MATH-035-09 - Fall 2006 CALCULUS I - 4.0 credits MTWF 01:15PM - 02:05PM REISS 112
See my index page for info on office hours Dec. 11 and over the break. Info on practice exam solutions later this weekend.
You can reach me by phone (or voice-mail) at 7-2703 or by e-mail kainen at georgetown.edu or drop by St. Mary's D315 during my office hours. See my index page.
**** Last updated ****
Dec. 8, 2006
**********************
Warning!! These are pdf files. Here are the url's for those practice exams:
I suggest that you try each exam, limiting yourself to 2 1/2 hours. The exam next Tuesday is not guaranteed to resemble these but there will likely be a good deal of overlap. I will post the annotated copies of the exam solutions sometime on Saturday late afternoon or Sunday by early afternoon to give you a chance to see how you did.
In fact, if a few of you send me e-mail asking me to post the answers to the 2005 exam first - maybe tomorrow in the early afternoon, I can do that. But be sure to try them all and wait to look up the answers untill you have made a good effort to do the problems - even if you need more than the allotted time.
Meanwhile, here's a neat little problem: Find the derivative of the log[x + sqrt(x^2 - a^2)]. Compare it to the derivative of arcsin(x).
Text is: Single Variable Calculus: Concepts and Contexts, 3rd Ed. by J. Stewart
Information is available on syllabus and course mechanics . I suggest you review this material which also explains homework, quiz policies, and the exam schedule.
Math assistance center, St. Mary's 3rd floor, Sun. - Thurs. evenings - open 6 to 10 pm
Review session on Sunday Dec. 3 at 5pm in Reiss 284.
Here are some practice problems for the review. Remember that you should always check your answers to indefinite integrals by taking the derivative to see if it gives you back the integrand. Don't forget that indefinite integrals need a "+ C" at the end.
1. If integral (from 2 to 5) f(x) dx = 7,
integral (from 3 to 5) f(x) dx = 9,
integral (from 2 to 3) g(x) dx = 80,
and h(x) = f(x) + (1/2)g(x),
what is the value of
integral (from 2 to 3) h(x) dx?
2. What is the value of
integral (from -7 to 7) [x^3/(x^2)cos(x)] dx?
3. If y''(t) = -4m/sec^2, y(0) = 10, y'(0) = 2, find t_0
such that y(t_0) = 0 and y(t) > 0 for t < t_0.
What is the formula for y'(t_0)?
If everything were the same except that y'(0) = -2,
should the corresponding time t_0 be larger or smaller?
4. Find the Riemann sum, using right-hand endpoints and
four subintervals (n=4)
R (x^2; [0,2])
4
Is this an over-estimate or an under-estimate of the
corresponding definite integral? Identify the integral,
make a drawing of its integrand, of the area under the
curve, and of the Riemann sum (as a set of rectangles).
5. Find the exact value of the integral in #4.
6. What is the indefinite integral
/ 2 dx
\ --------------
/ 3 + 5(x^2)
7. What is the area under the curve (2x)*e^(x^2) between
the vertical lines x=1 and x=2 and above the x-axis?
8. Find integral (from 0 to 2) [(x^2)/(1 + x^3)] dx
9. Make a substitution to find the value of
/ ds
\ -------------------------------
/ (s^2 + 2s + 2)[arctan(1 + s)]
10. Use IBP to find the antiderivative of x*sec^2(x).
You can also practice with the following:
Try-It Problems
5.5: #2,6,8,10,12,18,20,30,34,42; 5.6: #2,4,6,18,20
Here is the
Homework for Friday Dec. 1
5.5: #4,14,24,44; 5.6: #16,26
Note that for integration-by-parts (IBP) we are
only covering the indefinite case.
Here is how to do 5.5: #2
find I = integral x (4 + x^2)^10 dx. Let u = 4 + x^2.
Then du/dx = 2x so du = 2xdx and dx = du/2x. So
I = integral x * u^(10) du/2x = (1/2) integral u^(10) du
A simpler way to make the same conclusion is to note that
x dx = (1/2) du so I = (1/2) integral u^(10) du.
Once you have integral u^(10) du, it is clear this gives
(1/11) u^(11) so I = (1/22) (1 + x^2)^(11) + C.
Oh, since 5.6 involves "integration-by-parts", here
is what that means:
The product rule for derivatives (Leibniz rule) says that
(d/dx)(uv) = u'v + uv',
so d(uv) = v(du/dx)dx + u (dv/dx)dx = vdu + udv. Note that
du = u'dx and dv = v'dx.
Now integral dt = integral 1 dt = t. Hence,
uv = integral d(uv) = integral v du + integral u dv.
Hence, if I is an integral f(x) dx, and if f(x) can be
expressed in the form f(x) = u(x)v'(x) for some pair of
differentiable functions u = u(x) and v = v(x), then
I = uv - integral g(x)dx *** the IBP formula ***
where g(x) = v(x)u'(x). Sometimes this procedure converts
an otherwise intractable integral into something simple.
For example, consider I = integral ln(x)dx. Since ln(x) is
easy to differentiate, let's try u(x) = ln(x). Then for
v(x) = x, we have ln(x) = u(x)v'(x). By the IBP formula,
I = x*ln(x) - integral x(1/x)dx = x*ln(x) - x + C.
Check: (d/dx)(x*ln(x) - x) = x*(1/x) + ln(x) - 1 = ln(x).
For Monday, try 5.4: #26, 19, and 22 (in that order) For Tuesday, as homework to be collected 5.4: #8,12,14,18,20. I did #10 and 16 in class today. Using the Fundamental Thm of Calculus for both of them, #10 ans. is -tan(x); while for #16, also using Chain Rule, ans. is (-sin x)[(1 + cos^2(x))^10] - (cos x)[(1 + sin^2(x))^10] Note that for #10, since we can find an antiderivative of tan(x), it is possible to do the problem by actually finding the value of the integral with a variable endpoint and then taking the derivative of it. Since (d/dx) ln cos x = (-sin x)/cos x = -tan x, we have g(x) = int_x ^ 10 tan theta d theta (integral from x to 10 of tan(theta) dtheta) is equal to - ln cos 10 + ln cos x so g'(x) = -tan x. This second method is (1) much longer and (2) only available when we can find an explicit antiderivative. The fundamental theorem of calculus is much more powerful and much easier to use! Quiz on Tuesday on sections 4.9, 5.2, 5.3, and 5.4.
For week of Nov. 13 -- 17, try (4.9): #4,18,22, (5.1): #4,18, (5.2): #32,34,42, and (5.3): 4,8,16.
Here are the homework problems which were due on Friday 11/17: (4.9):#28, (5.1):#20, (5.2):#48,50, (5.3):#20,22.
The 2nd midterm was held on Wed. Nov. 8 at 9 pm in ICC 104. It covered 3.4 -- 3.7, 2.9, 3.8, 4.1--4.3, 4.5,4.6.
Practice problems ...
1. Find (d/dx)arccos x by using implicit differentiation.
2. If f(x) = 30x + 5 and g(x) = f(x) * e^(2x)
(a) find g'(x) and (b) determine g'(0)
3. For h(x) = (sin x)^3 * tan x,
find h''(x).
sin([pi/4]*x)
4. Let f(x) = x
find f'(1).
5. The sides of a square are each increasing at the rate
of 4 m/min at the instant of time when you measure the
side obtaining 20 m. What is the rate of increase of
area of the square at that same instant of time?
6. Show that the function g(x) given by the formula
g(x) = x^3 + 6x^2 - 42x - 1
must have a maximum and a minimum value on [0,1].
Find the max and min and show why they are as claimed.
7. Draw the graph of x^(1/2) * ln x justifying everything.
8. For an isoceles triangle with base 10 and height h(t),
let theta(t) be the angle at the apex (top) of the
isoceles triangle at time t. If at t = t_0, we have
h(t_0) = 5 and h-dot(t_0) = 3, find theta-dot(t_0).
Note that (1/2)theta = arctan(5/h).
9. Find lim(as x --> 0+) x ^ (sin 3x).
10. Suppose you choose a point P on the first quadrant of
the unit circle with angle theta between the radius
from the origin to P and the x-axis. We assume that
this determines a rectangular field w/ one corner at (0,0)
and the other at P. Find the angle which maximizes
the area of the field.
For the above problems, the answer for #8 is -6/10 rad/per unit time at time t_0. (Don't forget that you need to double the answer since the angle at the top is twice the arctangent. For #10, theta is pi/4.
Problems for Review 1 is available for first third of course. Some answers are now available online.
The 3rd midterm will be Monday Dec. 4 (instead of Wed. 11/30) - at 9pm. It will cover 4.9, 5.2 -- 5.6. I'll give a review session on Sunday Dec. 10 sometime in the evening.
For homework due 11/3/06: 4.3 10,22,30; 4.5: 8,24,28,38
For Monday, try 4.6: #2,6,12,18; for Tues.: 4.6: #4,10,16,22.
For #10, which I did in class yesterday, one got
x^2 * h = V and 4xh + x^2 = S for volume and surface area.
Since V is fixed and we are looking for a global (and hence
local min of S(x)), we can take both equations, differentiate
with respect to x, and set them equal to zero to find where
the minimum occurs. (This uses the fact that S(x) tends to
infinity as x tends to 0 and also as x tends to infinity, but
surface area is finite for any finite positve x so there must
be a global minimum which must occur at some positve finite x.)
Carrying out the differentiation, you get the equations
2xh + x^2 * h' = 0 and 4h + 4xh' + 2x = 0.
From the first one, h' = -2xh/(x^2) = -2h/x and
from the second one, h' = -(4h + 2x)/4x. Hence,
8h = 4h + 2x so h = x/2.
Now since we know that V = 32,000 and V = x^2 * h, we get
64,000 = x^3 so x = 40 cm.
4.6 #6 ans. Perim = 40*sqrt(10) meters
#12 The x-value (length of side of box), which minimizes
the value of C(x) = 24 xz + 10 x^2 of the box with square base,
no top, and volume equal to 10, is x = (12)^(1/3). This problem
can be done either way. So cost C is $30*(12)^(2/3).
#18 Ans. is 2ab.
I won't collect the latter set but I will ask some of you to
put them up on the board.
For next week (10/23 through 10/27), read 4.1 and 4.2. Homework for 10/27: 4.1: #12,18,26; 4.2: #36,38,44.
Here are some hints for homework problems from 4.1.
For #12 of 4.1:
Let x(t) denote distance (in ft) at time t from
where the man is standing to the lightpole, and
let y(t) denote the length of his shadow at time t,
measured from where he is standing. The problem
asks you to find the rate at which the tip of his
shadow is moving away from the lightpole - i.e.,
to find (d/dt)(x + y) which is the same as finding
x-dot + y-dot - when the man has distance 40 feet
from the lightpole. But in fact, the answer only
depends on the assumption that the man is moving
away from the pole at 5 feet per sec. at some time
t_0.
Since the lightpole is 15 feet high and the man is
supposed to be 6 feet tall, by similar triangles,
we have the equation
x + y y
------- = ---
15 6
so cross-multiplying
15 y = 6x + 6y
and solving for y in terms of x, y = (2/3)*x.
Now use the fact that x-dot(t_0) = 5 feet per sec.
to calculate y-dot(t_0) (the rate at which his
shadow moves away from his feet) and add x-dot(t_0)
to get the answer to the problem.
For #18 of 4.1:
Let x(t) be distance (in meters) of boat from dock,
measured horizontally, the height of the dock is
fixed at 1 m, and let z(t) be the length of the rope
between the bow of the boat and the dock, all at
some time t.
Then by the Pythagorean formula x^2 + 1 = z^2 and,
taking the derivative with respect to time of both
sides, we get the equation
2x*x-dot = 2z*z-dot, so x-dot = z*z-dot/x
Let t_0 be the time at which the boat has distance
8 m from the dock; i.e., x(t_0) = 8. Using the
Pythag. formula above, find z(t_0). You already
know that z-dot(t_0) = -1 m/sec, so
x-dot(t_0) = z(t_0)*z-dot(t_o)/x(t_0);
plug in the values and that's it. Since the problem
asks for the rate at which the boat is approaching
the dock, your answer should be given as a positive
value (i.e., it's like speed, vs. velocity).
For #26 of 4.1:
Let x(t) be the horizontal distance (in feet),
the height is constant (100 feet), and let z(t)
be the length of the string at time t. Let theta(t)
be the angle at time t between the string and the
horizontal. There are several formulas one can use
to relate these variables. For instance,
cos(theta) = x/z
could be used but when you take derivatives wrt t,
you would need to use the quotient rule, etc.
I suggest using instead
tan(theta) = 100/x
Take deriv wrt time of both sides (i.e., apply d/dt),
and evaluate what you get at time t_0, where z(t_0)
is 200 feet. Using our old friend, Pythagoras, we
find x(t_0) = sqrt(40,000 - 10,000) = 100*sqrt(3).
Hence, tan(theta(t_0) = 100/100*sqrt(3) = 1/sqrt(3).
To complete the problem, you'll need to use the fact
that sec^2(theta) = 1 + tan^2(theta). Of course,
you also need to use the fact that x-dot(t_0) = 8
feet per second. This will give you a formula for
theta-dot(t_0) which is what the problem asks for.
Prob. 22 of 4.1:
As in the similar example, let r(t), h(t), V(t), resp.,
be the height, radius of the top surface, and volume
of the liquid in the tank at time t. The formula for
volume in terms of radius and height (for cones) is
V = (1/3) pi (r^2) h
as in the example (or from the front/back pages of the
book which have many useful formulas from trig, geom.,
algebra, and for derivatives and integrals).
But by similar triangles (as in the example problem),
the radius of the tank is to the height of the tank
as the radius of the liquid is to the height of the
liquid; that is,
2 r
--- = ---
6 h
so r = h/3. Plugging this into the formula for V,
V = (1/3) pi [(h/3)^2] h
so V = (pi/27)*h^3.
Hence, since everything is a function of t, we can
differentiate with respect to t, getting
V-dot = (pi/9)*(h^2)*h-dot (writing f-dot for df/dt)
Let t_0 be such that h(t_0) = 2m = 200cm (careful here!).
By assumption, h-dot(t_0) = 20 cm/min. Hence,
V-dot(t_0) = (pi/9)(4 x 10^4 cm^2)(20 cm/min)
= (80pi/9)*10^4 cm^3/min (correct units)
But by assumption V-dot(t_0) = (c - 10^4) cm^3/min, where
c is the constant rate at which the liquid enters the tank.
Hence, c - 10^4 = (80pi/9) * 10^4 (leaving off the units now),
so c = (1 + (80pi/9)) * 10^4
(no need to go further but this is around 300,000 cm^3/min,
i.e., about .3 cubic meters per minute).
For 4.2, #48 f(x) = x - ln x on [1/2,2] has value
f(1/2) = 1/2 - ln (1/2) = 1/2 + ln 2 ~ 1.2
f(2) = 2 - ln 2 ~ 1.3
using the fact that ln 2 is approximately equal to .7, or a calculator. The critical numbers of f(x) are where the derivative is zero (since the function is differentiable everywhere). So df/dx = 1 - 1/x so x=1 is the only critical number. By the closed interval method, we need only check f(1) which is 1 - ln(1) = 1.
Hence, the Max of f(x) on the interval [1/2,2] occurs at the right-hand endpoint, The Min of f(x) occurs at x=1, which is an interior point. Moreover, you can see that f'(x) is negative for x in (1/2,1) and f'(x) is positive for x in (1,2), and this is another way to see directly that f(1) is the absolute minimum of the function on the interval.
For section 4.1: #2,4,6,10 (for Monday)
#14,22,34 for Tues/Wed
For section 4.2: #4,6,8,12,22,26,30 (For Tues)
#34,38 (for Wed)
The exercises for 4.2 are mostly shorter than for 4.1.
I'll select or add homework problems later in the week.
For 10/17 through 10/20, try the following: 2.9 #6,8,10,16,26; 3.8 #6,8,16,18,24. Homework for 10/20: 2.9 #10, 26; 3.8 #8, 20, 24b.
For homework for Friday 10/13 - for collection: 3.6 #18, 30, 40; 3.7: #12, 30.
Here are some additional try-it problems for 3.4 and 3.5, as well as those which will be due for homework on Friday. On Monday, we will also go over the exam.
3.4: #8, 24; 3.5: #2, 4, 10, 12, 26, 50, 54. Earlier, by e-mail, I suggested 3.4: 4,10,14. The reading for Monday is 3.4 and 3.5. And for homework due 10/6, 3.4: #16 (use the definition of derivative as a limit and mimic the argument we used to prove that the deriv of sin is cos), 18, 26; 3.5: #14, 20
For practice (i.e., try-it problems for Mon. Tues., and Wed. of next week) - 3.1: #4,8,22,38,42(a,b),46; 3.2: #6, 14,18,20,30,42a.
For all written assignments for collection: Write your answers clearly, show all work, and staple the sheets together. In addition to these, you ought to try a reasonable sampling of _all_ the problems, checking your answers to the odd-numbered ones using the key at the end of the textbook.
I suggest that as you work the problems for Chapter 2, if there are issues arising from more basic topics - e.g., graphing functions, meaning of trig functions, definition of functional composition, and so on - then try a few of the problems from appropriate sections of Chap. 1. The odd number problems have answers so you can check your understanding. Then if you still have questions, raise them in class!
For homework for collection on Friday 9/22. 2.7: #8,18,20; 2.8: #4,8,14,24. Additional problems and a list of topics for the midterm coming soon. Saturday problem review session 9/23 Reiss 284 5 pm till 6:30 pm. This will give you the chance to go over problems at the board. The room for all three of the midterms is ICC 104.
Here are some try-it problems for this week:
2.7: #4,6,16,22 (for Tuesday);
2.8: #2,6,10.16,24 (for Wed.)
Here's an extra: 2.8: #36.
Some of these problems require some algebra.
I'll remind you on Monday of how to do long-division
of polynomials. Once you know that, you will be able
to see that for #16 of 2.7,
x^2 + 1 5
-------- = x + 2 + -------
x - 2 x - 2
and the latter form is a lot easier to differentiate.
For Monday 9/11, please read sections 2.4. If you have time, look ahead also to section 2.5. Try the following problems from 2.4: #4,6,10,14,18,28,38. For Tues. 9/12, finish reading 2.5 and try 2.5: #4(a,c,e),6.
In class today (Tues.), we noted that
lim (x --> +oo) arctan x = + pi/2
lim (x --> -oo) arctan x = - pi/2
and I asked you to find two real numbers r and s so that
for g(x) = r + s*arctan(x)
we have
lim (x --> +oo) g(x) = +7
lim (x --> -oo) g(x) = +5
Here is the solution - I'll space it down a bit so that you
can first try this one on your own if you didn't already.
Or retry it later for review.
-- answer below --
-- answer below --
-- answer below, really! ----
Here it is:
For any real numbers r and s,
lim_(x --> oo) r + s*arctan(x) =
r + s*lim_(x --> oo) arctan(x) = r + s*(pi/2)
Similarly,
lim_(x --> - oo) r + s*arctan(x) =
r + s*lim_(x --> - oo) arctan(x) = r + s*(-pi/2)
= r - s*(pi/2)
To choose r and s so that the limits at + oo and - oo
are 75 and 5, resp., solve the equations
7 = r + s*(pi/2)
5 = r - s*(pi/2)
Adding them gives 12 = 2r so r = 6. Using either
the first or second equation and plugging in r = 6,
1 = s*(pi/2)
so s = 2/pi.
For Wed. 9/13, try 2.5: #16,22,30,40,46.
For homework for collection on 9/15 2.4: #16,22,26,30,40; 2.5: #4(b,d,f), 8,24,28,32.
For Tues. 9/5, try problems 2.2: #3 - 6 and 9 - 12 on pp. 106-107. Then read 2.3 and try 2.3: #2 - 22, 28 - 34.
Collected on Friday (9/8)
2.2: #12; 2.3: #4, 14 hint: (x-4) is a common factor of both numerator and denom, 34,36 (we did these in class on Friday).
For #4, there are many possible limit laws - but the theorem that states that limits of polynomials can be evaluated by "plugging in" is sufficient. For the later problems, don't forget that
|x| = x if x is non-negative
|x| = -x if x is negative
Homework is how you learn to _do_ calculus and so it is an essential part of the course. It is _not sufficient_ to just write down the answer. You must show your work. You will get full credit for homework which is clear and complete even if it is somewhat inaccurate. When you get homework back, you'll also get the correct answers and a hint about possible methods where needed. This gives you a chance to see if you can now get the right answer. After that, as necessary, we'll work them at the board. Writing lengthy answers won't improve the score. The goal is to be brief, clear, complete, and correct.