it seems as if your school has started now and thus you are flooding us with your Math homewoork problems! Well, obviously you're welcome to do that... and that's why we're here.
To help you and guide you. Now my attempt to clarify your mentioned concepts:
Aahh, you know how to differentiate and integrate? Well, basically, differentiation is a process to find out the gradeint / slope of a curve.
Integration is just the opp! (The thing to be used in your case.) (*even the procedure/method of integration is very much the opposite of the technique to find the differential-value!!!)
Integration is particularly used to find the eq. of a parabola.
Umn... the process varies from problem to problem. However, the basic thingee is as follows:
(OhM'Gosh! I'm getting late for my class! K9k, I promise I'll tell later.)
Till then, Pin~Jinx ******** 08-31-02, 09:13 AM silenteuphony Here are the standard equations for various parabolas with a vertex at (h,k):
(x-h)² = 4p(y-k) opens upward
(x-h)² = -4p(y-k) opens downward
(y-k)² = 4p(x-h) opens to the right
(y-k)² = -4p(x-h) opens to the left
where p is the distance between the focus and the vertex, or equivalently, the distance between the vertex and the directrix. ******** 08-31-02, 09:13 AM maiku I assume you mean to say that, besides knowing the graph you have is a parabola, you know some of the coordinates. If you know at least three of these, you can find the quadratic equation as follows:
In the equation (1) y = ax²+bx+c simply substitute 3 pairs of the coordinates x,y to get three simultaneous equations in the cœffiecients a,b, and c. Solve these, and substitute the results back into (1).
Here is an example. Suppose you know the graph passes through the points (0,-1), (1,4), and (-1,-2). Then substitution in (1) will give you the equations
c= -1 a+b+c=4,and a-b+c=-2. Solving these, you get a=2,b=3, and c=-1, so the quadratic you want is y = 2x²+3x-1.
[This message was edited by maiku on 08-31-02 at 09:31 AM.] ******** 08-31-02, 02:37 PM ko9000 Thanks alot guys big grin ******** 09-01-02, 06:28 AM Pin~Jinx ......but I guess I returned too late!
And SE and maiku have already done a better job than me!
So, just TAKE CARE!
Pin~Jinx ******** 09-01-02, 06:28 AM Pin~Jinx ......but I guess I returned too late!
And SE and maiku have already done a better job than me!
So, just TAKE CARE!
Pin~Jinx red face ******** 06-16-05, 01:40 PM Ballsy Hello,
I have the similiar question, the basis of which is to solve for the heights on a powerline - which I understand is a parabola opening upward. The challenge that I face is - besides being very rusty, - I will have two different elevations for the beginning and the end of the arc. As well I don't remember a lot of the jargon - focus, vertex, directric- it's all greek to me. ******** 06-16-05, 07:17 PM frankvan Ballsy, welcome to the pool. If memory serves, and it often doesn't, I seem to recall that a powerline assumes a curve called a "catenary", not a parabola. The formula for a catenary is y=coshx which is (e^x - e^-x)/2. One of these younger enthusiasts will probably confirm or deny, I hope. Wink ******** 06-16-05, 08:17 PM DorianGreyed I can confirm the "catenary" part. But I no longer remember enough to say that Frank's formula is correct. ******** 06-16-05, 10:03 PM Professor Yes, the hanging cable is a catenary which is described by the hyperbolic cosine function of a form given by Frankvan.
Of course it looks like a parabola, but in mathematics looks are not only deceiving but generally irrelevant. About all you can say they have in common is concave upward and bilaterally symmetric.
Gee, I never thought of myself as "one of the younger" contributors. Frank, did you personally know Archimedes? Big Grin ******** 06-17-05, 09:36 AM frankvan Well, no! But I did vote for FDR once. Wink ******** 06-17-05, 10:59 PM gerry A cable hanging under its own weight does take the shape of the catenary, but when it carries a uniform load across the entire span, such as snow or ice on powerlines, or such as the the bridge deck on cable suspension bridges like NY's glorious George Washington, it takes the shape of a parabola. For spans that are not very long and without 'steep' sags, even the catenary curve for cables hanging under their own weight may be approximated by the parabola. It's been a while since I did the math, but knowing the general form of the parabola y = ax2 + bx + c, then you need 3 known points to get the eqauation, per Maiku's response. For the powerline example, two of the known points would be the support points at either end of the span. You still need a third point, and this third point would depend on how tight the cable was pulled (it's tension, T), or, alternatively, how much sag (H)was in the span (sag and tension are related bt T = wL2/8H). The sag in the cable is the vertical distance between 2 parallel lines, one drawn between the two support points, and the other tangent to the parabola at the curve's low point. ******** 06-18-05, 08:51 AM gerry
quote: Originally posted by gerry: CORRECTION:The sag in the cable is the vertical distance between 2 parallel lines, one drawn between the two support points, and the other passing through a single point on the curve.
This message has been edited. Last edited by: DorianGreyed,
