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Exam Review
Things you should know
Math 1LT3
These are my notes for the Math 1LT3 exam review. Items in the Table of contents are clickable
for your convenience, as well as any equations that I refer back to. You are not required in any
way to read this whole document (it’s pretty long). Just use it to remind you how to do anything
that you’re having trouble with.
Contents
1 Review of 1LS3 things
3
1.1
Integration by Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
1.2
Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
1.3
L’Hopital’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
2 Differential Equations
9
2.1
Euler’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2
Separation of Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
2.3
Stability of equilibria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.4
The Logistic equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
3 Calculus of Several Variables
9
15
3.1
Domain and Range . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
3.2
Limits in higher dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
3.3
Partial derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
3.4
Critical Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
3.5
Gradient and Directional Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . 20
3.6
Differetiability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
4 Probability and Statistics
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Exam Review
Things you should know
Math 1LT3
4.1
Set Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
4.2
Discrete Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
4.3
Conditional Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
4.4
Binomial and Poisson Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
4.5
Continuous Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
4.6
The Normal Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
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Exam Review
1
Things you should know
Math 1LT3
Review of 1LS3 things
This section is dedicated to things you should remember from math 1LS3. These things are
probably not fresh in your memory though since you haven’t necessarily used them in a while. I’ve
included two examples in each topic with an explanation of what my thought process was, which
I hope is clear enough for everyone to follow. The problems using these techniques on your final
exam could potentially be a different difficulty level than what I’ve included here, but I’m sure
that if you understand what has happened here, you should be fine on the test.
1.1
Integration by Substitution
Integration by substitution is intended as the reverse process to doing chain rule for derivatives.
If you’re not sure how to do an integral, substitution is an excellent thing to start with. It doesn’t
normally take very much time to decide whether or not it will help, and can usually give some
pretty good insight on how the problem should work out. It does not work in every case however.
If you have something that looks like it could have came from a chain rule derivative, this is what
you want. In general, you will have something that looks like
Z
f 0 (g(x)) · g 0 (x)dx
(1.1)
Notice how that integrand is exactly the chain rule applied to f (g(x)). In order to correctly make
our substitution we will need to replace something in this integral with u. A general rule is that
you want the derivative of u to help get rid of other things. With that logic, the obvious choice
= g 0 (x). This derivative almost looks useful. In fact, we could go about
for u is g(x) since du
dx
replacing g 0 (x) with du
, but it will usually be easier in practice to replace just dx with something.
dx
Slightly abusing the notation, we can then solve for dx by pretending that the derivative on the
left hand side is a fraction.
let u = g(x),
then dx =
du
g 0 (x)
We then substitute 1.2 into 1.1 to get something that should be a little more palatable
Z
du
f 0 (u)
g 0 (x)
g 0 (x)
(1.2)
(1.3)
Notice how the g 0 (x)s cancel each other out? If this does not happen then we have possibly done
something wrong. Since we are now integrating with respect to u, u must be the only variable
remaining in the integrand. that means that if we have any x variables left, we have to replace
them with u somehow. In this nice case, we are left with
Z
f 0 (u)du = f (u) + C
(1.4)
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Exam Review
Things you should know
Math 1LT3
after which, for completion we should substitute our definition of u back into the final result
Example 1
Z
2
2xe−x dx
(1.5)
Since 2x is exactly the derivative of x2 this is a good indication that we should choose u = x2 .
du
du
= 2x ⇒ dx =
(1.6)
u = x2
dx
2x
We substitute 1.5 back into 1.6 to convert it to a more easily digestible form
Z
Z
du
−u
2xe
= e−u du = −e−u + C
(1.7)
2x
Once we have done the integration (remembering to add in the integration constant C), we remember that it is good manners to re-substitute our equation for u back into the final result so it
is in terms of x.
Z
2
2
2xe−x dx = −e−x + C
(1.8)
Example 2
Z
x
dx
(1.9)
x−3
In this example, there is nothing that looks like obvious as the derivative of something else. We can
still use substitution in order to make this look a little nicer though. In this case, the denominator
is really what’s giving us trouble, so we try substituting that out.
du
let u = x − 3, then
= 1 ⇒ du = dx
(1.10)
dx
by substituting our choices in 1.10 back into 1.9 we now have
Z
x
du
(1.11)
u
but sometthing looks wrong here. Remember back to earlier, we said that anytime you do a
substituion you need to get rid of all of the x variables. Luckily, our definition of u allows us to
easily solve for x in terms of u.
x=u+3
(1.12)
so we can replace x with 1.12 to complete the substitution.
Z
Z
Z
u+3
3
du = du +
du
(1.13)
u
u
These are integrals we can much more easily compute.
Z
Z
1
du + 3
du = u + 3 ln(u) + C
(1.14)
u
Now we should really substitute back the definiton of u so our final result is in terms of x. As a
side note, since the integration constant C is arbitrary, it has the ability to absorb other constants
like the 3 we will have left over. So we can write our final solution as
Z
xx − 3dx = 3 ln(x − 3) + x + C
(1.15)
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Exam Review
1.2
Things you should know
Math 1LT3
Integration by Parts
Integration by parts is the reverse process for the product and quotient rules in differentiation. We
use this rule when we have something that looks like its the product of two functions, or when all
our other methods have failed. In general you should have something that looke like
Z
f (x)g 0 (x)dx
(1.16)
This in itself doesn’t really look like it came from the product rule. If we remember way back to
when we learned the product rule we had the formula
(f (x)g(x))0 = f 0 (x)g(x) + f (x)g 0 (x)
(1.17)
The integrand in 1.16 is exactly the last term in 1.17. In fact, if we rearrange 1.17 to solve for this
term we have our basis for integration by parts
f (x)g 0 (x) = (f (x)g(x))0 − f 0 (x)g(x)+
which when we integrate, gives us the integration by parts formula
Z
Z
0
f (x)g (x)dx = f (x)g(x) − f 0 (x)g(x)dx
(1.18)
(1.19)
If you are having trouble remembering this formula, there is a clever little mnemonic for this. If
we replace f with u, g with v and their derivatives with du and dv respectively, 1.19 becomes
Z
Z
udv = uv − vdu
(1.20)
and just think of the words of duh to remember how the right side should be (Try to pronounce v
before duh if you can, it’ll help).
Knowing formula 1.19 (or 1.20 if you prefer) is really only half the battle. The other, and much
harder, half is figuring out how to decide what f (x) and g 0 (x) are in your integral. There really
is no steadfast method to do this everytime. What works in one case may not work in another.
There are, however, a few general principles that should helpf
i Whatever you choose g 0 (x) to be had better be something you know the antiderivative of
ii Try to choose f (x) so that taking its derivative will simplify something
iii f (x) · g 0 (x) must make up the entire integrand, you are not allowed to have left over bits.
iv If one combination doesn’t work, try a different one.
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Exam Review
Things you should know
Math 1LT3
Example 1
Z
x · e−2x dx
(1.21)
Since taking the derivative of x simplifies itself to 1, we should try letting f (x) = x. Then by
process of elimination, g 0 (x) is what’s left over
g 0 (x) = e−2x
−1 −2x
g(x) =
e
2
f (x) = x
f 0 (x) = 1
(1.22)
Once we’ve made our decisions, we need to substitute the things from 1.22 into the integration by
parts formula 1.19.
Z
−1 −2x
−1 −2x
x
e
−
e dx
(1.23)
2
2
The integral term here is something we know how to compute. So once we’ve computed that we’ve
gotten to our final answer.
Z
−x −2x 1 −2x
xe−2x dx =
e
− e
+C
(1.24)
2
4
Example 2
Z
x2 ln(x)dx
(1.25)
Normally, when I see a logarithm in an integral my first instinct is substitution. In this case,
substitution doesn’t get us anywhere. So when substitution fails, our next best friend is integration
by parts. Since we don’t know what the antiderivative of ln(x) is, (we could go about finding it
but that requires integration by parts anyway) we should let x2 = g 0 (x).
f (x) = ln(x)
1
f 0 (x) =
x
g 0 (x) = x2
1
g(x) = x3
3
(1.26)
Again, we’ve made our decisions, so its time to try them out in the integration by parts formula
Z
1 3
1 32 1
ln(x) x −
x dx
(1.27)
3
3 x
Miraculously, the integral simplifies a little bit, leaving something that is easily computed. Once
we do the final integration we achieve the final answer
Z
x3
x3
2
x ln(x)dx =
ln(x) −
+C
(1.28)
3
9
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Exam Review
1.3
Things you should know
Math 1LT3
L’Hopital’s Rule
L’Hopital’s rule can be an extremely convenient tool for computing limits. In order to be able to
use L’Hopital’s rule we need a direct substitution of the limiting value to yield one of the following
indeterminate forms
0 ±∞
,
, 0 · (±∞)
(1.29)
00 , (±∞)0 , 1±∞ ,
0 ±∞
Please notice that this list does not include anything like 01 . This is because 10 and things like it
are undefined as real numbers and blow up to ±∞. L’hopital’s rule says that if f (c) = 0 = g(c)
or f (c) = ±∞ and g(c) = ±∞ then
f 0 (x)
f (x)
= lim 0
x→c g (x)
x→c g(x)
lim
(1.30)
0
Notice the if part of that statement required that f (c) and g(c) be the same thing, that means ∞
∞
0
∞
1
is not a candidate, and neither is 0 . This is because ∞ is simply 0 while 0 is similar to 0 in that
it is regarded as infinity. In general, we use this rule to simplify limits. Taking derivatives is much
much easier most of the time than worrying about what happens as things approach inifinity or
zero.
Example 1
1 1
(1.31)
lim e− x
x→0 x
We should be able to convince ourselves that a direct substitution of x = 0 into 1.31 gives us
0
, which is one of the indeterminate forms. This means we can use L’Hopital’s rule. Since 1.31
0
1
appears to be in a form similar to 1.30, where f (x) = e− x and g(x) = x, let’s go ahead and take
derivatives
1 1
f 0 (x) = 2 e− x , g 0 (x) = 1
(1.32)
x
And therefore by L’hopitals rule we should have
1
1
e− x
e− x
lim
= lim 2
x→0 x
x→0 x
(1.33)
but wait, this looks worse than it did originally! Don’t panic, this sort of thing happens from time
to time. Usually it means that we need to change what we decided f (x) an g(x) should be. Let’s
try it backward then
1
1
1
f (x) = , g(x) = − 1 = e x
(1.34)
x
e x
Writing our limit this way doesn’t actually change what it is, it just moves it into a way that is
more easily manipulated. It now looks like
lim
x→0
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1
x
1
ex
(1.35)
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Exam Review
Things you should know
Math 1LT3
So taking our derivatives and applying L’hopital’s rule we should have
lim
x→0
−1
x2
1
x→0 −1 e x
x2
1
x
= lim
1
ex
(1.36)
After L’hopital’s rule, the limit simplifies a little, leaving us with only
1
lim e− x = 0
(1.37)
x→0
Which we should be able to justify through logic.
Example 2
lim xx
(1.38)
x→0
A direct substitution of x = 0 gives us 00 which is one of the indeterminate forms. Unfortunately
this looks nothing like one of the forms we know L’Hopital’s rule with. So how can we make it
look that way? In order to answer that question we need to remember an important property of
limits.
lim ln(f (x)) = ln lim f (x)
(1.39)
x→c
x→c
Based off of this property, we will want to concider the logarithm of 1.38. If we are going to
take the ln, we had better make sure that we don’t actually change anything by taking it to the
exponent as well.
x
eln(limx→0 x )
(1.40)
by using the limit property 1.39, we can rearrange the exponent in 1.40 to be
lim x ln(x)
(1.41)
x→0
Which is much closer to being something we can work with. In fact, if we let
f (x) = ln(x),
g(x) =
1
x
(1.42)
then we’re in good shape. Taking derivatives and applying L’Hopital’s rule we have
1
x
lim
x→0 −1
x2
= lim (−x) = 0
x→0
(1.43)
Substituting 1.43 back into 1.40 we have our final answer
lim xx = e0 = 1
x→0
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(1.44)
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Exam Review
2
Things you should know
Math 1LT3
Differential Equations
This section covers things about differential equations, starting with Euler’s method, then moving
on to separation of variables and then The Logistic map in its many forms. There will be about 16
marks on your test from this part of the course. For each of the 3 topics I’ve included an example
and some recommended problems.
2.1
Euler’s Method
Euler’s method is used for approximating the solution to first order differential equations when
we can’t or don’t want to find the solution. It may seem to be a bit tedious but the procedure is
not very complicated and is the sort of thing that a computer can do very quickly. The basic idea
comes from the difference quotient that we used in the definition of the derivative
f (x + ∆x) − f (x)
∆x
which, by rearranging to solve for f (x + ∆x) gives us the formula for Euler’s method.
f 0 (x) ≈
(2.1)
f (x + ∆x) = f (x)∆xf 0 (x)
(2.2)
where ∆x is the step size. We use this formula to approximate a function value when we don’t
actually know the function, but know the function value at a different value of x near by and know
the derivative. The approximations become better when our step size gets smaller, because then
we have a closer approximation of the derivative in 2.1.
Example Given
y 0 = exy
y(0) = ln(2) −
1
2
(2.3)
approximate the value of y(1) using 2 steps.
In this question we are given most of the information that we need, we just have to find out what
the step size is. Since we’re starting at xi = 0 and taking 2 steps to get to xf = 1 our step size
should be
x f − xi
1
∆x =
=
(2.4)
steps
2
Plugging this and our knowledge about y(0) into the Euler’s method formula 2.2 we can approximate y( 12 ).
1
1 1
y( 12 ) ≈ y(0) + y 0 (0) = ln(2) − + e0·y(0) = ln(2)
(2.5)
2
2 2
where we have gotten our information about y 0 from the differential equation in 2.3. We use this
approximation to approximate y(1).
1
1 1
y(1) ≈ y( 12 ) + y 0 ( 12 ) = ln(2) + e 2 ln(2)
2
2
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(2.6)
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Exam Review
Things you should know
Math 1LT3
We can simplify the exponential by using our logarithm rules
√
1 ln(√2)
2
y(1) ≈ ln(2) + e
= ln(2) +
2
2
(2.7)
♦
If you want some more practice with Euler’s method try questions 28-33 and questions 42-47 in
section 7.1 in your textbook.
2.2
Separation of Variables
The only method we learned for solving differential equations this year was separation of variables.
When we are given a separable differential equation and want to know what the solution is we can
use the separation of variables technique. A differential equation is separable if it can be written
in the form
dy
= f (x) · g(y)
(2.8)
dx
If this is the case, we abuse the notation on the left hand side, by pretending that the derivative
is a fraction. We then “separate” them to get all of our terms involving x on one side, and all of
our terms involving y on the other.
1
dy = f (x)dx
(2.9)
g(y)
It’s important that we end up with dy and dx in the numerator on their respective sides, since our
next step is to write an integral sign in front of both sides.
Z
Z
1
dy = f (x)dx
(2.10)
g(y)
For those of you who care, you could also just divide out the g(y) from the right and then integrate.
The integral on the left then just requires a substitution u = y. Once you integrate (remembering
the integration constant) its just a matter of solving for y (if it’s possible). Sometimes you will be
given an initial condition y(x0 ) = C. In that case you will need to use that information to figure
out what C is in this specific case.
Example
y 0 = yt − 2yt2
y(0) = e
(2.11)
The first thing we want to do here is to write the differential equation in a way that makes it look
dy
more easily seperable. We do that by writing y 0 = dx
and factoring the right hand side.
dy
= y(t − 2t2 )
dt
(2.12)
Now we do the separation, dividing both sides by y and multiplying both sides by dt
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Exam Review
Things you should know
Math 1LT3
1
dy = (t − 2t2 )dt
y
(2.13)
dy and dt look a little lonely here, so we pair them up with their best friend, the integral sign
Z
Z
1
dy = (t − 2t2 )dt
(2.14)
2
Once we’ve integrated we get an equation that only involves y, t and some costant C
1
2
ln(y) = t2 − t3 + C
2
3
Which we can then solve for y
1 2 2 3
−3t
y = e2t
eC
(2.15)
(2.16)
It is conventional that instead of writing eC we just write instead C. Personally, I think this can
be confusing, so how about we replace it with k instead, keeping in mind the fact that eC = k.
1 2 2 3
−3t
y = ke 2 t
(2.17)
We’re only really half way done the problem though. There was the initial condition given to us
at the beginning that we haven’t used (equation 2.11). From this, we know that when t = 0, y = 1
so we can substitute these two values into 2.17 to determine what C is supposed to be.
1 2 2 3
−30
1 = Ce 2 0
=C
(2.18)
So C = 1, and we have our final, particular solution.
1 2 2 3
−3t
y = e2t
(2.19)
♦
If you want more practice with separation of variables and separable differential equations, try
questions 14-17 and 25-28 in section 8.4 of your book.
2.3
Stability of equilibria
Consider the differential equation
dy
= f (x, y)
dx
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(2.20)
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Exam Review
Things you should know
Math 1LT3
An equilibrium solution is a constant solution to this differential equation (y = C). Since the
derivative of a constant is 0, this means that the function f (x, y) on the right hand side must be
equal to zero at that value of y and for all values of x as well (f (x, C) = 0 no matter what x is).
Equilibrium solutions come in three basic types, stable, semi-stable and unstable. For an equilibrium to be stable means that if our initial condition was close to, but not equal to the equilibrium,
the solution would eventually evolve to the equilibrium. An unstable equilibrium means that if we
start close to the equilibrium, the solution will move away from the equilibrium as time progresses.
Imagine a ball on a hilly landscape. An unstable equilibrium would be right at the very top of a
hill, we could balance the ball there, but if we don’t get it just right it will roll away. A stable
equilibrium would be at the low point of a valley. We can put the ball anywhere in the valley and
it will eventually just stop at that point anyway.
We have an excellent tool for determining the stability of equilibrium solutions in autonomous
differential equations. Remember autonomous means that the function on the right side of 2.20
only depends on y. An equilibrium solution C to an autonomous differential equation is stable if
d
f (y)
<0
(2.21)
dy
y=C
and unstable if
d
f (y)
> 0.
dy
y=C
(2.22)
If, when doing this calculation we get a result of 0, we can’t say whether it is stable or unstable.
Example Find all equilibrium solutions of
dy
= (1 − y 2 )
dx
(2.23)
and determine their stability.
Remember that we’re looking for constant values of y that make the right hand side zero.
0 = (1 − y 2 ) ⇒ 1 = y 2
(2.24)
Which has the solutions y = ±1. In order to determine their stability we want to look at the
derivative of the right hand side with respect to y, and figure out whether it is positive or negative
at these points.
d
(1 − y 2 ) = −2y
(2.25)
dy
When we substitute our equilibrium solutions into 2.25, we see that for y = +1 the derivative is
negative, which means that this equilibrium solution is stable, and when we substitute y = −1 in,
2.25 is positive.
If you want more practice with equilibrium solutions, and the stability theorem, try questions 11
to 14 and 21 to 24 in section 8.3 of your textbook, or try some stuff out with the logistic model
and it’s friends.
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12
Exam Review
2.4
Things you should know
Math 1LT3
The Logistic equation
The logistic equation, and all of its modifications, are a group of models used to model population
dynamics. The base logistic model is
P (t)
0
P (t) = rP (t) 1 −
(2.26)
K
r and K are parameters that can be changed to fit a specific situation. r is the rate at which the
population grows, while K is the carrying capacity, the maximum sustainable population within
an ecosystem. We should notice that if P = 0 then P 0 = 0 as well and the population will not
be able to grow. This is called the axiom of parenting, which makes sense, since if we have a zero
population, we can’t suddenly get a nonzero population; the new members of the population would
need to have parents. There is another equilibrium solution, and that is at the carrying capacity K.
This also makes sense since we expect the population to grow if it is below the carrying capacity,
and expect it to shrink if it is above. If you check, you can see that the equilibrium solution y = K
is stable, and the equilibrium solution y = 0 is unstable.
You may have noticed that the Logistic equation has a fatal flaw as a population model. It doesn’t
account for the fact that below a certain threshold population the population will eventually die off
due to inbreeding or a lack of suitable partners etc. This is called the Allee effect. The modified
logistic model deals with this fact by adding a new term
E
P
0
1−
(2.27)
P = rP 1 −
K
P
E is a minimum population that needs to be achieved, or else the population will die out, and is
called the Existential threshold. If P < E we say that the population is in existential crisis. You
can check that with this modification P = 0 is now a stable equilibrium, P = E is an unstable
equilibrium and P = K is a stable equilibrium.
The last of the logistic model’s variants that we looked at in class was a disease model. The disease
model keeps track of the percentage of the population infected with a certain disease, which we
call I.
I 0 = αI(1 − I) − µI
(2.28)
α and µ are parameters that describe the rate at which people are getting infected and the rate
at which people are recovering respectively. In this model, I = 1 is not an equilibrium since we’re
assuming that there will be some people recovering. The equilibria are I = 0 and I = 1 − αµ . The
stability of these equilibria is dependant entirely on the values of µ and α and will be the subject
of this section’s example.
Example People are being turned into zombies at a rate of 25% of people per year. After
Someone discovers a cure, we are able to help 10% of people per year, but they get no permanent
immunity. If we started with 50% of the population infected, will we be free of zombies eventually?
If not, what’s the maximum percentage of the population that will be infected?
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For this problem, we want to use the disease model. We have a rate of infection (.25) and a rate
of cure (.10) along with an initial condition. We need to check two things, if the “no zombies”
equilibrium (I = 0) is stable, and if it is, what side of the other equilibrium we are starting on.
If we’re on the far side of the other equilibrium, we won’t be able to pass it and we’ll forever
have zombies. If we are below the other equilibrium then we should converge to a world without
zombies.
First we should substitute our parameter values into the equation.
I 0 = 0.25I(1 − I) − 0.1I,
I(0) = 0.5
(2.29)
We want to check, using the stability theorem whether or not the equilibrium I = 0 is stable.
d
(0.25I(1 − I) − 0.1I)
= 0.25 − 2I − 0.1|I=0 = 0.15
(2.30)
dI
I=0
Since the derivative is greater than zero, that tells us that the equilibrium is unstable. This is not
good news for the world. On the bright side, we can still figure out what percentage of the world
will be zombies at its most infected. The other equilibrium point is at
I =1−
0.1
= 0.6
0.25
(2.31)
and since we start with a percentage of the population less than this infected, we can conclude
that this will be the maximum percentage of the population that are zombies since solutions to
autonomous differential equations don’t cross equilibrium solutions
♦
If you want more practice with the logistic equation and its friends, check out problems 39 and 40
in section 8.3, or 19 to 24 of section 8.2 in your textbook.
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Calculus of Several Variables
This section covers things from the middle section of the course, calculus in several variables. There
will be roughly 35 marks on the exam taken from this section. Again each subsection includes at
least one example with my thought process written out.
3.1
Domain and Range
Recall that the domain of a function is the set of acceptable input values for the function, and
that the range of a function is the set of possible output values for a function. In a function of
several variables this is no different. In most cases, you should be able to determine the domain
by recognizing what can go wrong with the function when inputing values. Determining the range
can be a little trickier, but you should be able to guess based off of what would happen for a
similar function of one variable.
Example 1
Find the domain and range of
f (x, y) = ln(1 − ex
2 +y 2
)
(3.1)
When determining the domain of the function, its good to start at the outside and work our way
in. The outermost part of this function is the logarithm. We know that we can only put positive
things into a logarithm,
0 < 1 − ex
2 +y 2
(3.2)
which implies that we need the exponential term to be less than one. Remember that an exponential
is less than one only if the exponent is negative. So we gain a new requirement
x2 + y 2 < 0
(3.3)
Since both x and y are squared here, this can’t ever happen. This tells us that there are no
acceptable input values for 3.1. So the domain and range for this function are both empty.
D = ∅,
R=∅
(3.4)
Example 2 Find the domain and range of
f (x, y) = arctan(x2 + y 2 − 1)
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(3.5)
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Like the last example, we want to work our way from the outside in. Arctan is a pretty nice
fuction, in that it will accept pretty much any real value, so since the innards of the arctan is a
polynomial, our domain is not restricted at all.
D = −∞ < x < ∞, −∞ < y < ∞ = {(x, y) ∈ R2 }
(3.6)
In order to find the range, we want to do the opposite from the domain. We want to work from
the inside out. The innermost part of this is the polynomial
x2 + y 2 − 1 = u
(3.7)
Since the x and y are both squared in this, no matter what we input, this polynomial will be
greater than or equal to negative one. So the smallest thing we can put into the arctan will be
minus one. If we turn our attention to the graph of arctan(u), our range should be every value it
takes for u ≥ −1. We can also notice that arctan is an increasing function and has a horizontal
arctan(x)
1.5
1
0.5
–10
–8
–6
–4
–2
2
4
6
8
10
x
–0.5
–1
–1.5
Figure 1: The graph of arctan(u)
asymptote at π2 . Since arctan(−1) = − π4 we can conclude that the range or our function is
R = z ∈ [− π4 , π2 ) = {z ∈ R : − π4 ≤ z < π2 )}
(3.8)
For more problems on domain and range try problems 7-12 in section 1 of the calculus of several
variables module.
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Things you should know
Math 1LT3
Limits in higher dimensions
We treat limits and continuity in higher dimensions basically the same as how we treated them
when we were doing calculus with a single variable. There are a few complications though. When
we were doing calculus with one variable, a limit existed when the function needs to approach
the same value from the left as from the right, or from both directions. Doing calculus in several
variables we need the same thing, but we have much more than just the two directions. We actually
have an infinite number of directions, although some of them differ by a tiny amount. Since there
are so many directions, its impossible to check them all, so we rely on some theorems to guide us
through the existence of limits.
i If the limit approaches two different values from two different directions, it does not exist
ii If the function is continuous at that point, then the limit at that point exists.
Remember that a function is continous at a point if the limit exists and is equal the functions
value at that point. That being said, we will generally only ask you to show that a limit does not
exist, or we will give you a function and ask whether it is continous of not. We could also possibly
ask you to explain why you can’t tell us if a limit exists or not.
Example 1 show that
x2 y
(x,y)→(0,0) 2x4 + y 2
lim
(3.9)
does not exist.
Since we’re asked to show that this limit does not exist, we need to find two different paths to
(0, 0) that have different end values. Let’s start by travelling along the x axis (x, 0). Since we’re
choosing to test this for y = 0 as x → 0 we can substitute y = 0 into the function
x2 0
=0
(x,y)→(0,0) 2x4 + 0
lim
(3.10)
We now need to find a second pathway that gives us a different result. We should notice that if
we put substitute y = x2 into the function, we will have x4 on top and bottom, so that seems like
a good path to choose.
(x,x
Since
1
3
x2 · x2
x4
1
=
lim
=
4
2
2
4
2
)→(0,0) 2x + (x )
(x,x )→(0,0) 3x
3
lim
2
6= 0 we have shown that the limit does not exist
(3.11)
♦
Example 2 show that
f (x, y) =
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cos(x2 + y 2 ) (x, y) 6= (0, 0)
1
(x, y) = (0, 0)
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is continuous
This is the second type of problem. Since cos(x2 + y 2 ) is continous for all values of x and y, we
know that the limit as (x, y) → (0, 0) exists, and is equal to cos(0) = 1. Since in the definition
of f (x, y) we see that this is indeed the function value at (0, 0), the function must be continous
there.
♦
If you want more practice with limits, try problems 8 to 21 in section 3 of the functions of several
variables module
3.3
Partial derivatives
Partial derivatives are an important tool in calculus of several variables, being used in determining
the gradient, in optimization and in determining directional derivatives.
Like derivatives for one variable, a partial derivative is defined as the limit of a difference quotient
f (x + h, y) − f (x, y)
,
h→0
h
fx = lim
f (x, y + h) − f (x, y)
h→0
h
fy = lim
(3.12)
In practice we find partial derivatives by pretending all the other variables are constant. What I
mean is that if you are taking the partial derivative with restpect to x, terms with y and z will
just come along for the ride. Some examples should help.
Example Find all the partial derivatives of
f (x, y, z) = x2 y + z sin(xy) − cos(z) + y z
(3.13)
We’ll start with the partial derivative with respect to x. We can treat any term that doesnt have
an x in it as if it were constant, and any y or z multiplied with an x as if they were just constants.
fx (x, y, z) = 2xy + zy cos(xy)
(3.14)
where we had to do chain rule on the sin function. The last two terms became zero since we were
treating them as constant.
fy (x, y, z) = sin(xy) + sin(z) + y z ln(y)
(3.15)
fz (x, y, z) = x2 + zx cos(xy) + zy z−1
(3.16)
♦
If you want more practice with partial derivatives, try 13 to 20 section 4 of the functions of several
variables module
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3.4
Things you should know
Math 1LT3
Critical Points
Like calculus with a single variable, we can use derivatives to find critical points, which will be
maxima, minima or saddle points. A critical point is any point for which all the partial derivatives
are zero. Also like calculus with a single variable, we can determine what type of critical point we
have by looking at the second derivatives. With calculus in several variables we need to look at
the second partial derivatives fxx , fyy , fxy . We have a result we can look at to determine which
type of critical point we have. We look at what is called the descriminant, although the name isn’t
important, so we’ll just call it D(x, y).
D(x, y) = fxx (x, y)fyy − (fxy )2
(3.17)
Remember, a point (a, b) is a critical point if fx (a, b) = 0 and fy (a, b) = 0. If D(a, b) < 0 then
this is a saddle point. If D(a, b) > 0 and fxx > 0 then it is a minimum, and if fxx < 0 then it is a
maximum. If D(a, b) = 0 then we can’t be sure what type of point this is and we could use some
other method, although we don’t really have any of those at the moment.
Example find and classify all critical points of
f (x, y) = (x − y)exy
(3.18)
The first thing we need to do is find our critical points. So we should be taking partial derivatives
of f (x, y).
fx (x, y) = exy + y(x − y)exy ,
fy (x, y) = −exy + x(x − y)exy
(3.19)
When we set these to zero we should get two equations for y in terms of x. By setting these
two equations equal to each other we will see where these two equations intersect (think back to
graphing with one variable). We should notice that we can factor out exy in both equations, this
will make life a little easier.
0 = exy (1 + yx − y 2 ),
0 = exy (−1 + x2 − xy)
(3.20)
Since exy is never zero, we can disregard it and focus on the polynomial terms. The second equation
is much easier to solve for y, so lets do that, and use the result in the first equation to solve for x.
−1 + x2
0 = −1 + x − xy ⇒ y =
x
We can now substitute this back the first equation in 3.20.
2
−1 + x2
1+
x−
x
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−1 + x2
x
(3.21)
2
=0
(3.22)
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We should multiply this equation through by x2 to kill off the x2 in the denominator of the last
term. This gives us the equation
2x2 + x4 − (1 − 2x2 + x4 ) = 0 ⇒ 4x2 − 1 = 0
(3.23)
which has the solutions x = ± 21 . We can resubstitute these back into the equation for y, and get
the critical points
(3.24)
( 21 , − 32 ) and (− 12 , 32 )
Now that we have our critical points, we want to figure out what type of critical points they are.
In order to do this, we need to figure out what our second partial derivatives are. We pretty much
do the same thing as with the first partial derivatives. for fxx we take the partial derivative of
fx with respect to x, for fyy we take the partial derivative of fy with respect to y, and for fxy we
either take the partial derivative of fx with resepect to y or we take the partial derivative of fy
with respect to x, it doesnt matter which.
fxx = exy (y +y 2 x−y 3 +y),
fyy = exy (−x+x3 −x2 y −x),
fxy = exy (2x+yx2 −y 2 x−2y) (3.25)
We need to figure out what these things are at our critical points. For the first critical point
( 12 , − 32 )
3
3
3
fxx = e− 4 ( 32 ) fyy = e− 4 (− 21 ) fxy = e− 4 ( 25 )
(3.26)
Which we substitute into the descriminant D(x, y)
D( 21 , − 23 )
3 3
= − e− 2 −
4
2
3
5
e− 2 < 0
2
(3.27)
This tells us that ( 21 , − 32 ) is a saddle. It should be safe to assume by symmetry that the other
critical point is also a saddle.
♦
If you want more practice with critical points, try problems 12 to 21 in section 10 of the functions
of several variables module
3.5
Gradient and Directional Derivatives
The gradient is a vector that points in the direction of the largest positive rate of change. It is
also a derivative, so the direction of the largest rate of change is in the immediate vicinity the
point we’re considering, meaning that we don’t really care if there are bigger hills far away. The
gradient vector is made up of the partial derivatives.
∇f (x, y) = (fx (x, y), fy (x, y))
(3.28)
Directional derivatives are scalar quantities (meaning they are just numbers). I think of them
as a sort of measure to how much the direction given points up hill. If they point up hill, we
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should get a positive number, if they point along the hill, neither up nor down, we should get 0,
and if they point downhill we should get something negative. Since we defined the gradient as a
vector pointing in the direction of greatest increase, we expect directional derivatives to be at their
largest when the direction is the same direction as the gradient vector. When finding directional
derivatives, we always use a unit vector, and we take the dot product of this unit direction vector
with the gradient.
D~u f (x, y) = ~u · ∇f (x, y)
(3.29)
Example Find the angle between ~v = 3ˆi − 4ˆj and the direction of largest increase of f (x, y) =
x2 + 2xy + y 2 at (2, 3).
Our first goal here is to find the gradient vector. Although it isn’t exactly required, we can factor
our fuction, and make our derivatives slightly simpler
f (x, y) = (x + y)2
(3.30)
∇f (x, y) = (2(x + y), 2(x + y))
(3.31)
specifically though, we are interested in the gradient at (1, 1)
∇f (2, 3) = (10, 10) = 10ˆi + 10ˆj
(3.32)
Our next step is to normalize the direction vector we’ve been given. This means that we want to
find a vector that is in the same direction as ~v , but has a magnitude of 1. To find this, we need to
divide ~v by the magnitude of ~v .
~v
(3.33)
~u =
||~v ||
In this case, the magnitude of ~v is
√
||~v || = 32 + 42 = 5
(3.34)
so the unit vector that points in the same direction as ~v is
3
~u = ˆi −
5
4ˆ
j
5
(3.35)
Now we just need to substitute all we have into 3.29 and use something we know about dot products
to find our final answer
D~v f (2, 3) = ~u · ∇f (2, 3) = ||~u||||∇f (2, 3)|| cos(θ)
(3.36)
θ is the angle between ~u and the ∇f (2, 3), so once we’ve solved for that we have our final answer. Recall, we compute a dot product by multiplying the two ˆi components together, the two ˆj
components together, and then adding the results together.
√
√
70 = 200 cos θ ⇒ θ = arccos 7 2 ≈ 2.28radians
(3.37)
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♦
If you want more practice with directional derivatives and the gradient, try problems 16 to 21 and
24 to 27 in section 9 of the functions of several variables module
3.6
Differetiability
We say that a function is differentiable at a point (a, b) if the partial derivatives fx and fy exist
and are continuous in a neighbourhood of (a, b). What this means is that the partial derivatives
need to exist and be continuous at (a, b) and that (a, b) can’t be at the edge of the domain of
definition. Another way to think about this is that you must be able to draw a circle around the
point (a, b) as small as you like, and the partial derivatives have to exist and be continuous in this
circle.
Example Is the function
f (x, y) =
sin(x + y) (x, y) 6= (0, 0)
0
(x, y) = (0, 0)
differentiable at (0,0).
First thing to check is if the function is continous at (0,0). If it isn’t then there is no way that it
can be differentiable there. A quick substitution shows us that the function is in fact equal to its
limit at that point.
The next step is to check the partial derivatives, and
cos(x + y)
fx (x, y) =
0
cos(x + y)
fy (x, y) =
0
whether they’re continuous.
(x, y) 6= (0, 0)
(x, y) = (0, 0)
(x, y) 6= (0, 0)
(x, y) = (0, 0)
In this case, the derivatives are not continuous at (0,0) since
lim
cos(x + y) = 1
(3.38)
(x,y)→(0,0)
which is not what the partial derivatives equal. Therefore the function is not differentiable
♦
If you want more practice with differentiability, try questions 11 to 14 and 37 in section 5 of the
functions of several variables module
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Probability and Statistics
This section contains things from the last section of the course, on probability and statistics. There
will be about 35 marks on your final exam taken from this section. Again, at least one example in
each subsection.
4.1
Set Theory
A Set is a collection of elements. Elements don’t necessarily have to be numbers, they can be
events, objects, letters, or really anything at all. For example R is the set of all real numbers.
{1, 2, 3, 4, 5}
(4.1)
Is a set containing 5 elements, the numbers 1 through 5.
There are 4 basic set operations that we need to worry about in this course. If A and B are two
sets then
i The union of two sets A ∪ B is the set of all things in A, in B or in both A and B
ii The Intersection of two sets A ∩ B is the set of all things that is in both A and B.
iii The complement of a set Ac is the set of all things that are not in A, but are in the universal
set
iv The subtraction of a set A \ B is the set of all things that are in A but not in B
Example Find the Intersection and Union of
A = (0, 1]
B = [ 21 , 2)
(4.2)
For the intersection we are looking for the things that are in common. It will be easiest to see if
we draw this on the number line.
0
1
( [ ]
2
)
Figure 2: Sets A and B on the number line, A is red and B is blue
The intersection of the two sets is exactly what it sounds like, the place where the two sets
intersect, or the set of things they have in common. So we can see that they overlap between 21
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and 1, including both end points. so the intersection is [ 12 , 1]. The union is also pretty easy to see,
it is just the whole of the two sets. The union then includes all the points between 0 and 2, but
does not include 0 or 2 because of the open brackets.
A ∪ B = (0, 2),
A ∩ B = [ 21 , 1]
(4.3)
♦
If you want more practice with set theory, try problems 11 to 23 in section 3 of the probability
and statistics module.
4.2
Discrete Probability
Probability is probably one of the least intuitive things that we will face in mathematics, and
we encounter it on an almost daily basis. The probability of an event happening is the expected
frequency of occurance during a number of trials. If each event has an equal chance of happening,
then the probability of an event x is
p(x) =
number of ways that x can happen
number of things that can happen
(4.4)
So in essence, discrete probability is just counting, but to be fair, it is the hardest counting you
will ever have to do.
Example
In texas holdem poker, you are trying to make the best hand you can out of the two cards you
have in your hand and the five cards that are lying on the table. If you have the hand seen below
(figure 3), what are your odds of getting a full house when the final card is turned up? (A full
house is when you have 3 of a kind and a pair).
In order to get a full house we need that last card to either be a 7 or a Jack. There are two 7s
that we are not accounted for and 2 Jacks, wich means there are 4 ways that we can get what we
need. There are normally 52 cards in a deck, but since we know what 6 of them are, we are left
with 46 unknowns. So that means that our probability should be
p(F ullHouse) =
Number of 7s and Jacks
4
2
=
=
cards left
46
23
(4.5)
This is one of the probability calculations that professional poker players are very good at. Other
ones involve conditional probability and also probabilities involving what their opponents could
have.
If you want more practice with discrete probabilities, try problems 23 to 36 in section 3 of the
probability and statistics module
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Figure 3: This could be a full house, but what are the odds?
4.3
Conditional Probability
Sometimes the probability of an event occuring can change based on an outcome already having
occured. When probabilities depend on other events like this we call them conditional probabilities.
In many cases, conditional probabilities still involve counting things. Often the questions for these
will be phrased like “given A, what is the probability of B”. We write such probabilites as
p(A|B) =
p(A ∩ B)
p(B)
(4.6)
which reads as “The probability of A given B”. Sometimes conditional probability can simplify
the problem, other times it can make it wildly more complicated. If the events are independent
then the probability will be
Example A study found that 15% of female huskies have blue eyes, while 25% of males do. If
there is an even split between female and male huskies in the population, what is the probability
that a given blue eyed husky is a female.
One important thing to notice here is that set of female huskies and the set of female huskies form
a partition of the set of huskies. This implies that we can use Bayes’ theorem. Recall that Bayes’
theorem says that if there is a set of Ei that form a partition of the sample space then
p(A|Ei )p(Ei )
p(A|E1 )p(E1 ) + p(A|E2 )p(E2 ) + ...p(A|En )p(En )
p(Ei |A) =
(4.7)
In our case, the partition is made by only two possibilities, male and female.
p(Female|Blue Eyes) =
p(Blue Eyes|Female)p(Female
p(Blue Eyes|Female)p(Female + p(Blue Eyes|Male)p(Male
(4.8)
Since we know that proability of having a blue eyed husky if the husky is a girl, and we know
the proportion of females in the population this is not so bad. We can actually just move on to
substituting all the things into Bayes’ theorem.
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p(Female|Blue Eyes) =
Math 1LT3
(.15)(0.5)
= 0.375
(0.15)(0.5) + (0.25)(0.5)
(4.9)
♦
If you want more practice with conditional probabilities, try problems 25 to 30 in section 4 of the
probability and statistics module
4.4
Binomial and Poisson Distributions
The binomial distribution is used when we are dealing with an experiment with only two possible
outcomes. That means that if A and B are the outcomes, then A and B form a partition of the
sample space.
p(A) + p(B) = 1
(4.10)
We generally call A a success and B a failure. The Binomial distribution is defined as
n
k
n−k
b(k, n, p) = p (1 − p)
k
In this equation, n is the number of trials, k is the number of successes and p is the probability of
a success.
n!
n
=
k
k!(n − k)!
is n choose k, which means the number of ways to pick k things out of n total things. The binomial
distribution has a mean of p and a variance of p(1 − p)
The poisson distribution is used when we have independent events and the probability is the same
over every interval of the same size. The Poisson distribution is
p(k; λ) =
e− λλk
k!
(4.11)
Where k is the number of occurances and λ is the expected nuber of occurances. The poisson
distribution has a mean of λ and a variance of λ.
Example 1
If the odds of winning roll up the rim are 1 in 5, what is the probability of winning at least once
with 10 coffees?
If the words “At Least” don’t stir up something deep within you, it should. There are really two
ways to go about this question, the long arduous awful way, and the quicker easier way. I’ll show
you the quicker easier way. Since there are only two possibilities, and we were not given any sort
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of interval, we want to use the binomial distribution for this problem. We are given the probability
of success, which is 0.20 ( 15 ). The probability of at least one win is the opposite of having lost all
times, so it is much easier to take 1 minus the probability of losing all than counting all the ways
we could win at least once, also 10 choose 0 is 1, so that’s nice. Therefore the equation we want is
b(k ≥ 1, 10, 15 ) = 1 − ( 15 )0 (1 − 51 )10 = 1 −
So about an 89.3% chance that we win at least one time.
410
≈ 0.893
510
(4.12)
♦
Example 2 The game board of Battleship is made up of a 10 × 10 grid, 17 squares of which
are covered by ships. Each player takes turns choosing squares to fire upon, if one of the squares
they choose is occupied by a ship, it is a hit, otherwise it is a miss, a ship is sunk if it is hit on
everysquare that it occupies. While a player has 5 ships remaining they may make 5 guesses per
turn, 4 when they have 4 left etc. What is the probability that you hit at least one ship on your
first turn?
In this problem we have an all or nothing scenario, like what’s required for the binomial distribution
to apply, but we are also given a probability over an interval (albeit in disguise), which is required
for the poisson distribution. It is actually possible to do this question with either, but much easier
for Poisson. This is usually the case when the sample size is pretty large. The probability over an
17
= 0.17 ships per box. The words “At Least one” should
interval comes from expecting E(x) = 100
indicate to us that it is more efficient to find 1 minus the probability of not hits, and since we’re
firing upon 5 boxes at once, we actually want to find out the expected number of ships in 5 boxes,
or µ = 5E(x) = 0.85. With this, and searching for the complement of 0 successes we are ready to
plug everything into the distribution.
e−0.85 0.850
≈ 0.5726
(4.13)
0!
So we have about a 57% chance of hitting something on the first shot, which is pretty good! ♦
p(x > 0, 0.85) = 1 −
If you want more practice with the binomial distribution, try problems 30 to 35 in section 10 of
the probability and statistics module
If you want more practice with the poisson distribution try problems 16 to 22 in section 12 of the
probability and statistics module
4.5
Continuous Probability
The main difference for dealing with continous random variables vs discrete random variables is
that we can no longer simply count things and add up the probabilities. There are an infinite
number of things to count, so we instead use something that has been developed to count up an
infinite number of little things, integrals. Dealing with continous probabilities we have one main
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tool that helps us to do everything, the probability density function or pdf for short. A pdf is a
function f (x) that satisfies two requirements:
f (x) ≥ 0 for all x
Z ∞
f (x)dx = 1
(4.14)
(4.15)
−∞
If a function does not satisfy both of these properties then it is not a pdf.
With continous random variables, we don’t look at the probability at exactly one x value, but
always instead over a range of x values. We write
Z b
f (x)dx
(4.16)
p(a ≤ x ≤ b) =
a
when we want the probability of x between a and b.
The mean of a continous random variable is
Z
∞
E(x) =
xf (x)dx
(4.17)
−∞
and the variance is
Z
∞x2 f (x)dx − (E(x))2
V ar(x) =
(4.18)
−∞
Example 1
Find k so that
f (x) =
kxe−4x
0
2
≥0
otherwise
is a probability function, or explain why no such k exists.
When faced with a problem like this, we should find out what k needs to be so that the integral
over the entire real line is zero.
Z ∞
Z 0
Z ∞
2
1=
f (x)dx =
0dx +
kxe−4x dx
(4.19)
−∞
−∞
0
The first integral is zero, while the second one can be done with the substitution 4x2 = u, which
means that du = 8xdx
∞
Z ∞ −u
e
−k −u 1=k
dx =
e (4.20)
8
8
0
0
When we substitute in our limits of integration we see that the upper limit goes to zero, while the
lower one gives just k/8.
k
1=
⇒ k=8
(4.21)
8
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Our final thing to check is that for all values of x, f (x) ≥ 0 with k = 8. This is indeed the case,
so this is a pdf if k = 8.
♦
Example 2 Find the mean and variance of
f (x) =
1
√
2 x
0
0<x≤1
otherwise
remember that we find the mean by first multiplying f (x) by x and integrating over the space that
it is defined on, and that we need the mean to find the variance. Since f (x) is defined as zero for
most x, its integral will be zero on most of x, so we can just integrate from 0 to 1.
3 1
Z 1√
Z 1
3
1
x
3x 2 =
dx =
(4.22)
E(x) =
x √ dx =
2
4 4
2 x
0
0
0
And the variance
Z
V ar(x) =
0
1
1
x2 √ dx − (E(x))2 =
2 x
Z
0
1
3
x2
dx −
2
2
3
9
11
5
=
= −
4
4 16
16
(4.23)
♦
If you want more practice with continuous random variables, try questions 17 to 23 in section 13
of the probability and statistics module.
4.6
The Normal Distribution
The normal distribution, or Gaussian distribution as it is sometimes called (also bell curve) is a
very common continuous probability distribution function. It has the form
(x−µ)2
1
f (x) = √ e− 2σ2
σ 2π
(4.24)
Despite its ugly appearance, it is actually very nice to deal with. All of the important quantities
we want are visible in the equation.
i The mean E(x) = µ
ii The Variance V ar(x) = σ 2
iii The standard Deviation Stdev(x) = σ
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It does not, however, have an antiderivative that we can write down, so when we are looking for the
probability of an event x happening between a and b we have to resort to numerical computations
that were done by a computer. In general you will have a chart for the standard normal distribution
x2
1
f (x) = √ e− 2
2π
(4.25)
which is the same as the normal distribution but with µ = 0 and σ = 1. In order to use this chart
properly we need to know how to convert between values on the chart, and values we have been
given. The following formula comes from the substitution you would do in the integral to convert
from the normal distribution to the standard normal distribution;
Z=
X −µ
σ
(4.26)
A reasonable number of problems will come from knowing how to use these conversions
Example
The wingspan of a blue jay is normally distributed with a mean of 39cm and a standard deviation
of 3 cm. What is the probability that a randomly chosen bluejay has a wingspan of greater than
42 cm?
The key to recognizing that the words “normally distributed” mean we need to use the normal
distribution (another common phrase to look out for is “approximately normal”). We have been
given the mean and standard deviation, which are the tools we need to make the conversion to the
standard normal distribution where we can use the chart (not included in this document). µ = 39,
σ = 3 and X − 42
(42 − 39)
=1
(4.27)
Z=
3
therefore we are looking for z > 1 in the chart. Since the chart gives us the value of the cumulative
distribution function at that point (the probability of less than 1), we want to take 1−F (1), which,
from looking at the chart gives us
p(wings > 42) = 1 − F (1) ≈ 0.159
(4.28)
♦
If you want more practice with the normal distribution, try questions 3 to 17 in section 14 of the
probability and statistics module
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