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Teaching Assistant: Claudine Myers, DuSable 370
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Homework problems
DATE DUE Friday, 8/29 NO QUIZ Homework #1: 1.1 #40,53,62 1.2 #12 1.3 #37 Homework #2: 2.1 #5,8 2.2 #6,22,33 Friday, 9/5 Quiz #1: Emphasis on 2.3, but covering everything we have done so far Friday, 9/12 Quiz #2: Section 2.4 #1  10 (bring a calculator) Section 2.5 Homework #3: 2.4 #29, 31 Friday, 9/19 Quiz #3: Sections 2.6, 3.1, 3.2 Friday, 9/26 EXAM I Covering sections 1.11.3, 2.12.6, 3.13.4 Note: no calculators Friday, 10/3 Quiz #4: Sections 3.5, 3.6 Homework #4: Appendix D #51, 56, 57, 74, 83
HONORS: The honors part of the course (Math 297) will use these notes: The Calculus of Polynomials. We will usually cover honors material on Tuesdays.
GENERAL ADVICE: The key ideas of calculus are not the main barrier in the coursethey can be explained rather intuitively. In my experience, students have trouble in this course either
TEXT: Calculus, 5th Ed., by James Stewart, McMaster University
COURSE OBJECTIVES: You are expected to acquire not only computational facility with the topics introduced, but also a basic understanding of the concepts and theory of calculus. Mathematics is a truly universal language; we want you to become more fluent in reading, writing, and using this precise language.
GRADING: Your grade will be based on 600 points as follows:
3 onehour examinations 300 points Final examination 200 points Homework/quizzes 100 points
FINAL EXAM: The comprehensive departmental final exam is scheduled as a mass exam on Friday, December 12, 9:009:50 A.M.
COURSE WITHDRAWAL: The last day to withdraw is Friday, October 17.
CALCULATORS: A graphing calculator is required for the course. with capabilities similar to those of the TI83 calculator. Graphing calculators will not be permitted on the final exam.
SYLLABUS: The course will cover Chapters 15, §6.1, and §8.7.
Tentative lecture schedule Monday Tuesday Wednesday Thursday Friday M Tu W Th F 1.11.4 2.1 2.2 Recitation 2.3 AUG 25 26 27 28 29 Holiday H1.1 2.3 " 2.4 SEP 1 2 3 4 5 2.4 H1.2 2.5 " 2.6 8 9 10 11 12 3.1 H1.3 3.2 " 3.3 15 16 17 18 19 3.3 H1.4 3.4 Review EXAM 1 22 23 24 25 26 3.5 H1.5 3.6 Recitation 3.7 OCT 29 30 1 2 3 3.8 H1.6 3.9 " 3.10 6 7 8 9 10 4.1 H Exam 4.2 " 4.3 13 14 15 16 17 4.4 H2.1 4.5 Review EXAM 2 20 21 22 23 24 4.6 H2.2 4.7 Recitation 4.7 27 28 29 30 31 4.9 H2.3 4.10 " 5.1 NOV 3 4 5 6 7 5.2 H2.4 5.2 " 5.3 10 11 12 13 14 5.3 H2.5 5.4 Review EXAM 3 16 17 18 19 20 5.5 5.5 Thanksgiving  holiday 24 25 26 27 28 6.1 H Exam 8.7 Review 8.7 DEC 1 2 3 4 5 FINAL EXAM Tuesday, 45:50 PM 8 9 10 11 12
ADVICE: The key to success in this course is to work as many problems as you can. The general rule in college is to spend at least two hours preparing for each class period. Above all, you can't afford to fall behind, since it is almost impossible to make up even a few days of lost time.
Recommended problems:
These homework problems are the key to your success in the course.
They represent the minimumdo as many additional problems as possible.
You should keep all of your homework problems in a notebook
so that you can study from them for the final exam.
Note: A *
indicates additional problems assigned in 229H.
Week Section Page Problems
8/25 1.1 p.22 2,79,19,21,24,26,29,35,38,40,48,53,62 1.2 p.35 2,3,6,10,12 1.3 p.46 4,10,12,17,18,24,27,32,36,37 8/25 2.1 p.69 1,35,8 2.2 p.80 5,6,12,14,15,22,23,25,33 2.3 p.89 1,46,10,12,13,15,21,31,37,46 9/1 *2.3 p.90 ***16,17,18,24,26,29,33,43,44,47,52,53,59 2.4 p.100 2,3,7,15 9/8 *2.4 p.101 ***29,31,37 2.5 p.110 3,4,6,11,12,15,16,20,33,38,39,41,43 2.6 p.119 7,8,12,15,17,18 9/15 3.1 p.132 3,4,7,8,13,1922,25 *3.2 p.142 47,17,24,31 ***39,40,43,44,45 3.3 p.154 4,6,12,14,17,2022,26,27,32,49,54,56,57,58,61,63,64 9/22 *3.3 p.156 ***6575,8088 3.4 p.166 1,2,7,8,11,14,18,20,25,26 EXAM 1 9/29 AppD p.A32 1,2,7,8,1315,1721,23,24,29,30,35,36,74,78,81 3.5 p.174 2,7,8,10,12,13,17,23,30,33,35,36,38,40,41,43,46 3.6 p.181 4,6,7,12,17,20,24,27,28,34,36,37,39,40,45,52,53 3.7 p.188 1,7,14,16,20,23,26,27,31,38,40 10/6 3.8 p.195 1,4,5,13,18,19,23,26,30,35,38,44,51,53 3.9 p.202 13,5,7,11,13,17,18,23,24,25,29,31 3.10 p.210 5,6,9,15,17,24,34,39,41,42 10/13 4.1 p.230 3,710,14,16,18,25,30,31,34,39,41,47,48,54,56 4.2 p.238 2,6,12,17,19,23,24 4.3 p.247 2,8,11,14,1618,22,24,25,28,32,34,38,50 10/20 4.4 p.260 4,710,13,16,20,32,39,40,47,48,52,54 4.5 p.270 3,5,7,13,14,25,26,31,37 EXAM 2 10/27 4.6 p.277 1,3,7,11,12,20,25 4.7 p.283 2,6,9,12,16,19,20,22,28 4.7 p.284 29,34,35,38,51,52,58 11/3 4.9 p.298 1,5,6,10,11,27,30,35 4.10 p.305 1,2,4,10,12,13,17,22,24,27,31,36,53,55,71 5.1 p.324 1,3,5,11,13,1720 11/10 5.2 p.336 1,2,5,11,18,20,2225 5.2 p.338 30,35,37,43,47,48,51 5.3 p.347 1,4,6,8,10,1921,25,26,28 11/17 5.3 p.348 30,31,33,34,45,46,53,54 5.4 p.356 1,2,513(odd),1838 (even),43,47,48 EXAM 3 11/24 5.5 p.365 131(odd),3751(odd) THANKSGIVING 12/1 6.1 p.380 515(odd),21,22,25 8.7 p.563 713(odd),29,33
Help on the WEB: The publisher's web site  Calculators (from NIU)  other WEB resources
Worksheets (in pdf format): Optimization problems  Antiderivatives  Integrals (by substitution)
The original motivation for studying derivatives came from physics, at a time when people were trying to understand things in motion. In this context, the derivative measures velocity, which we can think of as an instantaneous rate of change. The acceleration of an object measures how fast its velocity is changing, and so this can also be measured by a derivative (the second derivative).
When working with the graph of a function, the derivative can be very useful. For a straight line, the derivative is just the slope of the line, and a positive slope corresponds to a graph that moves up (from left to right), while a negative slope corresponds to a graph that moves down. For more general functions, just knowing the sign of the derivative is useful, since the graph goes up when the derivative is positive, and down when it is negative. This allows to find maximum and minimum values for a function.
The indefinite integral is the opposite of the derivative. It allows us to solve the following problem: given a formula that tells the rate of growth at any point in time, try to reconstruct the original function.
The definite integral is an averaging process. For example, the something is moving at a constant rate, the distance it travels is just the rate multiplied by the time it takes. But what if the rate is variable? That is where the definite integral can be used. As another example, in physics the work done by a force is defined to be the magnitude of the force times the distance through which it acts (provided the force is constant). But even in stretching a spring the force is variable, and this simple formula can't be used. The definite integral of the force finds the average force times the distance, and that is why physicists usually define work in terms of an integral. Finally, the integral allows us to find the area under a curve because it can find an average height, multiplied by the width of the interval.
There is a longer essay online, if you have the time to read it.
To take the derivative of the sum of two functions, we use the rule that the derivative of a sum is the sum of the derivatives. Unfortunately, the rule for taking the derivative of a product of two functions is more complicated. A simple example for which we already know the answer gives a good illustration of the difficulties.
Suppose that f(x) = x^{2} and g(x) = x^{3}. Then the individual derivatives are f '(x) = 2x and g '(x) = 3x^{2}. We know that the derivative of the product f(x)g(x) = x^{2}x^{3} = x^{5} should be 5x^{4}, but it is difficult to see how to combine f '(x) and g '(x) to get this answer. It turns out that we also have to use f(x) and g(x) in the formula.
One approach to understanding the formula for differentiating products comes from working with tangent lines. Remember that the equation of the line through the point (a,b) with slope m is
y = m(xa) + b.
At the point (a,f(a)) on the curve y = f(x), we have the ycoordinate b = f(a), and the slope of the tangent line at this point is given by m = f '(a). This allows us to calculate the equation of the line tangent to the curve y = f(x) at x=a asy = f '(a)(xa) + f(a).
In the same way, the line tangent to the curve y = g(x) at x=a isy = g '(a)(xa) + g(a).
To get the formula for the tangent line to y = f(x)g(x) at x=a, we could try multiplying the two formulas. We would get the following quadratic expression.f '(a)g '(a)(xa)^{2} + ( f '(a)g(a) + f(a)g '(a) )(xa) + f(a)g(a)
Because the tangent line has to be linear, this doesn't solve the problem, unless we can change it in some way. The tangent line is intended to be the line that gives the best approximation to the curve, near the point (a,f(a)). When x is very close to a, the factor (xa) is very small. In comparison, the factor (xa)^{2} is even smaller; in fact, it is very much smaller. For example, if xa = .1, then (xa)^{2} = .01; if xa = .01, then (xa)^{2} = .0001; if xa = .001, then (xa)^{2} = .000001. Because the tangent line is just an approximation to the curve, it turns out that we can ignore the quadratic term. Notice that the part we had to ignore contains the term f '(a)g '(a), which would have given us a "nice" formula.Conclusion: the tangent line to y = f(x)g(x) at x=a is
y = ( f '(a)g(a) + f(a)g '(a) )(xa) + f(a)g(a).
The general formula is best given using x instead of a.
d/dx ( f(x)g(x) ) = f '(x)g(x) + f(x)g '(x).
Besides remembering the product rule in the "visual" form given above, it is good to remember it in words, as an algorithm.
The derivative of the product of two functions is equal to the derivative of the first times the second, plus the first times the derivative of the second.
Example 1. Going back to the first example, we can now compute
d/dx ( x^{2} x^{3} ) = (2x)(x^{3}) + (x^{2})(3x^{2}) = 2x^{4} + 3x^{4} = 5x^{4}.
Our formula gives the expected answer, but has to include f(x) and g(x) along with f '(x) and g '(x).Example 2. It is interesting to see the formula for the derivative of the product of three functions. It clearly shows the pattern in which we add up terms where we differentiate one function at a time.
d/dx ( f(x)g(x)h(x) ) = f '(x)g(x)h(x) + f(x)g '(x)h(x) + f(x)g(x)h '(x)
The rule for taking the derivative of the composition of two functions is also complicated. Again, a simple example for which we already know the answer gives a good illustration of the difficulties.
Suppose that f(y) = y^{3} and u(x) = x^{4}. The formula for the composite function is f(u(x)) = (x^{4})^{3}. The individual derivatives are f '(y) = 3y^{2} and u '(x) = 4x^{3}. We know that the derivative of the composite function f(u(x)) = x^{12} should be 12x^{11}. In this case we just need to substitute y=u(x) into f '(y) and multiply by u '(x). We can show this by looking at the tangent lines.
The line tangent to the curve y = u(x) at x=a is
y = u '(a)(xa) + u(a).
At the point (b,f(b)) on the curve z = f(y), the slope of the tangent line at this point is given by f '(b). Thus the equation of the line tangent to the curve z = f(y) at y=b asz = f '(b)(yb) + f(b).
To compute the composite function f(u(x)), we substitute the formula for u(x) into the formula for f(y). If x=a, then we evaluate f(y) at y=u(a). This means that in the equation of the tangent line, we must take b=u(a). The equation becomesz = f '(u(a))(yu(a)) + f(u(a)).
To get the formula for the tangent line to z = f(u(x)) at x=a, we can substitute the formulay = u '(a)(xa) + u(a)
into the formulaz = f '(u(a))(yu(a)) + f(u(a)).
This gives usz = f '(u(a))(u '(a)(xa) + u(a) u(a)) + f(u(a))
orz = f '(u(a))(u '(a)(xa)) + f(u(a))
The derivative is the slope f '(u(a))(u '(a). The general formula is best given using x instead of a.
d/dx ( f(u(x)) ) = f '(u(x)) u '(x).
Besides remembering the product rule in the "visual" form given above, it is good to remember it in words, as an algorithm.
The derivative of the composite of two functions is found by taking the derivative of the first function and substituting in the second, the multiplying by the derivative of the second.
Example. 1 Going back to the first example, we can now compute
d/dx ( (x^{4})^{3} ) = 3(x^{4})^{2}(4x^{3}) = (3x^{8})(4x^{3}) = 12x^{11}.
Our formula gives the expected answer. It is important to remember to multiply the term f '(u(x)) by u '(x).Example. 2 To give another point of view, we can check the chain rule in a case where we can actually apply the product rule. If f(u(x)) = u(x)^{3}, we can write
f(u(x)) = u(x)^{3} = u(x)u(x)u(x),
and so from the product rule we getf '(u(x)) = u '(x)u(x)u(x) + u(x)u '(x)u(x) + u(x)u(x)u '(x).
Combining terms gives us the correct answerf '(u(x)) = 3 u(x)^{2} u '(x).
Past assignments for 229 H:
DATE DUE