Karl's Calculus Forum: Dec 2003 - Jan 2004

Archived Forum Discussion for December 2003 - January 2004

Please help me. For some reason I just do not get it. Kepler's law and using focus to figure the ellipse. I know that or maybe I do not but it seems like he used minor axis and major axis, but how did he figure what the 2th focus to fine the ellipse. The sun is the first one but the 2th outch. Does it have to do with the astronomical measureement. It seem that the 2th focus is empty, it is right in my face I just can not seet the tree for the forest. Please try not to laught to hard. airhead ron oh yes thank you if you do find the time someone.........
airhead

Jacksonville, Florida USA - Saturday, January 31, 2004 at 11:44:43 (EST)

Hi all, I've been having problems with a homework question and just wanted to find out if anyone knows how to do it. The question is:

1. A woman wants to make ice for a party she is giving. She fills her ice cube trays with cold water which comes out of her tape at 50 degrees. The freezer is set at 10 degrees. 30 minutes later, the water in the freezer cooled to 40 degrees. How long would it take for the water to reach 32 degrees from the point when it was put put into the freeze?

Any idea? Thanks.
Ray

Kansas Citiy, KS USA - Friday, January 23, 2004 at 12:45:37 (EST)

Does anyone know where I could find good information relating to Functional Models relative to Calc for Business, Econ, and the Social and Life Sciences on the Internet?? I am struggling with this particular portion of the class as it is based on "word" problems.
Gary Burchfield II

Lexington, Ky USA - Wednesday, January 21, 2004 at 22:16:52 (EST)

Reply to matthew: Your diff. eq. is separable and becomes:

    dy/(ky ln(L/y)  =  dx

You are correct to use the log identity to convert this to

    dy/(ky(ln(L) - ln(y))  =  dx

Note that ln(L) is constant. So integrating the left side, you can use the substitution:

    u  =  ln(L) - ln(y)

   du  =  -dy/y

and your equation becomes:

    -du/ku  =  dx

which is easy to integrate. The general solution (if I haven't made any mistakes) is -(1/k)ln(ln(L/y)) = x + C. Of course if you want y as a function of x, you still have a little work to do.
Karl <Click to Send Email to Karl>
USA - Tuesday, January 20, 2004 at 19:03:45 (EST)

Hi Karl, I was wondering if you could help me to find the antiderivative the "Gompertz Growth Model" Equation, which is (dy/dx = ky(ln L/y)) where k is a constant and so is L (but is also the population limit). I tried converting ln L/y to Ln L - ln Y... but it wouldn't work because tere was a Y multiplied to each term. Using separation of variables, how should I do this? -Matthew
matthew
USA - Sunday, January 18, 2004 at 16:49:16 (EST)

Reply to Matthew: Divide both sides by x to get

   y' + y/x  =  x^2y

   y' + y/x - x^2y  =  0

   y' + y(1/x - x^2)  =  0

which is now in the standard form to which you can apply the formula for linear first order diff. eq.
Karl <Click to Send Email to Karl>
USA - Thursday, January 15, 2004 at 13:20:05 (EST)

Reply to Matthew: Divide both sides by x to get

   y' + y/x  =  x^2y

   y' + y/x - x^2y  =  0

   y' + y(1/x - x^2)  =  0

which is now in the standard form to which you can apply the formula for linear first order diff. eq.
Karl <Click to Send Email to Karl>
USA - Thursday, January 15, 2004 at 13:17:32 (EST)

HELP SOLVE! SOLVE THE INEQUALITIES

2-ABS1-3YABS >ET 12

nikki







sacramento, USA - Monday, January 12, 2004 at 22:17:39 (EST)

Hi Karl! I am having trouble on the topic of first-order linear differential equations with the given equation xy' + y = x³y. And what I tried to do was to find the standard form of the equation as: y' + P(x)y = Q(x). But I wasn't sure what to do because there are two y's (one on the left side and one on the right). How do I attempt to do this? Thanks. -Matthew
matthew
USA - Monday, January 12, 2004 at 14:48:05 (EST)

Reply to Misty: You know that when x = 11, then f(x) = 11. That makes it easy to find f(14) because 14 is three more than 11.

    f(14)  =  f(11 + 3)  =  [f(11) - 1] / [f(11) + 1]  =  10/12

Now that you know it for 14, you can use the same method to find it for 17. Keep going for several more iterations, finding it for 20, 23, and so on. You will see a pattern. Once you know the repeating pattern, you can find f(x) for any x that is 2 more than a multiple of 3.
Karl <Click to Send Email to Karl>
USA - Sunday, January 11, 2004 at 23:25:39 (EST)

I am blanking on a problem. It reads: Given that f(11)=11 and f(x+3)=[f(x)-1]/[f(x)+1] solve for f(8) f(14) f(17) f(20) f(2000) How do I get this one?
Misty

USA - Sunday, January 11, 2004 at 15:25:28 (EST)

I've a problem with using the definition of a limit to show the following limit statement is true: Lim (x -> -1) x^2 = 1 I am missing a basic concept here and have gone astray....for I end up with a delta = to a negative fraction of epsilon???? Any help would be greatly appreciated.
Lloyd

Okeechobee, FL USA - Wednesday, December 31, 2003 at 21:12:40 (EST)

I am land surveyor in California and I'm studying more about geodesy. I am not very good with calculus - but I am getting better. The National Geodetic Survey (NGS) has a suite of programs that allow surveyors to perform geodetic calculations, datum transformations, and conversions (and much more). Most current geodetic datums use an ellipsoid of revolution (spheroid) with the axis of rotation coincient with the (mean) north and south poles to appriximate the shape of the earth. The major axis of the ellipse runs along the equatorial plane and the minor axis runs through the poles. They define geodetic coordinates for any point on the earth in terms of longitude, latitude, and ellipsoid height - where the longitude (angle in the equatorial plane) is measured east or west from Greenwich (England), the height is the distance above or below the ellipsoid (measured NORMAL to ellipse), and the latitude is the angle (north or south) resulting from the intersection of this NORMAL to the ellipsoid with the equitorial plane. This 'NORMAL' definition for latitude is necessary so that the latitude is INDEPENDENT of the height component. (I understand the difference between the ellipsoid and the geoid - this not an issue for this problem). They also define an 3-dimensional earth centered (geocentric) cartesian coordinate system with the origin (0,0,0) at the earth's center. The positive x-axis runs towards the prime meridian (0-degrees longitude) and the positive y-axis runs towards +90-degree longitude - both in the equatorial plane. The positive z-axis runs towards the north pole - perpendicular the xy-plane. Most of the United States has negative x and y cartesian values and a positive z value (northern hemisphere). One common calculation involves converting from cartesian (x,y,z) to geodetic (lat./lon./ell-ht.) coordinates. The longitude is easy to compute using trigonometry from the x and y values, but the latitude and ellipsoid height are much more difficult. We can reduce this to a 2-dimensional problem by 'cutting' the ellipsoid with a meridianal plane (perpendicular to the equator) through the point (x,y,z) and the z-axis. This produces an ellipse, still centered at (0,0,0) with the new x'-axis along the major axis and the new y'-axis (same as old z-axis) along the minor axis. The original point (x,y,z) may be represented by x' = SQRT(x² + y²) and y' = z - said point may fall above, below, or ON the ellipse. THE GENERAL PROBLEM IS: solve for the equation of the line going through point (x',y') AND NORMAL to the ellipse. This would be quite simple if the datum surface was actually a true sphere (the normal also going through the center of the circle), but it will involve calculus (and parametric equations, I think) to solve this for the ellipse. The major radius (a) and 1/f values are the usual defining values for geodetic ellipsoids - allowing us to calulate the minor radius (b) and location of the two foci (x=c and x=-c). Since the earth is ALMOST a sphere, the foci are rather close the center (origin) and x'>c is true except near the poles. We know that the formula for an ellipse is x²/a² + y²/b² = 1, with a>b. We know the sum of the distances from one focus --> any point on the ellipse --> the other focus is 2a. We also know that the NORMAL line will bisect the angle created by the lines connecting a point to both foci (and this line will, generally, NOT go through the center of ellipse). I'm not really sure where to start on this and any help would be appreciated.
Bill Hurdle

Tehachapi, CA USA - Thursday, December 25, 2003 at 15:31:27 (EST)

Hi - Is this correct?


lim     tan
x->0+

1 <---the answer?
Pete

USA - Sunday, December 14, 2003 at 19:58:12 (EST)

Please can someone help with this it is driving me crazy Determine the rate of change of m (to 4 significant figures) if M= 2ln (v/3) + w3 sec u and u = 3 cm, v = 3.7 cm and w= 4.4.cm, given that v is decreasing at 0.4 cm/s, u is increasing at 0.2 cm/s and w is increasing at 0.7 cm/s. Please help if you can many thanks Mark
Mark Hutton

Coventry, UK - Sunday, December 14, 2003 at 08:10:22 (EST)

Hey, I'm having a really tough time with this calculus problem, it states: A roman window is shaped like a rectangle surmounted by a semicircle. If the perimeter of the window is L feet, what are the dimensions of the window of maximum area?
Jen

Amherst, MA USA - Saturday, December 13, 2003 at 23:10:00 (EST)

Reply to Anthony S: Send me an email and I'll do this LaPlace problem out for you. If you want to try it yourself, here's a hint. Replace the sin function with the Euler expression for sine in terms of exponentials with imaginary arguments. Then do the resulting LaPlace integral by parts.

   infinity
   
   0

  t(e^iwt - e^-iwt)
                 e^-st dt
       2i

Multiply the e^-st through the numerator, remembering that the product of exponentials is the exponential of the sum. Remember also that i is a constant, and that i² = -1. Integrate by parts in the usual way, and you should get the result in the book.
Karl <Click to Send Email to Karl>
USA - Tuesday, December 09, 2003 at 08:34:11 (EST)

Reply to Matthew: Multiply top and bottom by cos(x). That will give you cos⁴(x) in the denominator, which is the same as (1 - sin²(x))². Now make the substitution, u = sin(x) and du = cos(x) dx. After that it becomes a 4th degree partial fractions problem.
Karl <Click to Send Email to Karl>
USA - Tuesday, December 09, 2003 at 08:12:55 (EST)

Hi Karl, I was wondering how does one solve the integral of sin x + 1/(cos x)³. I thought maybe you could use a trigonometric identity for the cos x, such as (cos x)²*(cos x), and then replacing (cos x)² with 1-(sin x)² but I'm not sure. Tell me what you think. Thank you!
Matthew
USA - Saturday, December 06, 2003 at 20:29:23 (EST)

Hi Karl. I had a question about an improper integral problem that has been bugging me. It is the integral from zero to 2 of the function: 1 dx/(x-1)^4/3. I broke the integral into two parts, one from 0->1 and the other from 1->2 where the 0->1 I took the limit from 1 to the left and the 1->2 I took the limit from 1 to the right. Am I doing this right? If so, doesn't the denominator reach a really small number so the number gets really big (for example like: 1/.000000000000000001)? Is it divergent? or Convergent? Thank you Karl! -Matthew
Matthew

Davis, CA USA - Tuesday, December 02, 2003 at 23:32:34 (EST)

Can you show the work on how to do this problem? f(t) = t*sin(w*t). It is a Laplace Transformation. The solution is: (2*w*s)/(s^2+a^2)^2
Anthony S
USA - Monday, December 01, 2003 at 20:08:47 (EST)

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