Math Is Fun Forum

As previously stated, I gave it an implicit plot. I determined the domain by finding where the radicand was zero and because I knew the "letter" would be in height some value close to its width I just made the range equal to its domain. In the case of the letter N I believe it was from x = 2..4.

The plot didn't look that great but it may be fixed by upping the number of points with the numpoint option. I did one letter because the equations are gigantic. I just thought it was a fun program. It would be more fun if you could copy and paste the math into something like Maple but all it yields is a picture of the equation.

Yes, thank you.

Are there any good books on set notation? I want to learn to write more formally in mathematics but the web sites I have looked at thus far are not so great.

Thanks again.

Actually you probably meant

but I knew what you meant.

Thank you! That is exactly what I was looking for.

Thanks. I'm getting there.

Ah, I remember. If the rational expression is a hyperbola that means it is a conic section; if a conic section then it is defined as the intersection of a plane with a double cone.

What is the equation of that plane that intersects the cone - for f(x) not the standard form of a hyperbola? I have read about it on wikipedia but I am not sure how to go about finding the equation for the plane that represents f(x) itself which, as I have shown, is rotated and shifted.

Thanks. And good luck with your gardening. You must not have too much more to do given the time of year unless you live in a warm part of the globe. Where I live we are guaranteed frost-free-ness until early October or so. Or so I have heard. I don't have a garden so I don't pay that much attention.

A "proof" is here given. I put proof in quotations as this is rather inductive and I still do not know how to go from one to the other. It could be very difficult given the thing is full of transcendentals but maybe there is an easier way than the following I have yet to see. As it is, graphing can show the following is apparently accurate.

That a rational expression of the form

is a rectangular hyperbola of the standard form

...

The value of "a" has been analytically found to be that of

and the center (h,k) is given by

both of which are values crucial to the problem. Please note that "a" in the hyperbola is not the same "a" in f. I went ahead and used the same symbol as we use the hyperbolic a very little. The difference between them is obvious based on the context of my work.

We can solve the equation for y,

but as it is unclear what to do with that we can switch to a parametric form of the hyperbola

The transformation matrix for the rotation of the parametric system by some angle is that of

such that its application for a 7π/4 rotation upon our vector curve <x,y> yields the vector-valued functions

Next, the curve, while the correct shape, is in the wrong location on the plane and simply needs shifted over and up by the original center of the hyperbola which yields the solution

I apologize for any errors made in my work.

Example: If

then f is a rectangular hyperbola. In standard form that hyperbola has the equation

Omitting all the steps already given in the general form above, the solution is

If f(x) is plotted so as to go from x = 0 to x = 5/3 the parametric equivalent is approximately t=-0.3398369094 to t=0.3398369094 or, if you prefer, the exact value is

The reason for choosing these bounds is completely by preference; it is where the domain is equal to the range such that f(5/3) = 5/3.

anonimnystefy: They are the same kind of graph only, as I have said, different by 45 degrees in a rotational sense. Suppose you want to get a rational function that is some hyperbola without knowing p, q, or r or anything about it such as its center, foci, vertices, etc. and all you have is a hyperbola in standard form in terms of a and perhaps b. How do you get the hyperbolic "standard" into the form of the rational function? That is the problem I am seeking help with.

Thanks to all who have and will reply. I am grateful for this forum and for your help.

Certainly. For instance, I believe its center is at

and the vertices, foci, etc. can be found.

Bob,

Thank you for your quick many-step response! I am working out what you have said and will get back to you.

I'm so sorry - I didn't see your questions from a couple of months ago.

The first-order differential equation is not separable because it is a subtraction problem and not one of multiplication or division - see how it has a minus sign. And so we need what is called an integrating factor. Here are the steps:

1) Move the part of the equation on the right hand side with the dependent variable in it to the left hand side.

2) Compute the integrating factor. That is, make up an equation where e is raised to the power of the integral of the dependent variable's maths without the independent variable itself. Here is a youtube video that teaches about the integrating factor:

http://www.youtube.com/watch?v=Et4Y41ZNyao

3) Once you have the integrating factor you can plug it in as shown in line 4 of my work you quoted.

Sorry again for not getting back to you. I am preparing for my final and am reviewing my old posts and just happened to see your questions. I hope this has helped you out. I can provide a few of my own examples with steps if you would like!