[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.