Karl's Calculus Forum: July/August 2003

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Oops, sorry, second given is (x-y)^2=2

Andre









 Ont Ca - Sunday, August 31, 2003 at 20:35:15 (EDT)









I have a question that im having a bit of a hard time answering,

Given

x+y=sqrt 10    and    (x-2)^2=2

solve x*y

This should also only take relatively few steps.

Thanks

Andre









 Ontario Canada - Sunday, August 31, 2003 at 19:24:00 (EDT)









Is there an equation for finding the mean arcradius of a spheroidal (oblate)
annular ring?
Let's say I asked for the mean arcradius of the perimeter of an ellipse. I would
be directed to extract it from the "elliptic integral of the second kind", which
finds the value *along* the ellipse.
What I want to know is the--for lack of a better description--mean "perpendicular
arcradius" of the ellipse.
For instance, take a spheroid, where the arcradius at the equator equals a: Its
"perpendicular arcradius" equals a * cos{Oz}^2 (Oz = acos{b/a}).
So, if a = 6378.135 and b = 6356.75, the arcradius along the equator = 6378.135,
but its perpendicular arcradius = 6336.4367 (i.e., radii of curvature in the
horizontal vs. vertical plane).
But now let's consider the mean arcradius of a meridian (i.e., "perimeter of an
ellipse").
Using the above spheroid, the mean arcradius equals about 6367.447. However, the
"perpendicular mean arcradius" equals somewheres between about 6388.85-6388.87.
I think I may have at least an approximation for
the mean arcradius of a given annular ring
(excluding the "sin{Ring}" factor):

          a * sec{Oz}^(.5*(1-3*cos{Ring}^2)) ???




     ~Kaimbridge~


Kaimbridge









Peabody, MA USA - Sunday, August 31, 2003 at 09:35:17 (EDT)









Reply to Nelson cabrera:
That's not really a calculus question but more of an engineering and
thermodynamics question. But I'll put my engineer's hat on and answer it for
you. In a turbojet engine, expanding hot gases from the combustion chamber
flow over the turbine, forcing it to turn. The turbine is on the same shaft
as the compressor fan(s), which compress the incoming air. The thermodynamic
cycle, as best I can remember, is that ambient gas undergoes adiabatic
compression at the compressor fan, isobaric heating as a consequence of
combustion of the fuel, then adiabatic expansion over the turbine and out
the nozzle. After that it undergoes isobaric cooling back to ambient temperature
to complete the cycle.

Karl <Click to Send Email to Karl>


 USA - Saturday, August 30, 2003 at 10:38:02 (EDT)









Igot problem with calculus I got a many question # 1 the main purpose of
the turbine in the turboje engineis to?
drive to compresor, compres the Air, increase the velosity of the
exhaust gases 0r rerduce the temperature of the exhaustgas

Nelson cabrera









silver spring, MD USA - Friday, August 29, 2003 at 18:31:55 (EDT)









Reply to John:
You can solve this problem by paying attention only to the crossection
of the roll and the crossection of the material. Unrolled any length of
the material has a crossection that is rectangular -- length by thickness.
The crossection of the roll is an annulus -- that is a circle with a smaller
concentric circle missing. Now let r be the core diameter and
R be the outer diameter. Then the crossectional area of the roll
is


   A  =  p(R2 - r2)



Presumably the thickness of the material is fixed. The area of the rectangle
is length times thickness. All you have to solve is for the length so that
length times thickness equals A. If s is length and
x is thickness, then


   s  =  A/x



Karl <Click to Send Email to Karl>


 USA - Wednesday, August 27, 2003 at 17:20:08 (EDT)









I have a formula for calculating the diameter of a roll with the following information available

SD= starting diameter (core),
mt= material thickness in mils (gauge),
ti= total inches of material run onto the roll,
the formula is : 
diameter=(PI*MT+SQR(PI*PI*(SD-MT)*(SD-MT)+4*PI*MT*TI))/PI

What I need now, is to calculate the total inches on a roll if
I know the thickness (gauge),
 and the final diameter,
 and the core diameter.

Can anyone help, Thanks
John


John









Tinley, IL USA - Monday, August 25, 2003 at 21:08:58 (EDT)









Reply to Lorrene:
First you need to know why


          sin(x)
   lim            =  1
  x -> 0    x



You can find that by clicking here. Once you
know that, to find


         sin(10x)
   lim           
  x -> 0    x



use substitution of variables. Let
 u = 10x.  Then you also have
 u/10 = x.  And clearly u goes
to zero whenever x goes to zero. So the substituted limit is


          sin(u)
   lim          
  u -> 0  (u/10)



which, because the limit of a product is the product of the limit, is the same as


             sin(u)
  10  lim          
     u -> 0    u



which you already know how to do.

Karl <Click to Send Email to Karl>


 USA - Friday, August 08, 2003 at 18:33:03 (EDT)









I'm having trouble finding the two closest points of the following
lines in free space.


x-1  =  y  =  z+1         &     x  =  y+1  =  z-1
---    ---    ---                     ---     ----
2       3      4                       2       -1




I know that the dot product of the vectors from the lines needs to
equal 0.  I just can't get started on the problem.  Thanks




Robert









Newport News, VA USA - Wednesday, August 06, 2003 at 19:16:42 (EDT)









I am suppose to use my graphing calculator on this but I'm unsure how!!
Here is the question:

lim{x->0} (sin(10x))/x
help me if you can:)


Lorrene









Indio, CA USA - Wednesday, August 06, 2003 at 17:18:03 (EDT)









Reply to Becky:
You didn't use any parentheses, so I can only guess at what you really mean.
I shall assume that the equation you want to solve is:


                  _
   6 csc(t)  -  4Ö3  =  0



Which is the same as"


                  _
   6 csc(t)  =  4Ö3



Dividing by 6:


              2  _
   csc(t)  =    Ö3
              3



And since


                1
   csc(t)  =        
              sin(t)



you can take the reciprocal of both sides to get


               _
              Ö3
   sin(t)  =    
               2



Now take the inverse sine of both sides and you're done.

Karl <Click to Send Email to Karl>


 USA - Thursday, July 31, 2003 at 13:18:55 (EDT)









having trouble getting this one:
solve the trig equation exactly for the indicated variable in [0,2pi]:

6 csc t - 4 sqrt3 = 0

please help

Becky









 USA - Wednesday, July 30, 2003 at 21:13:29 (EDT)










Looking to differentiate f(x) where

   f(x)      =  xlog x

Please see if you agree with this approach:

Say
   y         = xlog x

Take logs on both sides
   log y     = log (xlog x)

   log y     = log x [log x]

Differentiate wrt x on both sides
   1  dy              d                   d
   -  --     = log x  -- (log x) + log x  -- (log x)
   y  dx              dx                  dx


   1  dy              1           1
   -  --     = log x  -  + log x  -
   y  dx              x           x


   1  dy        2
   -  --     =  - log x
   y  dx        x


      dy        2
      --     =  - log x [xlog x]
      dx        x


      dy
      --     =  2xlog x - 1  log x
      dx









Saga









 NY USA - Tuesday, July 29, 2003 at 09:36:52 (EDT)









Reply to April:
Divide out the integrand before you integrate:




  x-1
       =  1  -
  x+1



   2
     
  x+1





Each of the summands on the right is easy to integrate.

You can't take the definite integral of this over any closed interval
containing the point,  x = -1.  Neither the
integrand nor the indefinite integral is defined at that point, and the indefinite
integral has no limit as you approach that point from either direction.

Karl <Click to Send Email to Karl>


 USA - Sunday, July 27, 2003 at 23:31:59 (EDT)









I can't integrate this for the life of me

1
  S (x-1)/(x+1) dx
-1

I intergrated by parts ans ended up with
[(1-x)ln(1+x)] + [(1+x)ln(1+x)] -1 -x +c

then with i slip in the definite intergral----
I get as a last step [0+2ln(2)-2]-[2 ln(0)+0]
and i don't know what to do with ln(0)??


april









 USA - Sunday, July 27, 2003 at 12:38:01 (EDT)









Reply to Melisa:
After applying integration by parts, you get




   



 x2 ln(x+1) dx  =



  1
   
  3



 x3 ln(x+1) -



  1
   
  3



 



  x3
     dx
 x+1





To do the second integral, divide the integrand out first using polynomial
long division:




   x3
       =
  x+1



  x2 - x + 1  -



   1
     
  x+1





which should be easy to integrate.

Karl <Click to Send Email to Karl>


 USA - Wednesday, July 23, 2003 at 12:36:38 (EDT)









Can't seem to get this one-

S x^2 ln (x+1) dx

(Integrate xsquared times ln(x+1))

I know you can use partial fractions, but even so, the answer seems to be 0/2 which can't be right!

These are my efforts Sx^2ln(x+1)=ln(x+1)(x^3/3)-S[(x^3/3)(1/(x+1)]
                  =ln(x+1)(x^3/3)- ln(x+1)(x^3/3) -Sx^2ln(x+1)dx
                  =Sx^2ln(x+1)dx + Sx^2ln(x+1)dx=0
                  =Sx^2ln(x+1)dx =0/2
Any ideas???

melisa









 USA - Wednesday, July 23, 2003 at 08:11:47 (EDT)









Reply to Melanie
Click here and then
scroll down to the subtitle for Inverse Hyperbolic Functions. There
you will find a similar derivation for arctanh(x). Follow that one
through and then see if you can apply the same technique to
arccoth(x).

Karl <Click to Send Email to Karl>


 USA - Tuesday, July 22, 2003 at 18:49:21 (EDT)









HelPPP!!!!! I need an emergency rescue for this question, i've been working on it for 4 days straight and it is driving me crazy, i just can't see it- e-mail me (by clicking the button below) with help, please please please


for x<-1 or x>1 show that

arccoth x = 1/2 ln [ x+1/ x-1]




melanie 









 canada - Monday, July 21, 2003 at 17:00:23 (EDT)









Reply to Ryan Swope:
First apply the
quotient rule
to this function to find f'(x). Numerator is


   u(x)  =  sin(5x)

   u'(x)  =  5 cos(5x)



The denominator is:


   v(x)  =  ln(x)

   v'(x)  =  1/x



Applying the quotient rule:


             v(x)u'(x) - u(x)v'(x)
   f'(x)  =                       
                   (v(x))2



Substitute the  u(x),  u'(x),  v(x), 
and  v'(x) 
given above into this equation and you have f'(x). Then substitute
x with e to get the final answer (remembering that
 ln(e) = 1).  For sine and cosine of e,
use your calculator.


Karl <Click to Send Email to Karl>


 USA - Monday, July 21, 2003 at 13:29:35 (EDT)











Any ideas on this?

If        sin(5x)
   f(x) = ------
          ln(x)

evaluate for f'(e).

           

Ryan Swope

 USA - Saturday, July 19, 2003 at 21:30:58 (EDT)




    











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