) is \frac{\sqrt{3}}{2}, - which is approximately 0.8660254038. - This is convincing evidence that - - \lim_{x \to 3} g(x) = \frac{\sqrt{3}}{2} - . -

- -

- As x \to 0, we observe that - \frac{\pi}{x} does not behave in an elementary way. - When x is positive and approaching zero, - we are dividing by smaller and smaller positive values, - and \frac{\pi}{x} increases without bound. - When x is negative and approaching zero, - \frac{\pi}{x} decreases without bound. - In this sense, as we get close to x = 0, - the inputs to the sine function are growing rapidly, - and this leads to increasingly rapid oscillations in the graph of g betweem 1 and -1. - If we plot the function - g(x) = \sin\left(\frac{\pi}{x}\right) with a graphing utility and then zoom in on x = 0, - we see that the function never settles down to a single value near the origin, - which suggests that g does not have a limit at x = 0. -

- -

- How do we reconcile the graph with the righthand table above, - which seems to suggest that the limit of g as x approaches 0 may in fact be 0? - The data misleads us because of the special nature of the sequence of input values \{0.1, 0.01, 0.001, \ldots\}. - When we evaluate g(10^{-k}), - we get g(10^{-k}) = \sin\left(\frac{\pi}{10^{-k}}\right) = \sin(10^k \pi) = 0 for each positive integer value of k. - But if we take a different sequence of values approaching zero, - say \{0.3, 0.03, 0.003, \ldots\}, then we find that - - g(3 \cdot 10^{-k}) = \sin\left(\frac{\pi}{3 \cdot 10^{-k}}\right) = \sin\left(\frac{10^k \pi}{3}\right) = \frac{\sqrt{3}}{2} \approx 0.866025 - . -

- -

- That sequence of function values suggests that the value of the limit is \frac{\sqrt{3}}{2}. - Clearly the function cannot have two different values for the limit, - so g has no limit as x \to 0. -

- - - - -

- An important lesson to take from Example - is that tables can be misleading when determining the value of a limit. - While a table of values is useful for investigating the possible value of a limit, - we should also use other tools to confirm the value. -

- - - -

- Recall that our primary motivation for considering limits of functions comes from our interest in studying the rate of change of a function. - To that end, - we close this section by revisiting our previous work with average and instantaneous velocity and highlighting the role that limits play. -

- - - - Instantaneous Velocity -

- Suppose that we have a moving object whose position at time t is given by a function s. - We know that the average velocity of the object on the time interval [a,b] is AV_{[a,b]} = \frac{s(b)-s(a)}{b-a}. - We define the instantaneous velocity - instantaneous velocity - at a to be the limit of average velocity as b approaches a. - Note particularly that as b \to a, - the length of the time interval gets shorter and shorter - (while always including a). - We will write IV_{t=a} for the instantaneous velocity at t = a, and thus - - IV_{t=a} = \lim_{b \to a} AV_{[a,b]} = \lim_{b \to a} \frac{s(b)-s(a)}{b-a} - . -

- -

- Equivalently, - if we think of the changing value b as being of the form b = a + h, - where h is some small number, then we may instead write - - IV_{t=a} = \lim_{h \to 0} AV_{[a,a+h]} = \lim_{h \to 0} \frac{s(a+h)-s(a)}{h} - . -

- -

- Again, the most important idea here is that to compute instantaneous velocity, - we take a limit of average velocities as the time interval shrinks. -

- - - -

- The closing activity of this section asks you to make some connections among average velocity, - instantaneous velocity, and slopes of certain lines. -

- - - - - - Summary -

+

  • - Limits enable us to examine trends in function behavior near a specific point. - In particular, - taking a limit at a given point asks if the function values nearby tend to approach a particular fixed value. + Algebra: factoring, multiplying out, common denominator, rationalizing

    • - -
  • -

    - We read \lim_{x \to a} f(x) = L, - as the limit of f as x approaches a is L, - which means that we can make the value of f(x) as close to L as we want by taking x sufficiently close - (but not equal) - to a. -

    -
    • -

  • - To find - \lim_{x \to a} f(x) for a given value of a and a known function f, - we can estimate this value from the graph of f, - or we can make a table of function values for x-values that are closer and closer to a. - If we want the exact value of the limit, - we can work with the function algebraically to understand how different parts of the formula for f change as x \to a. -

    -
    • - -
  • -

    - We find the instantaneous velocity of a moving object at a fixed time by taking the limit of average velocities of the object over shorter and shorter time intervals containing the time of interest. + Graphs: holes and jumps

-

+ + + + + + + + Additional Section Exercises + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + - + Foundational Exercises + + + + diff --git a/source/sec-1-3-derivative-pt.xml b/source/sec-1-3-derivative-pt.xml index 3140c86..07f8ad1 100644 --- a/source/sec-1-3-derivative-pt.xml +++ b/source/sec-1-3-derivative-pt.xml @@ -3,387 +3,59 @@
The derivative of a function at a point - +

  • - How is the average rate of change of a function on a given interval defined, - and what does this quantity measure? + Objective 1

  • - How is the instantaneous rate of change of a function at a particular point defined? - How is the instantaneous rate of change linked to average rate of change? -

    -
    • - -
  • -

    - What is the derivative of a function at a given point? - What does this derivative value measure? - How do we interpret the derivative value graphically? -

    -
    • - -
  • -

    - How are limits used formally in the computation of derivatives? + Objective 2

- - Introduction -

- The instantaneous rate of change - of a function is an idea that sits at the foundation of calculus. - It is a generalization of the notion of instantaneous velocity and measures how fast a particular function is changing at a given input. - If the original function represents the position of a moving object, - this instantaneous rate of change is precisely the instantaneous velocity of the object. - In other contexts, - instantaneous rate of change could measure the number of cells added to a bacteria culture per day, - the number of additional gallons of gasoline consumed per mile by increasing a car's velocity one mile per hour, - or the number of dollars added to a mortgage payment for each percentage point increase in interest rate. - The instantaneous rate of change can also be interpreted geometrically on the function's graph, - and this connection is fundamental to many of the main ideas in calculus. -

- -

- Recall that for a moving object with position function s, - its average velocity on the time interval t = a to t = a+h is given by the quotient - - AV_{[a,a+h]} = \frac{s(a+h)-s(a)}{h} - . -

- -

- In a similar way, - we make the following definition for an arbitrary function y = f(x). -

- - - -

- For a function f, the average rate of change - average rate of change - of f on the interval [a,a+h] is given by the value - - AV_{[a,a+h]} = \frac{f(a+h)-f(a)}{h} - . - Equivalently, the average rate of change of f on [a,b] is - - AV_{[a,b]} = \frac{f(b)-f(a)}{b-a} - . -

- - - -

- It is essential to understand how the average rate of change of f on an interval is connected to its graph. -

- - - - - - - The Derivative of a Function at a Point -

- Just as we defined instantaneous velocity in terms of average velocity, - we now define the instantaneous rate of change of a function at a point in terms of the average rate of change of the function f over related intervals. - - This instantaneous rate of change of f at a - instantaneous rate of change - is called the derivative - of f at a, - and is denoted by f'(a). -

- - - -

- Let f be a function and x = a an input value in the function's domain. - We define the derivative of f with respect to x evaluated at x = a, - derivativedefinition - denoted f'(a), by the formula - - f'(a) = \lim_{h \to 0} \frac{f(a+h)-f(a)}{h} - , - provided this limit exists. -

- - - -

- Aloud, we read the symbol f'(a) as either - f-prime at a - or the derivative of f evaluated at x = a. - Much of our work in Chapters 1-3 will be devoted to understanding, - computing, - applying, and interpreting derivatives. - For now, we observe the following important things. -

- - -

-

    -
  • - The derivative of f at the value x = a is defined as the limit of the average rate of change of f on the interval [a,a+h] as h \to 0. - This limit may not exist, - so not every function has a derivative at every point. -
    • - -
  • - We say that a function is differentiable - differentiable - at x = a if it has a derivative at x = a. -
    • - -
  • - The derivative is a generalization of the instantaneous velocity of a position function: - if y = s(t) is a position function of a moving body, - s'(a) tells us the instantaneous velocity of the body at time t=a. -
    • - -
  • - Because the units on \frac{f(a+h)-f(a)}{h} are - units of f(x) per unit of x, - the derivative has these very same units. - For instance, - if s measures position in feet and t measures time in seconds, - the units on s'(a) are feet per second. -
    • - -
  • - Because the quantity - \frac{f(a+h)-f(a)}{h} represents the slope of the line through - (a,f(a)) and (a+h, f(a+h)), - when we compute the derivative we are taking the limit of a collection of slopes of lines. - Thus, the derivative itself represents the slope of a particularly important line. -
    • -
-

- - -

- We first consider the derivative at a given value as the slope of a certain line. -

- -

- When we compute an instantaneous rate of change, - we allow the interval [a,a+h] to shrink as - h \to 0. - We can think of one endpoint of the interval as - sliding towards the other. - In particular, - provided that f has a derivative at (a,f(a)), - the point (a+h,f(a+h)) will approach (a,f(a)) as h \to 0. - Because the process of taking a limit is a dynamic one, - it can be helpful to use computing technology to visualize it. - One option is an interactive graphic in which the user is able to control the point that is moving. - For a helpful collection of examples, - consider the - work of David Austin - of Grand Valley State University, - and - this particularly relevant example. - For interactives that have been built in Geogebra - You can even consider building your own examples; - the fantastic program Geogebra is available for - free download - and is easy to learn and use. - , - see - Marc Renault's library - via Shippensburg University, - with - this example - being especially fitting for our work in this section. -

- -

- Figure - shows a sequence of figures with several different lines through the points (a, f(a)) and - (a+h,f(a+h)), generated by different values of h. - These lines - (shown in the first three figures in magenta), - are often called secant lines - secant line - to the curve y = f(x). - A secant line to a curve is simply a line that passes through two points on the curve. - For each such line, - the slope of the secant line is m = \frac{f(a+h) - f(a)}{h}, - where the value of h depends on the location of the point we choose. - We can see in the diagram how, as h \to 0, - the secant lines start to approach a single line that passes through the point (a,f(a)). - If the limit of the slopes of the secant lines exists, - we say that the resulting value is the slope of the - tangent line to the curve. - This tangent line - tangent line - (shown in the right-most figure in green) to the graph of - y = f(x) at the point - (a,f(a)) has slope m = f'(a). -

- -
- A sequence of secant lines approaching the tangent line to f at (a,f(a)). - -
- -

- If the tangent line at x = a exists, - the graph of f looks like a straight line when viewed up close at (a,f(a)). - In Figure - we combine the four graphs in Figure - into the single one on the left, - and zoom in on the box centered at (a,f(a)) - on the right. - Observe how the tangent line sits relative to the curve y = f(x) at - (a,f(a)) and how closely it resembles the curve near x = a. -

- -
- A sequence of secant lines approaching the tangent line to f at (a,f(a)). At right, we zoom in on the point (a,f(a)). The slope of the tangent line (in green) to f at (a,f(a)) is given by f'(a). - -
- - -

- The instantaneous rate of change of f with respect to x at x = a, - f'(a), - also measures the slope of the tangent line to the curve y = f(x) at (a,f(a)). -

- - -

- The following example demonstrates several key ideas involving the derivative of a function. -

- - - Using the limit definition of the derivative - -

- For the function f(x) = x - x^2, - use the limit definition of the derivative to compute f'(2). - In addition, - discuss the meaning of this value and draw a labeled graph that supports your explanation. -

- - -

- From the limit definition, we know that - - f'(2) = \lim_{h \to 0} \frac{f(2+h)-f(2)}{h} - . -

- -

- Now we use the rule for f, - and observe that f(2) = 2 - 2^2 = -2 and f(2+h) = (2+h) - (2+h)^2. - Substituting these values into the limit definition, we have that - - f'(2) = \lim_{h \to 0} \frac{(2+h) - (2+h)^2 - (-2)}{h} - . -

- -

- In order to let h \to 0, - we must simplify the quotient. - Expanding and distributing in the numerator, - - f'(2) = \lim_{h \to 0} \frac{2+h - 4 - 4h - h^2 + 2}{h} - . -

- -

- Combining like terms, we have - - f'(2) = \lim_{h \to 0} \frac{ -3h - h^2}{h} - . -

- -

- Next, we remove a common factor of h in both the numerator and denominator and find that - - f'(2) = \lim_{h \to 0} (-3-h) - . -

- -

- Finally, we are able to take the limit as h \to 0, - and thus conclude that f'(2) = -3. - We note that f'(2) is the instantaneous - rate of change of f at the point (2,-2). - It is also the slope of the tangent line to the graph of - y = x - x^2 at the point (2,-2). - Figure - shows both the function and the line - through (2,-2) with slope m = f'(2) = -3. -

- -
- The tangent line to y = x - x^2 at the point (2,-2). - -
- - - -

- The following activities will help you explore a variety of key ideas related to derivatives. -

- - - - - - - - - - Summary -

+

  • - The average rate of change of a function f on the interval [a,b] is AV_{[a,b]} = \frac{f(b)-f(a)}{b-a}. - The units on the average rate of change are units of f(x) per unit of x, - and the numerical value of the average rate of change represents the slope of the secant line between the points (a,f(a)) and - (b,f(b)) on the graph of y = f(x). - If we view the interval as being [a,a+h] instead of [a,b], - the meaning is still the same, - but the average rate of change is now computed by AV_{[a,b]} = \frac{f(a+h)-f(a)}{h}. +

    + Foundational knowledge or technique 1 +

    • - -
  • - The instantaneous rate of change with respect to x of a function f at a value x = a is denoted f'(a) - (read the derivative of f evaluated at a - or f-prime at a) - and is defined by the formula - - f'(a) = \lim_{h \to 0} \frac{f(a+h)-f(a)}{h} - , - provided the limit exists. - Note particularly that the instantaneous rate of change at x = a is the limit of the average rate of change on [a,a+h] as h \to 0, and that its units are also units of f(x) per unit of x. -
    • -
  • - Provided the derivative f'(a) exists, - its value tells us the instantaneous rate of change of f with respect to x at x = a, - which geometrically is the slope of the tangent line to the curve - y = f(x) at the point (a,f(a)). - We even say that f'(a) is the - slope of the curve - at the point (a,f(a)). -
    • - -
  • - Limits allow us to move from the rate of change over an interval to the rate of change at a single point. +

    + Foundational knowledge or technique 1 +

-

+ + + + + + + + + Additional Section Exercises + + + + + - + Foundational Exercises + + + +
diff --git a/webworkfiles/sec_1-2_prob1.pg b/webworkfiles/sec_1-2_prob1.pg new file mode 100644 index 0000000..113dca8 --- /dev/null +++ b/webworkfiles/sec_1-2_prob1.pg @@ -0,0 +1,105 @@ +## DBsubject(Calculus - single variable) +## DBchapter(Limits and continuity) +## DBsection(Finding limits using graphs) +## Institution(Univeristy of Utah) +## Author(Utah ww group) +## Level(2) +## TitleText1('Mathematical Applications') +## AuthorText1('Ronal J. Harshbarger and James J. Reynolds') +## EditionText1('8') +## Section1('.') +## Problem1('') +## KEYWORDS('limits') + +DOCUMENT(); # This should be the first executable line in the problem. + +loadMacros( + "PGstandard.pl", + "PGchoicemacros.pl", + "PGgraphmacros.pl", + "PGcourse.pl" +); + +TEXT(beginproblem()); +$showPartialCorrectAnswers = 1; + +$a=random(-3,3,1); +$b=random(-2,3,1); +$c=random(-3,2,1); +$m1=random(-1,1,0.5); +$m2=($b - $a)/2; +$m3=($c - $b - 1)/2; +$m4=random(-1,1,0.5); +@slice = NchooseK(3,3); + +@colors = ("blue", "red", "green"); +@sc = @colors[@slice]; #scrambled colors +@sa = ('A','B','C')[@slice]; + +$f1 = FEQ("${m1}*(x+1)+$a for x in [-2,-1) using color:$sc[0] and weight:2"); +$f2 = FEQ("${m2}*(x-1)+$b for x in (-1,1) using color=$sc[0] and weight:2"); +$f3 = FEQ("${m3}*(x-3)+$c for x in [1,3) using color=$sc[0] and weight=2"); +$f4 = FEQ("1+$a for x in [-1,-1] using color=$sc[0] and weight=2"); +$f5 = FEQ("${m4}*(x-3)+$c for x in (3,4] using color=$sc[0] and weight=2"); + +$graph = init_graph(-3,-6,5,6,'axes'=>[0,0],'grid'=>[8,12]); + +($f1Ref,$f2Ref,$f3Ref,$f4Ref,$f5Ref) = plot_functions($graph,$f1,$f2,$f3,$f4,$f5); + +$fig=image(insertGraph($graph), width=>200,height=>200,tex_size=>450); + + + +BEGIN_TEXT +$PAR +Let F be the function below. Evaluate each of the following expressions. $PAR + +$fig + +$PAR +Note: Enter 'DNE' if the limit does not exist or is not defined. $PAR + +a) \( \displaystyle \lim_{x \to -1^-} F(x) \) = \{ans_rule(4)\} +$PAR + +b) \( \displaystyle \lim_{x \to -1^+} F(x) \) = \{ans_rule(4)\} +$PAR + +c) \( \displaystyle \lim_{x \to -1} F(x) \) = \{ans_rule(4)\} +$PAR + +d) \( F(-1) \) = \{ans_rule(4)\} +$PAR + +e) \( \displaystyle \lim_{x \to 1^-} F(x) \) = \{ans_rule(4)\} +$PAR + +f) \( \displaystyle \lim_{x \to 1^+} F(x) \) = \{ans_rule(4)\} +$PAR + +g) \( \displaystyle \lim_{x \to 1} F(x) \) = \{ans_rule(4)\} +$PAR + +h) \( \displaystyle \lim_{x \to 3} F(x) \) = \{ans_rule(4)\} +$PAR + +i) \( F(3) \) = \{ans_rule(4)\} +$PAR + +END_TEXT + +$ap1 = 1 + $a; +$bp1 = 1 + $b; + +ANS(num_cmp($a, strings=>['DNE'])) ; +ANS(num_cmp($a, strings=>['DNE'])) ; +ANS(num_cmp($a, strings=>['DNE'])) ; +ANS(num_cmp($ap1, strings=>['DNE'])); +ANS(num_cmp($b, strings=>['DNE'])) ; +ANS(num_cmp($bp1, strings=>['DNE'])); +ANS(num_cmp('DNE', strings=>['DNE'])) ; +ANS(num_cmp($c, strings=>['DNE'])) ; +ANS(num_cmp('DNE', strings=>['DNE'])) ; + +ENDDOCUMENT(); # This should be the last executable line in the problem. +