Karl's Calculus Forum: Feb 2002

Archived Forum Discussion for February 2002

Please ignore my question below, I already have the answer. Thanks anyway
jim quigley toronto, on Canada - Thursday, February 21, 2002 at 09:47:04 (EST)

I can't make heads or tails of this question. Hopefully you can help. The question reads as follows: The integral U(sub n) = integral (from 0 to pie) of [cos (nx)/4cos(x) +5] is somewhat difficult to compute for various integer values of n. However the combined integral combination 2U(sub n+2) + 5U (sub n+1)+ 2U(sub n) is easier to compute, as is the integral 4U(sub 1) + 5U(sub 0). The steps below use these facts to evaluate U(sub n), for n> or = 0. a) Evaluate the integral U(sub 0)[hint: Use the transformation x=2tan^-1z] b) Evaluate the combined integral 4U(sub 1)+ 5U(sub 0), and hence determine U(sub 1) using part a) c) Show that the combined integral 2U(n+2) + 5U(sub n+1)+ 2 U(sub n) sums to zero using trig identities d)Solve 2U(sub n+2) + 5U(sub n+1)+ 2U(sub n) Thanks in advance
Jim Quigley Toronto, On Canada - Sunday, February 17, 2002 at 12:11:46 (EST)

Reply to Tom: This one isn't so bad. You do it by first completing the square, and then by trig substitution. Your integral is


     dp
  ________  =
 Öp(1 - p)


    dp
  ______
 Öp - p²

To complete the square on p² - p, you get (p - (1/2))² - (1/4). So let u = p - (1/2), du = dp, and your integral becomes


     du
  __________  =
 Ö(1/4) - u²


    2 du
  _________
 Ö1 - (2u)²

Now make the trig substitution, sin(v) = 2u, cos(v)dv = 2du, and you are nearly home. A quick trig identity in the denominator, and everything cancels except the dv. I'll let you take it from there.
Karl <Send Email to Karl>
USA - Friday, February 15, 2002 at 12:44:35 (EST)

I really need to integrate 1/"sd of a binomial distrbution" that is: Integrate f(p)= 1 ------------ (np(1-p)^0.5 with repect to p Any ideas???
Tom Cardiff, UK - Friday, February 15, 2002 at 11:40:33 (EST)

Reply to Jacob Rael: Let's integrate it using thin rectangles whose long axis are parallel to the i axis.

With each increment of dt you change t by, you change your v position by


   V₁ w cos(wt) dt

which is the width of the rectangle. The height of the rectangle is given by your equation for i. So the integral you have to do is (integrating the width times the height of the rectangles):




   V₁ w




 (I₁ cos(wt)  +  I₃ cos(3wt)) cos(wt) dt

Note that to integrate the general term, cos(nwt) cos(wt) dt (where n is an integer), do it applying integration by parts twice, which will get you back to an expression involving the original integral. Once you have it in the form of "original integral = expression involving original integral" simply solve for the original integral, and you have it.
Karl <Send Email to Karl>
USA - Thursday, February 14, 2002 at 10:24:36 (EST)

if there is anyone out there that can help iwould really appreciate it
tom spring hill , fl USA - Thursday, February 14, 2002 at 00:30:53 (EST)

if there is anyone out there that can help iwould really appreciate it
tom spring hill , fl USA - Thursday, February 14, 2002 at 00:30:48 (EST)

can anyone help: the solid lies between planes perpendicular to the x axis at x=-1 and x=1 the cross section is perpendicular to the x axis between the planes are squares with edges running from the semicircle y=the square root of 1-x squared to the semicircle y=the square root of 1-x squared
tom spring hill, fl USA - Thursday, February 14, 2002 at 00:19:41 (EST)

I have two parametric equations:


v = V1 sin(wt)

i = I1 cos(wt) + I3 cos(3wt)

If I plot v versus i I get a closed curve, if I3=0, it is an ellipse. How can I get the area of this curve? (If I3=0, then area = pi V1 g1 V1). Eventually I'll be adding more harmonic terms to i so I need a general answer. Thanks, jr
Jacob Rael Los Angeles, CA USA - Wednesday, February 13, 2002 at 13:58:02 (EST)

Reply to Helen: To find the tangent line to a function, you must first find the function's derivative. Your first function is:


   f(x) = (1/3)x³ + x² + x

so the derivative is


   f'(x) = x² + 2x + 1

So use algebra to solve for all the value's of x where x² + 2x + 1 = 1

Your other problem is asking for the line tangent to y = x² and passing through (4,15). This type of problem is covered in the online pages at section 4.5. Look down the page to the second example.
Karl <Send Email to Karl>
USA - Friday, February 08, 2002 at 17:22:46 (EST)

I'm really stumped on these two problems. I think they are basically the same problem in a different form #1 Find all points on the graph Y=1/3x^3+x^2+x where the tangent line has slope 1 #2 a space traveler is moving from left to right along the curve y=x^2. when she shuts off her engines, she will continue traveling along the tangent line at the pt where she is at that time. AT what point should she shut off the engines in order to reach the point (4,15) I have concluded that if the tangent line of the first pt goes to pt (4,15) then the share the same slope because the slope of a straight line is always the same so if I find the slope at pt (4,15) i then need to find the other pt that also shares the same slope. I took the derivative and plugged in the 4 and got a slope of 8 for the tangent line but i don't know how to find any other pts that share that same slope?
helen hates calculus la mesa, ca USA - Thursday, February 07, 2002 at 21:59:00 (EST)

Daniel Davis has done a great job showing how to apply L'Hopital's rule here. I would, however, be cautious about the rules of infinity given at the website, http://megadodo.com/articles/8R69.html. For example, that website states that infinity times zero is zero. Well not always. Karl's rules of infinity (which are the same as the ones taught in traditional calculus courses) are: if you have infinity divided by infinity, infinity minus infinity, or infinity times zero, then it is time to look deeper into what you've got and apply the theory of limits to resolve it. In each of those cases, when you work it through, depending upon how the limit pans out, you could have plus or minus infinity, any finite number, or zero. You never know until you have worked it through.
Karl <Send Email to Karl>
USA - Thursday, February 07, 2002 at 13:42:30 (EST)

the limit is neat because you can pretty freely move it all around and split it up all over the place, and if you know the rules of infinity, it makes it a lot easier. When I use "lim" below, assume it is lim x -> infinity. LH= is where I use L'Hopital's Rule

       x² - 4            lim x² - 4
lim   --------------  =  ------------------
      (x + 3)(x - 4)     lim (x + 3)(x - 4)

so you will end up with inf/inf which is 1.
an alternate way to do this is to use L'Hopital's Rule

       x² - 4                x² - 4
lim   --------------  =  lim ------------------
      (x + 3)(x - 4)         x² - x - 12


         2x
LH= lim -----  (still goes off to infinity, apply it again..)
         2x-1



         2
LH= lim ---  = 1 (same as we got above)
         2

Daniel Davis San Antonio, TX USA - Wednesday, February 06, 2002 at 12:09:27 (EST)

This: x² - 4 / (x + 3)(x - 4)
[R]ailGun
- Wednesday, February 06, 2002 at 08:19:27 (EST)

How do I find the limit of: x² - 4 --------------- (x + 3)(x - 4) as x --> infinity I'm having problems finding this one..
[R]ailGun
- Wednesday, February 06, 2002 at 08:18:15 (EST)

I just need a diagram of the unit circle with the radians listed! I would really appreciate it! Aghe Koondeh houston, tx USA - Monday, February 04, 2002 at 10:25:46 (EST)

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