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scorpeus
Average Member
USA
12 Posts 
Posted  08/12/2007 : 16:23:00

Hello all. My name is John, I am a high school student and have been teaching myself calculus using the internet for about eight months, and have made some exceptional progress from what I believe, but I have a few questions that I think will be very easy for any veterans to answer from experience.
1. I have been wondering this for a while. 1/t dt = lnt + C according to the what I've heard, but I thought it should be something like: (t/t)*lnt so that for f(t) = 1/t, for all t values less than zero, the area of that region would be negative as well, hence the absolute value of t over t (t/t). Should this be correct?
2. Another simple question. For the standard formula lim (f(x + h)  f(x)) / h = dy/dx h> I have seen h equal infinte or zero. Which one is correct, and why?
3. A third question, maybe a little off of the calculus subject, concerns e. According to my standard graphing calculator, e = 10^((43E6)/99,011,159) (E being 10 to the power of). In my calculator, just a standard graphing calculator, it shows that this is the exact value of e. Is this true? Or is it just really close?
Thank you to anyone willing to help out. 


Ultraglide
Advanced Member
Canada
299 Posts 
Posted  08/18/2007 : 20:15:09

Absolutely. 


msakowski
Average Member
USA
18 Posts 
Posted  08/20/2007 : 16:58:39

A third question, maybe a little off of the calculus subject, concerns e. According to my standard graphing calculator, e = 10^((43E6)/99,011,159) (E being 10 to the power of). In my calculator, just a standard graphing calculator, it shows that this is the exact value of e. Is this true? Or is it just really close?
Assume that e=10^(A/B), where A= 43E6 and B=99,011,159. Then LOG e = A/B = (LN e)/(LN 10) = 1/LN10 so A*LN10 = B which implies that an integer A multiplied by an irrational number LN 10 will equal an integer B  this is impossible. Good Question!
Your calculator uses a series approximation to calculate that value. The series is probably a modification of a Taylor Series  at least that is what some TI folks told me a number of years back. A series (on a calculator) only returns a finite number of places since it only uses a finite number of terms. No calculator can provide the exact decimal value of e (or pi) since these are irrational numbers. You will find discrepancies in all types of values that require a series approximation. For example, a 3rd root would use a series approximation and also any trig function would use a series (on a calculator that is). 
Edited by  msakowski on 08/20/2007 17:02:41 


scorpeus
Average Member
USA
12 Posts 
Posted  08/21/2007 : 21:50:29

thanks for the answers guys and to the suggestions of getting a calc book (which I have before from a local library) and yes, I did find the h>infinity a little wierd but like I said i've seen it written somewhere before which was why I felt like asking (but I still got the general concept of it). I also had the idea that my estimation was too good to be true but oh well, I just couldn't exactly find a precise enough calculator online to answer that for me and my graphing calculator just thought it was the same thing.
I have another question: a theorem of mine probably proven already (I have just yet to find it proven). It would be so much easier for me to make a diagram but I have no clue how to do it so i'm going to try and explain this as well as I can.
Take any parabola and put it on a graph. Get its focus and its directrix plotted. Then, put a point ANYWHERE along the parabola, which I'll call point C. sketch a straight line from the focus of the parabola to point C. Now, sketch a line from point C directly vertical to the directrix. From the focus, now, sketch a line vertical down through the vertex to the directrix. The area enclosed by the Focus to point C to the vertex back to the Focus we'll call A. The area enclosed by point C to the vertex down to the directrix along the directrix and back up to point C we'll call A. What I have calculated is that A = 1/2 A.
I realize that how I explained this can easily throw you off, but like I said I don't know how to post a diagram.
Is this true?
Thanx. 


msakowski
Average Member
USA
18 Posts 
Posted  08/22/2007 : 13:26:37

If I am reading this correctly, I am getting that the area of A1 is 1/2*xp if the parabola has vertex at (0,0), focus at (0,p), and with one point being (x,y).
The area enclosed by the xaxis, the directrix, and two vertical segments extending from directrix to origin and directrix and point will be xp. So this area is twice A1.
The area A2 you describe also includes the area bounded by the xaxis, the parabola, and the vertical segment down from the point (x,y). This extra bit of area would be 1/2bh = 1/2*xy 



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