Karl's Calculus Forum: Sep 2003

© 2003 by Karl Hahn 		KCT logo (click for Table of Contents)
Home Page Table of Contents How to email Math Notation Send Email to Karl 





    
Archived Forum Discussion for September 2003

    
return to current forum









I'm working on a problem and have run into a bump...

I just got the second derivative which I know is correct and is written as follows:

2(3X2 + 1)/(1-x2)3  (reads: 3 times x to the power 2 plus 1 over 1 minus x to the power 2 all powered by 3)

I need to find inflection points so I set my second derivative to equal zero, but I get that I need to take the root of -1/3....I don't think I did this right!

HELP!!!!!!!

Stephanie









OTown, Canada - Monday, September 29, 2003 at 13:46:59 (EDT)









Reply to John:

In either one of the integrals (but not both of them) substitute
 u = 1 - x, 
 1 - u = x,  and
 du = -dx.

Make sure you remember to substitute the limits of integration by the
same formula as well.

Finally observe that the variable that appears after the "d" in any
integral is a dummy variable. If you have two integrals in identical form
except for the choice of dummy variable, they are the same.


Karl <Click to Send Email to Karl>


 USA - Sunday, September 28, 2003 at 12:42:01 (EDT)









Reply to Mark:
The easy way is to use
L'Hopital's Rule.

The other way (which is a bit more work and is specific to this problem) is
to multiply numerator and denominator by


   (2+x)2/3 + 21/3(2+x)1/3 + 22/3



When you multiply out the numerator you'll get a whole lot of cancellations.
You will be left with something that has a common factor of x in
numerator and denominator, which you can also cancel. Then take the limit
of what's left.


BTW: This method uses the fact that
 un - vn  can always
be factored into


   (u - v)(un-1 + un-2v + un-3v2 + ... + vn-1)



for any whole number, n > 1.
Make sure you understand how I applied this to your problem.


Karl <Click to Send Email to Karl>


 USA - Sunday, September 28, 2003 at 12:34:45 (EDT)









I need help BADD!!!  How in the world do you do this, I dont even know where to start.

"Show that for any two positve real number 'a' and 'b',...{=sign of integration


{x^a(1-x)^b dx = {x^b(1-x)^a dx both limits of integration are defined as being between 0 and 1.

I have no idea where to begin or how to do any of it. Thanks for your help.

John


John









Knoxville, TN USA - Friday, September 26, 2003 at 23:36:10 (EDT)









need help with a limits problem!


Lim            (2+x)^(1/3) - (2)^(1/3)
x->0           -----------------------
                         x



( ^(1/3) = the third root of. )

I know the answer but i cant get to it.   im totally stuck.  ive tried the conjugation method but that didnt work.

btw the answer is:


    1
----------
3*(4)^(1/3)



mark

 USA - Friday, September 26, 2003 at 00:46:42 (EDT)









Follow Up to Follow up for Matt: Rather the limit is 0, not DNE.  Thanks for the help.  Again, I don't know where I got that from.

Matt









 USA - Monday, September 22, 2003 at 10:28:36 (EDT)









Follow Up From Matt: So, as x goes to 3 (x - 3) is 0, so the sin of 0 is zero which makes the answer, 0/3, which is not defined. So the limit does not exist?

Matt









 USA - Monday, September 22, 2003 at 09:55:33 (EDT)









Reply to Matt:
What is the limit of  x - 3  as x
goes to 3? Sine is a continuous function, so if the limit of
 x - 3  as x goes to 3 is some
value, L, then the limit of
 sin(x - 3)  as x goes to 3 is
 sin(L).  See if you can finish it using
those premises. [hint: the answer is NOT 1/3].

Karl <Click to Send Email to Karl>


 USA - Monday, September 22, 2003 at 00:08:51 (EDT)









I am having a hard time with the following.

lim x to 3    sin(x-3)/sqrt(x + 6)

I think, but don't know why, sin(3 - 3) = 1 and the sqrt (3 + 6) = 3
So the answer is 1/3, I think, but why is sin(3 - 3) = 1?

thanks,
Matt

Matt









 USA - Sunday, September 21, 2003 at 15:38:05 (EDT)









Reply to Tomas:
Numerator and denominator of your function have a common factor of
 x - 1.  Use
polynomial long division
to factor out the  x - 1  in the numerator
and the denominator. Cancel the common factor, then use the resulting function
to find the limit as x goes to 1.

Karl <Click to Send Email to Karl>


 USA - Thursday, September 18, 2003 at 12:17:47 (EDT)









Please help me with this TYPE of problem- as in just steps on how to complete it. Thank you!


Give a formula for the extended function that is continuous at the indicated point.

            x3-1
   f(x)  =  ---- ,     x=1
            x2-1




Tomas









 USA - Thursday, September 18, 2003 at 00:49:53 (EDT)









Reply to Lauren Leland:
Let  f(x) = x2 + 1. 
Take the reciprocal of that. Where's the discontinuity (removeable
or not removeable)?? Remember that it only takes one counterexample to make an
"always" statement false.

Karl <Click to Send Email to Karl>


 USA - Wednesday, September 17, 2003 at 15:59:12 (EDT)









hey guys I just had one quick question

True or False: A reciprocal of a function always has a removable discontinuity?? thanks sooo much!!! you can im me qtntx13 thank ya

Lauren Leland









arlington, tx USA - Tuesday, September 16, 2003 at 17:03:49 (EDT)









Reply to Zuber:
To integrate  x2sin(x),  your
instinct to do it by parts is correct. The first round of integration by
parts gives:




  -(1/a)x2cos(ax)  +  (2/a)



 



 x cos(ax) dx





and you of course have to do a second round, which I will let you do.
The final answer I got was:


  -(1/a)x2cos(ax)  +  (2/a2)x sin(ax)  +  (2/a3)cos(ax)



If you take the derivative of that you'll get lots of cancellations and
end up with the original integrand.

Karl <Click to Send Email to Karl>


 USA - Tuesday, September 16, 2003 at 12:42:54 (EDT)









Reply to Mel:

The first figure is entirely inside of the second. So take the difference
in area between them (the larger area minus the smaller area).


To find the area of a polar-coordinate figure, take the following
integral:




  1
   
  2



 2p
 
 0



 r2 dq





Where r is actually r(q), that
is it is your radial function of q.


Note that the integral might be only from 0 to
p if the second two quadrants of the
graph simply retrace the first two quadrants.

karl <Click to Send Email to Karl>


 USA - Monday, September 15, 2003 at 13:24:51 (EDT)









i stuck with this qeustion and i hope one of you guys will help me out. i did try to integrat by parts, but when i checked it by differentiating my result, i couldn't get the qeustion i started with.
   find the integration of

         x2 sine ax dx 

zuber









dallas, tx USA - Thursday, September 11, 2003 at 19:15:44 (EDT)









Don't even know were to start?

Find the are outside r=8 cos 0 and inside the cardiodic r=4 91+ cos 0)
The o in cos 0 are theda

mel









 USA - Thursday, September 11, 2003 at 19:00:43 (EDT)









hello dear friends.
would you mind helping me to solve this problem. if anybody could do it i would be very thankfull to him/her

the problem is to change the following sum into dm^3 (decimeter cube)

5200cm (centimeter)

3600cl (centilitre)


regards
Ahmed

Ahmed









oslo, norway - Tuesday, September 09, 2003 at 20:25:49 (EDT)









greetings intelligent people in calculus. i am a struggling student trying to figure the proof of the following theorem. i sure need your help folks. please send me a proof of the theorem that says :

if a function is continous at the closed interval [a,b], then it is integrable at [a,b]...

thanks!

way









 phils. - Friday, September 05, 2003 at 08:14:05 (EDT)









Reply to melanie:
To integrate the first one, complete the square in the denominator. So the
denominator becomse  (x+1)2 + 4. 
Substitute  u = x+1 and  du = dx.  That will put it into a form you've seen before, which
can be integrate by a trig substitution of
 tan(v) = u. 

The second one, just substitute
 u = x2  and
 du/2 = x dx. 

For the third one, use the same substitution.

The last one cannot be done without using advanced functions, sine integral and cosine integral, which are beyond the scope of
nearly every first year calculus class. So I suspect you've copied that one
incorrectly.

Karl <Click to Send Email to Karl>


 USA - Friday, September 05, 2003 at 00:21:20 (EDT)









Reply to Kaimbridge:
Finding the mean radius of curveture around some curve (say an ellipse) is
just a matter of integrating the radius of curveture formula along the length
of the curve. It's straightforward because the radius of curveture is a
defined scalar at every point on the curve (even though you have to use an
elliptic integral to do this one). But to find the mean radius of curveture
of a surface, you have to come to terms with the fact that at each point on
a surface there are two radii, one is the radius taken along one axis along
the surface, the other is taken along an axis (also along the surface) that is
normal to the first axis. So how do you define the mean radius?

Taking the product of the two radii at each point you can come up with a measure of total
curvature at each point. Integrate that product over the entire surface
to get the mean of that measure. Take the square root of that total to get
a mean radius. That's my best suggestion. Of course it doesn't work very
well if the mean measure comes out negative, but it won't if the surface
is an ellipsoid.



Karl <Click to Send Email to Karl>


 USA - Friday, September 05, 2003 at 00:01:59 (EDT)









Reply to Andre:
You have two equations in two unknowns:


               __
    x + y  =  Ö10

    (x-y)2 = 2



From the first equation you get


          __
   y  =  Ö10 - x



which you substitute into the second equation:


          __
   (2x - Ö10)2 = 2



Square out the binomial that's on the left of the equal, then subtract 2
from both sides, then use the quadratic formula to solve for x.
Finally back substitute x into the first equation to find y.

Karl <Click to Send Email to Karl>


 USA - Thursday, September 04, 2003 at 23:51:31 (EDT)









My exam is in one or two days...i need help as fast as you can. But even if it's after the exam, I have been struggling with these questions for a while so an answer would be heavenly:
Evaluate:

  positive infinity    ______1_________________dx
S                      x2 + 2x +5
  negative infinity
 










   positive infinity
S                     x e [-x2]  dx
   0











  positive infinity   ________________1____________________dx
S                    x square root of (x2 - 1 )
  1









  1    ___cos_x_________ dx
S      1- x 2
  1/2





Please Help--I have a feeling they will be on the exam cause they are so hard--

Thank you soo sosos so much

melanie









 canada - Thursday, September 04, 2003 at 22:46:21 (EDT)





    










Home Page


Table of Contents


How to email Math Notation











    


You can email me by clicking this button:
    

    
    
    
    
    
    
      Use your own emailer 
        Use form