Anyone else taking AP Calc tomorrow?

Yep, I'm taking the AB test tomorrow. Should be fairly easy, aside from--as you said--stupid mistakes here and there. The hardest part will be waiting a few months for the scores.

 

you got a 4.

 

I'm taking it on the late test date two weeks later due to a clerical error on my school's part. Not that I'm complaining :-)

 

I'm taking the BC in a few weeks. Should be pretty easy.

 

I'm taking the BC tomorrow. It shouldn't be too bad, better than the stat one today at least :)

 

Ha, I remember the AP tests. My school only taught Calc AB, so i took that test, and had an absolutely terrible headache immediately afterwards, was really stressed out, but got a 5 (probably did much better than I needed to get a 5 in fact).

The next year, I looked at the material tested for Calc BC and figured I just needed to learn a few things, so I decided to take the Calc BC test. I, however, hadn't done any Calculus for the entire year before exam and I put off the material until the last week, so I learned it in a few days, forgot to review some important things that I had learned like trig substitution, took the exam and only knew how to do about 3/4 of it (since there was a lot I hadn't reviewed. I knew how to do most of the new stuff and most of the simple stuff, but none of the difficult stuff from Calc AB) I wasn't at all stressed out during it because I figured I was going to do terribly anyway. Figured I only got a 4 (which wouldn't have done anything for me at my college since I already had a 5 in Calc AB), but I ended up with a 5 anyway.

Turns out that they don't really expect much in the Calc BC test. You can go in only remembering 3/4 of the material, and as long as you can do what you do know very well, you'll end up with a 5. And if you know all the material well, then you don't need to worry about it at all because they don't expect that much. (When I took it 3 years ago, they usually took an 80% as a 5)

 

Wow, my problem with AP tests have always been timing on the free response. I always take too long to answer #1 with this great, perfect answer. Unfortunately, that takes a huge chunk of time, and the person reading my #5 or something would think this kid is an idiot since he only wrote a single sentence in what little time he had left. On today's stat test, for example, I wrote paragraphs for the first few and spent my time formulating my answers before we were told we only had 10 minutes left, and I still had to do part of #4, #5, and #6 >.<. I still remember how on the chemistry test last year, I was rushing like crazy and somehow got an answer of ln(0) for reaction time or something, and I was like "huh?" but I put it down anyway since I ran out of time.

 

This is another thing that examinations are meant to test: the ability of the student to work well under pressure and to a deadline. One should know before entering the exam how much time is available for each question. Then, if one starts over-running by a large amount on the first question, one should then skip to the next.

Practising past exam papers under the same test conditions (i.e. same time limit) will enable you to better manage your timing during an exam.

Moved to academic guidance.

 

That's absolutely true for a lot of the math and science AP tests. For the AP Physics B test, my teacher told us the first day that we probably won't learn everything that will be on the test, but we will know plenty to do well. I think the standard most good AP teachers go by is, study 80% of the material 100%, and simply review the rest, and you will be okay.

 

I agree. Even if I do get a 5 on an AP test, I wouldn't use it unless I felt I was fully prepared(which I don't). I just look at the course as a good introduction to what I might see later in college. It would be idiotic to take an AP course just for the college credit. I will, however, take my AP English credits. So that eliminates at least one English comp, thank god.

 

Same, that's why I didn't take any AP tests(well that and the $80 cost.) I just went to the department and took one of the Profs. Calc II finals and got credit that way.

 

I thought it was pretty easy. The FRQ's were pretty straightforward.

 

Yeah, I rather liked the free response questions.

 

i teach at a state school, and have almost never had a student with credit for AP who should have accepted college credit for calc.

instead the best of them should have retaken beginning calc, but from an honor professor in an honors section, the hardest one they can manage.

students with AP credit are honors students. it is nonsense for them to accept colege credit for beginning calc, and then jump up into non honors sophomore courses in college for which they are ill prepared. instead they should take honors sections of freshman courses about which they know something.

i would guess our honors level freshman course in calculus, along lines of spivak's book, is challenging enough for almost any AP student with a 5. and if it isn't, by being in that class, the students ability will be noticed immediately, and he/she will be placed higher into perhaps our honors level several variables calc class.

in the event that the spivak class is actually too hard, which will be true of most AP students, they can drop back into our regular honors class or a regular non honors class like the one i taught this year.

most of the UNsuccessful calc students in my beginning non honors class have had AP calc in high school. and most of the sad stories in my fall second semester calc class, come from beginning freshmen who tried to skip the first semester with AP credit.

truth is the AP class simply does not have high enough standards to compare to a good college class. However, college standards have been long going DOWN as a direct result of trying not to flunk out all the incoming AP students, so be careful to see what the story is at your school.

 

e.g. here is a trivial little question from beginning calc:

if f is a differentiable function with f'(a) > 0, PROVE that on some interval a < x < a+e,

we have f(x) > f(a).

Using basic theorems from calc, PROVE that if f'(a) > 0 and f'(b) < 0, and f is differentiable on [a,b], (but do NOT assume the derivative is continuous), then f'(c) = 0 for some c: a< c < b.

If you cannot do this, that gives you an idea of the level of your AP class. If you can, it also tells you something.

 

Many schools no longer offer honors level single variable calculus because the entire audience for such a course would rather take a higher course their first year. For example, I (and everyone else who became a math major at my school) took a year long honors linear algebra and multivariable calculus sequence our first year.

We almost certainly covered less material than such a course would have covered if any of the incoming students had had a good knowledge of calculus. We learned how to answer questions like mathwonk posted above in this class rather than in the single variable calculus class where they probably belonged.

I believe that a lot of schools are going in this direction and eliminating honors single variable calculus classes.

 

Since [tex]f'(a)>0[/tex], there is some positive real number [tex]c[/tex] (i.e., the value of [tex]f'(a)[/tex]) such that for any given positive real number [tex]\epsilon[/tex], there is some positive real number [tex]\delta[/tex] such that [tex]|\frac{f(a+h)-f(a)}{h}-c|<\epsilon[/tex] whenever [tex]0<|h|<\delta[/tex]. Choose [tex]\epsilon=c[/tex]. Then [tex]-c<\frac{f(a+h)-f(a)}{h}-c<c[/tex] whenever [tex]0<|h|<\delta[/tex], so [tex]0<\frac{f(a+h)-f(a)}{h}[/tex] whenever [tex]0<|h|<\delta[/tex] and so [tex]f(a)<f(a+h)[/tex] whenever [tex]0<h<\delta[/tex]. This proves that for [tex]a<x<a+\delta[/tex], [tex]f(x)>f(a)[/tex], as asserted.

I haven't looked at the second one yet. Can I take my AP credit?

 

As an AP student, I agree that the typical AP calculus course doesn't bother with too much rigor. It seems more geared toward engineering, which I don't think bothers as much with proofs (correct me if I'm wrong, of course).

Since I don't have much experience in these types of proofs, I hope you guys don't mind me taking a stab at it. I have no idea what the best proof for each of these is, but hopefully I won't have any reasoning errors. Any pointers would be welcome :)

EDIT: I just noticed uman also posted, and his proof for #1 is similar to mine, so that's good. Formatting is a [censored].

Let's say [itex]f\,'(a) = L > 0[/itex]. Then, from the definition of the derivative,

[tex]f\,'(a) = \lim_{x \to a}\frac{f(x) - f(a)}{x - a}[/tex]​

So for all [itex]\epsilon > 0[/itex], there is a [itex]\delta > 0[/itex] such that if [itex]|x-a|<\delta[/itex], then

[tex]-\epsilon < \frac{f(x) - f(a)}{x-a}\,-\,L < \epsilon[/tex]​

thus

[tex](L-\epsilon)(x-a) < f(x)-f(a) < (L+\epsilon)(x-a)[/tex]​

The inequality doesn't switch because [itex]x-a[/itex] is positive. If we choose [itex]\epsilon < L[/itex], then

[tex]0 < (L-\epsilon)(x-a) < f(x)-f(a)[/tex]​

is true whenever [itex]x-a<\delta[/itex] (which exists by definition). So our interval is [itex](a,\,a+\delta)[/itex], which prevents us from choosing an interval [itex](a,\,b)[/itex] such that the slope of the secant line between a and b is negative or zero.

I'm not sure if this is the best way to do this, but we know by the first question that there exists x in (a, b) such that f(x) > f(a). By a similar proof, there is y in (a, b) such that f(y) > f(b). Thus, f(a) and f(b) are not absolute maximums, and with the extreme value theorem, we know f(x) must have an absolute maximum, f(M), for M in (a, b). To prove f'(M) = 0, we do the same thing that is used to prove Rolle's theorem: show that left and right hand limits for the derivative are always positive and negative, respectively. If they approach the same limit L, then they must be 0.

 

you used too many obscure symbols :rofl:

I wonder if engineering students would ever do these types of questions ... (It is obvious so why prove it :) )

But personally, I know many people who were doing extremely well in high school but started rolling down in the university... (I think this include all people I know )

High school standards are just way too low (you don't learn anything useful ><) - even in those enriched programs

 

I haven't read Tedjn's proof of the second question so I don't know if mine is similar.

Remark: Since [tex]f[/tex] is differentiable on [tex][a,b][/tex], it must be continuous on that interval.

Denote by [tex]\alpha_1[/tex] the least upper bound of the set of all real numbers [tex]x[/tex] in [tex][a,b][/tex] such that [tex]f'(x)>0[/tex] Denote by [tex]\alpha_2[/tex] the greatest lower bound of the set of all real numbers [tex]x[/tex] in [tex][\alpha_1,b][/tex] such that [tex]f'(x)<0[/tex]. If [tex]\alpha_1\not=\alpha_2[/tex], the derivative of [tex]f[/tex] is zero for every value between them. If [tex]\alpha_1=\alpha_2[/tex], call their value simply [tex]\alpha[/tex]. If there is some value of [tex]x[/tex] in [tex][\alpha,b][/tex] such that [tex]f'(x)=0[/tex], our work is done. If not, [tex]f[/tex] is strictly decreasing on this interval and so because (by continuity of [tex]f[/tex]) [tex]\lim_{x\to\alpha}f(x)=f(\alpha)[/tex], we must have [tex]f(x)<f(\alpha)[/tex] whenever [tex]\alpha<x<b[/tex], hence it is impossible for the derivative at [tex]\alpha[/tex] to be positive. The proof that it cannot be negative is similar. So it must be equal to zero. QED.

I think this proof could have been explained better but I'm tired and need to go study for my chemistry exam tomorrow :-/. Is it valid?

 

I like your proof better than mine, since I had to resort to the extreme value theorem, which I didn't prove. I suppose, to complete it and prove [itex]f\,'(\alpha) \geq 0[/itex], you would set [itex]\alpha_3[/itex] as the least upper bound of the set of real numbers x in [itex](a,\,\alpha\,][/itex] where [itex]f\,'(x) < 0[/itex]. If there is no such x, then f is increasing on [itex][\,a,\,\alpha\,][/itex], and we can complete the proof. If [itex]\alpha_3 \neq \alpha[/itex], then the interval [itex][\alpha_3,\,\alpha][/itex] is increasing, and we can complete the proof. If [itex]\alpha_3 = \alpha[/itex], then [itex]\alpha[/itex] is the least upper bound of a set of numbers with positive derivatives and of a set of numbers with negative derivatives, so it cannot be a member of either set (Just what was I thinking? ). Thus, [itex]f\,'(\alpha) = 0[/itex] since f is differentiable at that point.

Is that a valid conclusion?

I thought they were all right until I got to the point where I thought 8 < 6