There's plenty here:
But please make your own words and diagrams rather than copying directly.
I don't mind my post(with credits ) being copied. Or you could put a link from your forum to ours.
If I call the origin (0,0) point O and the lowest point on the curve L, then OL = FL (from the definition)
And I'll call the point on the y axis level with P, point Q.
Now it looks like you are saying PQ = 120. But where is 112 measured from and to. Most likely from Q but to F or L or even O ?
To use the equation accurately you'll need to work out f.
F (0,f) is the fixed focus on the y axis. D (x,0) is a moving point on the x axis directly below the general point on the parabola, P (x,y). The x axis has a special name for this parabola; it is called the directrix.
A parabola is defined as the locus of all points, P, such that the distance from P to the directrix is equal to the distance of P from the focus. Thus
differentiating with respect to x:
This is the gradient of a tangent AB to the curve at P.
On my diagram, the parabola is shown as a thick line. AB is the tangent to the curve at P. As PF = PD a circle may be drawn, centre P and radius PF = PD. DP extended cuts the circle again at C.
FD has gradient -f/x and so is at right angles to AB. So FPD is an isosceles triangle, bisected by AB.
So if angle FPD = 2 θ, then FPB = θ and using the angle properties of a circle, FCD = θ as well.
Let EP be at right angles to AB. Then CPE = 90 - θ = FPE . These angles are the angle of incidence and the angle of reflection of a light ray coming from C at right angles to the directrix, reflected at P towards F. Thus parallel rays perpendicular to the directrix reflect off the parabolic surface to go through the focus F.
And, making the rays start at F, a parallel beam of light is produced.
hi Morten Fredriksen
Welcome to the forum.
It is a property of the parabola that rays starting from the focus will reflect off the mirrored surface in parallel rays.** The reverse is also true: rays coming in towards a parabolic mirror will all go through the focus (hence the name). Reflecting telescopes utilise this property to produce focussed images of astronomical objects.
Remember the problems when the Hubble telescope was first launched. In essence, this was due to an incorrect shape for the mirror. https://en.wikipedia.org/wiki/Hubble_Sp … he_problem
If you had a perfect parabolic reflector and a point light source and were shining in a perfect vacuum, the light would go on for ever! In practice you'll never make a perfect mirror, the light source will not be a single point so not all light will originate from the focus, and finally, light will hit dust particles etc and so get scattered.
For a given specification of outer diameter and depth to the centre of the mirror, a parabola can be determined. Then it's a question of how good is your engineering? As well as worrying about the power of the light source, you need to make it as small as possible. And then take it to a high mountain away from cities in still air just after rain, to minimise the dust.
Sorry, I haven't read all those posts; too much this early in the morning (or any time come to think of it)
And your English is very good!
** I think I can produce a coordinate geometry proof of this if you want.
Q2. Parts 1 and 2
Initially, I'm assuming real values only
Here's a possible example to consider:
If values can be complex then the (x² +2) and (x^4+7) terms would not exist since these would also give rise to zeros, and we're told the only zeros are as given.
But, for example, f could have a factor (x+3)^2 and still satisfy the requirements.
This might be interesting to you. I certainly find it interesting.
Start with the Fibonacci sequence:
1, 1, 2, 3, 5, 8, 13, 21, 34, .....
Divide each term by the one before:
1/1 = 1
2/1 = 2
3/2 = 1.5
5/3 = 1.6666
8/5 = 1.6
The numbers converge on the golden ratio (phi) ; oscillating either side of that number.
If you compute 1/1, 1/2, 2/3, 3/5, 5/8, .......
the numbers converge to phi - 1, which is also 1/phi
Apparently, the sequence (of rationals converging on an irrational) has a special property (which I've forgotten for the moment) which means that a plant (such as a sunflower), growing a seed head from the centre, and with angular separation atan(phi), will be least likely to form cleavage planes which might cause the seed head to fall apart. So nature discovered the property first!
I googled the name and found that it is a Rubik cube with constraints that stop the faces from turning both ways. Since three clockwise turns of 90 is equivalent to one anticlockwise turn of 90, I assume the constraint has something more complex, to prevent this trick. I haven't actually tried one myself so I'm hoping the poster will enlighten us.
Welcome to the forum.
I had to google that name but it seems to be a Rubik cube.
I can tell you how but the explanation will take many posts.
EDIT: Just had a look at a u tube video. Seems you haven't got freedom to rotate a face any way you want. In which case, what are the constraints? This site has an explanation
hi Hannibal lecter
This diagram may help. The domain is the set of all t that the function, f, may be applied to. The range is the set of all C that result from the function.
I've also included another word, the codomain. The codomain is the set that contains the range. In some cases the range and the codomain may be the same anyway but for some questions it might be useful to be able to consider elements that are not mapped onto by the function.
For example if both the domain and the codomain are the full set of real numbers and the function f is t ---> t^2 then the range will only be those numbers that are zero or positive. Negative numbers lie in the codomain but no values map onto them, so they're not in the range.
When someone presents a function they should also tell you the domain so you know what values may be considered. Some questioners give a function and expect the student to provide the domain. I consider this to be poor practice since it is the responsibility of the questioner to specify this.
But with a function, a domain and a codomain a student should be able to work out the range for themselves.
To answer your specific questions:
where is the domain? and where is the range? The domain is the set of values of t that are to be considered. The range is the set of values of C that result from applying the function to values of t.
and where is the input and where is the output? The input is the possible values of t. The output is the resulting values of C.
I mean is C is a range or domain? or is it a domain.. Range and domain are sets. C represents a typical member of the range.
A further example: Let the domain be angles in degrees from 0 to 90 inclusive. Let f be the function sine.
If t = 30 then C = sine(30) = 0.5
If t = 60 then C = sine(60) = 0.866....
I can work out that the range is numbers from 0 to 1 inclusive.
Hope that helps,
Welcome to the forum.
Here's my thoughts on this: I'll refer to each little cube by using an whole number coordinate system; so the bottom left front cube is (1,1,1) and the one on its right is (2,1,1). Behind that is (2,2,1) and so on.
Consider the cube in position (2,3,4). If I extend its sides to the edges of the large cube I get a cuboid 2 by 3 by 4 so volume 24. The ones on a diagonal will have coordinates (a, a, a) where a is a number from 1 to 8 inclusive. The one at (4,4,4) will, when extended, make a cube 4x4x4. The one at (5,5,5) will make a smallest volume that is also 4x4x4 (go to the shortest outside edge).
Hope that helps
For number one the right hand side looks more complicated so I tried to make RHS into the LHS.
That gave me the clue to do it the other way round, ie. LHS => RHS.
Collect all the + terms together and factorise out an 'a'. Then all the - terms and factorise out a 'b'.
I think you can do the others similarly.
hi Zeeshan 01
You can always put in a value for x and test this for yourself. If the numbers do not agree then the algebra is wrong.
Let us examine what you have there. Two separate functions: first times x by 3 and second find the cosine of the result.
You cannot multiply together cos(3x) with another cos(3x) to make cos(9x^2). Functions don't work like that because the cos comes before the multiplying.
Now consider adding another function, 'square' to the sequence of functions.
You could start with x, square it, then times by 3, then find the cosine. .................................A
Or you could start with x, then times by 3, then find the cosine, then square the result. ..................................B
If you try these with different values of x, you will find they give different results. So it is important to show people which you mean when you write the algebra.
A is written like this:
You could write B like this:
but mathematicians have agreed to this shorthand for it:
Putting the squared immediately after the cos should be taken to mean work out the cos first and then square the result.
Welcome to the forum.
Have a look here for information on equations of straight lines:
the cost varies linearly with the distance
So there is a formula connecting cost and distance that looks like this:
You can substitute the given values to determine the constants b and m and then the rest should fall into place.
hi Monox D. I-Fly
Q3. P is (x,y), a point on an unknown curve C in the x-y plane. Form an expression for each gradient in terms of x, y, a, and b and multiply to make k.
For a circle the coefficients of x^2 and y^2 must be equal. That's enough to fix k.
Q4. What do you mean by 'meets'? I assumed this meant "has the x axis as a tangent at that point" but that doesn't give a sensible diagram. So do you mean 'goes through' the point?
Also is that y = (root3) times x or root(3 times x) ?
If the former then this is a straight line at 60 degrees so the angle bisector of this line and the x axis will be at 30 degrees and the centre lies on this bisector.
If the latter, it seems you need to differentiate as the circle and curve will have a common tangent at the point. That makes the question very complicated (well it is for me ). I've tried a diagram and cannot get such a tangent at all. Best I can get with my software is centre at about (2.78, -1.34) buit don't rely on that as accurate.
Q5. Again the word meets. Just touches or goes through ?
Later edit: Q5. If we assume that x+y+3 is a tangent to the circle at (2,1), then we can calculate the equation through (2,1) perpendicular to x+y=3.
The centre will lie on this line so you can get a second equation connecting a and b.
bobbym: The circle has centre (a,b) in conic section notation.
What I am trying to say is "It could be either". Someone had to decide and that's what they chose. Descartes https://en.wikipedia.org/wiki/Ren%C3%A9 … cal_legacy invented the x-y coordinate system and everyone has kept to his definition.