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I don't know what you mean.
Here is my attempt after looking in my old calc book.
And so my system of equations would be...
and
correct? Does this system solve for a = b? Is that how it should be integrated?
Okay... so... I guess the key is to integrate by partial fractions because it gives a friendler form for answering the question.
You would have
Let
I do not remember how to integrate by partial fractions though. What are the steps for integrating the left side using partial fractions?
So is B/3 supposed to be used in the differential equation, in the function A(t), or in a new derivative formed from the particular solution? It says to show that t = B/3 is where change occurs the quickest so it must have something to do with one of the two derivatives - the original or the derivative of A(t) as we have it.
I did not think it mattered since C is whatever it ends up being in the long run.
Is that better? I just skipped to the end.
No, that's just the way I typed it. The problem says jointly proportional to
.So the way it is typed in my work is correct to the problem. Good try though, that would have been better.
Hmmm...
Hey...
Well, I hadn't integrated yet. I separated the variables and had the equality
and you solved the left side... and it's the same answer I was getting out of Maple, so thanks for verifying it!
Now this needs solved for A(t):
I assume it should just be solved algebraically. Here is my effort:
And so that is A(t). That is so complicated, though. Am I even moving in the right direction? What is the question really asking in relation to this?
Thanks for always hanging around to answer questions.
"Let A(t) represent the area of a tissue culture at time t and let B be the final area of the culture when growth is complete. Most cell divisions occur on the periphery of the tissue and the number of cells on the periphery is proportional to the root of A(t). A reasonable model for the growth of tissue is obtained by assuming that the rate of growth of the area is jointly proportional to the root of A(t) and B - A(t). Formulate a differential equation and use it to show that the tissue grows fastest when t = B/3. Solve the differential equation for A(t). (Assume A(0) = 0.)"
In solving this problem I so far have
Separating produces
And solving this in a computer program yields a horrific answer and I don't think a computer is supposed to be used in the first place, though I suppose it is okay.
Do I have this set up correctly?
Thanks!
P. S. The solution of the ODE can be solved by hand. Here is how:
And solving for v(t) gives the general solution
Letting v(0) = 0 finds the particular solution
And you can factor out the g/k:
And so, in stating the answer as the original limit in terms of g and k,
In words: "The terminal velocity of a raindrop with mass m and velocity at time t is directly proportional to the acceleration due to gravity and inversely proportional to some constant k."
Thanks again for your help! Now I can see how it all goes together.
Neat! Now would this have been an easily-found solution by hand or no? I am still learning separation of variables and all that to do with natural logs, general solutions, and specific solutions.
13.
Sorry to keep withholding information. ![]()
The one you just solved in Mathematica is a good example.
In Maple I typed
to open the DETool box, then
In hopes of solving the differential equation. But I haven't really learned much about Maple's DE tools yet.
bobbym,
Thanks! I actually thought your first edit, the first three lines, was your final response and I worked out most of the rest myself. Which is good for me. But I do not know how to use Mathematica. I use Maple. Do you know anything about solving equations with generic constants such as g and c and k in Maple?
Thanks a lot for your help! I am glad I am now understanding this. I have not taken physics so all the stuff about gravity and proportioned mass to time kind of threw me off.
This is a problem from an introductory ODE course.
"When a raindrop falls, it increases in size and so its mass at time t is a function of t, namely m(t). The rate of growth of the mass is km(t) for some positive constant k. When we apply Newton's Law of Motion to the raindrop, we get (mv)' = gm, where v is the velocity of the raindrop (directed downward) and g is the acceleration due to gravity. The terminal velocity of the raindrop is the limit as t goes to infinity for v(t). Find an expression for the terminal velocity in terms of g and k."
For those in the wake, the full solution is shown in the posts below.
bobbym,
Thank you for getting back to me so quickly! I appreciate and understand your answer. However, in polar and spherical coordinates, are shapes not still based on coordinate axes? At least, in my last class, shapes were still drawn in the octanes, simply defined by functions converted from rectangular to spherical coordinates.
My question was with regards to situations where space curves might be redefined as linear lines on curved coordinate systems. Unfortunately I don't really know well enough of what I am talking about to explain it better than that so I could try giving an example instead.
Suppose you have some basic vector-valued space curve in rectangular coordinates and with some parameter t, such as
and for whatever reason you want to describe a coordinate system where r(t) - in 3D space - may be described as a straight line on a curved "2D" surface - what I mean is, simply a curved coordinate system. r(t) would be a straight line on a curved surface.
Sorry ahead of time if I am asking illogical questions. ![]()
What sort of "transformation" would r(t) undergo?
Hello. ![]()
I am interested in learning about curvilinear coordinates. I did a search on these forums and found only one use of the phrase and said incident was irrelevant to my question(s).
What I would like to know is, first of all, are curvilinear coordinates strictly a topic of tensor calculus? If so, need I first understand tensors in order to understand the topic of curvilinear coordinates?
My reason for asking is a curiosity in transforming a curve in space into a linear line on curve coordinate axes. I do not have any practical applications, I simply would like to learn about the concept.
How does ones "transform" a vector-valued space curve into a linear line on a curved coordinate system? Maple is one of my resources at hand.
I appreciate any and all answers. Thank you.
Yeah. I suppose I may have to settle for a half-accurate curve when dealing with graphic representations of 3D surfaces. I appreciate the value of splines, I just preferred a single function. It seemed cleaner.
With Maple you can plot a couple of points more along the curve than just the three linear points, as I said, I was trying out the mean values and quarter and three-quarter lengths of the arcs, but the resultant curves were really bad. I figured I could either fit a ton of points or try to figure something else out more logical.
Fitting a thousand different points seemed like cheating... and it created gigantic functions, anyway. I was hoping the solution would be something simplistic and beautiful in that simplicity.
I am a perfectionist. If there was something I could do over and over and get a better fit every time, I wouldn't stop doing it until I had it perfect. By near perfect I meant thousandths of a decimal off.
Exactly. A piecewise made into a single function h(x) or whatever we were calling it.
Part of my obsession with having one function instead of piecewise functions was experimentation I am doing in 3D surfaces which don't look so great when having to work with splines.
But, yes, a single function is the goal. One that matches the piecewise perfectly or at least extremely closely.
I will need some time to think.
Thank you in advance.
Correct - h(x), the unknown blue curve, is a piecewise function of f(x) from 0 to 75 and g(x) from 75 to 150. The ultimate goal is to fit a single curve defined by a single function.
The reason for using Thiele's Interpolation Formula was that it was a feature in Maple, yes. However, Maple also features polynomial, splines, linear, and so on but Thiele's formula does a perfect job on the regular curves shown in the majority of those pictures I uploaded.
The problem arises when the three primary points of the h(x) - the blue curve - are attempted to be fit for the points - (0,0), (75,25), and (150,50) - fit a perfectly linear line. But if more points from the original two functions f and g are chosen, say, the points of mean value, Thiele's formula does not fit very accurately if it fits at all for it is easy to hit vertical asymptotes as the formula divides by zero.
Polynomial interpolation does a terrible job of fitting the ideal h(x) perfectly.
I have read online about various interpolation methods but most of them are beyond me. One that at least sounded promising is found in this link:
http://en.wikipedia.org/wiki/Whittaker%E2%80%93Shannon_interpolation_formula
However it is quite beyond me. Do you happen to know if that is an idea worth trying? I am willing to learn anything new, I am just out of ideas on my end.
I will be taking differential equations starting next month and I know the course has something to do with finding unknown functions. Since you have probably already taken such a course, do you happen to know if there is anything similar to this in said class?
Thanks a lot for getting back to me as quickly as you did.
Alright, here are some images from Maple I made, color-coded in hopes to communicate effectively. On my monitor they look as though they exported with low detail to quality, but they are still readable.
Thanks!
bobbym,
1) Data? The two curves - f(x) and its inverse and on the stated boundaries - provide an infinite number of data points. Take your pick.
2)
Can I ask what you are trying to accomplish by using Thiele's Interpolation to begin with?
Seeing as I have already performed an application of it, it should be fairly obvious "what I am trying to accomplish".
3.
...you use Maple which places you ahead of lots of mathematicians.
My calculus class in junior college just happens to use Maple. It doesn't affect my amateurish status for good or not. Does it matter?
...
I apologize if my response is annoying, I simply didn't realize the question would be so difficult when it is the problem that has troubled me so. Perhaps it would help make the scenario, which is abstract in itself, easier for you if you imagine, say, a new road is being built that is half of an old function and half of the old function's inverse.
Perhaps I have simply made a mess of everything in attempting to explain it. An easier approach might be to simply start learning about new interpolation methods. I have tried reading about them online but it isn't the same as having someone explain it with the chance for questions and answers.
Also, I will try to get some images to support the question in the hopes that pictures will make the idea easier to understand. I will post again soon with said pictures.
Thanks...
Thank you in advance to any who offer their knowledge to my aid.
I am quite an amateur in all that I do and even then mathematics is not my strong point so please forgive any ignorant remarks on my part.
If one uses Thiele's Interpolation Formula to fit a curve that passes through three symmetrical points such as
the result will be the function
where the derivative of f(75) is exactly 1 and all derivatives on the interval x=[0,75] have a reciprocal set on x=[75,100].
My difficult problem is this: I need to fit a curve that is the above function f on the interval of x=[0,75] and then is the inverse of f on the interval [75,150] such that the domain of the function (in so far as we are concerned with) is [0,150], the range is [0,50], the derivative at x=75 is still 1 and all derivatives from opposite directions and toward x=75 are equal. That is, the derivative of this new function at x=0 is 1/9 and the derivative of this new function at x=150 is 1/9 and all derivatives from the two end points toward the middle at x=75 are also equal.
That may be confusing and so, to restate it, the function f, above, is the portion of a new function (say, h(x)) on its own interval x=[0,75] and at that point, where the derivative of the function is 1, the function needs to transition into its own inverse at the same point. The inverse of the function f, dislocated so as to connect to the function at the point of x=75, is
and the desired new function is, as stated, f on the interval from 0 to 75 and then its "dislocated" inverse on its interval from x=75 to x=150.
The complex problems arise right from the start when the first thing you try is to fit the points
which is a linear line.
I would provide you all with graphic representation to make clearer the objective but I do not have anywhere to host such a picture at this time. Plotting the function and its inverse and on said boundaries in Maple and using the multiplot feature display(p1,p2,etc) will show the desired curve.
I have no idea if this is something that has an easy solution or not. I am more than willing to learn new things in order to carry it out, however, and I welcome any and all advice. So far all I have managed to do is fit curves that are imperfect... and I am frustrated.
Thank you for your time.