|Week 1 (First day for U of I classes is Tuesday, January 17, 2012)|
|Jan 17 (Tue)||Go to your discussion-recitation section.|
|Jan 18 (Wed)||Read sections 1.1 and 1.2. In 1.1 do #4, 7, 8, 25, 34, 41, 47, 51, 54, 71, 73, 77. 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 – if you have not received a 70% or higher on the ALEKS math placement test some time between September 15, 2011 and January 23, 2012, then you will be automatically dropped from the course.|
|Jan 20 (Fri)||Read section 1.3. In 1.3 do #3, 8, 9, 10, 12, 14, 18, 19, 31, 32, 33, 38, 41, 43. We determined the domain for the following functions.
Students should know how to graph basic functions such as the following.
I used shifting techniques to graph y = 3(x − 2)2 + 5 by slowly modifying the graph of y = x2. Students should learn to shift any of the basic functions to obtain graphs of more complicated functions.
|Jan 23 (Mon)||For homework finish Trigonometry Worksheet #1 and read Appendix D.
In lecture I discussed some basic trigonometry. After today's lecture you should
|Jan 25 (Wed)||Skip section 1.4 and read section 1.5. For homework do #29, 30, 35, 36, 37, 38, 65, 67 in Appendix D. You will need a calculator for #35–38. Then do #2, 4, 11, 12, 13, 14, 15, 16, 17, 19, 20, 21, 22, 25, 29, 30 in section 1.5. There will be two quizzes next week – quiz #1 on Tuesday and quiz #2 on Thursday. Quiz #1 will cover sections 1.1, 1.2, 1.3 and problems on trigonometry (see worksheet and Appendix D). Quiz #2 will cover sections 1.5 and 1.6.
I 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 − 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.
I showed how to graph y = tan(x) and y = sec(x). You should use this approach or a similar approach to graph y = csc(x) or y = cot(x).
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.
If the x-value entries in a table of values are incremented by a constant amount, then the following holds.
|Jan 27 (Fri)||Read section 1.6. In 1.6 do #5, 7, 9, 10, 15, 17, 19, 21–26, 35–41, 51–54, 57, 58. Tuesday's quiz will be on sections 1.1, 1.2, 1.3 and problems on trigonometry (see worksheet and Appendix D which both now have solutions). Thursday's quiz will cover sections 1.5 and 1.6.
Students should have obtained the following formulas for functions which fit the data given in TABLE 1, TABLE 2 and TABLE 4 from last lecture.
TABLE 1: By plugging (x,y) = (0,3) and (x,y) = (2,12) into y = C*ax we obtained the formula y = 3*2x.
TABLE 2: By plugging (x,y) = (0,3) and (x,y) = (3,18) into y = mx + b we obtained the formula y = 5x + 3.
TABLE 4: By plugging (x,y) = (1,2) and (x,y) = (3,4) into y = C*ax we obtained the two equations 2 = C*a and 4 = C*a3. We saw two different ways to solve these equations and obtain a = sqrt(2) and C = sqrt(2) resulting in the formula y = sqrt(2)*sqrt(2)x. There are other ways to write this formula.
For y = C*ax we have exponential growth if a > 1 and exponential decay if 0 < a < 1. We looked at the graphs of examples y = 10*2x and y = 100*(1/2)x.
We looked at a population which is 50 in the year 1980 and doubles every 10 years after that. When does the population reach 600? A quick table shows that this occurs somewhere between 30 and 40 years after 1980.
To determine a more precise answer we plugged (t,P) = (0,50) and (t,P)=(10,100) into P = C*at to obtain the formula P = 50*2t/10. Setting P = 600 and solving for t requires logarithms. We get t = 10*ln(12) / ln(2) ≈ 35.8 years after 1980. Students should become comfortable using the rules of logarithms.
I introduced inverse functions by first looking at the following tables of values for y = x3 and y = x1/3.
The roles of x and y are reversed for inverse functions such as f(x) = x3 and f −1(x) = x1/3. Be careful with this misleading but standard notation since f −1(x) is not the same as 1 / f(x).
We saw that the graph of f(x) and f −1(x) are mirror images of each other across the line y = x.
Thinking about the relationship between a function and its inverse, one can quickly see that g(x) = x + 8 has inverse g −1(x) = x − 8, and that h(x) = 10x has inverse h −1(x) = x/10. For more complicated functions such as f(x) = (4x − 3)1/5 it isn't as immediately apparent that f −1(x) = (x5 + 3) / 4. We looked at two ways of obtaining the formula for the inverse of this function.
We 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) = x2 which are not one-to-one, we could restrict the domain so that it has an inverse.
I went over the basic definition of the logarithm for any base along with the various rules for manipulating logarithms. In particular
|Week 3 (Deadline to add this course is Monday, January 30, 2012)|
|Jan 30 (Mon)||For homework read sections 2.1 and 2.2. In 2.1 do #5. In 2.2 do #4, 7, 8, 11, 15, 17, 23, 24, 29, 31, 32, 39, 41. Quiz #1 on Tuesday will cover sections 1.1, 1.2, 1.3 and problems on trigonometry (see worksheet and Appendix D). Quiz #2 on Thursday will cover sections 1.5 and 1.6.
A function is called one-to-one if it never takes on the same value twice; that is, f(x1) ≠ f(x2) whenever x1 ≠ x2. The horizontal line test is a geometric way to understand the term one-to-one. All one-to-one functions have inverses. Since increasing functions are one-to-one, an increasing function always has an inverse. Since decreasing functions are one-to-one, a decreasing function always has an inverse.
The function f(x) = x3 + x5 is increasing so it has an inverse. Our technique to find a formula for the inverse did not work since we were unable to solve x = y3 + y5 for y. However, the function still has an inverse and we know things about it. For instance, since (1, 2) is on the graph of y = f(x), we know that the point (2, 1) is on the graph of y = f −1(x).
We spent a little time on logarithms with various bases. We spent more time on natural logarithms which have base e ≈ 2.71828...
The function f(x) = ex has inverse f −1(x) = ln(x). You should know the following definitions, identities and simplification rules.
We solved for x in the equation 6 + 2x = 4 + 2x + 2. Taking the logarithm of both sides didn't work directly since we do not have a rule for taking the logarithm of a sum. We had to manipulate the equation appropriately before taking the logarithm of both sides.
We used the informal definition of limit as found on the first page of section 2.2 to determine the value of the following limits.
|Jan 31 (Tue)||Quiz #1 on sections 1.1, 1.2, 1.3 and trigonometry will be given during today's discussion section.|
|Feb 1 (Wed)||Everyone should read section 2.3. 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, 25, 26, 37, 39. There will be a quiz tomorrow on sections 1.5 and 1.6.
I went over #7 from the homework in section 2.2. We used that limx → a f(x) = L if and only if limx → a− f(x) = L and limx → a+ f(x) = L.
We used the informal definition of limit as found on the first page of section 2.2 to determine the value of the following limits.
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) )
This can also be written as
• S ′(6) = limh → 0 ( (26 + h − 26) / (h) )
We made a table of values to approximate this limit. I mentioned that this S ′(6) notation for the rate 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. Most 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.
We next evaluated the following limits.
|Feb 2 (Thu)||Quiz #2 on section 1.5 and 1.6 will be given during today's discussion section.|
|Feb 3 (Fri)||Read section 2.5 and reread the material from section 1.6 on inverse trigonometric functions. In section 1.6 do #63, 64, 65, 68, 70, 71. In section 2.5 do #20, 45, 49, 51, 53.
You should know the following identities.
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).
We define 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 see Theorem 9 from section 2.5.
We found limx → 2 (3x2 + 5x − 4) = 18. Using the limit laws from section 2.3 it takes multiple steps to find this limit. However, knowing that polynomials are continuous everywhere it is much easier to simply plug in x = 2 to obtain the limit in one step.
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 quizzes and tests, I expect you to be able to carefully write out the statement of important theorems such as this one or The Squeeze Theorem. You should also be able to carefully show how and why you may apply these theorems to specific problems.
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 first of the following three quantities. For next time think carefully about the values of the last two quantities.
|Feb 6 (Mon)||Read section 2.6. In section 2.6 do #8, 15, 21, 24, 25, 29, 30, 33, 41, 43. There will be a quiz Thursday on sections 1.6, 2.1, 2.2, 2.3, 2.5 and 2.6. The test one week from Wednesday will cover 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.
The logarithm buttons found on most calculators are ln (natural logarithm with base e) or log (common logarithm with base 10), so for other logarithms it is useful to convert to one of these bases. We note that y = log2 13 ⇔ 2y = 13 so we expect y to be between 3 and 4. Taking the logarithm of both sides of this second equation and solving for y gives y = ln(13) / ln(2) ≈ 3.7. We can use this same technique to prove the change of base formula shown below.
We then discussed an approach for finding limx → ∞ f(x) or limx → −∞ f(x) where f(x) is the quotient of either polynomials or roots of polynomials. We can rewrite the quotient by dividing both the numerator and the denominator by xn where n is the highest power of x found in the denominator. We looked at the following examples. Think carefully about the last one.
Last time we defined f to be continuous at x = a if limx → a f(x) = f(a).
We looked at the piecewise function defined so that f(x) = x2 + C for x ≤ 2 and f(x) = 3 − x for x > 2. Which value for C makes f(x) continuous everywhere? Since continuity is defined in terms of limits, you should use limits in your solution.
It is important to know how to restrict the domains of the trigonometric functions so that they are one-to-one and thus have inverse functions. I showed the graph of sin(x) for −π/2 ≤ x ≤ π/2 along with the graph of sin −1(x). I discussed how the domain and range are related for the two functions. Students should work on their own to understand how to restrict the domains of cos(x), tan(x) and sec(x) to obtain cos −1(x), tan −1(x) and sec −1(x). We then evaluated the following quantities.
|Feb 8 (Wed)||Read sections 2.7 and 2.8. In section 2.7 do #5, 6, 7, 8, 9, 10, 13, 14, 27, 28, 29, 30, 31, 32. In section 2.8 do #4, 5, 6, 12, 16, 17, 18, 21, 23, 25, 27, 29. Prepare for tomorrow's quiz on sections 1.6, 2.1, 2.2, 2.3, 2.5 and 2.6. The test one week from today will cover 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.
I discussed the restricted domains for tan(x) and cos(x) so that they are one-to-one and have inverses cos −1(x) and tan −1(x), respectively. We also looked at the graphs of these inverse functions. We then evaluated the following quantities.
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 discussed how limits are used to determine horizontal and vertical asymptotes. We looked at graphs which showed why using limits is appropriate. We found limx → 2+ ( 3 / (x − 2) ) = ∞ so that the graph of f(x) = 3 / (x − 2) has a vertical asymptote at x = 2.
I showed how to choose which technique to use when finding the following limits.
f ′(x) = limh → 0 (f(x + h) − f(x)) / h
We saw how this definition gives us the slope of the tangent line at a point.
We used the limit definition of a derivative to show that f(x) = x2 has derivative f ′(x) = 2x.
|Feb 9 (Thu)||Quiz #3 on sections 1.6, 2.1, 2.2, 2.3, 2.5 and 2.6 will be given during today's discussion section.|
|Feb 10 (Fri)||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. The test will be given during your officially scheduled lecture period. 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 assigned homework problem. Detailed solutions to all odd problems and assigned even problems are available at Illinois Compass 2g. Solutions to the quizzes and the trigonometry worksheet are posted on the course homepage. You must be able to state and use the definitions of even functions, odd functions, continuity, derivatives. You must 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.
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 TAs will not answer any questions during the test.
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.
We saw graphically why the following three definitions of derivative are equivalent.
I mentioned that 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 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. One can use limits to find the velocity h ′(t) = −32t + 96. We set h ′(t) = 0 to determine that the ball reached its maximum height at t = 3 seconds. With this information we can determine h(3) = 304 feet to be the maximum height. We set h(t) = 0 to find how long it takes until the ball falls back to the ground. We plugged this value of t into the velocity formula to obtain the velocity at the moment the ball hit the ground. I discussed why we expected the velocity to be negative.
I mentioned the important theorem which states that if a function is differentiable at a point, then it must also be continuous at that point. I'll prove this quickly on Monday.
For most of Monday's lecture I will answer student questions in preparation for Wednesday's test.
|Feb 13 (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. The test will be given during your officially scheduled lecture period. 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 assigned homework problem. Detailed solutions to all odd problems and assigned even problems are available at Illinois Compass 2g. Solutions to the quizzes and the trigonometry worksheet are posted on the course homepage. You must be able to state and use the definitions of even functions, odd functions, continuity, derivatives. You must 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.
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 TAs will not answer any questions during the test.
Be sure to go to both discussion sections this week.
|Feb 15 (Wed)||Test 1 (given during lecture)|
|Feb 16 (Thu)||Discussion sections still meet today. Your homework is to read section 3.1 and do #3–30, 33, 35, 47, 51, 53.|
|Feb 17 (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, 47, 51, 53. In section 3.2 do the odd problems from #3–33. There will be a quiz Thursday on sections 3.1–3.2.
We discussed the results of the first test along with strategies for improving one's score on future tests. Students who wish to drop our 5 credit hour course, have the option of adding a second eight week course.
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, thinking about the derivative as a rate of change gives the same result. More formally, we use limits to prove these and the following derivative rules.
Next week we will use limits to derive short-cut methods for finding derivatives of all the basic functions. We will also discuss the Chain Rule for the derivative of a 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.
|Feb 20 (Mon)||Read sections 3.3 and 3.4. In section 3.3 do the odd problems from #1–23. Read the notes below for a few additional homework problems on deriving the derivative formulas for all trigonometric functions. There will be a quiz Thursday on sections 3.1 and 3.2.
Although one can directly find the derivative of a quotient using limits, it is easier to rewrite w(x) = f(x) / g(x) as w(x)*g(x) = f(x) and take the derivative of both sides using the product rule. Solving for w ′(x) gives us a simpler proof of 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.
We have shown earlier that limθ → 0 sin(θ) / θ = 1. Another useful limit is limθ → 0 (cos(θ) − 1) / θ = 0. We proved this result by first multiplying numerator and denominator by cos(θ) + 1.
Recall the trigonometric identities found in Appendix D.
Writing tan(x) = sin(x) / cos(x) we used the quotient rule to prove that ( tan(x) ) ′ = sec2(x). Writing sec(x) = 1 / cos(x) we used the quotient rule to prove that ( sec(x) ) ′ = sec(x)tan(x). For homework, students should rewrite cot(x) and csc(x) in terms of sin(x) and cos(x) to obtain the derivative of each of these functions.
Using limits the derivative of f(x) = ax is f ′(x) = limh → 0 ( (ax + h − ax) / h ) = limh → 0 ( (ax ah − ax) / h ) = ax limh → 0 ( (ah − 1) / h ) = ax f ′ (0). Thus the derivative of ax is ax multiplied by the slope of the curve y = ax at x = 0. This slope turns out to be ln(a) so that ( ax ) ′ = ax ln(a). In particular we get ( ex ) ′ = ex.
We can use 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.
|Feb 22 (Wed)||Read section 3.4. In section 3.4 do the odd problems from #7–55. Tomorrow's quiz is on sections 3.1 and 3.2.
The following table summarizes the derivative notation used in our textbook for first semester calculus. The prime notation is due to Joseph Louis Lagrange and Leibniz notation is due to Gottfried Wilhelm Leibniz. You should be comfortable using any of these notations for derivatives. For tomorrow's quiz you will be asked to correctly use Leibniz notation when computing derivatives. When using derivatives in other courses it may be worthwhile to see http://en.wikipedia.org/wiki/Notation_for_differentiation for other derivative notation due to Euler and Newton.
Using Leibniz notation the derivative of P = t3 is dP/dt = 3t2 and the derivative of w = r2 is dw/dr = 2r. Use the variables given in the problem instead of always using y and x.
If you wish to evaluate the derivative of x3 at 5 we have the following notation.
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.
Chain Rule: ( f(g(x)) ) ′ = f ′(g(x)) g ′(x)
The chain rule can also be written as dy/dx = (dy/du) * (du/dx) but we won't talk about this approach until next lecture.
Using the chain rule we found derivatives for the following functions.
|Feb 23 (Thu)||Quiz #4 on sections 3.1 and 3.2 will be given during today's discussion section.|
|Feb 24 (Fri)||Read sections 3.5 and 3.6. In section 3.5 do #5, 7, 9, 11, 13, 15, 17, 19, 29, 30, 31, 32, 49, 50, 51, 57. In section 3.6 do #3, 5, 6, 11, 13, 19, 31, 34, 39, 43, 45.
Here is one method for determining the derivative of sin−1(x).
I discussed logarithmic differentiation to obtain derivatives of the following functions.
We found the slope of the tangent line to the curve x2 + y2 = 25 at the point (3, 4) in two ways.
We 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.
|Feb 27 (Mon)||Read sections 3.7 and 3.8. In section 3.7 do #7, 8, 9, 10. In section 3.8 do #3, 4, 8, 9, 10, 11, 12. There will be a quiz Thursday on sections 3.3, 3.4 and 3.5. I have cancelled today's office hours, but will be available Tuesday 3-5:30pm in 121/123 Altgeld Hall.
We found the derivative of xcos(x) using logarithmic differentiation as follows.
See how the chain rule is used for each of the following problems.
I solved a couple of problems from the homework in sections 3.5 and 3.6.
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.
The following are examples of differential equations (equations which include derivatives).
We found formulas for y as a function of x to satisfy each of the following differential equations along with the given initial values.
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.
I did not have time for the following examples in lecture, but it is hopefully clear from our discussion why we obtain the given solutions.
|Feb 29 (Wed)||Read section 3.9. In section 3.9 do #6, 10, 13, 15, 20, 22, 24, 27, 28, 30, 31, 38, 41. Prepare for tomorrow's quiz on sections 3.3, 3.4 and 3.5. There will be two quizzes next week. 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.
I solved the first two problems on this related rates worksheet. Do the third problem on your own. Solutions will be provided soon.
Continuing our discussion of differential equations and exponential functions...
If dy/dx is given in terms of x (the independent variable), then we must think about the process of differentiation in reverse in order to determine a formula for y as a function of x. By adding an arbitrary constant C, we obtain a family of functions which have the given formula for dy/dx. If an initial value is given then we can plug this point into the formula for y in order to determine the value of C. Since we often use variables other than x or y, you must pay close attention to the variables before determining a solution to the differential equation. We have the following examples.
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". Since A = π r2 gives the area of a circle, we see that the area is proportional to the square of the radius. Since V = 4/3*π r3 gives the volume of a sphere, we see that the volume is proportional the cube of the radius. 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 know the meaning of the term half-life and be able to use exponential functions to help solve problems involving half-lives. Although y = C*ax and y = C*ekx are both valid formulas for exponential functions, we will begin using the second form more often. 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.
|Mar 1 (Thu)||Quiz #5 on sections 3.3, 3.4 and 3.5 will be given during today's discussion section.|
|Mar 2 (Fri)||For homework read section 4.1 and do #16–25, 30, 41, 43, 49–60, 63, 75 from that section. Quiz #6 on sections 3.6, 3.7 and 3.8 will be given during Tuesday's discussion section. Quiz #7 will be a take-home quiz on sections 3.9, 4.1, 4.3, 4.4 and 4.7. It will be due in lecture next Friday.
In lecture today I further discussed strategies for solving related rates problems. I solved problem #41 from section 3.9 and the third 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 critical numbers, absolute maximum, absolute minimum, local maximum and local minimum.
I discussed the Extreme Value Theorem and the Closed Interval Method.
|Week 8 (Deadline to drop this course without a grade of W is Friday, March 9, 2012)|
|Mar 5 (Mon)||Read sections 4.3 and 4.7. In section 4.3 do #10, 13, 17, 18, 33, 39, 43, 46, 48, 53, 86. In section 4.7 do #5, 6, 13, 14, 19, 21, 32, 34, 35, 38, 49, 54. Quiz #6 on sections 3.6, 3.7 and 3.8 will be given during Tuesday's discussion section. Quiz #7 will be a take-home quiz on sections 3.9, 4.1, 4.3, 4.4 and 4.7. It will be distributed in lecture on Wednesday and due in lecture on Friday.
We used the Closed Interval Method to determine the absolute minimum and absolute maximum values for the function f(x) = x3 − 6x2 + 5 on the interval [−3, 5].
On an interval we have the following.
We used information about the first and second derivative to obtain a graph of f(x) = 2x3 + 3x2 − 36x.
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? For our solution we drew the upper half of the circle of radius 2 centered at the origin. We noted that y = √(4 − x2) for each point on this semi-circle. If we use (x, y) as the coordinates of the point at the upper right corner of the inscribed rectangle, then the area of the rectangle is A = base*height = 2x*y = 2x*√(4 − x2). The next step is to maximize this area on the interval [0, 2] using the Closed Interval Method. That is, using the formula A = 2x*√(4 − x2), you will plug in the endpoints x=0 and x=2 as well as any points on the interval (0, 2) for which A ′ = 0 or A ′ does not exist.
|Mar 6 (Tue)||Quiz #6 on sections 3.6, 3.7 and 3.8 will be given during today's discussion section.|
|Mar 7 (Wed)||Read section 4.4. In 4.4 do #7, 11, 17, 18, 19, 21, 25, 33, 41, 45, 49, 50, 55, 57, 61, 62, 67. Next Wednesday's test will cover sections 3.1–3.9, 4.1, 4.3, 4.4, 4.7. Quiz #7 is a take-home quiz on sections 3.9, 4.1, 4.3, 4.4 and 4.7. It will be distributed on Friday and should be turned in at the beginning of Monday's lecture.
I briefly discussed inflection points, the first derivative test and the second derivative test.
We used l'Hospital's Rule to determine the following limits.
(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.
The following are considered indeterminate forms so it is helpful to have a technique such as l'Hospital's Rule for determining limits in these cases.
|Mar 8 (Thu)|
|Mar 9 (Fri)||Quiz #7 is a take-home quiz on sections 3.9, 4.1, 4.3, 4.4 and 4.7. It should be turned in to Mr. Murphy at the beginning of Monday's lecture.
Use the Test 2 Notes to begin your preparation for Wednesday's test.
|Mar 9 (Fri)|
|Mar 12 (Mon)|
|Mar 14 (Wed)||Test 2 (given during lecture)|
|Mar 15 (Thu)||Discussion sections still meet today. Your homework is to read section 4.9 and do #1–17, 20–22, 25–33, 41–43, 65, 69, 73–75.|
|Mar 16 (Fri)||Finish the homework from section 4.9. Read section 5.1 very carefully. Do #3, 4, 13, 14, 15, 18, 20 from section 5.1.
A function F is called an antiderivative of f on an interval I if F ′(x) = f(x) for all x in I. Since you have already memorized a lot of short-cut derivative rules, you can use this knowledge to quickly determine the following antidervatives.
It is straightforward to find an antiderivative for a constant multiplied by a function or the sum of two or more functions. From the more complicated short-cut derivative rules for products, quotients and the composition of functions, one should expect that it is not as straightforward to find antiderivatives for products, quotients and the composition of functions. We will discuss some of these later. For now if you see a complicated function, you should try to rewrite the expression using algebra or trigonometry before finding an antiderivative.
In particular we made the following simplifications before finding antiderivatives for #14 and #22 in section 4.9.
When the rate of change of a quantity is positive, we saw graphically that the total change in that quantity on some interval could be represented by the area between the function and the horizontal axis on that interval.
We looked at the simple problem where a car travels at a constant rate of 50 miles per hour for a 2-hour period. The total change in position is (50 miles/hour) × (2 hours) = 100 miles. If we graphed the horizontal line y = 50 and shaded in the area between the horizontal axis and this line on a 2-hour time interval, we obtain that the area of this rectangle is 50 × 2 = 100.
We next looked at an object which travels at v(t) = √(t) feet per second between times t = 1 and t = 3 seconds. By breaking the interval [1, 3] into smaller subintervals we could approximate the total change in position on each subinterval as the area of a rectangle. The total change in position would then be approximated by the sum of the areas of these rectangles. Using 4 subintervals of equal width and a right Riemann sum, we obtained the following approximation.
Total change in position ≈ √(1.5) × 0.5 + √(2) × 0.5 + √(2.5) × 0.5 + √(3) × 0.5 ≈ 2.98.
Graphically we see why the above approximation is an overestimate to the actual change in position. We discussed how one would obtain an underestimate. We also discussed the idea that using more subintervals usually leads to a better approximation. You may need a calculator for some of the homework problems where the computations are lengthy.
I used a few examples to introduce Σ notation for sums and relayed the story of Gauss quickly obtaining the sum of 1 + 2 + 3 + ... + 100 as a young boy. More generally we find that 1 + 2 + 3 + ... + n = n(n + 1)/2.
|Week 10 (Spring Break!)|
|Mar 26 (Mon)|
|Mar 28 (Wed)||Read section 5.2 very carefully. Start working on #2, 11, 18, 21, 22, 29, 33, 36, 37, 41, 48, 49, 52, 53, 55, 57, 59 from section 5.2. Quiz #8 on sections 4.9 and 5.1 will be given on Thursday.
Using Σ notation I wrote the sum 52 + 62 + 72 + ... + 202 in four different ways.
Using the second sum above, we found the sum from k = 1 to n of 5k2/n3 to be (5/n3)*n(n+1)(2n+1)/6. We then found the limit of this sum as n goes to infinity to be 10/6 = 5/3.
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]. Of course since the shape is just a trapezoid you can use geometry to find our answer more simply and should compare the two answers.
For another example we calculated the limit of right Riemann sums in order to determine the area between the x-axis and f(x) = x2 on the interval [2, 8]. This would be more difficult to check using geometry but you should at least have an approximate answer in mind to compare to the exact answer obtained in lecture.
|Mar 29 (Thu)||Quiz #8 on sections 4.9 and 5.1 will be given during today's discussion section.|
|Mar 30 (Fri)||Read section 5.3. In section 5.3 do #23, 24, 28, 31, 32, 33, 35, 39, 45, 48, 51.
Using Σ notation we again looked at the following sums.
I gave the definition found on page 372 for the definite integral including the term integrable for when the given limit exists.
Theorem: If f is continuous on [a, b] then f is integrable on [a, b].
We looked at the definite integral of 2x from 0 to 3. Since 2x is continuous on [0, 3] we know that it is integrable on [0, 3]. We can evaluate the definite integral in multiple ways.
Calculators have approximation techniques for evaluating definite integrals. For example the definite integral of 15x4 from x = 1 to 3 can be approximated in the following ways.
Suppose that a population is expected to increase at a rate of 6t + 2 people per year where t represents the number of years from now. What is the expected change in population between years t = 1 and t = 3? We began to solve this in two ways.
Method 1: After graphing 6t + 2 on the interval [1, 3], we saw that a limit of Riemann sums would give the exact change in population. Since by definition this limit is the definite integral, we saw that the exact change in population is the definite integral of 6t + 2 as t goes from 1 to 3.
Method 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) = 6t + 2. By solving this differential equation we obtain the formula P(t) = 3t2 + 2t + C as the population at time t. The exact change in population is then seen to be P(3) − P(1) = (33 + C) − (5 + C) = 28 people. Notice that C drops out here.
Our second method for this problem demonstrates the Net Change Theorem found on page 401 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) from section 5.3 which shows a wonderful shortcut for evaluating many definite integrals. We will discuss this more fully next week.
|Apr 2 (Mon)||Read section 5.4. In section 5.4 do #3, 5, 6, 15, 16, 17, 18, 27, 31, 37, 43, 53, 54, 64. Many of these are straightforward drill problems using antiderivatives instead of limits. If necessary you should do more than the assigned problems so that you become quick at obtaining these answers. Quiz #9 on sections 5.2, 5.3 and 5.4 will be given on Thursday.
For a given definite integral, I looked at how to write it as a limit of right Riemann sums as compared to a limit of left Riemann sums. Students should also think about how to write a definite integral as a limit of midpoint Riemann sums.
I solved #37 and #57 from the homework in section 5.2. For #37 we first wrote the definite integral as a limit of right Riemann sums. Since the limit appeared difficult to compute, we instead solved the problem geometrically by graphing the integrand and noting that the area of the region between the x-axis and the curve is simply the sum of the area of a rectangle and the area of a quarter circle. For #57 I used the minimum and maximum values of the function in order to approximate the definite integral of the function. Drawing a graph of the integrand and thinking about areas really helps here, but we are simply using the third comparison property listed below.
For integrable functions f(x) and g(x), I discussed the following comparison properties.
Method 1: We graphed 0.3t2 on the interval [0, 10] and then took the limit of Riemann sums to get the exact change in population. 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. When we computed this definite integral as a limit of Riemann sums, we obtained an answer of 100 people as the change in population.
Method 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 people.
Our second solution to this problem demonstrates the Net Change Theorem found on page 401 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) from section 5.3 which shows a wonderful shortcut for evaluating many definite integrals. We used this quick approach for the following problems.
I formally wrote out the Fundamental Theorem of Calculus (part 1) and the Fundamental Theorem of Calculus (part 2). Today we only discussed part 2 but students should read about part 1 in the book.
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 the last example above, the integrand x−2 = 1/x2 is not continuous on the interval [−2, 1] and it turns out that the function is not integrable on the given interval. We should not attempt to directly use the Fundamental Theorem of Calculus for this definite integral. However I'm still not quite sure why students came up with such a large number of different incorrect answers. It must have been due to incorrect arithmetic or incorrect antiderivatives.
The indefinite integral represents the most general antiderivative of the integrand on an interval. Students should know the table of indefinite integrals found on page 398 but can temporarily ignore the last two where the integrand is sinh(x) or cosh(x).
|Apr 4 (Wed)||Read section 5.5 and do #8, 16, 17, 18, 20, 21, 22, 23, 25, 28, 32, 39, 40, 41, 44, 46, 48, 54, 57, 59, 60, 61, 65, 66, 67, 69, 81, 82 from that section. There will be a quiz Thursday on sections 5.2, 5.3 and 5.4.
We began class by looking at the difference between definite integrals and indefinite integrals.
What is the definite integral of f(x) from a to b? If f(x) is integrable on the interval [a, b], then the definite integral of f(x) from a to b is equal to a finite number. This finite number may be obtained using the definition of a definite integral as a limit of Riemann sums. However it is often easer to find this finite number using the Fundamental Theorem of Calculus (part 2) or simple geometry.
What is the indefinite integral of f(x)? The indefinite integral is the most general antiderivative of f(x) on an interval. Thus it is a family of functions which all have f(x) for a derivative.
Students should know the table of indefinite integrals found on page 398 but you can ignore the last two where the integrand is either of the hyperbolic functions sinh(x) or cosh(x).
We looked at the connection between definite integrals and area with the following examples.
I evaluated the following integrals from section 5.3.
For the second problem I rewrote the integrand as 6x11 + 72x8 + 288x5 + 384x2 before finding the most general antiderivative.
Rewriting the integrand for the second problem was a little bit time-consuming. Rewriting the integrand for the third problem will be too time-consuming. We need a better way of handling problems like these.
In section 5.5 we see that the chain rule in reverse leads to the method of substitution. We used this method to evaluate the following integrals.
|Apr 5 (Thu)||Quiz #9 on sections 5.2, 5.3 and 5.4 will be given during today's discussion section.|
|Apr 6 (Fri)||Read sections 6.1 and 4.2. In section 6.1 do #1, 8, 11, 12, 13, 17, 23, 25, 27, 29, 50, 51.
From the graph of tan(x) that we should expect the definite integral from −1/3 to 1/3 of tan(x) dx to be equal to 0. In fact if f(x) is an odd function integrable on [−a, a], we always get that the definite integral from −a to a of f(x) dx = 0.
We found the indefinite integral of sin(2x) dx by solving it three different ways.
Here are some quick ways to check that your evaluation of an indefinite integral is correct.
For example I found the exact area of the following region.
|Apr 9 (Mon)||Read sections 4.2 and 6.2. In section 4.2 do #1, 2, 5, 9, 11, 15, 17, 20. Thursday's quiz will cover sections 5.5 and 6.1.
We sketched the region bounded by y = 2, y = 0, x = 0, and y = ln(x). We determined the area of this region in two ways – by integrating with respect to x and then by integrating with respect to y. Our answers were
Area = e2 − 1.
We solved #12 from section 6.1 which asks for the area of the finite region bounded by the curves 4x + y2 = 12 and x = y. By graphing the functions and finding the intersection points, we noted that we could draw a rectangle of area 72 around the region. Since the area of the region bounded the curves appeared to be about one third of the area of the rectangle, we expected the area of the region to be approximately 72/3 = 24.
We solved this problem exactly in two different ways.
We solved #23 from section 6.1 by integrating with respect to x to obtain
From section 4.2, the book introduces the Mean Value Theorem along with Rolle's Theorem which is a special case of the Mean Value Theorem. 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. 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 f(x) = x5 + 3x3 + 10x + 10 has exactly one real root (i.e. an x-intercept).
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 root for f(x) = x5 + 3x3 + 10x + 10.
|Apr 11 (Wed)||Read section 6.2. In section 6.2 do #2, 6, 7, 9, 12, 14, 16, 17, 33, 55, 56, 58. There will be a quiz Thursday on sections 5.5 and 6.1. There will be two quizzes next week.
We began lecture with the statement, visual interpretation, and finally a proof of both Rolle's Theorem and the Mean Value Theorem. I mentioned that the Mean Value Theorem is often used to prove other theorems. In particular, it can be used to prove the following theorem and its corollary.
|Apr 12 (Thu)||Quiz #10 on sections 5.5 and 6.1 will be given during today's discussion section.|
|Apr 13 (Fri)||Read sections 6.3 and 6.5. In section 6.3 do #3, 5, 9, 12, 14, 15, 17, 19, 20. In section 6.5 do #1, 2, 4, 5, 7, 9, 10, 13, 14, 17. Go back and try to solve each area problem (section 6.1) and each volume problem (sections 6.2 and 6.3) in two ways – once by integrating with respect to x and once by integrating with respect to y. There will be two quizzes next week. The quiz on Tuesday will cover sections 4.2, 6.2 and 6.3.
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.
We again looked at the region R between the x-axis and the graph of y = √(x) on the interval [1, 9]. We set up definite integrals for the volume of a solid obtained in each of the following ways.
We then found the volumes of two solids using the method of cylindrical shells.
|Apr 16 (Mon)||Read sections 3.10 and 4.8. In 3.10 do #6, 23, 24, 25, 26, 28, 31. Quiz #11 is an in-class quiz on sections 4.2, 6.2 and 6.3 which will be given during Tuesday's discussion section. I'll announce information about quiz #12 later this week.
We can 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.
I briefly discussed why it is good to use a polynomial (such as the tangent line) to approximate a given quantity. 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. Students who take MATH 231 (Calculus II) will learn more about these polynomials which are referred to as Taylor Polynomials. In this course we will restrict our attention to the use of tangent lines to approximate a function near a point.
I discussed five different methods for approximating the square root of 5.
|Apr 17 (Tue)||Quiz #11 on sections 4.2, 6.2 and 6.3 will be given during today's discussion section.|
|Apr 18 (Wed)||Read section 4.8. In section 4.8 do #11, 12, 13, 15, 18, 19, 29, 31. In section 7.2 do #1–8, 12, 14, 15, 17–31 and 34. When using Newton's Method in the homework from section 4.8, some initial estimates may lead to a sequence of estimates which converge very slowly to a root. Don't be too concerned about obtaining the level of accuracy they seek in the homework. Just make sure that you understand and can apply the iterative process correctly. Many of the integrals in section 7.2 require the use of some basic trigonometric identities along with substitution. You should already have these skills. However I will still discuss these more fully on Friday. Quiz #12 is a take-home quiz on sections 3.10 and 4.8. It will available Friday and due at the beginning of Monday's lecture.
I discussed Newton's Method for finding roots of a function f(x). This is equivalent to finding solutions to f(x) = 0. To solve this you decide upon a suitable first estimate x1. This iterative process then generates successive estimates x2, x3, x4, ... which hopefully converge to one of the roots. For n ≥ 1, the successive estimates are given by
xn + 1 = xn − f(xn) / f ′(xn)
You should know the algorithm as well as the graphical interpretation of how tangent lines are used to generate these successive estimates.
We used Newton's Method to solve the following problems.
For homework you will be doing a lot of computations on your calculator. Keep a lot of decimal places in each step until the last step when you finally round off your answer. I showed how to use the calculator in an efficient way so that after an initial set-up, each step in Newton's Method will only take only 3 calculator button pushes. It also keeps all of its decimal places so that you don't have to transcribe so many numbers. To do this on a TI-83 or TI-84, you should do the following set-up.
We evaluated the integral of each of the six trigonometric functions.
|Apr 19 (Thu)||Quiz #12 will be a take-home quiz on sections 3.10 and 4.8 due at the beginning of Monday's lecture.|
|Apr 20 (Fri)||Quiz #12 is a take-home quiz on sections 3.10 and 4.8. It should be turned in to Mr. Murphy before the bell rings at the beginning of Monday's lecture.
Use the Test 3 Notes to begin your preparation for Wednesday's test.
We looked back at the integrals for the six trigonometric functions. However for this test you will not be required to know the integral of sec(x) dx or the integral of csc(x) dx.
We discussed strategies for integrating sinn(x) or cosn(x) when n is odd.
|Apr 23 (Mon)|
|Apr 25 (Wed)||Test 3 (given during lecture)|
|Apr 26 (Thu)||Discussion sections still meet today. For homework read section 3.11 and do #1, 2, 3, 4, 23abcd, 31, 32, 33, 35, 38 from that section.
Section 3.11 includes a lot of details concerning hyperbolic functions. Here are the main details that you need to know for our course. This knowledge should be sufficient for answering the assigned homework questions.
|Apr 27 (Fri)||Finish the homework from section 3.11. Read the first five pages in section 5.3 on part 1 of the Fundamental Theorem of Calculus. In section 5.3 do #7, 8, 12, 16, 57. Physics and engineering students are encouraged to read section 6.4 and do #1, 2, 7, 8, 9, 13, 15, 18, 19, 20, 21, 24 from that section. Section 6.4 will not be on the final exam.|
|Week 16 (Last day for U of I classes is Wednesday, May 2, 2012)|
|Apr 30 (Mon)||On the final exam there will be a bonus problem on section 7.1 (Integration by Parts). For practice you may want to look at problems 1 – 36 in section 7.1.
My office hours this week are Tuesday/Thursday 4-5:30pm. The tutoring room will be definitely be open Monday, Tuesday and Wednesday. If there are additional hours I will post them here.
In lecture we used integration by parts to evaluate the following integrals.
|May 2 (Wed)||The cumulative final exam for section BL1 will be held Thursday, May 10, 1:30 PM – 4:30 PM in 314 Altgeld Hall. Here are the sections covered.
See the list of remaining tutoring/office hours for the semester.
|Final Exam Period (Friday-Friday, May 4-11, 2012)|
|May 10 (Thu)||Cumulative Final Exam from 1:30 PM – 4:30 PM 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