Karl's Calculus Forum: May 2003

Archived Forum Discussion for May 2003

My name is Ned Derksen. I repeat, I need help in calculus if you can help me in any way please e-mail me. If you think this is a joke, you are mistaken. I slept through my Trig classes last year and I barely remember how to take the square root of four but I still passed with a D-. I am failing Calculus and need at least a B on the final to pass. Please I only have three more days. Ned Derksen (Sir Quacksalot) P.S. A friend in need is a friend indeed.
Ned Derksen

Waupun, WI USA - Tuesday, May 27, 2003 at 10:34:07 (EDT)

I don't get anything in my Calculus class and I am in need of some major help. I realize that I am not the smartest person in the world, but this is ridiculus. If you can tutor me in any way please e-mail me at nedly_d@hotmail.com Thanx Ned Derksen (Sir Quacksalot)
Ned Derksen

Waupun, WI USA - Tuesday, May 27, 2003 at 10:25:31 (EDT)

Hi, I've finished "Teach Yourself Calculus" (in only 11 months!) and I feel like I've got a good foundation (Equivalent to Calculus 2, I think) for what's next. So...what's next? I'd like to get to Stochastic Calculus for Finance and Game Theory. I'm tempted to get a book on differential equations and/or partial derivatives. I think I'm lacking a big picture outlook of what's out there, how it's related, and where to go next. Thank you in advance for any advice.
Peter

Chatsworth, CA USA - Friday, May 16, 2003 at 19:16:46 (EDT)

Reply to Peter: Have you found your sign error yet? Look at where you factor out the x³dv/dx in last terms of

   x²v² + x²v - x²v² - x³vdv/dx + x³dv/dx = 0

When you add your ln(c) constant, add it as plus or minus ln(c). That way, one of the possibilities will give you the book's answer.
Karl <Click to Send Email to Karl>
USA - Wednesday, May 14, 2003 at 12:24:35 (EDT)

Here's another one. I'm guessing that I need to make c something other than ln(c) to get the answer in the book:

y² +(x² - xy)dy/dx = 0

x²v² + (x² - x²v)(v + xdv/dx) = 0

x²v² + x²v - x²v² -x³vdv/dx + x³dv/dx = 0

x²v -x³dv/dx(v + 1) = 0

-x³dv/dx(v + 1) = -x²v

((v + 1)dv)/v = -x²dx/-x³

Divide Out: (v+1)/v = 1 + 1/v

INT(dv) + INT(dv/v) = INT(dx/x)

v + ln(v) = ln(x) + c, let c= ln(c)

v = ln(x) + ln(c) - ln(v)

v = ln(cx/v)

e^y/x = cx²/y

y = cx² / e^y/x Book's Answer: y = ce^y/x
Peter

Chatsworth, CA USA - Tuesday, May 13, 2003 at 19:46:46 (EDT)

Stupid c! Why can't it commit to value. I get it now. I won't use c as a result to test against. Thanks!
Peter

Chatsworth, USA - Sunday, May 11, 2003 at 23:48:37 (EDT)

another person and i are supposed to teach a calculus lesson to the class that is fun/interactive and i dont have any ideas. it needs to apply things that we've learned in the course to a new concept. know of any suggestions?
nick

USA - Sunday, May 11, 2003 at 12:40:34 (EDT)

Reply to Peter: Your second to last equation is

   xv  =  c(v - 1)

Putting in y/x for v:

   y  =  c(x/y - 1)

   y - cy/x  =  -c

   1 - c/x  =  -c/y

   1/c  =  1/x  -  1/y

Can you see what happened? Remember that the constant, c, is arbitrary. So 1/c is just as valid as c, as long as c is not zero.
Karl <Click to Send Email to Karl>
USA - Sunday, May 11, 2003 at 12:31:50 (EDT)

Reply to Alex: On the first one, change the problem to two objects of the masses given in elliptical orbit about a common elliptical focus. The sum of the major axes of the two ellipses needs to be the same as the initial distance given in the problem. Use Kepler's laws figure out the ellipses. The length of the minor axes are unimportant. Use Kepler's laws to figure out the period of this system. The solution to your problem will be half of that period. Note that Kepler's laws requires that period is independent of geometric factors other than major axis of the ellipse. That's why this approach works.

For the electron problem, figure out the kinetic energy of each of the two electrons. Also figure out the potential energy function as a function of electrons' separation distance. Solve for the distance at which the potential energy is equal to the sum total of the initial kinetic energy of the electrons.
Karl <Click to Send Email to Karl>
USA - Sunday, May 11, 2003 at 12:21:51 (EDT)

Doh! I've had worked it out on paper a couple times without the sign error. But still:...

Solve Homogeneous Equation: y² - x²dy/dx = 0 Book's Answer : (1/x) - (1/y) = c My Answer: Let y/x = v, y=vx, dy/dx = v + xdv/dx

x²v² - x²(v + xdv/dx) = 0

x²v² - x²v - x³dv/dx) = 0

x²(v² - v - xdv/dx) = 0

v² - v = xdv/dx

dx/x = dv/(v² - v)

Decompose: 1/(v² - v) = (1/(v-1)) - (1/v)

INT(dx/x) = INT(dv/(v-1)) - INT(dv/v)

ln(x) = ln(v-1) - ln(v) + ln(c) Let c= ln(c)

ln(x) + ln(v) = ln(v-1) + ln(c)

xv = c(v-1)

xv/(v-1) = c

Let x= 2 and y=4, put in book's answer and get c= 1/4

Put in above answer (v = y/x = 2) and get c = 4

Where did I go wrong?

-Peter
Peter

Chatsworth, CA USA - Saturday, May 10, 2003 at 22:11:53 (EDT)

Just wondering if i could get a few thoughts on these questions: Both are very similar and are probably identical solutions besides the constants. 1)Two objects in space influenced by nothing other than the force of gravity are 1000 meters apart. One mass is 55kg and the other 65kg. How long does it take for the objects to hit one another. 2)Two electrons are fired at one another with an initial velocity of 3.5*10^6m/s. Calculate the minimum distance away from one another that they will reach. Thanks Alex
Alex

kitchener, canada - Saturday, May 10, 2003 at 19:47:18 (EDT)

Reply to Peter: You have no idea how close you are to the correct solution. The book's answer is correct. Here I'm taking the book's answer and solving for y.

   1/x  -  1/y  =  c

   1/x - c  =  1/y


      1
            =  y
   1/x - c


      x
           =  y
   1 - cx

If you do the same thing to your answer, you'll get something close. But, alas, you made a sign error following the line:

x²v² - x²(v + xdv/dx) = 0

Fix it and you'll have it right.

BTW the easy way to do this one is to multiply by the integrating factor: r(x,y) = 1/(x²y²). Try it.
Karl <Click to Send Email to Karl>
USA - Saturday, May 10, 2003 at 15:04:14 (EDT)

Reply to Philip:

Divide out e^2x / (e^x + 1) to get e^x - e^x / (e^x + 1)

A good example is at sos math:

http://www.sosmath.com/algebra/factor/fac01/fac01.html

Then the answer is: e^x - ln(e^x + 1)
Peter

Chatsworth, USA - Saturday, May 10, 2003 at 01:03:29 (EDT)

This is very puzzling. I need to find the integral of:

            e^2x
   f(x)  =  ----
            e^x+1

I have tried integration by parts using:

                e^x
   f(x)  =  e^x ----
                e^x+1

Is there a general procedure for solving problems of this form where the exponents in the numerator and denominator are different?
Philip Garza

McAllen, TX USA - Friday, May 09, 2003 at 18:00:02 (EDT)

I can't get my answer to match the book - so I'm sure the book is wrong! Solve Homogeneous Equation: y²2 - x²dy/dx = 0 Book's Answer : (1/x) - (1/y) = c My Answer: Let y/x = v, y=vx, dy/dx = v + xdv/dx

x²v² - x²(v + xdv/dx) = 0

x²v² - x²v + x³dv/dx) = 0

x²(v² - v + xdv/dx) = 0

v² - v = -xdv/dx

-dx/x = dv/(v² - v)

Decompose: 1/(v² - v) = (1/(v-1)) - (1/v)

-INT(dx/x) = INT(dv/(v-1)) - INT(dv/v)

-ln(x) = ln(v-1) - ln(v) + ln(c) Let c= ln(c)

ln(v) - ln(x) = ln(v-1) + ln(c)

v/x = c(v-1)

v/x / (v-1) = c

v/(x(v-1)) = c

(y/x)/(x((y/x)-1)) = c

(y/x)/(y - x) = c

y/(xy - x²) = c

Where did I go wrong?
Peter

Chantsworth, CA USA - Friday, May 09, 2003 at 16:57:51 (EDT)

Thanks Karl, but I think that's a better question to ask my Discrete Mathematics professor. I'm not familiar with convergence, which is probaly not somthing usually convered in a second level of Calculus. My Calculus professor would just say it's not covered in the course, and therefore wouldn't accept that. I do however still need a good problem for him.

Sorry to ask, but isn't this a public forum? I'm unclear on how to reply without having to make a new post. I would greatly appreciate any other suggestions (scroll down to my first post on the 8th). Thanks!
Martin

Bronx, NY USA - Thursday, May 08, 2003 at 22:56:37 (EDT)

Another note to Martin: On integrating

   dx
       
 1 + x⁴

I would not have bothered with that numerator munging thing. Instead I would have just factored the denominator and decomposed the result into partial fractions. You can see a similar integral arise as the partial result of an even harder integral by clicking here. You can also see the full solution using factoring and partial fractions at the same page.
Karl <Click to Send Email to Karl>
USA - Thursday, May 08, 2003 at 16:17:06 (EDT)

Reply to Martin: Your professor will probably get this one, but it's a neat problem anyway. Suppose you have a set of encyclopedeas with books numbered 1 through n. And suppose you have a bookcase for them with slots numbered 1 through n. Let b(n) be the number of ways you can insert the books into the bookcase so that no book's number matches the number of its slot.

            n!
   lim           =  ???
  n -> ¥   b(n)

This does not, at first, look like a calculus problem, but you are asked to prove the convergence value, and that is calculus. BTW I do know the answer and the proof. But I'm not telling because this is a challenge problem.
Karl <Click to Send Email to Karl>
USA - Thursday, May 08, 2003 at 16:08:39 (EDT)

Hello, I'm a college Calculus part II student here in NYC. Since the first day of class, my professor claimed that he can answer ANY Calculus II question, within the contents of the course material. He was even confident enough to challenge us by saying that he would give an A to any student who gave him a question, based on Calculus II, that he would be unable to answer.

Now I admit, he is a very knowledgeable professor, young even, but I doubt he can answer EVERY problem. Recently, he gave us an advanced problem, where he told us to integrate:

 1
----
1+x⁴ with respect to x.

No student could even figure out the correct first step, which was to multiply by a ½ and change the numerator to 1+x+1-x. We were all disappointed and discouraged as he used four whiteboards to answer this. Looked simple, but turned out to be very advanced.

I'm asking for all to help me find a Calculus II question that my professor will probaly not be able to answer. I've made searches for "most difficult calculus problem" and even "unanswered calculus questions," but both led me nowhere.

His terms are: (1) It has to be based on a topic covered in Calculus II, (2) You don't have to know the answer, but there has to be a way to check his answer right or wrong, (3) An answer must exist, (4) there must be a plausible way to do it, ie: an advanced triginometric integral with no limits.

That's it? There has to be someone who knows of a difficult problem that uses a mess of identities and properties that he probaly never even heard of! Any suggestions?

PS: Sorry if I seem to be using this forum for my own personal benefit, but maybe it can be a little fun. I personally enjoy this challenge, as a motivated math student as myself.
Martin

Bronx, NY USA - Thursday, May 08, 2003 at 11:45:43 (EDT)

Reply to Adonia: Solve it by separation of variables. Multiply by dx and then by (sin(y)/e^cos(y)):

  (sin(y)/e^cos(y)) dy  =  (1/x²) dx

  (sin(y) e^-cos(y)) dy  =  (1/x²) dx

The right is easy to integrate. On left, substitute u = -cos(y); du = sin(y) dy, and the left side becomes a piece of cake to integrate.
Karl <Click to Send Email to Karl>
USA - Tuesday, May 06, 2003 at 10:17:27 (EDT)

I need to solve this differential equation and I cannot seem to get the math right. dy/dx = (e^cos(y))/(x^2*sin(y)) Please Help!! Soon!!!
Adonia
Milwaukee, USA - Monday, May 05, 2003 at 22:41:12 (EDT)

Reply to Jo: To take a partial derivative, you hold all independent variables except one constant and then take the derivative with respect to the remaining independent variable only. You treat all the others as if they were constants. So you got the second one correct. The first one:

   f(x,y)  =  ln(y)/ln(x)

So finding the partial with respect to y first, that means treat x as a constant, so ln(x) is also treated as a constant.

  ¶f
      =  1/(y ln(x))
  ¶y

For the partial with respect to x, treat ln(y) as a constant and apply the chain rule:

  ¶f
      =  ln(y) (-1/(ln(x)²))(1/x)
  ¶x

Karl <Click to Send Email to Karl>
USA - Monday, May 05, 2003 at 19:58:08 (EDT)

Reply to Uncalcbum: On the first one, factor out the 3 and you are left with the sum of 1/n. This is called the harmonic sum. Notice that for n from 2 to 2, 1/n is greater than or equal to 1/2. For n from 3 to 4, 1/n is greater than or equal to 1/4, but you have two of them, so the sum over that range is greater than or equal to 1/2. For n from 5 through 8, you have 1/n greater than or equal to 1/8. And you have 4 of them, so what do you have as that partial sum? Indeed you can break all of the counting numbers up into such ranges, and the result is that the sum over each range is greater than or equal to 1/2. So you have an infinite number of halves. What does that tell you about the series?

On the second one, to find the power series, just do the polynomial long division. For the radius of convergence, how far is x=0 from the nearest x where the function is undefined?? That will be your radius.

For the last one, do the integral test. Remember that

  1/2^x  =  e^{-x ln(2)}

Use integration by parts to find the integral:

 x e^{-x ln(2)} dx

and determine whether the result converges or diverges as x goes to infinity.
Karl <Click to Send Email to Karl>
USA - Monday, May 05, 2003 at 19:47:18 (EDT)

Reply to matthew: In both cases the t is just the dummy variable. It will disappear when you do the definite integral. I'll do the first one for you:

   x²
   
   1

  dt
      =
 3+t

ln(3 + t)

x²
  =
1

   ln(3 + x²)  -  ln(3 + 1)

You can use the log identity to simplify this further if you like.
Karl <Click to Send Email to Karl>
USA - Monday, May 05, 2003 at 19:34:04 (EDT)

Reply to CalcChic: Let x be the number of trees. Then x - 1000 is the number of trees in excess of 1000. That means that revenue per tree is

   50 - 0.02(x - 1000)

To find total revenue, you multiply that by the number of trees, x:

   R(x)  =  50x - 0.02(x² - 1000x)

The profit is $10 per tree less than that. So subtract 10x from the above to get profit:

   P(x)  =  40x - 0.02(x² - 1000x)

Now take the derivative, dP/dx, and set it to zero:

   dP
       =  0  =  40 - 0.02(2x - 1000)
   dx

This is a linear equation in x, so you should be able to solve it pretty easily. That x will be the optimum number of trees.
Karl <Click to Send Email to Karl>
USA - Monday, May 05, 2003 at 19:26:55 (EDT)

Hi, I'm struggling through an on my own calculus course with James Stewarts book and I had a few problems I was wondering if someone could explain how to get the answers to. I'd GREATLY appreciate it! #1. Determine whether the series is convergent or divergent. If it is convergent, find its sum.

 inf.
(Sum) 3/n
 n=1

#2. Find the power series for the function and determine the interval of convergence.

f(x) = x/(4x+1)

#3. Determine whether the series is absolutely convergent.

 inf.
(sum) n²/2ⁿ
 n=1

If you can answer any or all of these, please email me! Thanks!
Uncalcbum

USA - Monday, May 05, 2003 at 13:02:39 (EDT)

Hi I am studying for the AP test for calculus and I was reading my review book on applications on the definite integral. However, the book had two problems that boggled me because the explanations were extremely paltry. I am hoping that someone can help me with these two problems.

Evaluate:
1) F(x) = integral from 1 to x^2 of [dt/3+t]
and
2) F(x) = integral from 0 to cos x of [(1-t^3)^(1/2) dt]

Both I thought were applying the idea of the fundemental theorem, with the direct substition of the varibles, but then "dt" was factored into solving the problem. Thank you and hope that someone will reply soon! -Matt
matthew

CA USA - Monday, May 05, 2003 at 01:19:49 (EDT)

HELP! I am working on a calc. problem dealing w/ optimization & maximizing profit. Can anyone help me? An apple orchard, of fixed size, produces an annual revenue of $50 per tree when planted with 1000 trees. Dure to overcrowding of the trees, the annual revenue per tree is reduced by 2 cents for each additional tree planted on the orchard. If the cost to maintain and pick each tree is $10 per year, how many trees should be planted in the orchard to maximize profit from the orchard? So far this is what I have: Revenue=$50,000 R(x)=(100+x)(50-.02x) R(x)=-.02x^2+30x=50000 R^1(x)=-.04x+30 x=750 Now I'm stumped with what to do for profit. Any help would be greatly appreciated. Please email me at tls1972us@yahoo.com Thank you!
CalcChic

Hartford, CT USA - Sunday, May 04, 2003 at 17:46:43 (EDT)

I was hoping that someone could give me a hand with some partial derivatives as I'm unsure that I'm on the right track.

We have to find f'(x,y) when f(x,y) = log_x(y)

I've said:

log_x(y) = log(y)/log(x)
df/dx = (0*log(x) - log(y)*1/x) / (log(x))^2
= -log(y) / x(log(x))^2
df/dy = (1/y*log(x) - log(y)*0) / log(x))^2
= 1 / ylog(x)
so f'(x,y) = (-log(y) / x(log(x))^2 , 1 / ylog(x))

Is this okay? Also,

f(x,y) = x(y^2 + 1)
df/dx = y^2 + 1
df/dy = 2xy

f'(x,y) = (y^2 + 1, 2xy)

Is this right?

Thanks,

Jo
Jo

Australia - Sunday, May 04, 2003 at 07:30:32 (EDT)

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