[Paleopsych] Danica McKellar: Mathematics
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Subject: Danica McKellar: Mathematics
Danica McKellar: Mathematics
http://danicamckellar.com and click on "mathematics."
[Okay, you mathematicians, go to it! She has an intriguing face, quite
beautiful, but just enough away from the stereotype to make her fascinating.
Are there any other math majors among celebrities?]
As some of you already know, Danica took a hiatus from acting for college, and
she graduated from UCLA with a Bachelor of Science in Mathematics! While she
was there, she even co-authored a math proof - new research proving an original
math theorem - highly unusual for an undergraduate. In fact, she was the only
undergraduate invited to speak at Rutgers University's biannual Statistical
Mechanics conference a few years back. Although she has returned to acting full
time, she still retains a passion for it, and likes to stay active.
Listen to a radio interview Danica did on the connection between mathematics
and the arts for Studio 360.
<http://danicamckellar.com/math/studio360.gif>
And for you hard core mathematicians, here is a PDF of Danica's published proof
Percolation and Gibbs states multiplicity for ferromagnetic Ashkin-Teller
models on Z2.
http://danicamckellar.com/math/percolation.pdf
Danica loves math so much that she wants to share it with you, and help you get
through some of your tough questions. Only a few questions each month will be
forwarded to her for answering, and you can submit your questions to:
math at danicamckellar.com
Says Danica: Let's face it; by and large math is not easy, but that's what
makes it so rewarding when you conquer a problem, and reach new heights of
understanding! I'll be answering questions ranging from middle school math to
Calculus and beyond, so skim along until you find something helpful or
interesting to YOU. I challenge all of you to embrace the mind-sharpening
qualities of practicing mathematics. Now let's roll up our sleeves and do some!
Doing the math
Q: Hi Danica, I heard a question from Mr. Feenie on a "Boy Meets World" episode
which he claimed to be unanswerable. After hearing that, I decided to figure it
out. If it takes Sam 6 minutes to wash a car by himself, and it takes Brian 8
minutes to wash a car by himself, how long will it take them to wash a car
together?
Danica Answers: Hm, unanswerable? That's TV for you. :)
Let's do it: This is a "rates" problem. The key is to think about each of their
"car washing rates" and not the "time" it takes them. Alot of people would want
to say "it takes them 7 minutes together" but that's obviously not right, after
you realize that it must take them LESS time to wash the car together than
either one of them would take.
So, what is Sam's rate? How much of a car can he wash in one minute? Well, if
he can wash one car in six minutes, then he can wash 1/6 of a car in one
minute, right? (think about that until it makes sense, then keep reading).
Similarly, Brian can wash 1/8 of a car in one minute. So just add their two
rates together to find out how much of a car they can do together, in one
minute, as they work side by side on the same car: 1/6 + 1/8 = 7/24 of a car in
one minute. That's their combined RATE. (Note: that's a little bit less than
1/3 of a car in one minute). From this point, the way you want to think of it
depends on your favorite way of dealing with fractions. You now have their
rate. It's 7/24 cars per minute. You can either just take its reciprcal and
say: 24/7 minutes for one car, and you're done.
Or, equivalently, you can think of the 7/24 cars/minute RATE as 24 minutes for
7 cars. (think about that until it makes sense, too) So just divide 24 by 7 to
find out how many minutes it would take to do just one car. You get around 3.42
minutes for one car, just a little less than 3 and 1/2 minutes. Done! Yes, I
think they should work together, it gets done much more quickly that way. :)
By the way, you said when you watched the TV show you decided that YOU would
figure it out, right? How did you do?
Q: I'm an adult trying to pass the GED test and you do alot of studying on your
own, and no matter what I do, I keep getting wrong answers. I know the
equations for parellelograms, the area of circles, etc. But I can't seem to get
the right answers. They only grade you on the final answer you get, not on the
work. I have no clue where I'm going wrong. I want to take this test so I can
go to school and become a drug and alcohol counselor.
Thanks so much!
-Pat
Danica Answers: Hi Pat- First of all, congratulations on having the fortitude
to go back and get your high school diploma! And I admire your goal of helping
people by becoming a counselor.
to you.
Now- Let me see if I can shed some light on your problem. Understanding and yet
getting the wrong answers usually means one thing: You're making a lot of
avoidable, careless mistakes. This is a VERY common problem these days, more
than ever before. But it's totally solvable.
Believe it or not, I think this is a symptom of a larger issue that our society
is facing, and it makes it that much more important that we deal with it. It
seems that in our ever-increasingly fast paced information age, we are more and
more often finding shorter and faster ways to do every day tasks: Abbreviating
whole phrases in emails to "lol" and "imo", our computers type in our own
addresses for us on websites, etc. We are accustomed to skipping steps. One of
the unfortunate results of this, is that students are more and more often,
skipping steps in solving math problems-- skipping the act of writing down
every step along the way. And no matter how well they understand the concepts,
they believe they cannot "do math" because they so often get the wrong answer.
You would be AMAZED, truly amazed, at how much more often you would get the
*right* answer, if you took the time to write down every step when you are
solving problems. We ALL make careless mistakes, but the chances of making
careless mistakes goes up at least tenfold when we skip steps.
Other ways to help avoid making careless mistakes, in addition to writing down
every step:
-Pay attention to WHAT is being asked for- you may have done a perfect job at
answering a slightly different question. Did they ask for the radius or the
diameter? etc.
-Restate the problem on your page; just write down all the info you're given.
This is especially helpful for word problems. Trust me, it works.
-If parentheses are used in the problem, keep them during the solution until
the last step. Especially if you've done any algebra, you know exactly what I'm
talking about. :)
-Don't try to squeeze too much onto one page. When numbers get scrunched and
small, mistakes are made.
-Once you've gotten the answer, then REREAD the problem to make sure you solved
for exactly what it was
g for.
So listen, the next time you solve a problem that you know you have the right
formula and understanding for, but you get the wrong answer-- do this:
Write down every step you can. Write down the formula. Then write down in
information you're given, etc. So if the book says that the radius of a circle
is 3, and you are supposed to get the area, then literally WRITE DOWN:
A=pi*r2
r=3
Then, beginning to solve it, don't do anything in your head without writing it
down:
A=pi*32
A=9pi
You'll get the right answer and you'll understand the value in making this a
habit. What takes a few extra moments now, will save you many minutes of
frustration and-- your grade. Another note on this: THIS IS HARDER TO DO THAT
YOU THINK. We all think to ourselves, "Oh, I'm smarter than this. I don't need
to write down every step. That's what you do in kindergarten, etc." So be
patient with yourself if you can't discipline yourself to do it right away--
but rest assured that when you choose to practice this method of not skipping
steps, your success will follow, I promise. :)
Q: I think you are great on "The West Wing"! Here's my current problem, it's in
advanced finite math (I'm a high school senior): At the height of the Beatles'
popularity, it was estimated that every popular music station played their
music 40% of the time. If you tuned through 10 such stations at any given
moment, what is the probability that at least *one* of the stations would be
playing a Beatles song? Thanks!
Danica Answers: A probability question! Okay, let's call "x" the probability
that "at least one of the 10 stations would be playing a Beatles song at that
moment." That's what we're asked to find. Then let' s call "y" the probability
that "none of those 10 stations would be playing a Beatles song at that
moment." Notice that x+y = 1, since the two situations are mutually exclusive,
but combined they make up ALL possible scenarios. And the probability that one
of "ALL possible scenarios" occuring is of course 100%, which equals 1.
Okay, so we' ll now determine the value of "y" which is much easier than going
through all the necessary calculations required to determine "x" directly. This
is a common strategy in probability.
So, what is the probability that NONE of the stations are playing a Beatles
song? That would be the (multiplicative) product each of the probabilities of
each station NOT playing the Beatles. We know from the statement of the
problem, that the probability of any given station, at any point in time,
PLAYING a Beatles song is 40% = .4. This means that the probability of a
station NOT playing a Beatles song is 60%, or .6. (After all, either a station
is playing a Beatles song or it's not: the two scenario's probabilities must
add up to 1 = 100%)
So, if the probability of 1 station NOT playing a Beatles tune at any
particular moment is .6, then the probability of all 10 stations NOT playing a
Beatles tune at that moment = [.6 raised to an exponent of 10]. Multiply .6
times itself 10 times and you will get a number like: .0060466176. This is our
"y." To get "x" we must solve x = 1-y, and we get that x = .9939533824.
Translated back into percentages, we get that, at any given time, there is a
99.39533824% chance that the Beatles are playing on at least one of the ten
stations. Wow! Did you expect it to be that high? :)
Q: Hi Danica, I am working on equivalent fractions. I forget the formula but
once I relearn it I'm usually ok. Do you know any tricks for this? Thanks!
Dena.
Danica Answers: Yes, there is a trick. It' s called "cross-multiplication." Say
you have the two fractions, a/b and c/d. To determine whether or not they are
equivalent, you can multiply "a" times" d," and also "b" times "c", and if
their products (ad and bc) are equal, then you have equivalent fractions. For
example, 2/3 and 8/12. You could "cross multiply" and see that 2 times 12 = 24,
and 3 times 8 = 24, so the two fractions must be equivalent.
Tricks are a good shortcut when you already understand the concepts, but tricks
can get you in trouble if you're fundamentally confused. So let's make sure you
understand the concept of equivalent fractions.
First of all, what does it mean for fractions to be equivalent? It means that
they represent the same VALUE. Just like if I wrote down the expressions "5-3"
and "10-8". They are *different* ways of writing the same VALUE. They both
equal "2." Equivalent fractions do the same thing. They are different ways of
writing the same number. 2/3 and 8/12 HAVE THE SAME VALUE. If you were to cut a
pie, and you said, "give me 2/3 of that pie" or if you said "give me 8/12 of
that pie", you'd end up with the *same* amount of pie. (That's a lot of pie!)
So the "real" way of determining if two fractions are equivalent would be to
determine if they represent the same value. After all, it is clear that 2/3
does NOT equal 24. We simply used a trick, and "24" has little to do with the
*value* of either fraction. A good way to determine if two fractions are
equivalent is to REDUCE them. That is, take out common factors in the numerator
and denominator. So, with 8/12, you might notice that both 8 and 12 have a
factor of "4" in them. So you can reduce the numerator and denominator by 4.
Then the fraction becomes 2/3. And certainly, we can see that 2/3 = 2/3. No
tricks needed. Hope this helped, and
Good luck!
Q: I think you are great on "The West Wing" and I just saw you on NYPD Blue!
This may be more of a physics question, but I was curious- a friend of mine was
talking about an outfielder who could throw a ball from the outfield, have it
go no higher than head height, and reach the catcher at home. It seemed
IMPOSSIBLE but then I started thinking about the viability of this being
possible. Can you help?
Here are the assumptions:
1. The outfielder (pt. A) and home plate (pt. B) are 180 feet apart (roughly
twice the distance from home plate to 1st base).
2. The ball is released from a height of 6 feet.
3. The ball travels along a curved path (pulled down by the force of gravity
(32 ft/sec2)).
4. It reaches home plate at ground level after not traveling at any time
above 8 ft above the ground (roughly head height for the tallest human).
5. An official baseball weighs 5 ozs. (although I'm not sure if that's
relevant.) Thanks!
Danica Answers: A Hi there! Alright, let's solve this. Some physics and algebra
knowledge is definitely needed to make it through this proof. I'm going to skip
most of the algebra steps, assuming you can do those on your own if you like.
So don't be discouraged if you don't follow it all- I answer all sorts of
levels of problems on this site. :)
First we will assume that there is no wind drag-- just to simplify things. You
are right that (with no wind velocity) the weight of the ball does not matter.
What we will do is find out what the velocity of the ball would have to be in
order for this hypothetical situation to be possible, and then see if a human
is capable of it. So, the way we do this is to first find out how long the ball
would be in the air. (it will be clear "why" later) I recommend drawing a
diagram just to make it clear to yourself. One thing to remember is that we can
treat the up/down component of velocity separately from the side to side
component of velocity.
First, looking only at the up/down motion: The ball gets thrown in the air from
6 ft, goes to 8ft, and then down to 0ft. (the ground). The equation for the
height change of the ball (when it starts with zero velocity) is:
H = (1/2)gt2
You can find this equation in any elementary physics book. g= the acceleration
of gravity, which is 32ft/sec2. (or 9.8 meters/sec2 ) First let's see the time
it takes for a ball to reach the ground, when dropped from a height of 8ft.
Since it starts with zero velocity, we can use this formula. Solving for "t"
when "H" = 8ft, using basic algebra, we get approx. .707 of a second. Since
when the ball is arcing across the baseball field, its up/down velocity is zero
at the point when it hits 8ft, this .707 of a second also represents the time
it takes the ball to go from the highest point of its arc, to the ground. If
that sounds completely foreign, there's a great lesson on this concept at:
http://www.glenbrook.k12.il.us/gbssci/phys/Class/vectors/u3l2b.html
So now we need the first part: the time it takes the ball to go from the
pitcher's hand to the highest point of its arc (8ft). Since the ball reaches an
up/down velocity of zero at its highest point (meaning that the vertical
component of its velocity is zero at that moment), and because the only
external force acting on the ball's up/down velocity is gravity, this would be
the same amount of time it would take for the ball to be dropped from 8ft, and
have it caught by someone at the height of 6ft.
So now, using the same above formula, we'd say that the change in height, H,
equals 2ft. And solving for "t" we get approx. .354 of a second.
So now we know, that in this hypothetical situation, the ball is in the air for
approx. 1.061 seconds and travels for 180 feet. So how fast would the ball have
to be going? It's a simple rt = d problem. Solving for r, the rate, we get
approx. 169.65ft/sec. Using the conversions 3600sec = 1 hour, and 5280ft = 1
mile, we get the rate of approx. 115.7 mph.
115.7 mph? Hm. And that's WITHOUT drag. If there were drag, the ball would be
slowing down throughout its journey, so the initial throw would have to be
FASTER than this. I've checked the Guiness Book of World Records and it seems
that the fastest anyone's ever thrown a baseball was 100.9 mph by Lynn Nolan
Ryan (California Angels) at Anaheim Stadium in California on August 20, 1974.
As was pointed out by one reader, (Thanks, Alan!) if we assume that the thrower
dropped his arm down as he let go of the ball at a height of 4ft, then the
thrower would only have to throw the ball with a speed of approx. 101.9 mph. I
would have assumed it to be harder to throw a ball with incredible speed from
far below shoulder height, but perhaps it's easier. I certainly know more about
math than I do about baseball! Of course, you could also run these numbers
using a very short (but fast) pitcher. Assume the guy (or gal) is only 5 ft.
tall; then a 4 ft release point becomes even more feasible. Experiment, and
have fun discovering what math can tell you!
Q: I have a calculus question for you: Gravel is being dumped from a conveyor
belt at a rate of 30 ft-cubed/min and its coarseness is such that it forms a
pile in the shape of a cone whose base diameter and height are always equal.
How fast is the height of the pile increasing when the pile is 10 ft high?
Danica Answers: Okay, this is a typical "related rates" problem, and it' s a
good problem to understand for ALL related rates problems in first year
calculus. We need the RATE of the changing height at a certain point in time.
We' re told the RATE of the changing volume (30 ft-cubed/min). So we will need
to "relate" the "rates" of the height and the volume. So we need to FIRST write
down an equation that determines:
1) The relationship between the VALUES of the heights and volumes, h and V.
And then we'll take the DERIVATIVE of this equation, which will then give us:
2) The relationship between the RATES of these values, dh/dt and dV/dt.
When determining this first, important, equation between the VALUES of height
and volume, always start with what you know.
Well, we know that for every cone,
V = (1/3)h(pi)r2.
Additionally, we are told that for THIS cone, the diameter, which equals 2r, is
always equal to the height. So we know that r = h/2. Plug this in for r, and we
get:
V = (1/12)(pi)h3.
This is our important equation #1 relating VALUES. Now, to get the #2 "related
rates" equation, we must take the derivative of the entire thing with respect
to time, t. Don't forget to use the chain rule!
dV/dt = (1/12)(pi)3h2dh/dt
Now remember that this equation, as it's written, is true for ALL moments in
time. And now let' s consider the moment in time that we were asked about: the
moment when the height = 10ft. So, at that moment, we can plug in h = 10. We
also know dV/dt; we were told in the problem that the "rate the volume is
increasing" is constant. It' s 30 ft-cubed/min. So we can certainly plug that
value in for this moment in time. Now the only variable left is dh/dt—the rate
that the height is growing. And when we solve for it with simple algebra, we've
solved the problem! (You should get dh/dt = 6/(5pi) ft/min.)
Q: How do you figure out percentages using word problems such as: To finish a
certain job, John made 35 parts while Allen made 15. So, how much MORE of the
job did John do than Allen? Express your answer in terms of percentages. Thank
you!
Danica Answers: Percentages rely on parts of a total. So what's the total
number of parts made? 35+15 = 50 total parts. How many more parts did John
make? 35-15 = 20 more parts than Allen made. So John did 20/50 more of the job
than Allen, which equals 2/5 = .4 = 40% more. The answer is: John did 40% more
than Allen. As a check, figure out what percentage of the job that each of them
did. So John did 35 parts total out of 50. That's 35/50th's of the job, which
equals .7 = 70% of the job. And Allen did 15/50th's of the job, which equals .3
= 30% of the job. Thus, you can see that John did 70%-30% = 40% MORE than Allen
did. It's always a good idea to check your work like this. Hope that helped!
Q: Danica, This is a problem for Calculus II
f(x)= x2+2x+3
Problem:Subdivide the function's domain into subintervals on which each
function has an inverse, and find the inverse function for each subinterval.
Now I found the critical #'s by first finding the derivative which is 2x+2 thus
the critical # is x= -1 Then I found the intervals which would be [negative
infinity, -1] & [-1, infinity]. But now I can't seem to find the inverse of
f(x).
So I'm asking;
What is the inverse of f(x)?
What are the two inverse functions for each interval?
Danica Answers: You started correctly! If f(x) = y, then you must solve for x,
in terms of y. That will give you the inverse function. [hint: currently the
function is in ax2 + bx + c form; subtract the “y” from both sides and make it
look like ax2 + bx + (c-y) and then solve it with the quadratic formula]
Then, you'll get an answer--a "function" that says "x = (something with Y in
it) with a "+/-" in it, and that will show you that you have two functions
represented. And then it is up to you to determine which of the functions is
applicable for which domain. [hint: substitute values and see what makes sense]
I highly recommend sketching a crude graph of the function. Then turn your head
sideways and get a feel for the inverse functions that will come from it. Just
do a simple drawing based on the critical value, and what the "zeros" of the
function are. Remember, the definition of a function includes the rule that for
each value in its domain, there must be a unique function value, not more than
one. Sketching graphs can be a GREAT aid in solving many of these problems, and
it's very fast to do once you get the hang of it! This function is especially
easy to get zero values from. (that means where x=0 on the graph of f(x)). A
grouping solution will show you that the zeros are 1 and -3. Good luck!
Q: Danica, I teach high school math and just read about your "Figure This!"
campaign to promote mathematics. How can I find out more about this?
Danica Answers: You can visit www.figurethis.org; it has all sorts of fun math
puzzles, and can give you more information about this organization. Best of
all, you can get ideas about how to bring math into the "real world" from this
site. Have fun!
Q: Hi Danica, my advanced calculus prof asked us to prove that the square root
of 5, Sqrt(5), is not a rational number. Any suggestions about where to start?
Also, I really loved your Wonder Years show, and I've seen you on West Wing,
too. Thanks a bunch.
Danica Answers: Thanks! Okay- as with most "disprove this" proofs, start by
writing down the hypothesis (as if the thing you are trying to disprove were
true) and then work with the equation until you get a contradiction. Here the
hypothesis is that the square root of 5 is a rational number, and we're going
to show that it's a faulty hypothesis. In "math language" this is equivalent to
saying that you can write the square root of 5 as a fraction of whole numbers;
that's in fact the definition of a rational number. We can assume that this
fraction looks like p/q where p and q do not divide each other; that is, they
share no common factors (except 1). In other words, we are assuming the
fraction is written in reduced form, and we shall also assume that q is greater
than 1. (And why can we do this? Here's a "mini sub-proof": This is really the
same as proving that Sqrt(5) itself is not a whole number; although this seems
obvious, we can easily prove it for those who like the detail: So- let's say
that q=1, then Sqrt(5) = p/1, where p is a whole number. This is the same as
saying that Sqrt(5) = p. But because Sqrt(5)>2 and Sqrt(5)<3 and no whole
number p satisfies those conditions, we arrive at a contradiction and we may
now assume that q is GREATER than 1 for the rest of the proof.)
Now it's time to work with the original expression and hope for a contradiction
to appear, that expression being: Sqrt(5) = p/q. Remember that p/q is a
fraction in reduced terms. That is, q does NOT share any factors with p. Since
every fraction can be written in reduced form, we are "allowed" to make this
assumption.
So, let's square both sides of the equation:
Sqrt(5) = p/q and we get: 5=p2/q2.
But if q does not share any factors with p, then q2 certainly cannot be a
factor of p2. Then p2/q2 cannot be a whole number, so it can't be equal to 5!
There's the contradiction we needed, which tells us that our original
hypothesis was false. We proved it!
Q: Hi Danica, my teacher (loves puzzles) has given us a problem, which, can
easily be solved using algebra. However, I have trouble grasping the question,
please help
here it is: "I am three times the age that you were when I was
your age. When you get to be my age, our ages will equal 63. How old will we
be? Thanks, Glynis, The Netherlands.
Danica Answers: This was not easy! Think of three timelines: Before, Now, and
Future. Let's call "I" Sam, and let's call "you" Daisy. I'll reword the
question here, with added timelines to help make the problem easier to
understand: "Sam is NOW three times the age that Daisy was (BEFORE) when Sam
(BEFORE) was Daisy's (NOW) age. When (FUTURE) Daisy gets to be Sam's (NOW) age,
THE SUM OF their (FUTURE) ages will equal 63. How old will they be?
Let's label these ages. For the BEFORE time, Daisy's age = y, and Sam's age =
x. Let's say there are m years between BEFORE and NOW, and there are n years
between NOW and FUTURE. Since NOW Daisy's age is Sam's age BEFORE, we know that
Daisy's age NOW = x, and that Sam's age NOW = 3y. (stop and go over this, make
sure you're following so far) We also know that y+m=x, and that x+m = 3y. This
is how Sam's and Daisy's ages have progressed from BEFORE to NOW. Let's say
that there are n years between NOW and FUTURE. Then their ages will be: Daisy
FUTURE = x+n, and Sam FUTURE = 3y +n. We also know that the future ages add up
to 63, which means that x+n+3y+n=63. One more thing we know is that in the
FUTURE, Daisy will be Sam's age NOW, so x+n = 3y. Put all of our known
equations together that will need to be reconciled:
y + m = x
x +m = 3y
x + n +3y + n = 63
x + n = 3y.
This is a set of four equations with four unknown variables, which can be
solved by regular algebra steps of substitution, etc. I highly recommend making
a chart with the timeline and ages data: Put across the top: BEFORE m NOW n
FUTURE and along the left write Daisy, and (below that) Sam. This is how I came
to better understand the age relationships. The hardest part of most word
problems is simply being able to translate English into Math!
By the way, you'll get that their future ages are 27 and 36 for Daisy and Sam,
respectively.
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