Math 1371: IT Calculus I. Fall '04, Lecture 040
Last changed: October 22,'04
The lectures are at 2:30pm MW, in SciCB 125.
Instructor: Max Jodeit ("Yodite") VinH 258, 5-3855, jodeit@math.umn.edu
Texts: Introduction to Differentiation by Chester Miracle
and
Calculus Early Transcendentals 5/e, by James Stewart
Fall 1371 Faculty, Workshop Leaders, Office Hours.
News and Announcements will go here
NEWS Oct 22: To prepare for the gateway exam, do exercises 1 - 48
on pages 270-271 of Stewart.
NEWS Oct 14: Classroom Support Hotline: 5-1086
Anent the Gateway Exam, see pages iii and iv in your Intro to Diff
pamphlet. It's on Oct 28 this semester.
Roger Sadlowsky's Problem-solving steps (my paraphrase):
1) Give variable names (x, y, t, p, q, etc.) to the quantities
the problem uses. Give names to the quantities the problem asks about.
2) State the given values and rates (x=25, (dp/dt)=3, etc.).
3) Write down what the unknown is (e.g., "Find dq/dt when so-and-so is true.").
4) Find equations that relate the relevant quantities.
5) In related-rate problems, DIFFERENTIATE the relevant equations,
substitute given values and solve for the unknown rate.
NEWS Sept 24: for Fall 1371 Faculty and Discussion Leaders,
follow the new link above
Yesterday in the Discussion Sessions it seemed that the biggest
troubles you had were with 6114, 2 and 3. In each case it's very useful to me to try to
imagine the physical situation. It's also important to think of the quantities you talk about
as variables instead of formulas! In 2, the problem was how to show work when you
said where the derivative was negative. There are two ways: geometric (picture) and
algebraic. To use the geometric approach you imagine the graph of the velocity, as a function
of time. Then you draw a t-axis, and it cuts thru the graph where the function it is the
graph of is zero. You figure out where the derivative is zero using algebra, your calculator or
the techniques you learned in High School Math. You somehow knew that the graph was a
vertex-down parabola, so you drew and showed that where the graph is under the
t-axis is where the derivative is negative.
To use the algebraic approach you draw a line and mark where t = 3 and
t = 7 are. Write t - 7 a couple of spaces above the line and write
t - 3 a couple of spaces below the line. Your line divides the t-axis into
three parts. Mark, above each part, the (algebraic) sign of t - 7 when t
is in that part (+ or -), and mark, below each part, the (algebraic) sign of t - 3
when t is in that part (+ or -).
The sign of the derivative is the product of those signs, in each interval. You should
have had this pattern: two -'s when t < 3, two +'s when t > 7,
and -+ when 3 < t < 7. These patterns tell you where the sign of the product is -.
Link to
Professor Carter's 1371 (recommended!) Webpage
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Jodeit's Home Page
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