What post is this again?

Anyway, I will now explain a few types of questions that in particular I got wrong on this week's AP, just to help me review my mistakes etc.

Number 83, Calculator Portion.
83. What is the area of the region in the first quadrant enclosed by the graphs of y=cos x, y=x, and the y-axis?

Okay so instinctively when I read this problem, I for some reason jumped to the conclusion that it was a volume problem and it would easy. Okay, I was completely wrong--it's an area problem. Other than that, my only advice here is that we/I need to be very particular in what we do when we read the problems. Identify what you are supposed to do, and double check that. Write it down next to the problem if need be...it might help you remember that "Okay, this is an area problem, not volume, so I won't be multiplying by Pi".

Number 21, Non-Calculator Portion
21. The limit as x goes to 1 of (x)/(ln x) is ?

Okay so this problem, I plugged in for 1 (because that is what you do for definite limits) and I got that it was 1 / 0. Okay so for whatever reason in my mind, I decided that this was a point in time where L'Hospital's rule applied; however, I was very wrong. L'Hospital's rule only applies when it comes out to be 0/0 or infinity/infinity. So, this answer is simply DNE.

Number 22, Non-Calculator Portion
22. What are all values of x for which the function f defined by f(x)=(x^2-3)e^-x is increasing?

Okay so for this problem, I identified "increasing" as to being my clue word. It hinted to me that I needed to take the first derivative, and set equal to 0 and solve for my critical points. Well, the only problem here was that the derivative was slightly...odd. It was different, but the trick is that you want to completely distribute everything in so that you can see what you can factor out. You end up being able to factor out the ugly e^-x, and you are left with a pretty polynomial to factor. The only tips on this problem is to watch your derivative, and remember that for first-derivative test, you have to plug into the first derivative, not the original.

Number 23, Non-Calculator Portion
23. If the region enclosed by the y-axis, the line y=2, and the curve y=sqrt(x) is revolved about the y-axis, the volume of the solid generated is

So the most important mistake that most people will make with this is that they won't realize it is being rotated about the y-axis. The reason this changes things is because you need to solve your equation for x. So, our equation becomes x=y^2. Now it is just a simple integral, however, you have to remember that since it is volume, it becomes squared. So the integral is of y^4, not y^2. Also, I guess it's important to note that the line y=2 is used as your bounds (drawing a picture will help immensely in figuring this out).

Have a great day :-).

Post #16

Calculus Week Number 16

So this week was a lot of different things. I'll start my post off by saying that I completely did not retain ANY of the information about the trig inverse functions and how to do all of those...I guess the only way I'll remember is by looking at my notes but I forgot my notebook for the weekend so this is all coming from my head...


Okay so the things I do remember

Volume by Discs

Basically you are given an equation, an interval, and what it is to be rotated about...either a line or one of axis's. Your basic steps are to first draw the graph. You need to know what you're looking at. So draw all the graphs as best as you can and determine in a somewhat 3D looking-way what the object or volume you will be finding is. Most of the time, with Volume by disks, you'll wind up with a sphere looking thing. Anyway, so a hint to remembering when to use this is when you have an equation and a line... if it was an equation with an equation you'd use volume by disks because you'd have a hole somewhere. Anyway, the way you do this is you take your equation, square it, and take the definite integral from a to b times by pi.

For a problem where you have to rotate the solid formed by two equations, you basically do it by washers. What you do is you do the top equation squares minus the bottom equation squared. You then take the definite integral from a to b and times that by pi to get your volume.

Both of those are really really simple...it's just a matter of getting it right. Sometimes you have to shift it because you always want it to be rotated about an axis so...what you do is just minus whatever you need from both equations, no big deal.

Anyway, I don't think anyone can really help me with the trig inverse stuff but if you think you can, have at it! :-P