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 #12

Calculus Post#12

I can't actually recall much of what we did this week. I was sick for Monday and Tuesday so I'm assuming we might of just reviewed... Anyway, other than that, we took a multiple choice test and free-response from all of derivatives in preparation to start integration soon. Turns out we suck pretty bad with limits...so...

Limit Rules:

When you take the limit as x goes to infinity, the rules are as follows
1) If the degree of the top is larger than the degree of the bottom, its infinity.
2) If the degree of the top is equal to the degree of the bottom, it is the leading coefficient of the top divided by the leading coefficient of the bottom.
3) If the degree of the top is smaller than the degree of the bottom, it is 0.

For limits that ask for the left or right side by using + or - after a number like as x goes to 9+, you either add or subtract (for positive or negative) .1, .01, and .001 and input those values into your calculator. This is used to approximate a value that it is approaching. Remember, you also have to do this if you get a division by 0 or anything like that whenever you plug in for finite limits. You would put it into your calculator and do both the - and + sides and make sure they match at a value, and if they do, that value is the limit.

Also, a trick for limit as x goes to 0 for

sin(ax)
-------
bx

Is simply a/b. This also works for

sin(ax)
-------
sin(bx)

Again, the answer is a/b.


Hope this helps some.