# Get Math Help from Me

My name is Stacey and I live in Corvallis, Oregon with my two kids.  I am passionate about math tutoring. Here are ways you can get help from me:

• If you have a quick question, I can help you on Twitter.
• I tutor people privately both in person and online.  See my Facebook tutoring profile page to learn more about me.
• I can tutor you on Skype — if you learn well from me, you can hire me.
• Browse through examples and useful tips here, or post a new question for me to answer.

You can contact me via Twitter or Facebook if you are interested in setting up a tutoring session:

# Venn Diagram Word Problem

Q:  Superburger sells hamburgers with the choice of ketchup, mustard and relish. One day they sold 256 hamburgers; 140 had mustard, 140 had ketchup, 84 had ketchup and relish, 62 had mustard but no relish, 68 had ketchup and mustard, 38 had all three condiments and 20 had none.

(a) The number sold with relish only is?

(b)The number sold with no relish is?

Read More »

# Integration Example: Disk (Washer Method) vs. Shell Method

Here is an example to help you understand and visualize the difference between the disk method and the shell method.  I will do the same problem twice: once using disk method and once using shell method.  Keep in mind: not all problems are equally as easily solved with both methods — that’s why we have multiple methods!  Some problems may be easy to do with the shell method but nearly impossible with the disk method, or vice versa.

If you haven’t read the blog posts that discuss the basic differences between the disk method and the shell method, read those first:

To get the most out of this problem, grab a pen and paper and do the problem along with me: draw each picture and write each equation.

Q:  Find the volume obtained by rotating the area contained by y = √(x), y = 0, and x = 4 around the y-axis.

A:  First, draw a picture of the area that is to be rotated.

Visualize the rotation.  Actually give the shape a name to help solidify it in your mind.  I think of this shape as a cinder cone:

Obviously this isn’t a perfect visual match, but giving a real image to an abstract shape really helps with mentally processing the steps.

Method 1:  Disk (washer) Method.  Remember, the disk and washer method are the same thing.  In the disk method, we visualize stacked circles (or pancakes, as I like to say).  The washer method is the same stacked pancakes with holes (like washers or donuts).

I like to complete every problem with the same thought process.  Follow these steps for all volume/integration problems and you will get the hang of it!

Step 1.  Determine if this is a dx or a dy problem.

Which way are we stacking our pancakes?

Pancakes are being stacked vertically, so this is a “dy problem”.  This means the limits of integration and the equations used will all be in terms of y.

?? ____?____  dy

Step 2.  Find the limits of integration.

Where does the pancake stacking start?  Where does it end?  Remember — in terms of y since this is a dy problem.

Pancakes start at y=0 and end at the place where y = √(x) and x = 4 intersect (the top of the cinder cone).  These bounds intersect at the point (4, 2).

So, pancake stacking starts at y=0 and end at y=2.

02 ____?____  dy

Step 3.  Find the equation of the areas that are being stacked.

We’ve pre-decided that we are going to use the disk (washer) method.  There are definitely washers happening, because there is a big outer circle minus a hole (to create the cinder cone).  So, we need to stack outer pancakes minus inner pancakes (big circles minus small circles):

Area = πR2 – πr2

Let’s start with the area of the big circles, which I’ve called πR2.  What is the radius, R, of the big circles?  Is it changing throughout the problem or is it constant?

The radius of the larger circles is constant.  The larger circles have a radius of 4 throughout the entire cinder cone:  R = 4

Now look at the smaller circles.  Is the radius, r, constant or changing?

Notice the radius of the smaller circles is changing.  This radius is a horizontal distance, starting from the y-axis and moving on out.  The radius is x: r = x.

BUT WAIT…. remember, this is a dy problem.  All equations need to be in terms of y!  So, using r = x will not help.  We need to find a way to represent x in terms of y.  Fortunately, we have the equation to help: y = √(x), so, x = y2

Good.  So, R = 4 and r = y2

Our equation is now complete and ready to solve:

02 π(4)2 – π(y2)2 dy

I will leave the solving to you… but as a final answer I get 128π/5 (approx = 80.425)

Now… Are we ready to solve this same problem using the shell method?

Method 2:  Shell method.  Instead of visualizing stacked pancakes to create our cinder cone, we will visualize stacked “Russian Dolls” (cylinders).  We are going to stack these cylinders so tightly together that the lateral area of the cylinders will stack to create volume.

Back to our steps:

Step 1.  Determine if this is a dx or a dy problem.

The dolls are being stacked inside of each other and on outward, expanding along the x-axis.  This is a “dx problem”.  This means the limits of integration and the equations used will all be in terms of x.

?? ____?____  dx

Step 2.  Find the limits of integration.

Where does the cylinder stacking start and where does it end?  In other words: what is the radius of our smallest cylinder and what is the radius of our largest cylinder?

The smallest cylinder has a radius of x=0 and the largest cylinder has a radius of x=4

04 ____?____  dx

Step 3.  Find the equation of the areas that are being stacked.

We are stacking lateral areas of a cylinder, which has equation: 2πrh (r is the radius of the cylinder, h is the height).

Let’s figure out the radius, r, of a random cylinder/doll in our shape.  The radius does change, so it is a variable.

But, it does not necessarily depend on the functions.  The radius is simply an x-value that continues to grow until we hit the wall of x = 4.

So, r = x.

Now, let’s figure out the height, h, of a random cylinder/doll in our shape.  The height is changing and it is definitely affected by the functions.  The height is the y-value of the bounding function: h = y…. BUT WAIT… remember, this is a dx problem — no y’s allowed. So, use the equation: y = √(x).  Therefore, h = √(x)

So, we have:

04  2π(x)(√(x)) dx

I will leave the solving to you… but as a final answer I get 128π/5 (approx = 80.425) — no coincidence that this is the same answer obtained by method 1!

Same shape, two methods, same answer.  Phew.

# When to use Disk Method versus Shell Method, Part 2

To start, read: When to use Disk Method versus Shell Method, Part 1 to get a general visual.  This is a little more detailed:

Volume should be thought of as infinitely stacked area. In the disk method, you are infinitely stacking circles (think pancakes). In the shell method, you are infinitely stacking lateral surface areas of cylinders (think “Russian Dolls” that stack inside of each other)

The Disk Method: Since you are stacking pancakes, the general formula that you will be integrating is pi*r2. If the radius of each disk is changing throughout the shape, the radius, r, will be a function, dependent on either y or x, depending on how you are rotating.

The Shell Method: Since you are stacking lateral areas of cylinders, the general formula that you will be integrating is 2*pi*r*h (lateral area of cylinder formula). The radius, r, will be a simple function involving an x or a y. The height, h, will depend on the functions that are being rotated.

Now that you have an understanding of the concept, view this example… I solve a volume problem using the disk method.  Then, I solve the exact same problem using the shell method:   Integration Example: Disk Method vs. Shell Method

# When to use Disk Method versus Shell Method, Part 1

Q:  When should I use a disk / washer method versus a shell / cylinder method for integration?

Answer:

First, the visual difference:

The disk / washer method is used when you can think of your shape as “stacked pancakes” (the washer method is just removing any center “holes” from these pancakes).

The shell / cylinder method is used when you can think of your shape as “stacked Russian Dolls” (you know, those dolls that stack inside of each other, and you keep opening them up to find a small doll inside, etc..)

A mathematical difference:

Example 1:  Rotate around the x-axis to create a volume.

If you rotate around the x-axis and use the disk method:  You will be stacking your pancakes horizontally (with respect to x).  Therefore, your limits and functions of integration will be in terms of “x” (it will be a “dx” problem).  The radius of your disk will need to be in terms of x.

If you rotate around the x-axis and use the shell method:  You will be stacking your cylinders vertically.  Your limits and functions of integration will all be in terms of “y” (it will be a “dy” problem).  The height of your cylinder will need to be in terms of y.  The radius of your cylinder will most likely just be “y” itself – though not guaranteed.

Example 2:  Say you are rotating an area around the y-axis to create a volume.

If you rotate around the y-axis and use the disk method:  You will be stacking your pancakes vertically (with respect to y).  Therefore, your limits and functions of integration will be in terms of “y” (it will be a “dy” problem).  The radius of your disk will need to be in terms of y.

If you rotate around the y-axis and use the shell method:  You will be stacking your cylinders horizontally.  Your limits and functions of integration will all be in terms of “x” (it will be a “dx” problem).  The height of your cylinder will need to be in terms of x.  The radius of your cylinder will most likely just be “x” itself – though not guaranteed.

Which method to pick?

Take into account the visual of your shape.  Does it look like pancakes stacked on top of each other or cylinders nested inside of each other?

Take into account the math.  Is it easier to put the functions in terms of y or in terms of x?  That will guide you on which to pick (depending on your axis of rotation).

This was a start to understanding.  Next read: When to use Disk Method versus Shell Method, Part 2.

Also check out this example…  I do one problem using the disk method.  Then, I do the same problem using the shell method:  Integration Example: Disk Method vs. Shell Method

# Determining the End Behavior of a Function

How do you determine the end behavior of a function?  And, what does this mean?

When looking at a graph, the “end behavior” is referring to what is happening all the way to the far left of the graph and all the way to the far right of the graph.  Your goal is to analyze the y-value (height) of the function when x is really large and negative, and then again when x is really large and positive.  What is the pattern on each end?  What is the “end behavior”?

Notationally, we are thinking:

1. As x → -∞, y → ?
2. As x → +∞, y → ?

OK, so let’s try this on a polynomial example:

Q:  What is the end behavior of the function y=5x3+7x2-2x-1

A:  OK.  Let’s look at the left end behavior first:

As  x approaches -∞, what is the function (y-value) doing?

Imagine x=-1000000 (some super large and super negative number, like the idea of -∞), we have:

y=5(-1000000)3+7(-1000000)2-2(-1000000)-1

Don’t do the actual math.  Just think:

Is this number large or small?

Is it positive or negative?

I can look at the x3 term and see that it dominates this function. x2 and x are small peanuts compared to x3. So, in reaity, in polynomials, I can focus on the term of the largest degree:

y=5(-1000000)3+7(-1000000)2-2(-1000000)-1

y=5(-1000000)3

This number gives y = negative and super large.

So, I can jump to conclusions here…

As x → -∞, y → -∞

(As x approaches negative infinity, y approaches negative infinity).

Now, let’s look at the right end behavior:

As  x approaches +∞, what is the function (y-value) doing?

Imagine x=+1000000 (some super large and super positive number, like the concept of +∞), we have:

y=5(+1000000)3+7(+1000000)2-2(+1000000)-1

And, by the same reasoning, we can focus on the term of largest degree:

y=5(+1000000)3+7(+1000000)2-2(+1000000)-1

y=5(+1000000)3 = super large and super positive

So, as x → +∞, y → +∞

(As x approaches positive infinity, y approaches positive infinity)

Note: in this example, y behavior mimicked x behavior, this isn’t always the case!

# Visualizing limits vs values

Look at the function f(x) in orange below:

We are going to answer 4 questions about this graph.  They are all related to each other, but different questions.  Seeing the difference will help us sort out the difference between a function value and a limit.

Q1:  Find f(1)

Q2:  Find  limx→1 f(x)

Q3:  Find  limx→1+ f(x)

Q4:  Find  limx→1 f(x)

OK….. Try to answer these questions with what you know… Then continue reading to see the answers and explanations!

Read More »

# Basic concept of a limit

Here is a brief example on the concept of a limit:

Look at the function f(x) in orange below:

Question 1.  Find f(3)

Explanation of question 1: Find the value of the function when you plug in 3.  What is the height of the function at the exact moment when x=3?

Answer 1:  The function is undefined at x=3.  There is a hole when x=3.

So, f(3) is undefined.

Question 2:  Find limx→3f(x)

Explanation of question 2:  We are being asked to find what the function is doing around (but not at) 3.  What is happening to the path of the function on either side of 3?

In order to find limx→3f(x), we must confirm that limx→3+  f(x) and limx→3–  f(x) both exist and are equal to each other.

So, let’s find limx→3+  f(x).  What is happening to the function values as you approach x=3 from the right-hand side?  Literally run your finger along as if x=4, then x=3.5, then x=3.1.  What value is the function getting closer to?

The function is approaching a height of 4.

Let’s find limx→3–  f(x).  What is happening to the function values as you approach x=3 from the left-hand side?  Literally run your finger along as if x=1, then x=2, then x=2.9.  What value is the function getting closer to?

The function is also approaching a height of 4.

So:

limx→3+  f(x) = 4

limx→3–  f(x) = 4

Since, the left-handed limit at 3 and right-handed limit at 3 exist and are equal, this gives:

limx→3f(x) = 4.

So, to summarize, here are 4 different things we found.  They are related, but not necessarily the same:

f(3) is undefined

limx→3+  f(x) = 4

limx→3–  f(x) = 4

limx→3f(x) = 4

Are you ready to try one on your own? Click here! (Don’t worry, I’ll walk you through the solutions too)

# Limit Example

Q:  What is limx→2 (x-2)/|x-2|

A:  This question is not too bad if you know what the function (x-2)/|x-2| looks like graphically.  But, let’s say you don’t.

We are going to “talk our way” through this problem to help solidify the concept of a limit.

If you plug in 2 to the function, you are finding the value of the function when x=2.  This is important, and related, though it is not the limit.  This is even sometimes a skill used to help us find the limit, but it is still not the limit.  Sometimes the function value is equal to the function limit, which can also be confusing, but not all the time.  Let’s find the value of the function when x=2:

(2-2)/|2-2| = 0/0 = undefined.

Okay.  This function is undefined when x=2.  This means there is a hole, or an asymptote, or a break or a jump or some disruption in the continuity of the function.

So, let’s talk about the limit of the function as x approaches 2:

To find the limit as x approaches 2, we need to make sure the left-handed limit and the right-handed limit both exist and are equal.

I’m going to start with the right-handed limit (remember, this means we are getting closer and closer to 2 from the top-side).  We are going to do this by clever analysis:

When x=3, we get (3-2)/|3-2| = 1

When x=2.5, we get (2.5-2)/|2.5-2| = 1

When x=2.01, we get (2.01-2)/|2.01-2| = 1

See the pattern here?

When x>2, the function ALWAYS equals 1 (tricky function huh?).

So this tells us about the right-handed limit at 2:

limx→2+ (x-2)/|x-2| = 1

Now, let’s trying the left-handed limit at 2, again by analysis:

When x=1, we get (1-2)/|1-2| = -1

When x=1.5, we get (1.5-2)/|1.5-2| = -1

When x=1.99, we get (1.99-2)/|1.99-2| = -1

Again, a pattern: When x<2, the function ALWAYS equals -1.

This tells us about the left-handed limit at 2:

limx→2- (x-2)/|x-2| = -1

Since the right-handed limit at 2 ≠ the left-handed limit at 2, “the limit at 2” does not exist.  The right-handed limit must equal the left-handed limit to have a “complete limit” so to speak.  The concept:  If I walk the path from the left, and you walk the path from the right, we better think we are going to the same place.

Here is a picture of the function just for fun.  See what is happening as x approaches 2 from the right?  See what is happening as x approaches 2 from the left?  And, see what happens exactly at 2?

# One-sided Limit Example

Q:  Find the one-sided limit (if it exists):

limx→-1–     (x+1)/(x4-1)

A:  So we need to find the limit of this function (x+1)/(x4-1) as x approaches -1 from the left.  Remember, from the left means as x gets closer and closer to -1, but is still smaller.

The concept: What is happening to this function as x = -2, x = -1.5, x = -1.1, x = -1.0001, etc…

We test first and plug -1 into the function: (-1+1)/((-1)4-1) = 0/0

Whenever you get 0/0, that is your clue that maybe you need to do “more work” before just plugging in or jumping to conclusions.

So, let’s try “more work” — usually that means simplifying.  I see that the denominator can factor.  We have:

(x+1) / (x4-1) = (x+1) / [(x2-1)(x2+1)]

Let’s keep factoring the denominator:

(x+1) / [(x-1)(x+1)(x2+1)]

Now, it appears there is a “removable hole” in the function.  This means, we can remove this hole by reducing the matching term in the numerator with the matching term in the denominator:

(x+1) / [(x-1)(x+1)(x2+1)]

= 1 / [(x-1)(x2+1)]

Notice that hole exists when x = -1 (and it was removable! This is good news for us since we are concerned with the nature of the function as x approaches -1)

Now that we have removed that hole, let’s once again try to plug in -1 to see what we get.

1 / [(-1-1)((-1)2+1)]

1 / [(-2)(2)]

-1/4

So, after removing the hole at (x+1), we found the function value when x=-1 is -1/4.

Due to the nature of this function, this means:

limx→-1(x+1)/(x4-1) = -1/4

Since the limit exists as both the right-handed and left-handed limit, it follows that limx→-1–     (x+1)/(x4-1) must also be -1/4.  It ended up not being necessary that we only do a “one-handed limit analysis.”

# Integration Example: Sines an Cosines

Q:  ∫cos5(x)sin4(x)dx

A: Some books have tables and charts to memorize on how to integrate these types of problems: what to do if the power is odd on one but even on the other, etc (called reduction formulas)… Boring… Who has time, space or desire to memorize formulas? Let’s solve the damn problem.

First, I see that there are sines and cosines, and we know that one is the derivative of the other (more or less).  This tells me that u-substitution is likely going to come up.

Remember: whenever you see a function and its derivative present in a problem, you want to be thinking u-substitution!

I also know that by using the Pythagorean Identity, sin2(x)+cos2(x)=1, I can convert an “even number” of cosines to sines and vice versa.

So, now that I know u-substitution is most likely, I want to leave behind one function to be “du” and the rest should be converted to “u’s”.

Cosine is literally the “odd man out”.  There is an odd number of cosines, so I will leave one cosine behind to eventually serve as du and convert the rest like so:

∫cos5(x)sin4(x)dx

∫cos(x)*cos4(x)*sin4(x)dx

∫cos(x)*(cos2(x))2*sin4(x)dx

∫cos(x)*(1-sin2(x))2*sin4(x)dx

Now, I can do a fairly clean u-substitution:

Let u = sin(x)

Then, du = cos(x)dx

∫cos(x)*(1-sin2(x))2*sin4(x)dx = ∫ du*(1-u2)2*u4

Or, rewrite it: ∫(1-u2)2*u4du

Simplify and clean up:

∫(1-2u2+u4)*u4du

∫u4-2u6+u8du

Integrate to get:

(1/5)u5-(2/7)u7+(1/9)u9+C

And, substitute back in u = sin(x):

(1/5)sin5(x)-(2/7)sin7(x)+(1/9)sin9(x)+C

This is only 1 way to solve this problem.  I could’ve picked a different u-substitution and I could’ve likely picked different trig identities to help me along the way.