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.

Deriving the volume of a sphere solution

note* I'm assuming you are already familiar with calculus. If you're not, it's probably not going to make any bit of sense.

I will be doing this using integration techniques. Remember learning about solids of revolution from calc. I? That's what I'll be using. I'm not really sure if it's the best way, but it works.

Consider the graph of a basic semi-circle. If you were to revolve it around the x-axis, the solid generated would be a sphere whose cross sections are circles.

Imagine slicing through a sphere. You'll notice that the cross sections are circles of varying sizes. What I'll essentially be doing is creating a summation of disks that each have a very small thickness of dx.

To start off, I'm going to create a function of the cross sectional area and integrate it from one end of the semi circle (-r) to the other (r). I'll need the formula for the area of a circle. The radius of each cross section will vary as a function of x, and so I need to make the appropriate substitution for the radius. Through integration, the disks of varying cross sectional area will be summed together to create a sphere.

There's nothing all that tricky about solving this integral. If you work it all out you should end up with the correct formula for the volume of a sphere.

Deriving the volume of a sphere

For this next one I want to derive the volume of a sphere using calculus. I don't have much more to add, but I feel like I should at least post a picture of something...

problem 6: the derivative of y=x^x^x

When I saw the solution to this, I thought it was really cool. Check it out at khanacadamy.org. This website really helped me get through calc. I.
https://www.khanacademy.org/math/calculus/differential-calculus/implicit_differentiation/v/calculus--derivative-of-x--x-x

problem 5: solution to the slightly harder integral

I must say, this was a bit more challenging, but nothing I can't do.

Using properties of logarithms, I can rewrite the integral like this:

It simplifies to this:

Now I just need a little u-substitution. 

For u-substitution to work, I'll also have to rewrite the integral as follows:

After doing u-substitution, it should look like this:

The integral of ln(x) requires integration by parts. I'm not going to go through the process now because I did it in a different post. It should like this this:

Well, it's simple from here. Just plug back in (x^2+1) in place of u. I'm going to leave it in terms of u because it's long and messy in terms of x.

problem 5: A slightly harder integral

My brother said this problem was on one of his Calc. II exams and challenged me to do it. Challenge accepted

Problem 4 solution: The ACT math problem.

Well, this is what we started out with. It doesn't look like there's much we can do, but the coordinates of point C can be found by dividing the figure into multiple 30-60-90 triangles.

 Based on the proportions of a 30-60-90 triangle, we can figure out the length of each side in terms of w. The shortest leg is w, the adjacent leg is w√3, and the hypotenuse is 2w.  Put it all together and you should get (w√3,2w).

problem 4: An ACT math problem

I was taking a practice test in my Princeton review book and I came across this problem. ACT math is usually pretty boring, but this one had an interesting solution.

Triangle ABC, shown above in the standard (x,y) coordinate plane, is equilateral with vertex A at (0,w) and vertex B on the x-axis as shown. What are the coordinates of vertex C?

Problem 3 solution

With a little bit of ingenuity, this problem can easily be solved. I can't recall any sort of rules to follow when trying to integrate this sort of function, but we can rewrite it using properties of logarithms.

Looks familiar right? Now we can rewrite the integral like this:

To help out more, we can move the ln(3) outside of the integral since it's a constant. Now all we really need to do is think about how to integrate ln(x).

In order to integrate ln(x), we need to use the integration by parts formula. 

 Now we can use the integration by parts formula to rewrite it as follows:

Simplify it and this is what you get:

Well, that took a bit of creativity, but it was totally doable.

Problem 3: Integrals of Logarithms other than base e

I recently got through AP calculus AB and let me just say that there was not one problem that involved logarithms other than the natural log of...something. I don't know why we never ended up using logarithms other than base e, so why not do one right now?

Lets start with something simple. Got any ideas?

problem 2 solution: The weird integral solution

So this was the one I made up. I'm kind of proud of it because so far, no one in my family has been able to figure it out :). The solution is actually not a hard one.

In fact, it's really easy to do once you recall the derivative of  inverse tangent.

looks familiar doesn't it? All I really did was multiply the derivative of arctan(e^x) with the function itself. And I love how no one was able to figure it out.

Solution

We just have to follow the standard procedure for u-substitution and we're good.

When we differentiate both sides it looks as follows. Notice how it looks almost exactly like the function we started out with.

We now have all the ingredients for u-substitution.

Problem 2: weird integral I just made up

So I was messing around with some numbers and I came up with this. I hope you think it's somewhat cool...

It looks really ugly at first glance, but the solution is actually fairly simple.

Problem 1 solution

Not too bad right? Not too bad at all. The height of the cylinder can be calculated by constructing a right triangle as shown in the figure with the diameters of the sphere and the cylinder as shown. Now all you have to do is use the Pythagorean theorem to find the length of the third leg (the height).

10²-6²=h²

64=h²

h=8