DAILY ASSIGNMENTS
Week 1 (First day for U of I classes is Tuesday, January 18, 2011)
Jan 18 (Tue) Go to your discussion section if it meets Tue/Thu.
Jan 19 (Wed) Read sections 1.1 and 1.2. In 1.1 do #2, 5, 6, 21, 30, 35, 41, 45, 48, 63, 65, 69. In 1.2 do #5, 10, 16, 18. You should complete these before your next discussion-recitation section meeting. Be sure that you are prepared for this course – you must have received a 70% or higher on the ALEKS math placement test some time between September 15, 2010 and January 24, 2011. Otherwise you will be automatically dropped from the course. Here are some notes taken by one of my TA's last semester. Since I gave roughly the same introduction this semester, you may find them useful.
Jan 21 (Fri) Read section 1.3. In 1.3 do #3, 8, 9, 11, 12, 13, 17, 18, 21, 31, 32, 33, 38, 41, 43. I spent time showing how to obtain the domain for functions like f(x) = x2 + 3, y = 1 / (2x – 5), g(t) = sqrt(8 – 2t), h(x) = sqrt(x2 – 4x + 3). I then gave the definition for even and odd functions and we looked at examples such as f(x) = x2, g(x) = x3, h(t) = 3t5 + 5, w(x) = (3x2 + 1)3, v(t) = 3cos(t) + t4, p(x) = 2sin(x) + x5, including looking at the graphical approach for the first three of these examples. I then showed how to graph basic functions like x, x2, x3, x4, ..., 1/x, 1/x2, 1/x3, 1/x4, ..., x1/2, x1/3, x1/4, ..., sin(x), cos(x), ex, ln(x). From the reading, students should learn to shift these basic functions to obtain graphs of more complicated functions. I then introduced the definition of cosine and sine in terms of the coordinates of points on a unit circle. We used this to obtain cosine and sine of special angles. I have not yet discussed shifting of graphs or composition of functions, but you should still do the relevant homework problems based on your reading.
Week 2
Jan 24 (Mon) Today I discussed shifting of graphs and graphed y = 3(x – 2)2 + 5 by slowly modifying the graph of y = x2. Students should learn to shift basic functions to obtain graphs of more complicated functions. I then discussed composition of functions. The domain of (f o g)(x) is all x in the domain of g for which g(x) is in the domain of f. The key thing to remember here is that (f o g)(x) = f(g(x)) and we evaluate the inside function g(x) first. For example if f(x) = x2 + 3 and g(x) = sqrt(x – 2) then (f o g)(x) = f(g(x)) = f(sqrt(x – 2)) = (sqrt(x – 2))2 + 3 = x + 1. Even though we can plug any x-value into the expression x + 1, the domain of (f o g)(x) is not all real numbers. The domain of (f o g)(x) is [2, ∞) since we need to evaluate g(x) first. Finally I discussed some basic trigonometry. After today's lecture you should
• Know the definition of cos(θ) and sin(θ) as the x and y coordinates of points along the unit circle.
• Be able to easily switch between degrees and radians.
• Be able to evaluate cos(θ) and sin(θ) for special angles.
• Be able to easily recall these or similar trigonometric identities by thinking about the definition of cos(θ) and sin(θ) in terms of the unit circle.
• cos(–θ) = cos(θ)   (thus cosine is an even function)
• sin(–θ) = –sin(θ)   (thus sine is an odd function)
• cos(θ + 2π)=cos(θ)
• sin(θ + 2π)=sin(θ)
• cos(π – θ) = –cos(θ)
• sin(π – θ) = sin(θ)
• cos(π/2 – θ) = sin(θ)
• sin(π/2 – θ) = cos(θ)
• sin2(θ) + cos2(θ) = 1
• Be able to graph y = sin(x) and y = cos(x). Know the x-intercepts along with the domain and range for both functions.
• Know the other trigonometric functions in terms of sine and cosine. That is,
• tan(θ) = sin(θ) / cos(θ)
• cot(θ) = cos(θ) / sin(θ)
• sec(θ) = 1 / cos(θ)
• csc(θ) = 1 / sin(θ)
• Be able to evaluate tan(θ), cot(θ), sec(θ) and csc(θ) for special angles by first writing everything in terms of sine and cosine.
• Be able to graph y = tan(x). I will discuss the graphs of cot(x), csc(x) and sec(x) later but try to determine the shape of these graphs yourself by rewriting everything in terms of sine and cosine.
• Know the identities tan2(θ) + 1 = sec2(θ) and cot2(θ) + 1 = csc2(θ). These are easily derived by dividing both sides of the equation sin2(θ) + cos2(θ) = 1 by either cos2(θ) or sin2(θ).
Finish Trigonometry Homework #1 and read Appendix D tonight. Homework is never turned in. Quiz #1 will be given by your TA in your first discussion period next week. It will cover sections 1.1, 1.2, 1.3 and trigonometry. We may also have a second quiz on sections 1.5-1.6 in your second discussion period next week. Homework solutions from sections 1.1 — 1.3 are available on Illinois Compass.
Jan 26 (Wed) Skip section 1.4 and read section 1.5. For homework do #35, 36, 37, 38 in Appendix D using a calculator. Then do #7, 8, 9, 10, 11, 12, 13, 15, 16, 17, 18, 21, 25, 26 in section 1.5. In a right triangle where θ is one of the acute angles we label the length of the side opposite θ as opp, the length of the side adjacent θ as adj, and the length of the hypotenuse as hyp. In lecture I used similar triangles and the definition of cosine and sine on the unit circle to obtain the following relationships in a right triangle.
• sin(θ) = opp / hyp
• cos(θ) = adj / hyp
• tan(θ) = opp / adj
• csc(θ) = 1 / sin(θ) = hyp / opp
• sec(θ) = 1 / cos(θ) = hyp / adj
• cot(θ) = 1 / tan(θ) = adj / opp
I used right triangle trigonometry to solve problems like the following:
• Given that 0 < θ < π/2 and tan(θ) = 3, determine the value of cos(θ).
• Given that 0 < θ < π/2 and sin(θ) = x/2, determine the value of tan(θ).
I then talked about recognizing linear versus exponential functions from a table of values, finding formulas for the linear functions in the form y = mx + b and formulas for the exponential functions in the form y = C*a^x. In general we get exponential growth if a > 1 and exponential decay if 0 < a < 1. We only looked at exponential growth so far and graphed the exponential function y = 3*2^x. We then looked at a population which was 50 in the year 1980 but doubled every 10 years after that. We sought to determine when the population would reach 600. A quick table showed that this would occur somewhere between 30 and 40 years after 1980. To determine a more precise answer we found a formula for the population (P) as a function of the number of years after 1980 (t). We then plugged in P = 600 and used logarithms to solve for the time t. Students should know the basic rules of logarithms. Since students had difficulty doing this in class, an additional homework problem is to find a formula for an exponential function which has the following table of values.

x y
1 2
3 4
5 8
7 16

Quiz #1 will be given by your TA in your first discussion period next week. It will cover sections 1.1, 1.2, 1.3 and trigonometry. Homework solutions from sections 1.1 — 1.3 are now available on Illinois Compass. Solutions to the trigonometry problems are posted on the course home page. Quiz #2 on sections 1.5 and 1.6 will be given by your TA in your second discussion period next week.

Jan 28 (Fri) Read section 1.6. In 1.6 do #33–39, 47–50, 7–10, 15, 17, 19, 21–26. Prepare for two quizzes next week (see above for topics and dates).

In lecture I plugged some of the points from the table shown last time into the formula y = C*a^x to arrive at the formula y = sqrt(2)*sqrt(2)^x. There are other ways to write this formula.

I sketched graphs for a couple of exponential functions. In the formula y = C*a^x, we get exponential growth if a > 1 and exponential decay if 0 < a < 1.

We then discussed inverse functions. This included the important features of a table of values, a graph, and a formula for a function and its inverse. We saw multiple ways to obtain the formula for the inverse of a function. I also discussed the concept of one-to-one functions and the horizontal line test as a graphical way to see whether or not a function is one-to-one and thus has an inverse. We saw that for functions like f(x)=x^2 which are not one-to-one, we could restrict the domain so that it has an inverse.

I went over some of the basic rules for working with logarithms of any base including base e for the natural logarithm. We used these rules to solve the equation 3 = 24x – 1 for x.

Week 3 (Deadline to add this course is Monday, January 31, 2011)
Jan 31 (Mon) For homework read sections 2.1 and 2.2. In 1.6 do #53, 54. In 2.2 do #4, 7, 8, 12, 13, 15, 21, 22, 25, 27, 28, 33, 35. Prepare for both quizzes this week in your discussion section. Quiz 1 will be on 1.1, 1.2, 1.3 and trigonometry. Quiz 2 will be on 1.5 and 1.6.

You should know the following identities and simplification rules.

• sin(2θ) = 2sin(θ)cos(θ)
• cos(2θ) = cos2(θ) – sin2(θ) which can also be written in the following two ways.
• cos(2θ) = 2cos2(θ) – 1
• cos(2θ) = 1 – 2sin2(θ)
• ln(AB) = ln(A) + ln(B)
• ln(A/B) = ln(A) – ln(B)
• ln(Ak) = k ln(A)
• eln(x) = x for all x > 0
• ln(ex) = x for all x
• ln(1) = 0
• ln(e) = 1
• e0 = 1
In lecture today, we used the informal definition of limit as found on page 88 to investigate the value of the following limits.
1. limx → 3 ( x2 )
2. limx → 4 ( (x2 – 2x – 8) / (x – 4) )
3. limx → 0 ( sin(x) / x )
4. limx → π/2 ( tan(x) )
5. limx → 7 ( (sqrt(x + 2) – 3)/(x – 7) )
I graphed the first four functions and illustrated graphically what is meant by a limit. Even though many students immediately expected the limit in (1) to be 9, it was harder to guess what the limit in (2) would be. We factored the numerator and cancelled the x – 4. This resulting in us expecting the limit in (2) to be 6. We could not easily simplify the expression in (3) so we made a table of values for y = sin(x) / x and plugged in values of x getting closer to 0 (I used 0.1, 0.01, and 0.001). We then saw a trend and guessed the limit. Since students got more than one answer for (3), I mentioned the need to put the calculator in radian mode to get the limit of 1. For the limit in (4) we saw from the graph of tan(x), that we got different limits depending on whether we approached π/2 from the left or the right. So we introduced the proper notation (using – or + as shown in the book) to approach π/2 from one particular side. For the limit in (5) we multiplied the numerator and denominator by (sqrt(x + 2) + 3) / (sqrt(x + 2) + 3) to rewrite the expression in a different form before guessing the appropriate limit. We next looked at an example where S(t) = 2t gives the size of a tumor in cubic millimeters t months after its discovery. In order to determine how quickly the size of the tumor is increasing precisely 6 months after its discovery, we ended up taking the following limit.

• S ′(6) = limt → 6 ( (2t – 26) / (t – 6) )

We made a table of values to approximate this limit. I mentioned that this S'(6) notation for the speed at which the tumor is growing at precisely t = 6 is called the derivative. Since some students have learned a little about derivatives before, I asked if anyone knew a formula for S'(t) so that we could simply plug t=6 into this formula. Nearly all of the students that have seen derivatives before obtained an incorrect formula for S'(t). I used this example to point out the need for understanding the method I am using with tables of values and other techniques before trying to use short-cut techniques.

Feb 2 (Wed) University classes are cancelled today due to weather conditions. I have also cancelled my office hours. If you have a Tues/Thurs discussion section, you should still be prepared for quiz #2 on Thursday. See homework solutions at Illinois Compass.
Feb 4 (Fri) Everyone should read sections 2.3 and 2.5. Math majors or those who want a better understanding of proof techniques for limits should read section 2.4. In section 2.3 do #11, 13, 15, 17, 18, 20, 21, 26, 35, 37. There will be a quiz in your second discussion section next week.

In lecture today I began by proving limx → 0 ( sin(x) / x ) = 1 by starting with the unit circle and drawing a small triangle, the sector of a circle, and a large triangle. We then compared areas and used The Squeeze Theorem. I then stated this theorem more formally as given on page 105.

We next evaluated the following limits.

1. limx → 3+ ln(x – 3)
2. limx → 4 (x + 2) / (x – 4)
3. limx → 0 ( 3 + 2e–x )
4. limx → ∞ ( 3 + 2e–x )
5. limx → –∞ ( 3 + 2e–x )
6. limw → 1 (w4 – 1)/(w – 1)
7. limx → 0 x2 sin(1 / x)
I talked about the numerical approach which can sometimes provide evidence to suggest a particular limit. However for these problems we mostly used a graphical approach to quickly provide answers for limits (1), (3), (4) and (5). For limit (2) we rewrote (x + 2) / (x – 4) = (x – 4 + 6) / (x – 4) = (x – 4) / (x – 4) + 6 / (x – 4) = 1 + 6 / (x – 4) and then used the graphical approach. For limit (6) we factored the numerator and cancelled terms in order to obtain the limit. For limit (7) we saw that –1 ≤ sin(1/x) ≤ 1 so that –x2 ≤ x2sin(1/x) ≤ x2. This limit was then easily found using The Squeeze Theorem.

Although it is good to use common sense, we saw from a few examples that what we think of as common sense may not actually be correct in a given situation. For instance, even though limx → 0 ( x2 ) = 0, we do not immediately know the value of limx → 0 ( x2 * f(x)). We looked at f(x) = 5x + 2, f(x) = 1/x6, and f(x) = (6x + 8) / x2 to see that limx → 0 ( x2 * f(x)) might equal 0, ∞, 8, or some other value depending upon the choice for f(x).

Does common sense help us determine limx → ∞ ( 1 + 1/(2x) )x ? Many students expect the limit to be either 1 or ∞ depending upon whether they concentrate on the term in parentheses or the exponent. We will see later that surprisingly this limit is equal to the square root of e.

Finally I defined f to be continuous at x = a if limx → a f(x) = f(a). We looked at a few graphical examples which demonstrated when f was not continuous at a particular point.

Week 4
Feb 7 (Mon) Read section 2.6 and reread the material from section 1.6 on inverse trigonometric functions. In section 1.6 do #59, 60, 61, 64, 66, 67. In section 2.5 do #18, 41, 45, 47, 49. In section 2.6 do #5, 6, 15, 16, 19, 20, 25, 28, 29, 39, 41. We will have an in-class quiz in your last discussion section this week. It will cover sections 1.6, 2.1, 2.2, 2.3, 2.5 and 2.6.

In lecture we looked carefully at why we restrict the domain of f(x) = sin(x) in order for it to have an inverse function f-1(x) = sin-1(x) = arcsin(x). Note that sin-1(x) is not the same as 1/sin(x). Students should also understand how the domains of cos(x), tan(x), and sec(x) are restricted in order to have inverse functions. We then evaluated the following quantities.

• sin-1(-1/2)
• tan-1(sqrt(3)/2)
• sin(sin-1(1/5))
• sin-1(sin(3π/4)) (be very careful with this one — the answer is not 3π/4)
• cos(sin-1(1/3))
• sec(sin-1(x))
Although we could use identities to answer these last two questions, it is a bit easier to use the approach of drawing a particular right triangle as shown in lecture.

I introduced some of the limit laws from section 2.3 but explained why we will rarely need to use them once we understand continuity a little better. We defined f to be continuous at x=a if limx → a f(x) = f(a). Nearly every function we will use in calculus is continuous on its domain so we can often just plug in a particular x-value to determine a limit. In particular, polynomials, exponentials, logarithms, roots, trig functions, inverse trig functions and rational functions are all continuous on their domains. We can also combine continuous functions by adding, subtracting, multiplying or dividing and obtain another continuous function. We can even do composition of functions to get another continuous function. For dividing, just make sure that the denominator is not 0. For composition of functions we have to be a little bit careful but I'll explain why next time.

We then discussed the very important Intermediate Value Theorem. We used it to explore the location of the roots (i.e. x-intercepts) for f(x) = x3 – 3x2 – x + 5. For homework, quizzes and tests, I expect students to very clearly show how and why they may apply theorems such as The Intermediate Value Theorem or The Squeeze Theorem.

Feb 9 (Wed) Read sections 2.7 and 2.8. In section 2.7 do #5, 6, 7, 8, 9, 10, 13, 14, 25, 26, 27, 28, 29, 30. In section 2.8 do #4, 5, 6, 12, 14, 15, 16, 19, 21, 23, 24. Prepare for this week's quiz on sections 1.6, 2.1, 2.2, 2.3, 2.5 and 2.6. Prepare for next Wednesday's test on sections 1.1, 1.2, 1.3, 1.5, 1.6, 2.1, 2.2, 2.3, 2.5, 2.6, 2.7, 2.8, and the trigonometry material discussed in lecture.

Today we discussed limits which are of the indeterminate forms ∞ / ∞ or ∞ – ∞ and used algebra in order to evaluate the following limits.

• limx → ∞ (2x2 + 1) / (5x2 + 2x + 3) = 2/5
• limx → ∞ (6x + 1) / (5x2 + 4x + 10) = 0
• limx → ∞ (8x3 + x - 6) / (4x2 + 2) = ∞
• limx → ∞ (2x2 + 5) / sqrt(9x4 + 2x + 6) = 2/3
• limx → ∞ (sqrt(4x2 + 3x) - 2x) = 3/4
We then found sin(2*tan-1(1/5)) by letting θ = tan-1(1/5) or tan(θ) = 1/5 and drawing a right triangle with opp / adj = 1 / 5. With Pythagorean's Theorem we found the hypotenuse to be sqrt(26). This resulted in

sin(2*tan-1(1/5)) = sin(2*θ) = 2*sin(θ)*cos(θ) = 2*(1/sqrt(26))*(5/sqrt(26)) = 10/26 = 5/13.

I then defined the derivative of f(x) as

f ′(x) = limh → 0 (f(x+h) - f(x)) / h

I discussed the interpretation of a derivative as a rate of change as well as the slope of the tangent line. We used the limit definition of a derivative to show that f(x) = x2 has derivative f ′(x) = 2x. We also used the limit definition of a derivative to investigate the form of f ′(x) given that f(x) = 2x.

Next we used the graph of a function y = f(x) to determine a reasonable graph for the derivative function y = f ′(x).

Feb 11 (Fri) Wednesday's test will be given during your lecture period and will include all sections in chapter 1 and 2 except 1.4 and 2.4. It will also include the trigonometry material covered in lecture. You must present your student ID in order to turn in your test. The cover page on your test will include a seating chart. When you pick up your test on Wednesday, one of these seat numbers will be circled and that will be your assigned seat for the first test. Try to arrive early to obtain your assigned seat. Unless you happen to sit at the end of a row, you will be required to stay for the full 50 minute testing period. The TA's will not answer any questions during the test. Check the list of office hours for me and the TA's. It may be updated to include extra hours so check back regularly.

In lecture today we saw graphically why the following three definitions of derivative are equivalent.

1. f ′(x) = limh → 0 (f(x + h) – f(x)) / h
2. f ′(x) = limw → x (f(w) – f(x)) / (w – x)
3. f ′(a) = limx → a (f(x) – f(a)) / (x – a)
We used the 1st approach to find that the derivative of f(x) = sqrt(x) is f ′(x) = 1/(2*sqrt(x)). We used the 2nd approach to find that the derivative of f(x) = x3 is f ′(x) = 3x2. For homework you should use the 1st approach to see that you obtain the same formula for the derivative of f(x) = x3. Usually one just uses whichever approach seems easiest for a particular function.

I again mentioned how the derivative, the rate of change, and the slope all represent the same quantity. We looked at a population P(t) = 2000 + 3t2. We calculated P(5) = 2075 and P ′(5) = 30 to show that 5 years later the population is 2075 people and increasing by 30 people per year. We then looked at the height of a ball thrown upwards from an apartment window h(t) = –16t2 + 96t + 160. We saw that h(0) = 160 feet is the height of the window. We graphed h(t) to see it had a slope of 0 when the ball reached its maximum height. Then we found the velocity h ′(t) = –32t + 96. We set h ′(t) = 0 to determine that the ball reached its maximum height at t = 3 seconds. We then found h(3) = 304 feet to be the maximum height. I mentioned that you could set h(t) = 0 to find when the ball fell back to the ground. We could then plug this value of t into the velocity formula to obtain the velocity at the moment the ball hit the ground.

We discussed some of the different notation to refer to the 1st derivative, 2nd derivative, 3rd derivative, etc. We used f ′(x), f ′′(x), f ′′′(x) as well as Leibniz notation dy/dx, d2y/dx2, d3y/dx3. Students should be comfortable with the notation for evaluating derivatives at a point as in f ′(5) or dy/dx|x=5. Students should also be comfortable using different variables. Using Leibniz notation the derivative of P = t3 is dP/dt = 3t2 and the derivative of w = r2 is dw/dr = 2r.

I showed why a function is not differentiable at any point where there is a sharp corner on a graph.

I proved the important theorem which states that if a function is differentiable at a point, then it must also be continuous at that point.

Finally we used the graph of a function y = f(x) to determine a reasonable graph for the derivative function y = f ′(x). We saw that graph of the derivative of sin(x) looks like cos(x) and the graph of the derivative of ex looks like ex. Later we will prove that these are actually the correct formulas for these derivatives.

Week 5
Feb 14 (Mon) Prepare for Wednesday's test on sections 1.1, 1.2, 1.3, 1.5, 1.6, 2.1, 2.2, 2.3, 2.5, 2.6, 2.7, 2.8, and the trigonometry material discussed in lecture. No calculators or notes are allowed, and you should bring a student ID. See the notes written here after each lecture to get a summary of some of the important material. It is expected that you now know how to solve every homework problem. Detailed solutions to all odd problems and assigned even problems are available at Illinois Compass. Solutions to the quizzes and the trigonometry worksheet are posted on the course homepage. You will need to be able to state and use the definitions of even functions, odd functions, continuity, derivatives. You will need to be able to state and use the Intermediate Value Theorem, Squeeze Theorem, and the theorem which says If f is differentiable at a, then f is continuous at a. You will definitely have one problem where you will be asked to find the derivative of a function using limits. Use proper notation as you show all the appropriate steps. You may want to look over old tests and quizzes from previous semesters that I taught a Calculus or Calculus I course. Check the available tutoring hours. No appointment is necessary. Be sure to go to both discussion sections this week.
Feb 16 (Wed) Test 1 (given during lecture)
Feb 17 (Thu) Discussion sections still meet today. Your homework is to read section 3.1 and do #3–30, 33, 35, 49, 51, 53.
Feb 18 (Fri) Read sections 3.1–3.2 to learn the short-cut methods for finding derivatives. In section 3.1 do #3–30, 33, 35, 49, 51, 53. In section 3.2 do the odd problems from #3–33. There will be a quiz Thursday (Wednesday for Merit sections) on sections 3.1–3.2.

In lecture we saw from the graphical interpretation of a derivative as a slope that the derivative of a constant is 0 since the graph of a constant function is a horizontal line which has slope 0. We also used this approach to see that the derivative of f(x) = mx + b is f ′(x) = m. In addition to the graphical approach, we could also have used limits to prove these derivative rules.

I then showed how limits can be used to prove the following derivative rules.

• ( xn ) ′ = nxn-1     (True for all real numbers n, our proof with limits was only for positive integers n)
• ( cf(x) ) ′ = cf ′(x)     (Do a proof with limits on your own)
• ( f(x) + g(x) ) ′ = f ′(x) + g ′(x)     (A proof with limits was given in lecture)
• ( f(x) – g(x) ) ′ = f ′(x) – g ′(x)     (Do a proof with limits on your own)
• ( f(x)g(x) ) ′ = f ′(x)g(x) + f(x)g ′(x)     (A proof with limits was given in lecture)
• ( f(x) / g(x) ) ′ = ( f ′(x)g(x) – f(x)g ′(x) ) / ( g(x) )2     (Try a proof with limits on your own.
These last two rules are referred to as the Product Rule and the Quotient Rule, respectively. Don't be too worried if you have trouble proving the product rule or quotient rule on your own, but make sure that you understand and can duplicate the proofs of the derivative rules for f(x) + g(x), f(x) – g(x), and cf(x).

Next week we will use limits to derive short-cut methods for finding derivatives of all the basic functions we have discussed in calculus including the composition of functions. Here are the basic derivative rules I plan to discuss. It will be helpful for you to quickly memorize (or be able to derive) these rules. I plan to give more frequent quizzes to check that you are keeping up.

• ( c ) ′ = 0     (c is any constant)
• ( xn ) ′ = nxn–1
• ( ex ) ′ = ex
• ( ax ) ′ = axln(a)     (a > 0)
• ( ln(x) ) ′ = 1 / x
• ( sin(x) ) ′ = cos(x)
• ( cos(x) ) ′ = – sin(x)
• ( tan(x) ) ′ = sec2(x)
• ( cot(x) ) ′ = – csc2(x)
• ( sec(x) ) ′ = sec(x)tan(x)
• ( csc(x) ) ′ = – csc(x)cot(x)
• ( tan–1(x) ) ′ = 1 / (1 + x2)
• ( sin–1(x) ) ′ = 1 / sqrt(1 – x2)
• ( cos–1(x) ) ′ = – 1 / sqrt(1 – x2)
• ( sec–1(x) ) ′ = 1 / (x*sqrt(x2 – 1))
Week 6
Feb 21 (Mon) Read sections 3.3 and 3.4. In section 3.3 do the odd problems from #1–23. There will be a quiz in your last discussion section this week on sections 3.1 and 3.2. There will be two quizzes next week.

In lecture today I rewrote w(x) = f(x) / g(x) as w(x)*g(x) = f(x) and took the derivative of both sides. By then solving for w ′(x) we were able to derive the quotient rule. If you search YouTube for quotient rule or quotient rule song you will likely find many different mnemonic devices or songs to remember this rule.

I used limits to find the derivative of sin(x) and mentioned the similar method for finding the derivative of cos(x). I then wrote tan(x) = sin(x) / cos(x) and used the quotient rule to obtain the derivative of tan(x). I mentioned that this same technique could be used for the other trig functions.

I also used limits to show that the derivative of f(x) = ax is f ′(x) = f ′(0) ax where f ′(0) = limh → 0 ( (ah – 1)/h ). This limit turns out to be ln(a) so that f(x) = ax has derivative f ′(x) = ln(a) ax. In particular we get that f(x) = ex has derivative f ′(x) = ex. The rest of the class was spent on examples and answering homework questions.

Feb 23 (Wed) Read section 3.4. In section 3.4 do the odd problems from #7–55. There will be two quizzes next week. On Tuesday (Monday for Merit sections) there will be a quiz on section 3.3. On Thursday (Wednesday for Merit sections) there will be a quiz on section 3.4.

In lecture we used the product rule to show that ( f(x) g(x) h(x) ) ′ = f ′(x) g(x) h(x) + f(x) g ′(x) h(x) + f(x) g(x) h ′(x). This generalizes naturally to the product of more terms. Since ( f(x) )2 = f(x) f(x), ( f(x) )3 = f(x) f(x) f(x), etc., I used this generalized product rule to obtain the following derivatives

• ( f(x)2 ) ′ = 2 f(x) f ′(x)
• ( f(x)3 ) ′ = 3 ( f(x) )2 f ′(x)
This generalizes to the following derivative rule.
• ( f(x)n ) ′ = n ( f(x) )n–1 f ′(x)
This was our first hint of the more general chain rule for derivatives of the composition of functions which says that
• ( f(g(x)) ) ′ = f ′(g(x)) g ′(x)
The chain rule can also be written as dy / dx = (dy / du ) * (du / dx).

I then worked through many examples. We saw some that resulted from a chain of two functions, but others that resulted from a chain of 4 or 5 functions. We saw examples that required a combination of the product rule, quotient rule, and chain rule. We saw some complicated functions for which it was much easier to simplify before taking the derivative.

Finally we found the derivative for y = ln(x) by rewriting it as ey = x, finding dx / dy = ey, and then writing dy / dx = 1 / (dx / dy) = 1 / ey = 1 / x.

Feb 25 (Fri) Read sections 3.5 and 3.6. In section 3.5 do #5, 7, 9, 11, 13, 15, 17, 19, 27, 28, 29, 30, 45, 46, 47. Prepare for next week's two quizzes on sections 3.3 and 3.4.

We saw another way to obtain the derivative of ln(x) by writing eln(x) = x and taking the derivative of both sides. I then discussed the change of base formula logb(x) = ln(x)/ln(b) to obtain that the derivative of logb(x) is 1/(xln(b)).

We looked at explicitly defined functions versus implicitly defined functions. We found the slope of the tangent line to the curve x2 + y2 = 25 at the point (3,4) in two ways. First we solved for y explicitly, took the derivative, and plugged in x=3. Next we just took the derivative of both sides of the original equation to get 2x + 2y*dy/dx = 0, solved to get dy/dx = -x/y, and plugged in x=3 and y=4 to obtain the same answer. This 2nd approach is called implicit differentation and you have to think carefully to see that we used the chain rule to obtain that the derivative with respect to x of y2 is 2y*dy/dx. We next found the derivative dy/dx for the following implicitly defined functions. For the first example we also plugged in (x, y) = (2, 4) to find the slope of the curve at that point.

• x3 + y3 = 9xy   (this curve is called the folium of Descartes)
• y2 + x5y + 2x = 0
• 1 + x = sin(xy2)
For the second example above, we could have used the quadratic formula to obtain that y = ( –x5 ± sqrt(x10 – 8x) ) / 2 and then found the derivative, but implicit differentation is an easier approach. We then discussed the ease of finding roots of polynomials of degree 2 using the quadratic formula, the difficulty of finding roots of polynomials of degree 3 and 4, and the impossibility of finding an explicit formula for the roots of an arbitrary polynomial of degree 5 or higher.

I next showed how to rewrite y = sin–1(x) as sin(y) = x and differentiate each side with respect to x in order to eventually obtain dy/dx = 1/sqrt(1 – x2). For additional homework, students should use this same approach to discover the derivative of tan–1(x).

Week 7
Feb 28 (Mon) Read sections 3.6, 3.7 and 3.8. In section 3.6 do #3, 7, 11, 13, 16, 19, 31, 33, 37, 41, 43. In section 3.7 do #7, 8, 9, 10. In section 3.8 do #3, 4, 8, 9, 10, 11, 12. On Tuesday (Monday for Merit sections) there is a quiz on section 3.3. On Thursday (Wednesday for Merit sections) there is a quiz on section 3.4. We will also have two quizzes next week.

In lecture I showed how to obtain the derivative rule for tan–1(x) and mentioned the other inverse trigonometric derivative rules. I did #15 and #46 from section 3.5. I then discussed logarithmic differentation and its use in finding derivatives of the following functions.

• y = ln( x5 ex3 (x4 + 3)6 )   (simplify using rules of logarithms before differentiating)
• y = x3 sqrt(x4 + 3) / (5x + 1)8   (take the natural logarithm of both sides, simplify, then use implicit differentiation)
• y = xcos(x)   (take the natural logarithm of both sides, simplify, then use implicit differentiation)
Students should use this approach to find derivative of y = xx.

I had intended to find the derivative of y = xn where n is any real number but ran out of time. The derivative rule is exactly what you would expect ( dy/dx = nxn–1 ), but we never proved the formula to be valid when n is any real number. Interested students should prove this short-cut derivative rule by taking the natural logarithm of both sides before using implicit differentiation.

My only applications from section 3.7 are the ones concerning position, velocity and acceleration. Since I have already discussed these concepts, I will expect students to already have the tools needed to solve these homework problems. I then introduced section 3.8 by talking about differential equations. We guessed then checked potential solutions to the following differential equations and initial values.

• dy/dx = 3ex, y(0) = 8   (solution: y = 3ex + 5)
• dy/dx = 10x, y(2) = 30   (solution: y = 5x2 + 10)
• dy/dx = 8x, y(0) = 20   (solution: y = 4x2 + 20)
• dy/dx = 8y, y(0) = 20   (solution: y = 20e8x)
These last two look very similar. Carefully check the variables used to see why they lead to two different solutions. We obtain in general that the differential equation dy/dx = ky where k is a constant has solution y = Cekx. If an initial value is given then we can plug in that particular point to solve for unknowns like C.
Mar 2 (Wed) Read section 3.9. In 3.9 do #6, 10, 13, 15, 20, 23, 24, 27, 28, 30, 31, 39. There will be two quizzes next week. Quiz #7 on sections 3.5, 3.6, 3.7 and 3.8 will be a take-home quiz posted online by 5:00pm this Friday. It will be due at the beginning of your first discussion period next week. I'll give details about quiz #8 later. The test is in two weeks so don't fall behind. Sections 3.9 and 4.7 include a lot of word problems so be prepared to work hard if this is an area of weakness for you.

In lecture I continued our analysis of differential equations with initial values such as the following.

• dP/dt = 12t, P(0) = 25   (solution: P = 6t2 + 25)
• dP/dt = 12P, P(0) = 25   (solution: P = 25e12t)
We could check the first problem directly to make sure that our answer of P = 6t2 + 25 satisfied the differential equation (yes, dP/dt = 12t) and the initial value (yes, P(0) = 25). Since it satisfied the differential equation and had the correct initial value, we know that we have found the correct solution. For the second problem, we can quickly see that our answer of P = 25e12t does give the correct initial value of P(0) = 25. To check that the differential equation is satisfied, we compare the left side of the equation to the right side of the equation. Using the function P = 25e12t, for the left side we get dP/dt = 25e12t*12. Using the function P = 25e12t, for the right side we get 12P = 12*25e12t. The left side equals the right side so dP/dt = 12P. Now we know that we have found the correct solution.

In general the differential equation dy/dx = ky where k is a constant has solution y = Cekx. For example dy/dx = 5y with y(0)=4 has solution y = Ce5x. By plugging in (x, y) = (0, 4) we find that C = 4 so that our solution becomes y = 4e5x. A second example is dy/dx = 3y with y(5) = 2 which has solution y = Ce3x. By plugging in (x, y) = (5, 2) we find that C = 2/e15 so that our solution becomes y = (2/e15)e3x or y = 2e3x–15.

We looked at two populations. One population is currently 100 and increasing by 5 people per year. Another population is currently 100000 and increasing by 5000 people per year. We define relative growth rate as (dP/dt) / P. Since 5/1000 = 5000/100000 = 0.05, we see that these two populations are growing at the same relative growth rate of 5%.

If a population is currently 200 and growing at a constant relative growth rate of 3%, then this leads directly to the differential equation dP/dt = 0.03P with P(0) = 200. Solving this differential equation gives us the following formula for the population: P = 200e0.03t.

If a quantity A is proportional to B, then this means that A = k*B where k is a constant. That is, you can translate "is proportional to" to "equals a constant times". Note then that if a population is growing at a rate which is proportional to the population size, this translates to dP/dt = k*P which has solution P = Cekt.

Students should also know the meaning of the term half-life and be able to use exponential functions to help solve problems involving half-lives. The example solved in class was to determine how long it takes for 100mg of caffeine in the bloodstream after a cup of coffee to be reduced to 10% of that amount. The half-life of caffeine in the bloodstream is about 4 hours for most people but closer to 10 hours for pregnant women.

I then solved the first and third problems on this related rates worksheet. Try to do the second one on your own. Solutions will be provided soon.

Mar 4 (Fri) For homework read section 4.1 and do #16–25, 31, 41, 43, 49–60 from that section. Quiz #7 is a take-home quiz on sections 3.5, 3.6, 3.7 and 3.8. It should be turned in at the beginning of your next discussion section meeting.

In lecture today I further discussed strategies for solving related rates problems. I solved problem #39 from section 3.9 and the second problem on the related rates worksheet.

We obtained a graph of the basic shape of f(x) = x4 – 4x3 + 16x – 16 by first looking at its derivative f ′(x) = 4x3 – 12x2 + 16. Since the derivative factors as f ′(x) = 4(x + 1)(x – 2)2 we can quickly see which x-values cause the derivative to be positive, negative or zero. This tells us where the graph of f(x) is increasing, decreasing or level.

We used this same approach to obtain a graph of f(x) = 5xe–2x = 5x / e2x. We found f ′(x) = (5 – 10x) / e2x and noted that f ′(x) > 0 for x < 1/2, f ′(x) = 0 for x = 1/2, and f ′(x) < 0 for x > 1/2. Thus the graph of f(x) is increasing for x < 1/2, level at x = 1/2, and decreasing for x > 1/2. Even though the graph of f(x) is decreasing for x > 1/2, we see from the formula for f(x) that the y-values never become negative. We used this along with limx → ∞ f(x) = limx → ∞ ( 5x / e2x ) = 0 to get a better graph for this function.

The derivative function f ′(x) tells us the shape of the graph of f(x) but not the y-values. When actually graphing a function f(x) we should plug specific x-values into f(x) to obtain the corresponding y-values. Next week we'll see how the second derivative gives us further information about the shape of a graph.

I introduced terms such as absolute maximum, absolute minimum, local maximum and local minimum. We looked at a few different graphs to determine whether or not a given function had absolute or local maxima or minima. We looked in particular at the function f(x) = tan–1(x) along with its graph to see that it had no absolute maximum and no absolute minimum.

Week 8 (Deadline to drop this course without a grade of W is Friday, March 11, 2011)
Mar 7 (Mon) Read sections 4.3 and 4.7. In section 4.3 do #10, 13, 17, 33, 39, 43, 45, 50, 53. In section 4.7 do #11, 12, 17, 19, 24, 30, 32, 33, 36. Turn in Quiz #7 at the beginning of your first discussion section meeting this week. Turn in Quiz #8 on sections 3.9, 4.1 and 4.3 at the beginning of your second discussion section meeting this week.

In lecture today we discussed the terms increasing, increasing, concave up and concave down. In particular we obtained a graph of f(x) = 2x3 + 3x2 – 36x. We discussed critical points and discussed the fact that a continuous function on a closed interval always attains an absolute maximum and an absolute minimum. The closed interval method guarantees that for such continuous functions on closed intervals, we can find the absolute extreme values by plugging in the critical points (points where the derivative is 0 or undefined) and the endpoints.

We next solved the following optimization problems.

• A farmer wishes to enclose a rectangular pen with area 100 square feet next to a road. The fence along the road is to be reinforced and costs \$34 per foot. Fencing that costs \$16 per foot can be used for the other three sides. What dimensions for the pen will minimize the cost to the farmer? What is that minimum cost?
• A rectangle is to be inscribed in a semi-circle of radius 2. What is the largest possible area and what are the dimensions that will give this area?
Mar 9 (Wed) Read section 4.4. In 4.4 do #7, 11, 16, 17, 19, 29, 31, 40, 43, 44, 48, 52, 55, 59, 60, 65. Next Wednesday's test will cover sections 3.1-3.9, 4.1, 4.3, 4.4, 4.7. Quiz #8 is a take-home quiz which should be turned in at the beginning of your last discussion section meeting this week. In lecture today we discussed limits for indeterminate forms such as
• 0 ÷ 0
• ±∞ ÷ ±∞
• 0 × ±∞
• ∞ – ∞
• 1
• 00
• 0
You should think about why forms like 0, ∞1 and 1 ÷ 0 are not indeterminate. For example, we can directly obtain limx → ∞ ( 1 / x )x2 = 0 and limx → 0 ( cos(x) / x2 ) = ∞.

We used L'Hospital's Rule to determine the following limits.

1. limx → 0 ( (3x – sin(x)) / x ) = 2
2. limx → 0 ( (sqrt(1 + x) – 1 – x/2) / x2 ) = –1/8
3. limx → 0 ( (1 – cos(x)) / (x + x2) ) = 0
4. limx → ∞ ( ln(x) / (2sqrt(x)) ) = 0
5. limx → ∞ ( e2x / (x2 + 3x + 5) ) = ∞
6. limx → ∞ ( x2 / ex ) = 0
7. limx → 0+ ( x ln(x) ) = 0
8. limx → 0 ( 1 / sin(x) – 1 / x ) = 0
9. limx → ∞ ( (1 + 1 / x)x ) = e
For examples (4), (5) and (6) we saw how orders of growth to infinity should immediately give us answers without the need for L'Hospital's Rule. Specifically as x approaches ∞ the following functions approach ∞ from slowly to quickly in the order shown.

(slowly)   ln(x), ..., x1/3, x1/2, x, x2, x3, ..., ex   (quickly)

If you need to take the limit of the ratio of two such functions, then the slowness or quickness of growth toward ∞ should be enough to tell you if the limit of the ratio is 0 or ∞. For example one should immediately see that limx → ∞ ( (4x1000 + 5x50 + 10) / (0.001e2x) ) = 0 since the numerator approaches ∞ slowly while the denominator approaches ∞ quickly.

Mar 11 (Fri) In lecture we discussed the Second Derivative Test to determine when a function has a local maximum or local minimum. We looked at g(x) = 200 + 8x3 + x4, f(x) = (x – 1)4 and y = x1/3, and answered questions dealing with the terms critical point, increasing, decreasing, local maximum, local minimum, concave up, concave down and inflection point.

I answered homework questions including #36 from section 4.7. For this problem we used similar triangles and Pythagorean's Theorem to obtain a formula for L2, the square of the length of the ladder L, as a function of the distance x between the base of the ladder and the bottom of the fence. Although we could take the square root of both sides to obtain a formula for L and then take the derivative, we found it easier to immediately take the derivative of both sides with respect to x to get 2L*dL/dx on the left. Now setting dL/dx = 0 results in a simpler equation.

Use the Test 2 Notes to begin your preparation for Wednesday's test. No new homework was assigned today. Come prepared with questions on Monday.

Week 9
Mar 14 (Mon) Prepare for Wednesday's test on sections 3.1, 3.2, 3.3, 3.4, 3.5, 3.6, 3.7, 3.8, 3.9, 4.1, 4.3, 4.4 and 4.7. See the Test 2 Notes for further details. No calculators or notes are allowed and you should bring a student ID. See the notes written here after each lecture to get a summary of some of the important material. It is expected that you now know how to solve every homework problem. Detailed solutions to all odd problems and assigned even problems are available at Illinois Compass. Solutions to the quizzes are posted on the course homepage. You may want to look over old tests and quizzes from previous semesters that I taught a Calculus or Calculus I course. Check the available tutoring hours. No appointment is necessary.
Mar 16 (Wed) Test 2 (given during lecture)
Mar 16-17 (Wed-Thu) The TA's will introduce new material in discussion section. For homework read section 4.9. In 4.9 do #3-15, 18-20, 23-31, 39-41, 63, 67, 71, 72, 73.
Mar 18 (Fri) Finish the homework from section 4.9. Read section 5.1 very carefully. Do #3, 4, 11, 12, 13, 16, 18 from section 5.1. Quiz #9 on sections 4.9 and 5.1 will be given in your last discussion period the week after Spring Break.
Week 10 (Spring Break!)
Week 11
Mar 28 (Mon) Read section 5.2 very carefully. Do #9, 18, 21, 23, 29, 33, 35, 36, 41, 48, 49, 53, 55 from section 5.2. Quiz #9 on sections 4.9 and 5.1 will be given in your last discussion period this week.

In lecture today we used summation notation for various finite and infinite sums. I relayed the story of Gauss quickly obtaining the sum of 1 + 2 + 3 + ... + 100 as a young boy. I used what was probably his approach to find a formula the sum from k = 1 to n of k. I also gave formulas for the sum from k = 1 to n of a constant, the sum from k = 1 to n of k2, the sum from k = 1 to n of k3.

We looked at area as a limit of Riemann sums. I talked about right Riemann sums, left Riemann sums, midpoint Riemann sums, and sums where an arbitrary xk* was chosen on each interval [ xk–1, xk ] in order to generate f(xk*). We then used this limit approach with right Riemann sums to evaluate the area between the x-axis and f(x) = 2x on the interval [1, 5]. Since the shape is just a trapezoid we also checked our answer with simple geometry.

I also defined the definite integral to be the limit of Riemann sums as shown on page 366.

Mar 30 (Wed) There is no new homework – get caught up with the current homework and prepare for this week's quiz on sections 4.9 and 5.1.

In lecture today I discussed section 5.2. I wrote the definition of a definite integral as the limit given on the first page of this section. Students should be able to set up this limit for any given definite integral and should be able to evaluate the limit for some particular functions such as polynomials. By the text given just prior to theorem 4 in section 5.2 we see that we can choose whether to use a left Riemann sum, right Riemann sum, or midpoint Riemann sum in the evaluation of the limit. I suggest that you use a right Riemann sum as shown in theorem 4. It makes the calculations a bit simpler. If you are not using limits and are just approximating the value of a definite integal, then you should be prepared to compute left Riemann sums, right Riemann sums, or midpoint Riemann sums.

I used limits to solve #23. I then used the fact that the region is a semi-circle to solve #36 without limits. For #53 I used the minimum and maximum values of the function in order to approximate the definite integral of the function.

I then discussed the various properties about definite integrals found on pages 373 and 374.

Apr 1 (Fri) Read sections 5.3 and 5.4. In section 5.3 do #24, 25, 26, 31, 32, 35, 37, 43, 46, 49. In section 5.4 do #3, 6, 15, 16, 17, 18, 23, 28, 37, 41, 51, 52, 62. These are the straightforward drill problems using antiderivatives instead of limits. If necessary you should do more than the assigned problems over the weekend so that you become quick at obtaining these answers. You should know the table of indefinite integrals found on page 392 but you can ignore the last two where the integrand is sinh(x) or cosh(x).

Suppose a population is currently 2000 and is expected to grow by 0.3t2 people per year where t represents the number of years from now. What do we expect the population to be in 10 years? We solved this in two ways.

1. We graphed 0.3t2 on the interval [0, 10] and then computed a Riemann sum to approximate the change in population. To get the exact change in population, we took the limit of Riemann sums. Since by definition this limit is the definite integral, we saw that the exact change in population is the definite integral of 0.3t2 as t goes from 0 to 10. Computing this definite integral as a limit of Riemann sums takes many steps but gives an answer of 100 as the change in population.
2. If we let P(t) represent the population at time t, then P ′(t) represents the rate of change of population at time t. Thus P ′(t) = 0.3t2 and P(0) = 2000. By solving this differential equation we obtain the formula P(t) = 0.1t3 + 2000 as the population at time t. The exact change in population is then seen to be P(10) – P(0) = 2100 – 2000 = 100.
Finally to answer the question we see that the population in 10 years is equal to the current population added to the change in population and so equals 2000 + 100 or 2100 people.

Our second solution to this problem demonstrates the Net Change Theorem found on page 394 which basically says that the definite integral of a rate of change gives the total change.

More generally we have the Fundamental Theorem of Calculus (part 2) (FTC) from section 5.3 which shows a wonderful shortcut for evaluating many definite integrals. We used this quick approach for many definite integrals in lecture.

Recall that f(x) is integrable on an interval [a, b] if the appropriate limit of the Riemann sums exists (i.e. the definite integral of f(x) from a to b exists and is finite). Theorem 3 in section 5.2 states that if f is continuous on [a, b], then f is integrable on [a, b]. For example, since functions x2, ex, cos(x) and ex2 are all continuous on [1, 3], we know from this theorem that the limit of the appropriate Riemann sums and hence the definite integral of each of these functions from a to b exists and is finite. Note that these limits exist regardless of whether or not we are able to personally determine the limits. Prior to section 5.3, you have learned the skills necessary to evaluate the definite integral from 1 to 3 of x2 as a limit. From today's discussion of the FTC we saw quick ways to evaluate the definite integral from 1 to 3 of x2, ex or cos(x) without using limits. None of this helps in determining the definite integral of ex2 from 1 to 3. Even though ex2 is integrable on [1, 3] our only option is to estimate the value of the definite integral using an approximation technique such as a Riemann sum with a lot of subintervals.

Note that the FTC can be used when the integrand is continuous. We saw in lecture that the FTC gave us an obviously incorrect answer when applied to the integrand x–2 on the interval [–2, 1]. However x–2 = 1 / x2 is not continuous on [–2, 1] so we never should have tried using the FTC. We'll learn later how to deal with functions that have discontinuities.

Week 12
Apr 4 (Mon) Read section 5.5 and do #8, 12, 14, 19, 21, 23, 25, 26, 30, 32, 35, 36, 37, 39, 44, 46, 51, 53, 57, 58, 60, 65, 67, 77, 78 from that section. There will be a quiz during your last discussion section this week on sections 5.2, 5.3 and 5.4.

I began lecture by answering about 4–5 questions from the homework in sections 5.3 and 5.4.

I then introduced section 5.5 where we saw that the chain rule in reverse leads to the method of substitution. In particular we looked at the following examples.

1. The indefinite integral of 6x2(x3 + 4) dx.
2. The indefinite integral of 6x2(x3 + 4)3 dx.
3. The indefinite integral of 6x2(x3 + 4)10 dx.
4. The definite integral from 1 to 2 of 6x2(x3 + 4)3 dx.
5. The indefinite integral of 1 / (3x + 10) dx.
6. The indefinite integral of x4cos(x5) dx.
7. The definite integral from –1/3 to 1/3 of tan(x) dx.
8. The indefinite integral of tan(x) dx.
For example (1), it was easiest to just multiply through before integrating. For example (2), we multiplied through before integrating but it was starting to get just a little bit messy since we first had to correctly cube the binomial. For example (3), it would be unreasonable to multiply through beforehand so I used this as our first example of the benefit of substitution. We saw that when using substitution for a definite integral such as example (4), we should be sure to change the limits of integration. We then used substitution for each subsequent example, but also saw that you have the option of just thinking carefully about the correct antiderivative without formally using substitution. For example (7), we saw from the graph of tan(x) that we should expect the definite integral to be equal to 0. In fact if you have a definite integral from –a to a of f(x) where f(x) is an odd function, you should understand why the value of the definite integral is necessarily 0. For example (8) we rewrote the integrand as sin(x) / cos(x) before making the substitution u = cos(x).
Apr 6 (Wed) Read sections 6.1 and 4.2. In section 6.1 do #6, 8, 10, 13, 16, 20, 21, 23. On Friday I will assign homework from 4.2 and possibly 6.2. Prepare for this week's quiz on sections 5.2, 5.3, 5.4.

In lecture today I solved #35, 36, 58 and 77 from section 5.5.

If f(x) ≥ 0 for a ≤ x ≤ b, the area of the region between the graph of y = f(x) and the x-axis for a ≤ x ≤ b is given by the definite integral from a to b of f(x) dx. Similarly if g(y) ≥ 0 for c ≤ y ≤ d, the area of the region between the graph of x = g(y) and the y-axis for c ≤ y ≤ d is given by the definite integral from c to d of g(y) dy. We saw how this comes directly from the definition of a definite integral as the limit of a Riemann sum.

We then saw how a limit of Riemann sums reveals how to find the area between curves. If f(x) ≥ g(x) for a ≤ x ≤ b, then the area of the region between these curves for a ≤ x ≤ b is the definite integral from a to b of (f(x) – g(x)) dx. When we say to find the area between curves, we usually mean the area of the finite region bounded by the curves. This often involves finding intersection points.

As examples I found the exact area for each of the following regions.

• The region between y = 1 / (x2 + 1) and the x-axis for 0 ≤ x ≤ sqrt(3).
• The region bounded by f(x) = x + 3 and g(x) = 5 – x2.
• The region bounded by y = 2, y = 0, x = 0, and y = ln(x).
We solved this last problem two ways by first integrating with respect to x and then integrating with respect to y. Our answers were
• Area = 2 + the integral from 1 to e2 of (2 – ln(x)) dx.
• Area = the integral from 0 to 2 of ey dy.
For the first answer, we broke it up into two areas. For 0 ≤ x ≤ 1 we have a rectangle with area 2. For 1 ≤ x ≤ e2 we used a definite integral. At this point we are stuck since we have not yet learned an antiderivative for ln(x). For this reason the second answer is better and gives us a final answer of

Area = e2 – 1.

Apr 8 (Fri) Read sections 4.2 and 6.2. In section 4.2 do #1, 2, 5, 12, 13, 15, 17, 20. There will be one quiz given during your last discussion section meeting next week on sections 5.5, 6.1, and possibly 4.2. The following week we will have two quizzes.

I solved #20 from section 6.1 in the following two ways.

• Area = the integral from –6 to 2 of (x – (– sqrt(12 – 4x))) dx + the integral from 2 to 3 of (sqrt(12 – 4x) – (– sqrt(12 – 4x))) dx.
• Area = the integral from –6 to 2 of (3 – 0.25y2 – y) dy.
For this problem, we see that integrating with respect to y is much easier to set up and also easier to evaluate the definite integral. I suggest you try to solve other area problems both ways to get a sense of when one approach will be easiest.

I solved #23 by integrating with respect to x to obtain

• Area = the integral from 0 to π/6 of (cos(x) – sin(2x)) dx + the integral from π/6 to π/2 of (sin(2x) – cos(x)) dx.
We see that it is essential to have a graph here (and on the previous problem) to see the relative position of the curves and to see which intersection points we need.

I stated the Mean Value Theorem along with Rolle's Theorem which is a special case of the Mean Value Theorem. I proved Rolle's Theorem today and plan to quickly prove the Mean Value Theorem on Monday. Students should be able to properly state each of these theorems correctly. I drew pictures of functions which were not continuous or not differentiable to see why we need the conditions on continuity and differentiability before stating the conclusion for each theorem. Students are encouraged to think about the pictures I drew on the board which demonstrate the conclusion for each theorem. The most important aspect of these theorems is their use in proving other important theorems. In particular I will use the Mean Value Theorem to prove the Fundamental Theorem of Calculus next week.

Using Rolle's Theorem in the following way, we are able to prove that the equation x5 + 3x3 + 10x + 10 = 0 has exactly one real solution.

Proof: Let f(x) = x5 + 3x3 + 10x + 10. Since f(–1) is negative, f(0) is positive, and f is continuous everywhere, the Intermediate Value Theorem implies that there is a real root between –1 and 0. What would happen if there were two distinct real roots c1 and c2? Since f is continuous and differentiable everywhere with f(c1) = f(c2) = 0, we would apply Rolle's Theorem to find a value of c between c1 and c2 with f ′(c) = 0. However, this is impossible since f ′(x) = 5x4 + 9x2 + 10 ≥ 10 for all x. Thus there cannot be two (or more) real roots. This shows that there is exactly one real solution to f(x) = 0.

Week 13
Apr 11 (Mon) Read section 6.2 and do #2, 7, 9, 12, 14, 17, 32, 57, 60 from that section. There will be a quiz on sections 5.5 and 6.1 given in your last discussion section meeting this week.

We began lecture with a proof of the Mean Value Theorem. I then used the Mean Value Theorem to prove the following theorem and its corollary.

• Theorem: If f ′(x) = 0 for all x in the interval (a, b), then f(x) = C (a constant) on the interval (a, b).
• Corollary: If f ′(x) = g ′(x) for all x in the interval (a, b), then f(x) = g(x) + C on the interval (a, b) where C is a constant.
We found the indefinite integral of sin(2x) dx by solving it three different ways.
1. Using the substitution u = 2x to obtain (–1/2)*cos(2x) + C
2. Writing sin(2x) = 2sin(x)cos(x) and using the substitution u = sin(x) to obtain sin2(x) + C
3. Writing sin(2x) = 2sin(x)cos(x) and using the substitution u = cos(x) to obtain –cos2(x) + C
These answers look different but are all correct. Through basic trigonometric identities we see that (–1/2)*cos(2x), sin2(x), and –cos2(x) all differ by constants. Thus when adding an arbitrary constant, we arrive at the same family of functions.

We saw how to calculate volume as V = limn → ∞ Σ A(xk*) Δ x where A(xk*) is the cross-sectional area at xk*. Since the limit of this Riemann sum is by definition a definite integral we obtain

• Volume = the definite integral of cross-sectional area
I graphed the function y = sqrt(x) and shaded in the region R between the x-axis and the curve on the interval [1, 8]. We set up definite integrals for the volume of a solid obtained in each of the following ways.
• Revolving the region R around the x-axis: We get volume = the definite integral from 1 to 8 of π (sqrt(x))2 dx.
• Revolving the region R around the line y = 3: We get volume = the definite integral from 1 to 8 of (π (3)2 – π (3 – sqrt(x))2) dx.
• Letting R form the base of a solid for which cross-sections perpendicular to the x-axis are squares: We get volume = the definite integral from 1 to 8 of (sqrt(x))2 dx.
• Revolving the region R around the y-axis: We get volume = the definite integral from 0 to 1 of (π (8)2 – π (1)2) dy + the definite integral from 1 to sqrt(8) of (π (8)2 – π (y2)2) dy.
For this last problem I mistakenly wrote 2 instead of sqrt(8) at the board during lecture, so be sure to change your notes to the correct definite integrals listed here.
Apr 13 (Wed) Read section 6.3 and do #3, 5, 9, 12, 15, 17, 19 from that section. Prepare for this week's quiz (quiz #11) on sections 5.5 and 6.1. There will be two quizzes next week – one in each of your two discussion section meetings. Quiz #12 will be on sections 4.2 and 6.2. One of the problems on the quiz will either ask that you precisely state the Mean Value Theorem or Rolle's Theorem. You need to know both. You may also need to apply these theorems as you did on the homework from section 4.2. Quiz #13 will be on sections 6.3, 6.5, and possibly another section.

In lecture today I proved the Fundamental Theorem of Calculus (part 2) without relying on part 1. Students are not responsible for knowing the proof but I encourage math majors to make sure that they understand each step in the proof.

I looked again at the final example from the last lecture. We obtained

• volume = the definite integral from 0 to 1 of (π (8)2 – π (1)2) dy + the definite integral from 1 to sqrt(8) of (π (8)2 – π (y2)2) dy.
I used this example to illustrate another approach called the Cylindrical Shells method. With this approach we obtained
• volume = the definite integral from 1 to 8 of 2π x sqrt(x) dx.
Note that the integrand is simply the cylinder's surface area 2 π r h where for this example the radius r = x and the height h = sqrt(x).

We then found the volumes of two solids. We discussed integrating with respect to x versus integrating with respect to y – sometimes one approach will be easier than the other and sometimes one approach will be impossible. The two solids we worked with are:

• A region is bounded above by y = 9 – x2 and below by the x-axis on the interval [–3, 3]. This region is revolved around the vertical line x = 5 to form a solid.
• A region is bounded above by y = 4x – x2 and below by the x-axis on the interval [1, 4]. This region is revolved around the y-axis to form a solid.
Apr 15 (Fri) Read section 6.5 and do #1, 2, 4, 5, 7, 9, 10, 13, 17, 18b from that section. There will be two quizzes next week – one in each of your two discussion section meetings. Quiz #12 will be on sections 4.2 and 6.2. One of the problems on the quiz will either ask that you precisely state the Mean Value Theorem or Rolle's Theorem. Although Rolle's Theorem is just a special case of the Mean Value Theorem, you need to be able to state both theorems. You may also need to apply these theorems as you did on the homework from section 4.2. Quiz #13 will be on sections 6.3, 6.5, and possibly section 3.10 or section 4.8 which we will discuss next week.

In lecture today I derived the formula for the average value of a function and showed the geometric interpretation which helped to obtain an approximate value for the average before applying the formula. I then answered homework questions including #57 from section 6.2 where the cross sections are triangles.

Week 14
Apr 18 (Mon) Read sections 3.10 and 4.8. In 3.10 do #6, 23, 24, 25, 31. There will be two quizzes this week – one in each of your two discussion section meetings. Quiz #12 will be on sections 4.2 and 6.2. One of the problems on the quiz will either ask that you precisely state the Mean Value Theorem or Rolle's Theorem. Although Rolle's Theorem is just a special case of the Mean Value Theorem, you need to be able to state both theorems. You may also need to apply these theorems as you did on the homework from section 4.2. Quiz #13 will be on sections 6.3 and 6.5.

In lecture today we sought to approximate e0.2 without the use of a calculator. To do this we compared the function f(x) = ex to some other functions as follows.

1. Graph f(x) = ex and P0(x) = 1 on the same coordinate axes. Note that f(0) = P0(0) (i.e. same y-value). From the graphs we see that ex ≈ 1 for x near 0. Thus e0.2 ≈ 1.
2. Graph f(x) = ex and P1(x) = 1 + x on the same coordinate axes. Note that f(0) = P1(0) and f ′(0) = P ′1(0) (i.e. same y-value, same slope). From the graphs we see that ex ≈ 1 + x for x near 0. Thus e0.2 ≈ 1 + 0.2 = 1.2.
3. Graph f(x) = ex and P2(x) = 1 + x + 0.5x2 on the same coordinate axes. Note that f(0) = P2(0), f ′(0) = P ′2(0) and f ′′(0) = P ′′2(0) (i.e. same y-value, same slope, same concavity). From the graphs we see that ex ≈ 1 + x + 0.5x2 for x near 0. Thus e0.2 ≈ 1 + 0.2 + 0.5(0.2)2 = 1.22.
Polynomials are easier to deal with than some other functions like ex, sin(x), cos(x), etc. When you plug a value into a polynomial, you end up using only addition, subtraction, multiplication and division. This means we can do this more easily by hand. It also means that computers and calculators can more easily perform these calculations. A quick check with a calculator shows our approximation e0.2 ≈ 1.22 is a very good approximation. Students who take Calculus II will learn more about these polynomials which are referred to as Taylor Polynomials. In this course we will concentrate on the second one listed which is simply the tangent line to the curve at a given point.

So for section 3.10 we will find the equation of the tangent line to the graph of f(x) at a particular point. If we call this tangent line L(x), then from the graphs of f(x) and L(x) we see that f(x) ≈ L(x) for x near the point of tangency. We used this approach to approximate the following quantities without a calculator.

• To approximate sin(0.1), we used the tangent line to the graph of sin(x) at x = 0 to obtain that sin(x) ≈ x for x near 0. Thus sin(0.1) ≈ 0.1.

• To approximate sin(315), we used the tangent line to the graph of sin(x) at x = 100π to obtain that sin(x) ≈ x – 100π for x near 100π. Thus sin(315) ≈ 315 – 100π ≈ 315 – 100(3.14159265358979323) = 0.840734641020677. Of course you don't need to use this many decimal places for π.

• To approximate sqrt(4.06), we used the tangent line to the graph of sqrt(x) at x = 4 to obtain that sqrt(x) ≈ 0.25x + 1 for x near 4. Thus sqrt(4.06) ≈ 0.25(4.06) + 1 = 2.015.

Apr 20 (Wed) Read section 4.8 and do #5, 13, 18, 19, 29, 31 from that section. Prepare for quiz #13 on sections 6.3 and 6.5.

In lecture today I talked about five different methods for approximating the square root of 5.

1. Use the Intermediate Value Theorem to find the positive root of the function f(x) = x2 – 5. Since f(2) = –1 < 0 and f(3) = 4 > 0 we know the root lies between 2 and 3. Now since f(2.2) = –0.16 < 0 and f(2.3) = 0.29 > 0 we know the root lies between 2.2 and 2.3. Now refine further to get the accuracy you desire.
2. Use this algorithm which was often taught to high school students 30 or more years ago.
3. Use a linear approximation. Find the tangent line to the graph of f(x) = sqrt(x) at x = 4 to be y = 0.25x + 1. Thus sqrt(x) ≈ 0.25x + 1 for x near 4. We conclude that sqrt(5) ≈ 2.25.
4. Use Newton's Method to obtain successive estimates for the positive root of the function f(x) = x2 – 5. See section 4.8.
5. Use a calculator.
The second method is mentioned only for historical interest – students will not be tested on this algorithm. Students are not allowed the use of a calculator on quizzes, tests or the final exam, but can use one to answer the homework questions from section 4.8. Students should fully understand methods 1, 3 and 4 which apply generally to many different types of functions.

We ended class by approximating the x-value where the graphs of y = x3 and y = x + 1 intersect. We began by setting the equations equal to each other to get x3 = x + 1. Next we rewrote this as x3 – x – 1 = 0 in order to apply Newton's Method to the function f(x) = x3 – x – 1 with an initial estimate of x1 = 1.

Apr 22 (Fri) In section 7.2 do #1, 2, 3, 4, 5, 6, 15, 16, 17, 18, 21, 23, 25, 27, 29, 30, 31, 33, 36. You will need to use basic trigonometric identities and substitution to do these homework problems. Prepare for Wednesday's test which will cover sections 3.10, 4.2, 4.8, 4.9, 5.1, 5.2, 5.3, 5.4, 5.5, 6.1, 6.2, 6.3, 6.5, 7.2 (parts involving substitution and basic trigonometric identities). See the test #3 notes for further details.

In lecture we integrated the following functions.

• sin(x) [ You should already know this. ]
• cos(x) [ You should already know this. ]
• tan(x)   [ Rewrite as sin(x) / cos(x) and use the substitution u = cos(x). ]
• cot(x)   [ Rewrite as cos(x) / sin(x) and use the substitution u = sin(x). ]
• sec(x)   [ Rewrite sec(x) = 1 / cos(x) = cos(x) / cos2(x) = cos(x) / (1 – sin2(x)) and use the substitution u = sin(x) to obtain the integral of 1 / (1 – u2) du. Since 1 / (1 – u2) = 1 / ((1 + u) (1 – u)) = 1/2 (1 / (1 + u) + 1 / (1 – u)), we get 1/2 (ln |1 + u| – ln |1 – u|) + C. Writing this in terms of x and then using a lot of simplification we obtain that the integral of sec(x) dx is ln |sec(x) + tan(x)| + C. Most people just memorize this since it takes so many steps to derive. ]
• csc(x)   [ An approach similar the one for sec(x) gives that the integral of csc(x) dx is ln |csc(x) – cot(x)|. ]
• sin3(x) [ Rewrite as sin2(x) sin(x) = (1 – cos2(x)) sin(x) and use the substitution u = cos(x). ]
• cos5(x) [ Rewrite as sin4(x) sin(x) = (1 – sin2(x))2 cos(x) and use the substitution u = sin(x). ]
• sin2(x) [ Rewrite using the identity sin2(x) = 1/2 – cos(2x)/2 and then use the substitution u = 2x. ]
• sin3(x) cos5(x) [ Method 1: Rewrite as sin2(x) cos5(x) sin(x) = (1 – cos2(x)) cos5(x) sin(x) and use the substitution u = cos(x). Method 2: Rewrite as sin3(x) cos4(x) cos(x) = sin3(x) (1 – sin2(x))2 cos(x) and use the substitution u = sin(x). ]
• tan(x) sec4(x) [ Method 1: Rewrite as tan(x) sec2(x) sec2(x) = tan(x) (tan2(x) + 1) sec2(x) and use the substitution u = tan(x). Method 2: Rewrite as sec3(x) sec(x)tan(x) and use the substitution u = sec(x). Method 3: Rewrite as sin(x) / cos5(x) and use the substitution u = cos(x). ]
• tan2(x) [ Rewrite as 1 – sec2(x) and integrate term by term. ]
Week 15
Apr 25 (Mon) Prepare for Wednesday's test which will cover sections 3.10, 4.2, 4.8, 4.9, 5.1, 5.2, 5.3, 5.4, 5.5, 6.1, 6.2, 6.3, 6.5, 7.2 (parts involving substitution and basic trigonometric identities). See the test #3 notes for further details.
Apr 27 (Wed) Test 3 (given during lecture)
Apr 29 (Fri) Read section 5.3 about the Fundamental Theorem of Calculus (part 1). Do #7, 8, 12, 14 from section 5.3. Read about hyperbolic functions in section 3.11.

• Fundamental Theorem of Calculus (part 1): If f is continuous on the interval [a, b], then the function defined by g(x) = the definite integral from a to x of f(t) dt for a ≤ x ≤ b is continuous on [a, b], differentiable on (a, b), and g ′(x) = f(x).

Defining a function as a definite integral with a variable as the upper limit of integration is sometimes referred to as an accumulation function. For positive integrands, as x increases you can think of this definite integral as accumulating area under a curve. Defining a function as a definite integral may seem strange at first, but it is often done in physics, chemistry, statistics and other fields. In particular we looked at the following examples.

• g(x) = the definite integral from 0 to x of t3 dt has derivative g ′(x) = x3. We see this immediately from FTC (part 1), but also checked it by first evaluating the definite integral and then differentiating.
• g(x) = the definite integral from 0 to x of (1 + t2)1/2 dt has derivative g ′(x) = (1 + x2)1/2. We see this immediately from FTC (part 1). I noted that it is more difficult to evaluate the definite integral and then differentiate since you may not immediately know an antiderivative for (1 + t2)1/2. Those going on to take MATH 231 will learn to make the substitution t = tan(θ) and dt = sec2(θ) dθ followed by some more careful work to find an antiderivative.
• S(x) = the definite integral from 0 to x of sin(π t2 / 2) dt has derivative S ′(x) = sin(π x2 / 2). We see this immediately from FTC (part 1). It turns out that it is impossible to find an elementary function which is the antiderivative of sin(π t2 / 2) so we must rely on FTC (part 1) if we need to know the derivative of this function. This particular function S(x) is called the Fresnel function which first appeared in Augustin Fresnel's theory of diffraction of light waves, but more recently has been applied to the design of highways.
We next looked at the following two examples which require us to use the chain rule.
• g(x) = the definite integral from 1 to x3 of cos(t) dt has derivative g ′(x) = 3x2cos(x3). We did this one by first evaluating the definite integral and then differentiating.
• y = the definite integral from 1 to x3 of et2 dt has derivative dy/dx = 3x2ex6. Since it is impossible to find an elementary function which is the antiderivative of et2, we must rely on FTC (part 1). To do this we make the substitution u = x3 which has derivative du/dx = 3x2. With this substitution, the definite integral is now y = the definite integral from 1 to u of et2 dt which has derivative dy/du = eu2 by FTC (part 1). By the chain rule we now get dy/dx = (dy/du)(du/dx) = (eu2)(3x2) = 3x2ex6.

I introduced the following two hyperbolic functions.

• cosh(x) = (ex + e–x) / 2   (note that cosh(x) and cos(x) are different functions)
• sinh(x) = (ex – e–x) / 2   (note that sinh(x) and sin(x) are different functions)
The graph of y = cosh(x) is called a catenary. It appears in natural forms such as a spider web or as a hanging chain.

We took derivatives to find that

• ( cosh(x) ) ′ = sinh(x)
• ( sinh(x) ) ′ = cosh(x)
Week 16 (Last day for U of I classes is Wednesday, May 4, 2011)
May 2 (Mon) For homework do #1, 2, 3, 4, 23abcd, 31, 32, 33, 35, 38 from section 3.11.

The final exam is cumulative and will take place in 314 Altgeld Hall from 1:30pm to 4:30pm on Monday, May 9th. There will be 50 points of test 1 material, 50 points of test 2 material and 50 points of test 3 material. The test 2 material may include questions from section 3.11. The test 3 material may include questions from section 5.3 about the Fundamental Theorem of Calculus (part 1). You have not had quiz questions on this material so be sure to read the sections and do the assigned homework. From section 3.11, be sure that you know how sinh(x) and cosh(x) are defined in terms of ex and e–x. Also be sure to know the graphs and the derivatives for sinh(x) and cosh(x).

The best way to study for the final exam is to print blank copies of each test and quiz from this and past semesters. Take them in the allotted time and compare your answers to the online solutions. Additionally you should look at the notes from each lecture listed on this page, the handout of test 2 notes, and the handout of test 3 notes. For material that you struggle with, look back at the specific section in the book or stop by during office hours to see me or a TA.

May 4 (Wed)
Final Exam Period (Friday-Friday, May 6-13, 2011)
May 9 (Mon) Cumulative Final Exam from 1:30pm to 4:30pm in 314 Altgeld Hall Department of Mathematics College of Liberal Arts and Sciences University of Illinois at Urbana-Champaign 273 Altgeld Hall, MC-382 1409 W. Green Street, Urbana, IL 61801 USA Department Main Office Telephone: (217) 333-3350 Fax (217) 333-9576