An example of an infinite series.

Damian has two identical cups, cup 1 and cup 2. Cup 1 is initially full of water, and cup 2 is initially empty. He begins a procedure in which he exchanges the water among the cups. For the first turn of the process, he pours half of the water in cup 1 into cup 2. Next, he pours half of the water in cup 2 back into cup 1, and then half of the total water in cup 1 back into cup 2 (see illustration below). He continues this process for some time.

If Damian were to continue this process indefinitely (for an infinite amount of turns), how much water would remain in each cup?

Set up

Cup 1 is initially full, so the total volume of water it holds is considered to be 1. A “turn” is each time Damian pours half of the water in one cup into the other. Let’s create a table to see how the volume of water changes after a certain number of turns.

It seems that the volume of water in cup 1 is approximately 1/3 for odd numbered turns and 2/3 for even numbered turns, and the volume in cup 2 is 2/3 for odd numbered turns and 1/3 for even numbered turns. This gives us a pretty good idea that if this process were continued indefinitely, the volume of water in each cup should oscillate from 1/3 and 2/3, but how can we actually prove this?

What to do in a nutshell

We’ll have to figure out a formula for the volume of water in each cup after some number of turns and take its limit as the number of turns approaches infinity.

Getting started

We’ll have to express the volume of water in each cup in a different manner. It looks more complicated, but a very interesting pattern becomes evident. Closely examine the way the volume in each cup is expressed on the chart below:

 Because the total volume for both cups must always be 1, we couldn't end up with more or less than what we started out with. After we've expressed the volume of water in one cup, we can simply subtract it from 1 in order to have an expression for the volume of water in the other cup.  In addition, the exponent in the denominator represents the number of times the volume of water in each cup has been halved.

Let us from now on focus on even numbered turns of cup 1. Consider the pattern seen in turn 6, an even turn. The first term that appears in the numerator (2^n-1 ) is raised to a power that is one less than the number of the turn (turn n), and this is true for the denominator as well. The exponent also lessens by 1 as we progress through the series. Upon analyzing this pattern, we can write volume of water in cup 1 for the nth even term as the following series V_n:

If we distribute the negatives and remove the parentheses, we get the following alternating series:

*First of all, let me point out why the quantity (1-1/2) is negative. Look at the chart above. If we examine the expression of the volume in cup 1 for even turns, we’ll notice that there are an odd number of negatives that get distributed, ultimately leaving the quantity (1-1/2) as negative. We can conclude that for even turns, this portion of the series will always be negative, and because V_n applies specifically to even turns of cup 1, its final term will also be negative; however, we will soon see that this portion of the series can be ignored regardless of its sign.

Since that’s out of the way, let’s go ahead and break up V_n into two parts:

Let’s go ahead and divide the terms of the series by 2^n-1:

This is where it gets exciting. Notice that the terms of the series share a fixed ratio of 1/4; it’s a geometric series! It was pretty evident earlier that this is a geometric series, but doesn't it look so much cleaner now? To make V_n even more compact, we will rewrite it using sigma notation. Below I've provided the formula for the nth term of a geometric sequence because we will use it to rewrite the sum. The term a₁ is the first term of the sequence and r is the common ratio

This looks significantly better. This takes us back to the quantity (1-1/2) that became 2^-n after dividing by 2^n-1. Remember that we are working with an infinite series. As n becomes very large, 2^-n approaches 0. That’s why this term of the series is ignored, and it doesn’t matter to us whether this quantity is positive or negative; it’s zero either way.

We are interested in calculating the sum of this geometric series, so we can apply the following formula:

Notice that the geometric series we are dealing with has a common ratio that has an absolute value less than 1, and when such a number is raised to the power of some positive number greater than 1, the output decreases. For example,

0.5²=0.25 and 0.5³ is even smaller. If we take the limit of S_n as n approaches infinity, we will get the following formula:

This is actually the standard formula for an infinite geometric series. Note that it is only valid if the absolute value of the common ratio is less than 1.

We are now ready to apply this formula to V_n. We know that a₁ is equal to 1 and r is equal to 1/4:

This is our final solution, and it seems that our conjecture from the very beginning is correct! The volume of water held by cup 1 after many even turns is 2/3. From this, we can conclude that the volume held by cup 2 must be 1/3 for even turns. Well, what about odd turns? Look at the chart below:

The volume of water held by each cup will oscillate from 1/3 and 2/3 over and over and over and over…

Deriving the Volume of a Cone with Integration

We start by drawing an arbitrary line in the coordinate plane with y-intercept of some radius r and x-intercept of some height h. These will be dimensions of the cone we will use later on.

We will create an equation of the line using r and h as shown above

Notice that the region closed by the and the x and y axes is a right triangle . If we revolve this region around the x-axis, then a cone is formed. Just try to picture it in your head

Notice that the cross sections of a cone are circles. We could break the cone up into a bunch of circular disks and add the volumes of the disks together.This will only be an approximation of the volume, but as the number of disks approaches infinity, we will have an exact value for the volume of the cone.

We can create a bunch of rectangles in the region bounded by the line and the axes. Revolving the rectangles will create disks whose radii vary as a function of the line. That is, the radius of each disk is equal to the y value of the line at its specific location on the graph.

The volume of each disk represents a small change in the volume of the cone as shown by the expression below.

As delta x approaches zero, the number of disks that can fit in the region bounded by the lines approaches infinity. We can now write the equation as follows.

We will create a summation of these disks by integrating this equation from 0 to h. We must also rewrite y in terms of x. If you work it all out correctly, you should come up with the right formula for the volume of a cone.

Using u-substitution makes the integral A LOT less messy.

A physics problem in my book that I thought was hard

A particle starting from rest slides down a frictionless ramp. It is to attain a given horizontal displacement in a minimum amount of time. What is the best angle for the ramp? What is the minimum time?

Usually I'll solve something and realize that it wasn't that bad, but I still think this is hard...sort of.

Start out by drawing a diagram of the particle sliding down a ramp or incline. This is just to give an idea of what we're dealing with. The arrow I've drawn indicates the direction of the particle's motion.

Draw another diagram of the particle that shows its acceleration vectors. The vector pointing straight down is the acceleration due to gravity (g). The other two are components of g, one of which is in the direction of the incline and the other of which is perpendicular to the incline. We will focus on the acceleration vector that points in the direction of the incline because this is the one that causes the particle to accelerate down the plane.

As the particle moves down the plane, it experiences a displacement in the direction of the ramp denoted by delta z; however, we are interested in the horizontal displacement. We will break up the displacement vector delta z into a horizontal and vertical component denoted by delta x and delta z respectively. Now we should be able to write an equation of the horizontal displacement in terms of delta z and theta.

Keep this equation in mind because we will use it later. Right now we will focus on writing the displacement of the particle in the direction of the plane (or in the direction of delta z) in terms of the change in time and the angle theta.

Now this is where the equation above comes in: Once we figure out and equation for the displacement in the delta z direction , we will multiply the whole thing by cos(theta) to obtain an expression of the displacement in the horizontal (delta x) direction.

Also Recall from the acceleration diagram that the acceleration in the direction of the plane is equal to g*sin(theta). This is what we'll use in our kinematics equation.

we have now written the horizontal displacement in terms of theta and the change in time.

Want we want to know is what angle of inclination will cause a certain horizontal displacement in a minimum amount of time. In other words, for what angle does the particle experience the greatest displacement?

It may have struck you that this problem becomes similar to finding the relative maximum of a function. It almost is. The tricky thing here is that the equation is in terms of two variables, not one.

In this case when we solve for the relative maximum of the equation, we will treat the change in time as a constant and differentiate the horizontal displacement with respect to the angle theta. If you think about it, the change in time is really irrelevant in this situation because it's not what causes a change in the vector components of the particle's acceleration. We are interested in the variable that causes a direct change in the particle's acceleration components, so we differentiate with respect to theta.

Now that we have the angle that will cause a horizontal displacement in a minimum amount of time, we simply plug it back into the original equation (the one we differentiated) and solve for delta t.

Solution to the physics problem I found

You may have noticed that as object A moves to the left there are several variables that change with respect to time. The vertical height y increases, the horizontal length x decreases, and the angle alpha increases. This is really no different from an ordinary related rates problem, so we'll treat it like one.

Start out by using the Pythagorean theorem to write the vertical length y as a function of the horizontal length x.

recall that the derivative of displacement with respect to time is equal to velocity. So in order to create a velocity equation, we will differentiate both sides of the equation above with respect to time.

We're almost ready to solve the problem. All we have to do now is a little bit of substitution and manipulation of the problem.

important note* dx/dt is the velocity of A. It's important to know that the velocity of A is going to be negative because as A is moving to the left, the length of side x gets smaller. This means that the rate of change of x with respect to time (dx/dt) is negative. It is thus substituted with negative v.

at this point it's just a matter of rearranging parts of the equation and making it look nice. It should simplify to something fairly simple where we can plug in the angle alpha.

Isn't it mind blowing? We started out with a nasty looking velocity equation and simplified to something simple.

A cool physics problem I found

I found this problem in my physics book. It really stumped me for a while, but I finally figured it out after some time. It looked intimidating, but it's not as bad as I thought. I tried to redraw the diagram as closely as I could and I almost copied the problem down word for word.

Two objects, A and B, are connected by a rigid rod that has a length L. The objects slide along perpendicular guide rails, as shown in the figure. If A slides to the left with a constant speed v, find the velocity of B when the angle, alpha equals 60 degrees.

Solution to the cool integration by parts problem

This problem requires a bit more ingenuity.

Start out using integration by parts as you normally would.

You may notice that the new integral acquired still poses the same issue we started out (but with cosine instead of sine). Let us do integration by parts once more and see what happens

Now I need to substitute this back in. I need to also make sure to distribute the negative.

Now we have the exact same integral we started out with on both side. all we really need to do is add the integral of e^(x)sin(x) to both sides of the equation. Then we'll have 2 times the integral of e^(x)sin(x) on one side. Simply divide by two and the whole thing will be solved.

Is your mind blown or what?

A cool integration by parts problem

Here's a cool integral that requires integration by parts. It's a little tricky, but the solution is mind blowing.