For Spring '04, class meets on MTWF at 2:15; Reiss 264. Calc II (Math 036).
You can reach me by phone (or voice-mail) at 7-2703 or by e-mail kainen@georgetown.edu or drop by Reiss 248 during my office hours. I have a new office in Preclinical Sciences Building LR3 room 26 from TBD ...
**** Last updated ****
May. 8, 2004
**********************
Other general information, including for other courses, on the index (classroom) page .
Check out information on the syllabus and course mechanics from time to time. I suggest you review this material which also explains homework, quiz policies, and the exam schedule.
There will be an extra review class at 9pm on Monday May 10 in Reiss 283.
For some _new_ problems (in pdf format) as review for the final, see review problems . (As of May 8, with a few additional by tomorrow night.) If you find any misprints, please let me know - and you'll get credit for noticing ;-)
Try the following for Monday: 8.7 #48,50,52,54; the answer for 50 is e^(3/5).
As promised here are a few more. I won't collect them but we will discuss some of them on Mon. and Tuesday so be prepared. Also, they provide a review of some useful material.
Converge or Diverge - with justification!
sqrt(2 + 7j + j^5) + sin(e^j)
(1) series sum (j=1 to oo) -------------------------------
cuberoot(1 + j^2 + j^11)
(2) sequence (2, 3/2, 4/3, 5/4, 6/5, ...)
(From now on, I'm not going to specifically remind you which are series.)
(3) sum (j=1 to oo) [1 + (1/j)]
(4) sum ( " ) [(-1)^j + (1/j)]
(5) sum ( " ) [-1 + (1/j)]^j
(6) sum ( " ) (-1/j)^j
The following three series converge. Give two different arguments,
when possible, for convergence:
(1 + 1000*n + 2*n^4) 2^(3n)
(7) sum (n=0 to oo) --------------------- * --------
(7 + n^4) (27)^n
(8) same as in (7) except replace 2*n^4 by 2*n^3.
(9) sum (n=0 to oo) (-1)^u(n) c_n r^n, where
(a) u(n) is any integer-valued function,
|c_(n+1)|
(b) lim (as n goes to oo) ---------- = 1,
|c_n|
(c) |r| < 1.
Find the value of the following three series:
(13)^n
(10) sum (n=5 to oo) (-1)^n ----------
5^(2n-7)
(11) sum (n=1 to oo) 3^(-n)/(n-1)!
(-1)^n (pi/4)^(2n+1)
(12) sum (n=0 to oo) ---------------------
(2n)!
(13) If a series converges absolutely, it must converge.
The converse claim (that convergence implies absolute convergence)
is false; give an example to demonstrate this.
(14) Give an example of an alternating series which does not converge
even though its terms decrease monotonically in absolute value.
(15) Can you state the integral test from memory?
(16( Give an example of a sequence c_n which satisfies 9(b).
(17) Derive the following "q-test":
sum (n=2 to oo) 1/[n * (ln n)^a] converges if and only if a > 1.
(18) Derive the comparison test for series from the comparison test for
improper integrals.
(19) If the sequence (a_n) is increasing and bounded below, must it converge?
(20) If the sequence (a_n) converges and the sequence (c_n) diverges,
what can you say about the sequence (a_n -c_n)?
(21) What is the radius of convergence and the center of the power series
sum (n=0 to oo) (n^3 + 7n -3) * (3/2)^n * (x - 14)^n?
(22) Same question but replace (x - 14)^n by (x - 14)^(3n)
(23) Find the interval of convergence for
(2n)!
sum (n=3 to oo) a_n x^n, where a_n = --------
5^(5n)
For Fri. May 30, please do 8.6: 18,20,24; 8.7: 18,20,22. Also, there will be a special review on May 10 (Mon.) at 9pm till 10:30 pm (room TBA) where we can review practice material for the final - I'll give out some problems on the last day of class (May 4).
For tomorrow (Fri. 23rd) as I said in class on Wed. please do 8.7 #4,6,10 for collection. For Monday the 26th, here are some problems: 8.7 #9,11,12,13,16 (hint: use thm 8 and exercise 13),20,39,40.
Also, we can skip (for now) the so-called ``telescoping'' series, like sum_(n = 1 to oo) 1/n(n+1) = 1.
Proof. Note that 1/n(n+1) = 1/n - 1/(n+1) (as in Partial Fractions). Hence, for each k,
sum_(n = 1 to k) 1/n(n+1) = 1 - (1/2) + (1/2) - + ... - (1/(k+1))
= 1 - (1/(k+1)) (all the other terms cancel)
Hence, the sum of the infinite series = the limit of the partial sums = 1.
Here's one to try: What is the interval of convergence of
sum_(n=0 to oo) n^(-1/2) * (-27/8)^n * (x+4)^3n?
The answer follows a bit further down the page. But try it first! Make a guess as to what you think the answer is and then check it by writing out the details. What do you think happens if n^(-1/2) is replaced by n^(-3/2)? What about if it is replaced by n^2?
Method 1: Use the Ratio Test. The RT-limit for this series is lim_(n --> oo) (sqrt(n)/sqrt(n+1))* (27/8) |x+4|^3 and this is < 1 when |x+4| < 2/3 so R = radius of convergence is 2/3 and the center of the interval of convergence is -4 (NOT +4). Checking the endpoints, we get IOC = (-4 - 2/3, -4 + 2/3] using the p-test (p = 1/2) to exclude the left-hand endpoint and using the AST to include the right-hand endpoint.
Method 2: Use substitution to rewrite the series as sum c_n t^n, where t = (-27/8)(x+4)^2. The t-series has IOC [-1,1) using again the AST and p-test to include and exclude, resp., the left and right endpoints. Now use the reverse mapping to translate into x-coordinates: x = -4 + (-2/3)t^(1/3). The cube root leaves the interval [-1,1) unchanged; multiplying by -(2/3) rescales and reverses so the interval is now (-2/3,2/3]; and adding -4 gives the same interval of convergence as in Method 1.
x = -4 + sqrt[(-9/25) * t]
maps [-1,0] to [-4 + 3/5,0]
For Monday 19 April, in addition to the problems listed below, try the following: 8.6: #12,14,16,22. And on p. 641, #2,4,6,10,12,14,16,18, 30,32,34. Don't forget that the solutions manual for the odd-numbered problems is available in the Reiss lib.
Meanwhile, of the ones below for 8.5, here are a few selected answers: #4 R = 1, IOC = (-1,1] (when x = -1, the series becomes the harmonic series so diverges, while when x = 1, the series becomes the alternating harmonic series and so converges by the AST), #10 R = 10, #14 R = oo.
Also, try the following from 8.6: #3 through 10; selected ans. #6 f(x) = sum (n from 0 to oo) ((-1)^n)*(3^2n)*(x^2n), IOC = (-1/3,1/3).
For tomorrow, Tues. 13 April, please do 8.5: #3 - 15, including both odd and even. Of course, the odd ones have answers so you can use them to check your methods. To determine the interval of convergence, i.e., to test the endpoints, you will need to use methods from earlier sections, as shown in the examples 1 through 5 in this section. For now, we aren't told where these power series come from. But in the two subsequent sections, we see that one can build power series to represent a given function f. This can come from knowledge of the derivatives of f at some particular point, or it can come from certain special forms for f, as we will cover in 8.7 and 8.6, respectively.
The problems just given in 8.5 are not (yet) for collection but be ready to try them at the board!
For Monday, April 5, here are some problems to review sections 8.1 through 8.4: p. 641:# 2-20, and 28 (of which 8, 20 and 28 are the harder ones). We will also go over some of the earlier homework problems. If you have time, take a look at section 8.5 on power series.
For Fri Apr. 2 for collection do 8.3: #24, 26(a only); 8.4: #2,4,8,14.
As I said in class on Friday, for Monday the 29th you should read the rest of section 8.3 and also section 8.4. Try the examples in the text and a few of the easier problems. For Tuesday, try 8.4: #2,4,6,8,12,14,16; and for 8.3: #22,23,24,26(a),38.
For Friday March 26: 8.1#10, #38; 8.2:#14,16; 8.3:#16,20.
For Friday (to be collected): pp. 558--559, #6,8,10,12,16,18.
For hw on Friday 2/27, as announced in class, find the standard deviation of the uniformly random distribution u_a,b with parameters a < b, (see problem #12 of 6.7 where the same problem is posed for the exponential distribution) and for 7.3, do problems #4,14,24, and 34.
For homwork due on Feb. 20, please do 6.4: #6,10, and 6.7: #6,12.
For Wed. Feb. 25, try 7.3: 2,4,6,8,14,16.
Here are answers for the problems you just did for homework in 6.7:
#6 (a) (i) P(0 < X < 200) = 1 - exp(-1/5)
(ii) P(X > 800) = exp (-4/5)
(b) The median m = ln(2)*mu, where mu is the mean.
So here m = ln(2)*1000 which is approximately 693.
#12 Hard but some of you did get it! The standard deviation
for the exponential distribution turns out to be the same
as the mean; that is, sigma = mu for this distribution.
(Try, e.g., integration by parts and l'Hospital's rule).
and for 6.4:
#6 (a) Writing for f_average,
for f(x) = ln x on the interval [1,3],
= (1/2)* integral of ln x from 1 to 3
= (1/2)*(x ln x - x) evaluated from 1 to 3
= (3/2)*ln 3 - 1
(b) Suppose c satisfies f(c) = . Then
ln c = (3/2)* ln 3 - 1 so ln (ce) = ln (3^(3/2)),
hence, c = (1/e)*(3^(3/2)).
#10 We want b such that 3 = (1/b)*int_{0 to b} f(x) dx,
so 3 = 2 + 3b - b^2 and solving the quadratic equation,
b is (1/2)*(3 plus or minus sqrt(5)). Both of these
roots are positive and so valid.
================================================================
Some additional practice problems:
From the chapter review exercises for Chapter 6, pp. 500--501:
#4 answer: pi*[(25/12) + (1/4e^4)], #6 (a), (b), and (c),
#12 answer is 64/15, #14 (a) and (b), #24, #29.
From the corresponding exercises for Chap. 5, pp. 439--441:
#8,12,14,16,24,26,28,30,32,34,42,54,58 (answers to some of them later)
answers: for #8(a): -1 + exp(-pi/4), for #14: 146/15, for #26: (1/3) x sin 3x + (1/9) cos 3x + C,
for #30: (1/2) arcsin(x^2) + C, for #58: the integral from 0 to 2/3 of 1/(2 - 3x) diverges; hence,
so does the integral from 0 to 1 for the same integrand.
For Monday the 23rd, we'll review in class. Then on Monday evening, we have our first midterm
For Monday Feb. 9 you tried: 6.2 #2,4,6 (ans = 2*pi*(-1 + tan 1) #8,10 (ans = 2*pi*[(1/3) + ln3] and read the section.
For collection on Fri. Feb. 13 , 6.2: #4,8,28,46; 6.3: #24. Hint on #24: When you calculate what should go under the square-root sign (formula 1 p. 468), you can greatly simplify the expression in the first quadrant. The graph of the parametric function has a four-fold symmetry. It looks like a baseball diamond centered at the origin, except that the "basepaths" are bowed in toward the center of the diamond so that the four corners of the diamond are now cusps. Hence, what you get for the first quadrant can be multiplied by 4 to get the length for the entire figure. Don't forget that "a" here is just a given constant (aka parameter).
For Friday Feb. 6, homework to be collected, 6.1:#6,8,10,12,16.
For Monday Feb. 2, we will cover section 5.9 on approximate integration.
The idea is to replace int_(from a to b) f(x)dx (a kind of infinite sum) with
_finite_ sums, where the finite sum is computed in a systematic way (called
an "algorithm") and where the error is explicitly bounded above. If the
definite integral is I and the finite sum is A_n (i.e., A can be M,T, or S),
where n is the number of subintervals, then for a suitable pos. integer s
K (b - a)^(s+1)
|I - A_n| = E_A <= ----------------
c_A n^s
where K >= f^(s)(x), x in [a,b] is an upper bound on the s-th derivative
of the integrand f on the interval [a,b] and c_A is a constant depending
on A. If A is M, for instance, c_A = 24 and s=2. When the s-th deriv. of
f is continuous on [a,b], it does have a maximum value.
Since for a given function f and interval [a,b] everything in the upper bound
is constant except the power of n, we can make the error as small as desired
by taking n sufficiently large.
Try the following for practice: int_(from 1 to 2) dx/(1+x^2) using n=4 for all three methods (note that this can be exactly computed but find the approximations and see if they have the claimed closeness to the exact value). Also, do examples 1 and 4 with n=4. Then do 5.9: #18 (but be sure to utilize example 7).
Homework due on Friday for collection on Jan. 30 :
(1) indefinite integral of (3x+4)/(x)(x-1)(x-2) (2) int_(from 0 to 2) dx/(x^2 + 2x + 5) (3) 5.7 #30 (see hint below) (4) Review of chapter 5 (p 440) #28 (5) " " " " ... #30
Homework was due Fri. Jan. 23 in class is the following: 5.10: #6,14,22,26,32. Also, please read section 5.7, focusing first on the technique called ``partial fractions''.
For Monday Jan. 26, try the following problems. I may call on some of you to put them up at the board. 5.10:#30, and the following:
int_(from 1 to 4) dx/(x+2)(x+5) (hint: following the general form we did in class, a would be -2 and b = -5, so A is 1/(a-b) = 1/3, and B = -1/3, etc.) int_(from 0 to 3) dx/(x^2 - 16) (ans. is (1/8)ln 7 - unless I made a misprint!) int_(from 1 to -1) dx/(9 + x^2) int_(from -3 to 3) (x^5 - 3x^3 + 19x)dx/(1 + x^8) The last two of these problems don't involve PF but they are examples of rational functions which we now can integrate. Here are a few more from 5.7: #18,20,24,26(no PF required for this one),30. For #24, use Example 5 as a model; for #18,20,30, follow Example 4. For #18, the correct answer is ln (27/32). For #20, you need three linear factors for the denominator. For #30, try the substitution u=sqrt(x+2). Also, look at the earlier part of 5.7 which involves trig substitutions. We'll discuss it in the next class after a bit more PF practice. You can try #2 (ans. 8/15), 4,10,14 (ans. 40/3).
On Mon. Jan. 12, I suggested that you review 5.5:26-32 and 42-52 even numbered problems especially. We worked 22,24,30 from 5.5 in class on Tues. and I did 29 and 37, also from 5.5, as well as #15 from 5.6. We noted that for 5.5 the answers to #38 is 26 and for #48 the ans. is 10/3 (with the sub. u=(1 + 2x). For 5.6, I went over how to find the integral of ln x and of xlnx using IBP, and we also did the integral of sin^2(x).
HW was due on Fri. 1/16 (for collection) 5.5 #50, 5.6 #20, 26, 28. If you can't be here, have someone else hand in your homework. I did give a quiz. But those who come by to see me before class on Wed. 21st can still take it. My office hours on Tues. are usually 1 to 2 pm but I'll come in at noon. Then I'll be around on Tues. from 6 to 7pm as well. Look for me in Reiss 256, 258 or 248, as indicated in the office hours section of my index page. In general, there will _not_ be make ups of quizzes so try to be there when I give one!
On Fri. 16 Jan., I reviewed improper integrals and we saw that the integral of e^(-t) from 0 to oo is 1.
For Tues. 1/20, please read the section on Improper Integrals (5.10) and try 6-24 of 5.10 (evens).
Rough syllabus for Calc II in Spring 04
We will cover Stewart's textbook (second edition) "Single-Variable Calculus: Concepts and Contexts" including sections on integration by parts, trig substitution, approximate integration, improper integrals, area, volume, and other applications, as well as separable differential equations, the case of exponentially increasing and decreasing functions, sequences, series, power series, and Taylor's formula. This corresponds to the material in Stewart from the middle of Chapter 5 through most of Chapter 8. In addition, I (and most other instructors) review the definition of the integral and its elementary properties. When time permits, some additional topics (not part of the final) are considered such as the Gamma function. As for grading methods, if you look at the web page given above, you'll see (in the page on "course mechanics") the grading scheme for 035 which is similar to what I will use in 036.
Ignore the stuff below; it will be re-edited shortly.
Here are some selected answers to the practice final. More may appear later. An extra "+" for correcting any misprints ... This stuff is from the Fall semester and will be moved shortly ...
2a) e^3
2b) done in the review
2c) use l'H
2d) 4/11
3a) 8x^7 - 4x^3 + 14x + (1/2)x^(-3/2)
3b) either quotient rule or product rule applied to x^(-2) sin(x)
3c) 2 cos (sqrt theta) * (-sin (sqrt theta)) * (1/2) (1/sqrt theta)
= =cos(t)sin(t)/t, where t = sqrt theta
3d) -A (omega^2) cos (omega * t)
3e) done in the review
4) slope is 2(2) = 4 so (y - 4)/(x - 2) = 4; i.e., y = 4x - 4
5) 2 + (.01)(1/12)
6) 3/2 ft/sec
7) h^2 = 4 and h=2 so s=sqrt(6). (see below "SPOILER" for how to do this.
Try it yourself first - there are different methods which work
but I suggest ID.)
8d) skipping the positive constant, the numerator of g''(x) can be factored
as (x - sqrt 3)*x*(x + sqrt 3) so numerator < 0 if x < - sqrt 3, num > 0
if - sqrt 3 < x < 0, num < 0 if 0 < x < sqrt 3, num > 0 if x > sqrt 3
(since x^2 + 1 is never zero)
8e) no vertical asymptotes since g(x) is defined for all x; h.a. is 1 since
lim as x goes to plus or minus infinity of g(x) = 1
9a) (1/18)x^18 - 3cos(x) + C
9b) (-2) exp(-x^2/4) + C
9c) 78
10) y(t) = (-5/2)t^2 + 60t + 200
11a) Riemann sum _approximation_ just means "Riemann sum"
(it's an approximation since the integral is the limit of
the Riemann sums as n goes to infinity so calculating one
of the sums with a small value of n gives an approximation)
In our notation, this problem asks for M_4(1/(1+x^2); -3/2, 5/2).
The midpoints are -1,0,1,2 and Delta x is ((5/2)-(-3/2))/4 = 1.
Hence, the ans is 1*(1/2 + 1 + 1/2 + 1/5) = 11/5.
SPOILER: sol'n to #7
Use implicit differentiation. First, describe the shed as having
side s and height h (both in meters). Then 12 = (s^2)h and one is
to minimize the cost C = 12h + 4s^2 (in dollars); the 12h is the
cost of the posts and 4s^s is the cost of the roof. Now differentiate
these two formulae with respect to s. Since 12 is a constant, the
first formula becomes 0 = 2sh + (s^2)h' (where h' is dh/ds). Hence,
h' = -2h/s. When s is chosen so as to minimize C, dC/ds will be zero.
So the second formula gives 0 = 12h' + 8s, yielding h' = -2s/3.
Solve the problem by equating these two formulae for h' so s^2 = 3h.
But now the volume is 12 so 12 = (3h)h.
h^2 = 4 and h=2 so s=sqrt(6).
Here's a problem: Suppose you know that f'(x) = -xf(x) and also f(0)=2. Find f(x).
Here are some additional problems:
Graph x ln(x) on the interval (0,1]. What is the maximum value and
where does it occur? Where is the minimum achieved?
** The max part is easy. Since ln(1)=0 and ln(x)<0 for x in (0,1),
clearly x ln(x) is negative except at x=1 where the function
is zero so the max value is zero and it is achieved at x=1.
Now if the function were just ln(x), then there would not be
a minimum value on (0,1] but x goes to zero as ln(x) --> oo
so x ln(x) needs to be evaluated using l'H rule as x --> 0.
On the other hand, (1/e)ln(1/e) = -(1/e) < 0 so the function
can be negative. Where is the minimum (and what is its value)?
Use the usual technique.
Show that (d/dx)(2x + 5)^(1/2) is (2x + 5)^(-1/2) using the limit definition
of derivative (not by using the power and chain rules but directly).
Give the Riemann sum M_3(x^2;1/2,13/2) for the function f(x)=x^2 on the
interval [1/2,7/2] using a partition into 3 subintervals and choosing
the midpoint of each subinterval.
Consider the parametric curve y = sin(2t), x = sin(t). For what values
of t is the tangent line to the curve horizontal? Hint: Recall that
if y = g(t) and x = h(t), then dy/dx = (dg/dt)/(dh/dt). What is the
slope of the curve at the origin (0,0) which clearly is on the graph?
** Hint: For instance, t=pi/4 is a solution. What are the other solutions?
Suppose f(x) is differentiable on the real line, f(0) = 2 and |f'(x)| <= 2
for all x in (0,5). Make a reasonable estimate for the value of f(5) (i.e.,
find an interval which must contain f(5). Justify your answer.
Hint: You need to use a theorem! (I hate to be mean ...)
Approximate the value of the definite integral of sin(x) between pi/12 and
5*pi/12 and justify your answer. Hint: Try a Riemann sum but pick one where
you can evaluate the terms of the sum explicitly - no half-angle formulas
are needed ;-)
** Try n = 2 and midpoints. Is there another possibility?
From time to time, I'll add a few answers here so you can check your work.
On the review problems due Wed. 3 Dec.: Ch. 4 review: #30 ans. is 1, #32 ans. is 1/e, #34 ans is 8/9, #38 Hint: you get a 4th degree polynomial p(x) = x^4 - 3x^3 - 64 (when you minimize the square of the distance as usual) for which you want to find the roots (in the interval) and one can see that x=4 is indeed a root (i.e., p(4)=0) so one can divide p(x) by x-4 and the result is a cubic polynomial with all positive coefficients which can't have any other roots for x>0 (why not?). For Ch. 5 review, #70 take the derivative of both sides wrt x. Show your work on all of the problems!
For Friday, Nov. 21, 5.3: #10,14,16,22; 5.4: #6,8,10,12,16,18,24,26(a).
In class, I mangled the argument for #24 of 5.4 since I was evaluating the integral of 2 - x from 0 to 2 instead of 1 to 2. In the latter case, there are extra terms which make the arithmetic work properly. So the function g(x) is defined in the interval of x from 1 to 2 by g(x) = (1/2) + 2x - (1/2)x^2 - 2 + (1/2) = 2x - (1/2)x^2 - 1. Now g(1) is 1/2 as it should be, and g(2) is 4 - 2 - 1 = 1, also correct. Try the same thing for f(x) which is 0 for x < 0, sin x when 0 <= x <= pi/2, 2 - (2/pi)x for pi/2 <= x <= pi, and 0 for x > pi. What is the value of g(4) if g(x) = integral from 0 to x of f(t) dt?
For Tues. Nov. 25, 5.4: #20; 5.5: #2,6,10,14,22,40,42,44,46. For practice: 5.5 #48 (ans. is 10/3 using substitution u = 2x + 1) and #52 (ans. is pi^2/72 "pi squared divided by 72") and the integrand is of the form f f'.
No class on Wed. Nov. 26! As review, the last homework set is due on Wed. Dec. 3: Ch. 4 review (pp. 337-339) #2,4,6,8,10,26,28,30,32,34,36,38,44. Review of Ch. 5 (pp. 439-441) #8,12,16,22,28,30,34,40,54,70. The latter set only covers the parts of Ch. 5 we have (or will) cover in class - which is sections 5.2, 5.3, 5.4 and 5.5. Though I'll talk briefly about integration by parts and improper integrals later in class, we won't do homework problems on IBP or improp integ. nor will they be on the final.
The last homework set (long but one extra day and the long holiday weekend available for doing it) will be graded (probably by the 9th) and I'll put it in the box outside Reiss 248 for you to pick up.
Note: We'll have a review session on the evening of the 16th of Dec. in Healy 105 from 8:30 to 10:30 pm. I will also give you some problems on the 2nd or 3rd of Dec. which we'll go over at the review.
For Tues. Nov. 18, 4.9: #4,10,16,20,26,40,44; 5.2: #40,42,46,48. Section 4.9 is about antiderivatives, which we have already seen. But it does more with them, as we will need shortly (shown in class today). There are also a couple of problems about gravity, which are special cases really of the situation where the second derivative of some function is known to be constant. Assuming the independent variable is time, then the second derivative of the position function is acceleration and ordinarily we view gravity as the force corresponding to a constant acceleration. If you look at problem 38 (not assigned), the quadratic function (in t) shown there will have constant second derivative a as required since the coefficient on the t^2 term is (a/2), and the first derivative will be (a/2)*t + v_0 so for t=0, we get v_0, etc. That is, if you can do #38 (the generic form) you can just plug in the numbers on the others. Gravity corresponds to the case that the constant acceleration is negative (in the usual choice of coordinates), with the constant being 32 ft/sec^2.
For 5.2, you should focus on (1) the concept of the definite integral as the limit of Riemann sums, and (2) on the basic properties of such definite integrals (there are 8 of them) on pp. 364-366. Don't worry about evaluating them yet. Note that just as with derivatives, there is an analytic definition (a limit) which can be geometrically described.
For Tues. Nov. 11, 4.1: #6,8,10 (done in class ;-), 12,14; 4.6: #10,14
By popular demand, I'm extending the due date to tomorrow, the 12th.
For problem 14 of 4.6, you will get a cubic polynomial P(x) equal to zero.
Solving such an equation can be tricky but here there is a simple way
to proceed: Find a value of the variable (say x = a) such that
P(a) = 0 using trial and error. In this case, if you try the small
values of x - say x = 0,1,-1,etc., you'll soon discover a value which
works. Now P(a) = 0 if and only if (x-a) divides P(x) so you can then
find the other two roots (if they are real). For instance,
P(x) = x^3 + x^2 - x - 1 = 0
has x = 1 as a solution so (x-1) divides P(x). In fact,
P(x) = (x-1)(x^2 + 2x + 1) = (x-1)(x+1)(x+1),
so x = -1 is also a root (in fact, it is a root of multiplicity two,
illustrating that the three roots may not all be distinct.
Once you have the roots of the cubic equation, you can finish the problem.
For Fri. Nov. 14, 4.1: #30; 4.6: #16,18,20,22,24,34. Remember that rate means deriv wrt time. Use x-dot (x with a dot above it) to mean dx/dt, and write t_0 for the moment of time when the conditions in the problem hold. Use (usually) either the Pythagorean Thm or similar triangles to analyze the relationship between variables and let the variables stand for simple geometric aspects of your drawings. Try to make the choices so that the algebra is easier and check to make sure that plus and minus signs seem correct!
For number 22 of 4.6, I suggest you use trigonometric functions of the acute angle theta between the ladder and the ground (the ground is assumed to be horizontal!) also using similar triangles. The two given data (height of fence and distance from the building) allow you to calculate the length of the two portions of the ladder (inside and outside the fence) in terms of the angle. Thus, you end up by expressing the length L in terms of trig fcns of the angle. At a minimum or maximum, dL/dtheta should be zero and there is no maximum (why?). Hence, showing a unique theta where dL/dtheta is zero, it must be where the minimum length occurs. Now you still have to calculate the length corresponding to this angle, and this is a matter for trig.
I showed in class that to solve x = tanh(y) for y in terms of x, one can use algebra to see that
x = 1 - 2/(1 + e^(2y))
which you can easily solve for y in terms of x. Once you have the explicit formula for y = arctanh(x), use some trig identities to show that arctanh(sin x) = ln |sec x + tan x|. This is not part of the homework assignment but is a nice chance to explore algebra and trig.
For Friday, Nov. 7, I gave the following assignment in class today: 4.5 #36 and the following challenging but, I hope, interesting assignment designed to help you discover _why_ the antiderivative of sec(x) is as given in our book.
The hyperbolic functions are introduced after section 3.7.
sinh(x) = (1/2)(e^x - e^-x)
cosh(x) = (1/2)(e^x + e^-x)
sinh(x) e^x - e^-x
tanh(x) = ---------- = ------------
cosh(x) e^x + e^-x
1
sech(x) = ----------
cosh(x)
A useful identity is cosh^2(x) - sinh^2(x) = 1; this follows from
the algebraic identity (a + b)^2 - (a - b)^2 = 4ab by taking
a = e^x, b = e^-x, so 4ab = 4.
Part (1): Find the derivatives of sinh(x), cosh(x), and tanh(x).
In finding the deriv. of tanh(x), use the quotient rule since you
already will know the derivs of sinh and cosh, and use the identity.
The argument is just like finding the deriv of tan(x).
Part (2): Find the derivative of the inverse function arctanh(x)
to tanh(x). Note that if h and g are inverse functions so that
(*) g(h(x)) = x,
then h'(x) = 1/g'(h(x)) by the chain rule applied to (*).
In fact, (d/dx) arctanh(x) = 1/(1 - x^2). The argument is exactly
analogous to what we did to show that (d/dx) arctan(x) = 1/(1 + x^2).
Part (3): (Easy!) Show that arctanh(sin(x)) is the antiderivative
of sec(x) - that is, (d/dx)(arctanh(sin(x)) = sec(x). This is just
the chain rule applied to Part (2). Note that two of these are
trignometric functions and only one is hyperbolic - i.e., it is
not a misprint that I wrote "sin" here.
Part (4): (Only for extra credit) Show that the answer obtained
in Part (3) agrees with the answer given in the back of the book:
(**) arctanh(sin(x)) = ln |sec(x) + tan(x)|.
You can first identify the inverse function arctanh(x) explicitly,
using the method given in our text - i.e., to find the inverse fcn
h to g, let x = g(y) and then solve for y in terms of x so that y = h(x).
Doing this when g is the tanh function requires a couple of algebraic
manipulations. Then consider the explicit function arctanh(sin(x))
and rewrite the expression using trig.
An alternative way to show (**) is to note that both sides of
the equation have the same derivative, and hence they must differ
by a constant. But when x=0, both left and right sides are 0 so
the constant must be zero! This is faster but less satisfying ;-)
For Tues. Nov. 4, please do 4.3: #12,16,18,22,24,30; 4.5: #6,8,10,12,14. Thanks to Alejandro for reminding me to post them since I'd put them on the board. Don't forget also that other exercise in algebra which I won't recopy now.
For Friday Oct. 31, please read section 4.2 and do #10,12,16,20,24,28,30,34,38,40; also read 4.3 and do #8,10. I'll post problems for next Tues. shortly.
Due on Tuesday Oct. 28 Ch. 3 Review (pp. 258--259): #2,6,10,16,18,20,28,32,34,42. For problem #2 only, also find y'(pi/4).
Here's a tough one - can you find a function f with f' = sec? What is the antiderivative of the secant? Hint: You can look at the hyperbolic functions defined at the end of the chapter. Or you can look up an antiderivative in the "backpiece" of the text. Of course, you should _check_ that it is the antiderivative ...
For Tuesday Oct. 21, 3.6: #38,42,46,48,52; 3.7: #16,22,32,42; 3.8: #4,12,16; also read 3.8 and go back also to read 2.9. I'm aware that you guys are beginning to get a bit overloaded with new material so we'll pause for a while and make sure everyone is up to speed. In particular, I'll help you with some of the homework problems on Monday. We may also have a quiz either Mon. or Tues.
For Fri. 3.5: #32,38,46,48,50,56,62,68; 3.6: #2,16,24,44,50; 3.7: #2,8.
On Friday, I sketched part of the following in one section but
not the other. Here is a repeat: Consider two families of equations:
x^2 + y^2 = r^2
y = mx,
where r and m are ``parameters'' (that is, they are constants but
we must consider the equation for all possible values of r and m).
Since the first family only involves r^2, we can and do assume r >= 0
(" >= " means "greater than or equal to"). The first equation
corresponds to a circle centered at the origin with radius r,
while the second corresponds to a straight line through the
origin with slope m. It is obvious geometrically that these are
orthogonal families, but it would be nice to see how this works
analytically. Using implicit differentiation, the first eq'n
gives 2x + 2yy' = 0 so y' = -x/y, while the second clearly gives
y' = m. It is left to you to _check_ that where the two curves
intersect, these two expressions for y' represent the slopes of
orthogonal (i.e., perpendicular) lines.
For Tues. Oct. 14 (remember that the 13th is a holiday ;-): 3.5: #2,4,6,14,16,26. Also, mimic what I did in class on Friday to find the derivative of ln(x). (In case you missed it: take the function x = e^(ln x) and differentiate it. On the one hand, the derivative of x (with respect to x) is 1 but on the other hand, using the chain rule, (d/dx) e^(ln x) is ... so (d/dx) = ... Your job on this last exercise is to fill in the dots ... If you forgot the last exercise for Friday (i.e., the one involving the power rule for x to a negative power), you can hand it in, too, on Tuesday.
For Fri. Oct. 10 (a light assignment ;-): read section 3.4 (skip 3.3 for now). Don't worry about the "proofs" there; just read what the derivative of the trig functions are and use the information to do 3.4:#2,4,16,28, plus the following (which goes with the power rule):
Assume the result of exercise 3.2:#42 and the power rule for x^n when n is a positive integer to show that the power rule also holds when n is a negative integer.
For Fri. Oct. 3, section 3.1: #6,10,22,44,50,52,54; for Tues. Oct. 7, section 3.2: #4,14,18,30,34,36,42.
Answers to 3.2 assignment - as it was not graded:
#4 [(1/2)x^(-1/2)e^x](2x+1) (a product with 4 factors) 14 (in class) 18 (ad-bc)/(cx+d)^2 (lots of stuff cancels ;-) 30 -2.5 34 (in class) 36 f''(x) < 0 when x in (-2 -sqrt(2), -2 + sqrt(2)) 42 (in class) Some other probs to try: 3.2 #6,8,28. Here are the answers to 28: (a) -1, (b) 8, (c) -8, (d) 8 Also try 3.2:#32(a) and (b), and of course any odd numbered problems that seem interesting. I suggest you study by looking at the old homework - especially the ones which were straightforward but not trivial. Speaking of old homework:
The old homework and quizzes (including a couple I couldn't find before) are now outside Reiss 248 in a box - so pick yours up at your convenience.
Some of you have asked about additional practice - here's one I mentioned in class. Try to find the derivative of the cube root of x. OF course, using the power rule, it should be (1/3)x^(-2/3), but the exercise is to show this using the limit definition of derivative - either of the two equivalent forms will do. It takes a bit more algebra than the problem I did in class - the deriv of the square root of x. There are lots of problems in the chapter reviews and you can use those also as opportunities to practice. Check this space later during the weekend - I may add some other problems.
Note that sometimes the book doesn't indicate which variable is supposed to be differentiated and which ones are constants. Usually, k and c are used to denote constants.
old homework below ...
Homework for Tues. 23 Sept. section 2.4: #6,12,14,20,26,30; section 2.5: #6,8,14,16,18,20; section 2.6: #10,12. We'll go over some similar odd-numbered problems on Monday so bring your favorites ;-)
Here are some future assignments. I might slightly modify them as their due dates approach, depending on how much we cover in class.
For next Friday (9/26), here is the assignment: Read section 2.7 and do #4,6,8,18,30, and read section 2.8 and do #4,6,8,36,38.
Also for Friday the 26th though not for collection (but do try it!): Show that D(cf)=cD(f), where D means the derivative operator and c is a constant. In other words, show that if a function f has a derivative f', then cf has the derivative cf'. I did a similar problem in class by showing that (f+g)' = f' + g', using the fact that the limit of a sum is the sum of the limits. You'll need a corresponding property of limits for the exercise.
For Tues. 9/30, read section 2.10 (skip 2.9 since we'll cover this idea later) and do 2.10: #4,6,8,10,12,16,22,26, and chap. review (concept check), p. 181, #4,8,12,14. I've avoided giving you certain problems that involve a lot of rather pointless computation since our soon-to-be-developed tools for derivatives will do them much more efficiently. If you want to look even further ahead, try to prove the following power rule:
For n a positive integer, the deriv. of x^n is n*x^(n-1). (Note that "*" means "multiply" and "^" means "raise to the power of".) The proof involves a new technique, called Induction, which is described in the text (and which I'll review in class briefly), and the proof also uses a formula for the derivative of the product of any two differentiable functions (fg)' = fg' + f'g, which is the so-called product rule.
Homework for Tues. 16 Sept. In addition to exercise 1 below, pp. 108-110 (section 2.2): #4,6,10; pp. 117-119 (section 2.3): #2,4,6,8,12,18,38.
Given in class: Suppose that lim (as x --> a) f(x) = L and let c be any nonzero constant. Show that lim (as x --> a) cf(x) = cL. Do this by showing that the delta which makes cf(x) within epsilon of cL can be expressed in terms of the delta which makes f(x) within epsilon of L.
There is an assignment for Friday 12 Sept. which I gave in class. Ordinarily I also put it on the web as well but this week there was not sufficient time to do it. I did explicitly say in class that these problems were due on Friday - here they are:
Appendix B: (pp A16-17) #18,22,32,34,42 Appendix C: (pp A27-29) #12,30,36,42,44a
For Monday, review Chap 1 as needed - especially to do the homework below. Have a nice weekend!
For Tues. Sept. 9 (to be collected) do Exercise 1 given in class (see below for a repeat) and the following from the book: section 1.3 (pp 48-49): #32,38,52,56,58; section 1.5 (pp 63-64): #16,18,24a,b; section 1.6 (pp 73-74): #12,14,18.
Ex. 1 for Tues. Show that |a + b| is at most |a| + |b|. Hint. Consider two cases: (i) a and b have different algebraic signs (i.e., one is positive and the other is negative), and (ii) the other possibility that a and b do _not_ have different signs (i.e., one is positive and the other is nonnegative, or one is negative and the other is nonpositive). Remember that |x| is just the distance from x to the origin 0.
For Wed. Sept. 3, look at the graph of sin(x) and find an interval where sin(x) is increasing. This is useful since functions which are increasing on some domain are 1-1 there - while sin(x) is not 1-1, some of its restrictions are. Also, figure out the domain of tangent of x - for what values of x does tangent make sense?
Due on Friday, pp. 85-87: #6,8,16,18,22; p. A7 (appendix): #18,32. Bring some questions to class on this stuff or anything else in Chap. 1 so we can review it. Once we've finished reviewing Chapter 1, we'll be following along in the textbook by various sections. But in Chapter 1 we are skipping around to pick up the points that need review.
For Tues. Sept. 2, there are two problems: (1) Show that for any two distinct rational numbers, there is another rational number which is strictly between them. (2) Using the result from (1), show that in fact there must be infinitely many rational numbers between any two distinct rationals. Today (9/2) I added hints - for (1), try the midpoint, and for (2), use an argument by contradiction.
For Friday Aug. 29, please read through Chapter 1 (review) and bring some questions to class. Regarding homework, if for any reason you can't be in class, ask one of your classmates to hand in the homework for you - or put it in my mailbox in the math dept office in Reiss 256. Show your work and staple the pages together; you can include a second better draft but let me see the rough try where you work stuff out - or at least enough of it to get the idea. Your grades on homework include both effort and accuracy.
There were two homework problems assigned for 8/29. I asked you to (1) find a formula for the difference of two successive cubes (check it for a concrete case, too).
Solution: (n+1)^3 - n^3 = 3*n^2 + 3*n + 1. So, e.g., for n=3, the formula gives 27 + 9 + 1 which is 37, and this is correct since 64 - 27 = 37.
Let N be the set of natural numbers - i.e., of all nonnegative integers. (2) Show that the function s from N to N defined by s(n) = n + 1 is a 1-1 function; that is, s(n) is not equal to s(m) whenever n and m are not equal.
Using the contrapositive form, if s(n) = s(m), then n+1 = m+1 so (adding -1 to both sides of the equality) n=m.
To see the old homework assignments and other material from this page, you can go to the Old Calculus Page (not yet containing anything ...).