So, I think
hopefully everybody knows the distance
formula, if
not I made it just on one slide.
So, distance equals rate times time.
This just means if I'm say driving at a
certain speed say
30 miles per hour for three hours, I'm
going to go 90 miles.
it's actually amazing how often this is
the correct answer to
a question, or the corre, correct way to
solve a question.
if you can recognize what you're going to
think of as distance, what you're going
to think of as rate, and what you're going
to think of as time.
This form of the distance formula really
only works
when the rate is constant across the
entire time.
But now that we can integrate, we're
going to be
able to allow that to change a little bit.
So, it provides intuition for a lot of
problems, and
all that we're really doing is imagining
rate on one
axis, time on another axis, and then the
distance you're
going to cover is going to be equal to
this area.
So that's just the definition of the area
of a rectangle is height times width.
So let's look at something I call the, the
job problem, which
seems to give a lot of people quite a lot
of trouble.
so, I take my motivation quite often from
the Simpsons.
So, I said if Lenny can do a job in 6
hours and Carl can do
the same job in 4 hours, how long does the
job take if they work together?
a lot of times when you ask
people this question, they will say
something the first
number that pops into their head is 5
hours.
which seems to make sense, because it's
just the average of 4 and 6.
But that would actually mean that by
helping
out Carl, Lenny is making the job take
longer.
So, you know, clearly that's not the right
way to think about the problem.
And the right way to think about this
problem is
that it's the rates that are additive.
So, Lenny, if he can do a job in 6 hours,
he can do 1/6th of the job per hour.
And Carl, can do 1 quarter of the job per
hour.
So if they're working together, their rate
is going to be 1/6th
plus 1/4th, and that ends up being 5/12ths
of the job per hour.
And so, here, the job is what's going to
end up
being the distance and the time is what
we're asking for.
How long does it take them to do the job
if they're working together?
So, solving for time, I just get distance
divided by rate.
And then I've worked out what these
numbers are, we have one job.
The rate that they're working is 5 over 12
jobs per hour.
I can multiply the top and bottom of that
by 12 over 5 and get 12/5th.
So the job is going to take 2 hours and 24
minutes if they work together.
But we've still made the assumption, that
Lenny
and Carl are working at a constant rate.
So, suppose instead, that Lenny works at a
rate L
of x and Carl works at a rate C of x, then
my
condition that I started out with, I could
rewrite
this in terms of definite integrals.
So I know
that if I let Lenny work for 6 hours, he's
going to finish one job.
So that means that the integral of L of x
from 0
to 6 has to be equal to 1, and then oops,
now I've
confused L's and my C's.
So I think this should be the C
and the integral from 0 to 4 of c of x
should also be equal to 1.
And that's just the way of mathematically
writing down the condition, or the
statement that Lenny can do one job in 6
hours and Carl can do one job in 4 hours.
And so now my job question is, can I find
a capital T,
so that's a, a time, so that when I add
these rates together I get 1.
And
so now, this is not quite as nice
a problem, because there are actually a
whole
bunch of different rates that I could add
together that would, that would give me
this condition.
So for instance I could just say well, I
guess I can't.
So I have some constraints to, so I am
bound by the constraint that Carl works at
this rate.
So the integral of that one has to be over
0 to
4 to be 1 and the integral of L of x over
0 to 6 also has to be 1.
So I just invented some functions to, to
try and show, demonstrate this.
So I chose 12 minus x divided by 54.
And all this is going to, all this is
going to do,
is when we're at time zero.
So when x is zero, I'm going to have 12
minus zero divided by 54.
And at the end of the job, so when x
is equal to 6, I'm going to have 12 minus
6, which is going to be 6 divided by 54.
And we can forget about the bottom,
because that's the same.
So at the beginning, we have 12 over
something
and at the end, we have six over
something.
Six is half of 12,
so all this is saying is that Lenny is
working twice as hard
at the beginning of the time period as he
is at the end.
So he gets tired while he's working, so by
the
end of the 6 hours, he's going only half
as fast.
And then I just have to put this 54 down
here so that the, the condition,
the constraint that he finishes the job in
6 hours is met.
And then I'm going to do
exactly the same thing for Carl.
So initially he's working at 8 over 24, at
the
end of his 4 hours, he's working at 4 over
24.
So he works faster, so he's also getting
tired faster.
And
now, how long is it going to take them if
they, if they work together?
So, would it be more or less than the 2
hours and 24 minutes?
So basically what I want to do is find T,
so that when I integrate this
function, L of x plus C of x from zero to
T, I get my one job finished.
And so I'm just going to start by plugging
in the two functions.
So L of x was 12 minus x over 54 and C of
x was 8 minus x over 24.
And then I don't like having these numbers
in the denominator.
So I'm going to multiply both sides of
this by 216, that's going to be
4 times 54.
So the 4 times 50 is 200, 4 times 4 is 16,
216.
Or it'll be 9 times 24.
You can do that in your head pretty easily
too, because 10 times 24 would be 240,
and that's 24 more than 216 so that's got
to be 9 times 24.
And so
that's going to get everything out of the
denominator.
So, mostly the reason I wanted to do that,
it makes
the lines a little skinnier so it all fits
on one slide.
But it also means that I can now add these
together, so my integral is going to
simplify itself a bit before I actually
have to compute an anti-derivative.
And so what I end up having to compute
the anti-derivative of is this function
here, 120 minus 13x.
So I can just use my anti-power rule to
get that.
And that's going to give me 120x minus 13
over 2 x squared from 0 to T.
And now, even though the variable T is on
this side, I, I treat it exactly the
same way, so I'm just going to evaluate my
anti deriv, oops, my anti derivative here.
I evaluate it at T and then subtract it,
evaluate
it at 0, so I get 120T minus 13 over
2 T squared.
And then luckily on this side, since I've
got that zero, when I plug in zero to the
to, to this anti-derivative, I just get 0,
so that makes things a little bit simpler.
So I end up with 13 T squared.
So I, I multiplied both sides of this by,
by 2 as well, 13T squared minus 240T plus
216, oops, I
think I forgot to multiply that bit by 2.
And
then you can solve that by using the
quadratic formula.
And that gives 57 minutes.
Although I think it gives 57 minutes for
this one that
I forgot to multiply by 2, so I'll have to
check
that.
And then another application where this is
going to
turn of for those of you that are taking
461.
in the next couple of weeks, we're
going to
learn about something called a probability
density function.
So, I'm going to denote a probability
density function for a random variable x.
It says f sub X, and its argument is going
be little x, and I'm
going to use it to assign probabilities to
a random variable being in an interval.
So I'm reading this thing on the left hand
side.
It's the probability that my random
variable is in the interval
[a, b].
It's just going to be the integral of my
probability density function over the
interval [a,b].
The support of a random variable is just
going to be
the, the set of points, so generally it's
an interval.
a set of points S, where the probability
density function is greater
than or is greater than zero, when x is in
S.
And so, some of the profi-, properties of
a probability density function
are, if f sub X is a probability, so I'm
going to eval, abbreviate that
as pdf, then I have to have f of x is
greater than or equal to 0 for all real
numbers.
And when I integrate this, so sometimes
instead of
instead of writing an interval as just
from a
to b, I might want to write a set that
might be made up of more than one
interval density function.
And so, in that case, when I've defined S
to be a set.
So, here S is the set of points where this
thing strictly greater than 0.
I can write the integral sign, and I'll
just
put the name of the set at the bottom.
And the integal has to be equal to 1.
So the exercise that'll show up quite
often is, can you
prove that a certain function is actually
a probability density function?
And so I'll go through one example of that
pretty quick.
So the PDF of a beta 2-2 distribution, is
defined to be 6x
times 1 minus x, when x is in the interval
[0, 1], and its defined to be 0 otherwise.
So if I want to verify that f of x is
greater than or equal to zero, that should
be pretty straightforward.
If I have a number between zero and one,
then it's greater than or equal to zero.
So 6x has to be greater than or equal to
zero.
And one minus x, so the smallest this
could be
is when x is equal to 1 and that's zero.
So, it's pretty easy to verify the first
condition.
The, the work is usually verifying that
the
PDF is going to integrate is going to
integrate
to one over it's support.
So here going to have the, the integral,
so S is going to be 0 to 1.
So, it doesn't actually include the
endpoints, but
those end up not mattering for the
integral.
So, I'm going to say this is just an
integral from
0 to 1 of 6 x times 1 minus x.
Oops, again that typo.
dx, so this x, shouldn't be here.
Hopefully that's going to go away.
[COUGH]
And
so, using some of the properties I've
listed in the first set of slides.
I have, I can think of x times 1 minus x
as
my f of x, and so I'm just multiplying
that by a constant.
So I can take that constant outside of my
integral sign to make things a
little bit simpler, and then I'm just
going to just go ahead and do the
multiplication.
So I have x times 1 gives me x.
And then x times x gives me the x squared
here.
And then I can calculate the
anti-derivative of that, just using my
anti power rule on each term.
So the first one came from 1 half x, 1
half x squared, and the second term came
from 1/3rd x cubed.
And now that I've written down my
anti-derivative here,
I've moved my limits to the right hand
side.
So it means this notation again, just
means evaluate
my anti-derivative upper end point and
subtract
the anti-derivative evaluated at the lower
end point.
So, again, because I have an x in each
term, the bottom
endpoint is just going to be zero, that
simplifies things quite a bit.
So, I end up with just 1 half minus 1/3rd.
Oops, and that's equal, so 1 half is, I
can write that as 3 over 6.
1/3rd, I could think of as 2 over 6, so 3
over 6 minus 2 over 6 is 1 over 6.
And I
have this 6 out in front and that cancels
each other out.
So, I verified that this function here, fx
of x is a valid probability density
function.