Karl's Calculus Forum: September/October 2004

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Archived Forum Discussion for September/October 2004 (and 2 days of November)

    
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Reply to Vinay:
To integrate this one, subsitute u = x3.
Then du = 3x2 dx. Observe that x5 dx = (1/3)u du. When you've made
the substition, integrate by parts, then back-substitute.



Karl <Click to Send Email to Karl>


 USA - Tuesday, November 02, 2004 at 13:07:32 (EST)









Reply to Gene:

To expand sqrt(x3 + 1), the easy way to
do it is to let u = x3, and expand
y = sqrt(u + 1). Here are the first few derivatives
of y with respect to u:


  y' = (1/2) (u + 1)^(-1/2)

  y" = -(1/4) (u + 1)^(-3/2)

  y"' = (3/8) (u + 1)^(-5/2)

  y"" = -(15/16) (u + 1)^(-7/2)



And so on. I think you should be able to discern the pattern here. The
numerators of the coefficient in each case except the first are the product
of the odd numbers. The signs alternate. The denominators are powers of 2.
Evaluate each at u = 0. Then write the Maclaurin
expansion of sqrt(u + 1). Finally, replace u
with x3. Do 1/sqrt(x3 + 1)
the same way.


For x/sin(x), first observe that this is an even function, so
you expect that all the odd powers of the expansion will be zero. This function
is the same as x csc(x). Find the first four derivatives
of csc(x), then apply Leibniz method for finding the nth
derivative of x f(x). The method says that the
nth derivative will be


   x f(n)(x)  +  n f(n-1)(x)



When you have all that, you will have to take the limit as x goes
to zero of each of the derivatives to get the coefficients of the Maclaurin
series for this. If any of the limits don't exist, then the series doesn't
exist.


Karl <Click to Send Email to Karl>


 USA - Tuesday, November 02, 2004 at 13:03:08 (EST)










I need some help with a problem from stewarts 5th edition section 8.5 problem 45.

this is the problem



integral { (x)5 (e)-x3 ) 

Vinay









Union City, CA USA - Sunday, October 31, 2004 at 22:15:45 (EST)









Hi, I'm learning Taylor series right now and I'm having trouble evaluating the indefinite integral sqrt(x^3+1)dx and 1/sqrt(1+x^3) as an infinite series.

And also, if possible I'd like to get some help on using multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for the function y=x/sinx.

Thank you so much for your help.

Gene









 USA - Sunday, October 31, 2004 at 22:12:30 (EST)









Reply to Kyle Schmitz:
The problem is over-specified and has no solution.



Observe that the zeros of a parabola (if they exist) must be symmetric
about its vertex. This problem is asking you for a parabola with a vertex
at  x = 3,  but zeros at
 x = 1  (2 away from the vertex) and at
 x = 4  (1 away from the vertex).
For a solution to exist, both zeros must be the same distance from
the vertex.

Karl <Click to Send Email to Karl>


 USA - Tuesday, October 26, 2004 at 01:37:15 (EDT)









My professor gave me this local maximum problem.  I have tried several times to solve it, but cannot get the right values.  Can anyone help!





Let f(x) = ax2 + bx + c. 



Find the numbers a,b,c so that f has a local maximum f(x) = 2 for x = 3, and f(4) = f(1) = 0.






I tried using the three equations:

f(4) = 16a + 4b + c = 0

f(1) = a + b + c = 0

f(3) = 9a + 3b + c = 2



and get:

a = -1

b = 5

c = -4



Which is a parabola which does include all three points, however the point (3,2) is not a local maximum of the function.  I also tried setting the f'(x) = 0 where x = 3, solving for b, then replacing b into f(1) and f(4); but using this method I get:

a = 0

b = 0

c = 0



This is obviously wrong.  



Does anyone else have any ideas?  I feel like either the problem is flawed, or the professor is really just trying to say that the local maximum of 2 is simply the local maximum over the closed range x = 3, but does not actually refer to the critical point (which would be at x = 2.5 using the first method I described above) 

Kyle Schmitz









Norman, OK USA - Monday, October 25, 2004 at 09:59:16 (EDT)









I have a conic sections question i cannot do. can you please help?
A conic has parametric equations x=2sec(theta)  and y=3tan(theta)
so its a hyperbola,
find the equation of the tangent at the point where theta= pi/4

i have found the point of intersection to be (2root2, 3) but dont know what to do from there.

Alex









Auckland, New Zealand - Saturday, October 23, 2004 at 21:35:12 (EDT)










Reply to Sumant:
Since this is a problem that involves complex numbers, I am going to assume
that you are familiar with them and with Euler's formula. First, this integral
is the same as





  



(xn / (x2n + 1)) dx





The denominator has roots in the complex plane at
 e±ijp/(2n), 
where j goes from 1 to n, and where
i is the square root of -1.
Once you have factored the denominator, you need to break it into
partial fractions.
This is the hard part. You can use Heaviside or solving a matrix to
do this. For any particular n, this isn't so bad, but for general
n it is harder. Email me if you need more help with that.
Finally take the integrals of each of the fractions, then recombine
the results. They will all be complex logs. But when you recombine them
into logs and arctangents, everything will come out as real numbers.


Karl <Click to Send Email to Karl>


 USA - Friday, October 15, 2004 at 13:59:52 (EDT)










_____________1____________.dx
 xn + x-n





sumant

 India - Thursday, October 14, 2004 at 08:46:38 (EDT)









i have been trying to do this calc problem for awhile, and somebody recommended me to these forums.  can someone help me do it?  thanks.

------estimate the largest value of the absolute value of y1(x)    (y1= the derivative of y),
for y = the square root of 1 - x, inside 0 is less than or equal to x and x is less than or equal to 0.8

thank you very much if you can help.

kyle tently









 USA - Sunday, October 10, 2004 at 13:36:25 (EDT)









As you are driving around a curve on a road your headlights shine on a deer off the road at (x,y). What is your location (a,b) on the road?

  I have already solved for the equations of the two tangent lines passing from curve to (x,y). Now how do I find the (a1,b1) (a2,b2) two points on the curve.

Brian









Pensacola, FL USA - Thursday, September 30, 2004 at 14:38:56 (EDT)









Hi, in my Calculus class we've started to talk about the chain rule with, jumping straight into sin, cos, log, and ln functions.  I began to have trouble from minute 1.  The teacher explains the chain rule with inside and outside functions (example, f(g(x)), where f(u) is outside and g(x) is inside).  The biggest problem comes here, I can never figure out what is what.  Is the inside function always the function that includes the variable?  And then what all does the inside function include?  And then what about the outside function?

I'm also having a bit of trouble remembering the derivitives of such things as sin(x), cos(x), etc.  But I guess I can always find that straight off the web.

Thanks to anyone who can offer any help,
Dave

David Herald









Alexandrie, KY USA - Wednesday, September 29, 2004 at 00:18:15 (EDT)









Reply to loreal
First find where the two curves intersect. You have
 f(x) = x2 - 4x,  and
 g(x) = x - 4.  So you need to find where
 f(x) - g(x) = 0.  So you solve:


   x2 - 4x - x + 4  =  x2 - 5x + 4  =  0



That quadratic is easily factored into


x2 - 5x + 4  =  (x - 4)(x - 1)



So  x = 4  and  x = 1.  These solutions form the limits of your integration.
You will integrate the difference of
 f(x) - g(x)  or
 g(x) - f(x),  whichever is positive
between the limits you solved for above.




  4
  
  1



 -x2 + 5x - 4 dx





which gives




  



-(1/3)x3 + (5/2)x2 - 4x



4

1



  =  (16/6) - (-37/6)  =  53/6






Karl <Click to Send Email to Karl>


 USA - Sunday, September 26, 2004 at 13:23:52 (EDT)









I'm stuck

Determine the area of the region bounded by the graphs of y = x^2 - 4x and y = x -4.

thank you

loreal









 USA - Sunday, September 26, 2004 at 00:14:53 (EDT)









Reply to BoyHowdy:
First click on
this, and
read what it has to say about functions. The basic idea of a function is
that you have two groups of things -- a domain group and a range group
(the two groups may indeed be the same or have elements in common, or not).
A function is a rule by which a unique thing in the range group is assigned
to each thing in the domain group. Note that the same range thing can be
assigned to several things in the domain group, but not vice-versa. Each
thing in the domain group must have one and only one thing from the range
group assigned to it.

Karl <Click to Send Email to Karl>


 USA - Friday, September 10, 2004 at 12:56:59 (EDT)









Reply to Jack:
To prove that


   lim     f(x)  =  L
   x -> a



using delta-epsilon, you must show that for any positive value of epsilon,
no matter how close it is to zero, you can always find a positive value
for delta in order to make the following true:


   |f(x) - L|  <  epsilon



whenever  |x - a| < delta. 


So a delta-epsilon proof should read like a recipe, where if somebody
gives you a positive value for epsilon, your recipe can always turn it into
a positive value for delta that makes the above inequality true.

Karl <Click to Send Email to Karl>


 USA - Friday, September 10, 2004 at 12:44:08 (EDT)









Can someone give me an analogy I can use to determine what exactly a function is?  Or a simple definition of a "function".

BoyHowdy









New York, NY USA - Friday, September 10, 2004 at 03:31:52 (EDT)









I have kind of a fundamental question.  Is there a proper way to write a proof?  I was working on some delta-epislon proofs of limits, and I was wondering if there's a proper way of phrasing or writing them out.  Thanks for any advice.

Jack









 USA - Wednesday, September 08, 2004 at 17:25:31 (EDT)









Find the x- and y- intercepts for the graph of...

4x2+3y2+6xy-24=0

Kris









 VA USA - Thursday, September 02, 2004 at 20:56:46 (EDT)









Please help me solve this problem-
Solve the following for z:

6x2+10yz-5y=3xz



Kris









 Va USA - Thursday, September 02, 2004 at 20:51:19 (EDT)









Hi Mai!

Try this step-by-step calculator:


          http://www.calc101.com/webMathematica/derivatives.jsp

Ä

Xanadu,   USA - Thursday, September 02, 2004 at 07:33:07 (EDT)









Can someone help me find derivatives, not just answers but tell me how to find them for:


y=xlnx






y=xsec-1x-.5lnx






y=zcos-1z-rad.(1-z2)






y=csc-1(secx)
[0,2pi]
for this one i thought it might be  csc-1(sec4x)cotxtanx
but i wasn't sure how the [0,2pi] affects it






y=lnxx






y=logbase5(t-7)   (i think i have to use change of base)




thank you anyone who helps

mai









 USA - Wednesday, September 01, 2004 at 13:20:04 (EDT)
















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