Some problems:
- Page 223: #1,2,3,5,7,9,25,27,30,31,33,35,38,39,40
- Page 238: #5,6,7,8,12,13
- Page 262: # 10,21,22,23.25.26
Some answers to the quiz (November 5) 1. A spherical ball 8" in diameter is coated woth a layer of ice of uniform thinkness. If the ice melts at a rate of 10 cubic inches per minute, how fast is the thinkness of the the ice decreasing with it is 2" think?
Let x be the thinkness of the ice. The radius of the iron ball is 4, so
that the total volume is:
V = (4/3) pi (x + 4)^3
Differentiating gives:
dV/dt = 4 pi (x + 4)^2 dx/dt
Plugging in numbers:
-10 = 4 pi (2 + 4) ^2 dx/dt
or
-10/(4 pi 36) = dx/dt
3. A balloon is 200 ft off the ground and raising vertically at a constant
rate of 15 ft/sec.. An automobile passes beneath it traveling at 66 ft/sec.
How fast is the distance between them changing 1 second later?
Let r be the distance between them, h the height of the balloon, and x
the distance the car travels. Then
r^2 = h^2 + x^2
Differentiating,
2 r dr/dt = 2 h dh/dt + 2 x dx/dt
One second later x = 66, h = 200 + 14 = 215, so r = Sqrt(215^2 + 66^2).
Plugging in numbers gives:
2 Sqrt(215^2 + 66^2) dr/dt = 2 215 15 + 2 66 66 or
dr/dt = (6450 + 8712)/ (2 Sqrt(215^2 + 66^2))
which is about 33.7
Next quiz November 5 Related Rates, some curve sketching.
Problems from Related Rates (3.1) and Curve Sketching (3.2,3):
First Computer Projects:
These are Maple files. If you have enough memory you can set (in
preferences) that Maple is the helper app for files with suffix ".mws".
Then when you click on the following, Maple will start up with the
assignment. Save the file on your own computer, and start....
http://www.math.gatech.edu/~carlen/1507/notes/CalcIMap.mws . (This is
an introduction to Maple), and
http://www.math.gatech.edu/~cain/calculus/assn3.mws (Although
Georgie doen't say this, this project is an adaptation of an original
project by A. D. Andrew and T. D. Morley.)
1. Suppose the f'(x) = f(x), and that f(g(x)) = x. Find g'(x)
3. The radius of a certain sphere is increasing at a rate of 4 inches
per minute. If currently the radius is 5 inches, how fast is the volume
increasing?
4. Find the equation of the norml line to the curve x^2 y^2 + xy = 5
at the point x =1, y =2.
Some problems on the chain rule, power rule, trig functions:
Solutions to Quiz #2. For compatibility and for ease of loeading,
I am attempting to do these in plain text. When I get mathematica
working again, I will begin to post solutions to quizes in html + gifs.
Problem 1: (10 points) Calculate the derivative of (3x)^(1/2) by the definition.
Problem 2: 5 points each. I'll just give the answers.
Problem 3: Find the tangent line to the curve sin(x)/6x at the
point x = pi/3.
WIll cover up through 1.5,1.6,2.5,2.1,2.2
Things to know for the quiz:
As I said yesterday, I will be arround this afternoon from
about 3 to about 4 pm. In my office (Skiles 263) or if I'm not there,
check the door
of my office. This will say where I am.
Suggested problems for the material covered Wed Oct 7
and Fri Oct 9, Mon 12.
First Quiz: Tuesday October 6, 1998
Will cover up through 1.2
Suggested problems for the material covered Chapter 0. and 1.2
Sone odd numbered problems to work yourself:
Material coverered from Chaprter 1. On September 28, ew did example
4 on page 56, and example 7 (page 58 and page 161).
Book Calculus, by Stanley I Grossman (Fifth edition)
All tests a quizes will be open book and notes
Office Skiles 263.
TA: Honor Hutton
gt2366b@prism.gatech.edu
Class (TM) is at MWF Skiles 270: Time: 8:05-8:55 am
Recitation: (CC): T Th Skiles 270
All tests and quizes will be open book and notes. Calculators
allowed.
Regrading policy: Quizes will be handed back
in recitation or lucture. We will go over the quizes the day that
they are handed back. If there is any quiestions
about the grading, you must write the quiestions
on the quiz and return the quiz before you leave the classroom.
No exceptions. OK?
The mechanics of calculation, by hand, calculator, or computer are
still important, but not in and of themselves. They are tools.
There is no such thing as a stupid question. Yell them out!
If something
confuses you, raise you hand and shout out to the world that the
person sitting next to you doesn't understand! :-)
If you have not had Calculus before, that is also OK. Rest assured that
this course does not presume any knowledge from AB or BC calculus.
Return to Tom Morley's
Home Page
email:
morley@math.gatech.edu
Or
The Calc I
Swiki . An editable web page that is yours. Under construction.
Please add whatever you like -- When you get there, just click on "Edit
this page". This might be a place to hook up with a partner for the
computer projects.
Some solutions to quiz #3:
Solution:
Take f(g(x)) = x, and differentiate useing the chain rule, giving:
f'(g(x)) g'(x) = 1
But f'(junk) = f(junk), so:
f(g(x)) g'(x) = 1
Using f(g(x)) = x, we get:
x g'(x) = 1
Giving g'(x) = 1/x.
V = (4/3) pi r^3, so differentiating with respect to t, we get:
dV/dt = 4 pi r^2 dr/dt.
Plugging in r = 5, and dr/dt = 4, we get the answer
dV/dt = 4 pi 5^2 4
= 400 pi
Differentiating the euqation gives:
2 x y^2 + 2 x^2 y dy/dx + y + x dy/dx = 0.
Solve for dy/dt, giving:
- 2 x y^2 - y
dy/dt = --------------
2x^2y + x
Plug in x = 1, y = 2 and get:
- 2 1 2^2 - 2
dy/dt = --------------
2 1^2 2 + 1
= -10/5 = -2
So slope of the normal is -1/(-2) = 1/2.
Equation is therefore (y-2) = (1/2) (x-1)
Solution: We compute the limit:
(3(x + h))^(1/2) - (3x)^(1/2)
lim ---------------------------
h->0 h
Multiply the numerator and denominator by (3(x + h))^(1/2) + (3x)^(1/2):
(3(x + h))^(1/2) - (3x)^(1/2) (3(x + h))^(1/2) + (3x)^(1/2)
lim --------------------------- --------------------------- =
h->0 h (3(x + h))^(1/2) + (3x)^(1/2)
(3(x + h)) - (3x)
lim ------------------------------ =
h->0 h (3x + h)^(1/2) + (3x)^(1/2)
3h
lim ------------------------------ =
h->0 h (3x + h)^(1/2) + (3x)^(1/2)
3
lim ------------------------------ =
h->0 (3x + h)^(1/2) + (3x)^(1/2)
3
------------------------ =
(3x)^(1/2) + (3x)^(1/2)
3
------------- =
2(3x)^(1/2)
Solutions:
a) 4 x^3 sin(x) + x^4 cos(x)
b) (4 x^3 + 6 x^2 + 4) (3x^2+x^6) - (x^4 + 2 x^3 + 4 x)(6x + 6 x^5)
------------------------------------------------------------------
(3x^2+x^6)^2
c) (4 x^3 + 6 x^2 + 4)(sin(x) + 42) - x^4 + 2 x^3 + 4 x)(cos(x))
-------------------------------------------------------------
(sin(x) + 42)^2
Solution: Let f(x) = sin(x)/6x, therefore the tangent line goes through
the point x = pi/3, y = f(pi/3) = Sqrt(3)/2.
f'(x) = cos(x) 6 x - 6 sin(x)
---------------------
(6x)^2
So f'(pi/3) = (1/4 pi) - (3/4) Sqrt(3)/pi^2
So we get
y - Sqrt(3)/2
-------------- = (1/4 pi) - (3/4) Sqrt(3)/pi^2
x - pi/3
This can be simplified, if you like.
Second Quiz: Tomorrow!! Thursday October 15, 1998
The grade will be based on
The quiz grades will be added together, with
one quiz dropped. They count for 55% of the grades. Group Projects 10%,
Homework 10%, and Final exam 25%.
Topic
Sections
Appoximate Days (lecures)
Review of some algebra
0.1-0.5
3
Limites and Derivatives
1.1-1.8
5
More about derivatives
2.1-2.7
6
Applications of the Derivative
3.1-3.7
7
Conic sections and polar coordinates
8.1-8.3, 8.5-8.6
3
Complex Numbers
A5
3
Total
27
General comments: I realize that some of you all
have have calculus before. This is a good, right and proper thing,
and my sister Mary mmorley@ets.org
would approve. There will be considerable overlap between AP calculus
and this course. But the pupose of this course is not the same as AP
calculus:
Need to do an integral? Try this.
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the Math Lab is open on M - Th from 11 to 4 in Skiles 257. It is staffed
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taking the above classes.
To get a hold of me:
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