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