studiot

Perhaps you will remember the James Bond mnemonic OHMS But I don't see the point of the circle, you can do just as well by drawing a triangle if you have a protractor. If you don't have a protractor, but just have the angle drawn you can still get the sine but you won't know what the angle is then.

Friction and cohesion are entirely different forces with entirely different causes and properties. What was the rest of your post about please?

Always good to do it yourownself. Hope you also enhanced your knowledge of the properties of circles and how these properties all fit together as a result.

Parabolas are often used in highway engineering particularly for valley curves as they provide a natural slowing down near the bottom. I think, but I am not an expert in fairground rides, that they are used here as well as they give a much better roller coaster rush. I have not seen catenaries used in this respect. Some arched bridges have been constructed with a catenary profile in modern times however.

The rate law is determined experimentally and so is the 'correct' one in the sense that it is what actually happens. However that does not mean that the simple expression obtained from the stochiometric equation is useless. In fact comparison with the experimentally determined rate law tells us when something more complicated, such as intermediate reations, is going on. It also depends whether you insist on whole numbers in your stoichiometry. Take, for instance, a reaction that requires two different molecules A and B to meet and react and form products C and D. Then the rate of reaction will depend upon the presence of both A and B so the assumption is that A + B = C + D (stochiometry) v = k [A] ie is second order. But suppose that the reaction is actually a decomposition 2N2O5 = 4NO2 + O2 experimentally rate equation is v = k[N2O5] not v = k[N2O5]2 This is only first order. But of course the dinitrogen pentoxide molecule does not need to meet another to decompose. The stochiometric equation could in fact be written in terms of a single reactant molecule and fractions for the products.

One thing Imatfaal didn't seem to mention is that his alpha is twice gwiyomi's theta as shown here. Edit just found the hidden meaning in post#5 - spoiler.

You don't seem to have used my suggestion. Your original idea has taken you away from the radius which is a constant for a given circle and given you an expression in terms of the half chord length, which is variable. You really want to do the opposite ie work towards an expression that puts the area in terms of the circle constants and theta alone. Then you can differentiate and set to zero. Here is some help towards this. My final expression can be further simplified, in particular look for a simplification of (1+tan2)

A chord forms the base of your triangle. Angle theta is the angle subtended from the ends of the chord to the circumference to form your triangle at its apex. I suggest you use 2 theta rather than theta and work on half the triangle. Consider the angle subtended to the circumference on the other side of the chord to your isoceles triangle. this also forms an isoceles triangle. The half chord is a side common to both triangles and allows calculation of the division of the diameter.

20 is not the repeated root in Imatfaal's solution, y is 2.5 at this x.

Don't be, I had some fun with this question trying out different quadratics in my excel spreadsheet for comparison. I also looked at matching derivatives. A surveyor or engineer would probably not use direct junction but introduce 'transition curves' between the parabolas. I hope the extended posts have also helped dawoodr with his (her) English. As to the < and > sign I find that sometimes complicated TeX can be built up by placing adjacent multiple statements. I use the [math] [\math] tags in this forum as TeX doesn't seem to take. Finally IMHO we need to encourage more thought provoking questions like this one and less rubbish.

Like most forums this one has an incomplete Tex implementation. I get the the line LaTeX Error all too often. Anyway I'm glad it worked out. Don't forget there are many solutions. The one by Imatfaal has the advantage that it does not go below zero.

Ah. Thanks for the clarification.

There are infinitely many quadratics that pass through the point p ={10,10}. Imatfall has chosen a different one that sits on the x axis for its minimum, as given by this condition here "dy/dx = 0 when y =0 (3)" This is not unreasonable if y= 0 represents ground level. But this is not the reflection of the first curve, it is a different one. As a matter of interest was this a typo or can I not see the decimal point here "dy/dx(10) =-3 (2)" However your second equation appears to work It may interest some to know that the expression [math]a{x^2} + bx + c[/math] may also be written [math]a\left[ {{{\left( {x + \frac{b}{{2a}}} \right)}^2} - \frac{{{b^2} - 4ac}}{{4{a^2}}}} \right][/math] which has an extreme value when [math]x = \left( { - \frac{b}{{2a}}} \right)[/math] The value is [math]y = \frac{{4ac - {b^2}}}{{4a}}[/math] and is a maximum if a is negative and a minimum if a is positive.

OK it is more complicated than I first thought. So the general parabola is y = ax2 + bx + c The first parabola is as I already stated, but I will rework it in the new format. There are three points available on this parabola {x=-10, y=10} ; {x=0, y=25} ; {x=10, y=10} This leads to three simultaneous linear equations that we can solve for a1, b1 and c1 [math]100{a_1} - 10{b_1} + {c_1} = 10[/math] [math]0 + 0 + {c_1} = 25[/math] [math]100{a_1} + 10{b_1} + {c_1} = 10[/math] leading to [math]{a_1} = ( - 0.15)[/math] [math]{b_1} = 0[/math] [math]{c_1} = 25[/math] Thus the first polynomial P1 is [math]y = ( - 0.15){x^2} + 25[/math] If we turn this curve upside down and require it to pass through P ={10,10} it has a minimum of 15 units below (y=-5) and x displaced by 10 units along (x=20). By symmetry must also pass through the point x=30, y=10 So as before [math]100{a_2} + 10{b_2} + {c_2} = 10[/math] [math]400{a_2} + 20{b_2} + {c_2} = (-5)[/math] [math]900{a_2} + 30{b_2} + {c_2} = 10[/math] leading to [math]{a_2} = 0.15[/math] [math]{b_2} = ( - 6)[/math] [math]{c_2} = 55[/math] Thus the second polynomial is P2 = [math]y = 0.15{x^2} - 6x + 55[/math]

So you have two parabolas 1) y = ax2 + b 2) y' = a'x2 + b' The first one satisfies the points {y=25, x = 0} and {y=10, x =10} so you can calculate a and b by substitution, Leading to y = -15/100x2 + 25 So dy/dx is zero at x=0 so the max of this parabola is on the y axis at y=25. As Imatfaal says in theory you need another another point on the second one, say the value of x when y=0. If, as he says you choose an inverted form of this parabola, the constant b needs to be adjusted as in my sketch. The max of parabola 1 is (25-10 = 15) units above P. So the min of parabola 2 needs to be 15 units below P ie at y = -5 Edit the original offering had the wrong second parabola since The second parabola second also need to be shifted to the right so the minimum is not on the y axis. I will think some more about that tomorrow.

Well I did offer you 2 alternative proofs originally, but you ignored them both. If you must pursue your analysis you should carry your trigonometric manipulations through to the end. [math]\frac{{1 - \sin t + {{\cos }^2}t}}{{\cos t(1 - \sin t)}}[/math] [math] = \frac{{1 - \sin t + {{\sin }^2}t - 1}}{{\cos t(1 - \sin t)}}[/math] [math] = \frac{{\sin t}}{{\cos t}}[/math] [math] = \tan t[/math] and [math]\frac{{1 + \sin t - {{\cos }^2}t}}{{\cos t(1 - \sin t)}}[/math] [math] = \frac{{{{\sin }^2}t + \sin t}}{{\cos t(1 + \sin t)}}[/math] [math] = \frac{{\sin t}}{{\cos t}}[/math] [math] = \tan t[/math]

Are you asking about the force on the wire or the Hall effect, they are different.

So do you ,mean the rate of progress along the curve is constant, or do you mean the rate of progress horizontally and/or vertically is constant? The required curve is then a solution of some differential equation = this constant rate of progress. Which differential depends upon the meaning of steady shift.

You are asking about the Lorenz force http://en.wikipedia.org/wiki/Lorentz_force

Yes this is correct. No. the speed of light is a constant (in a given medium). 'c' is the (constant)speed in a vacuum. This implies that any change to the frequency must be accompanied by a change in the wavelength to maintain that constancy. What's the issue here? c is constant so hc is constant so the energy is proportional to the frequency and inversely proportional to the wavelength.

There are several articles in this week's New Scientist about the existence (or non existence) of time. http://www.newscientist.com/section/physics-math

Yes, but once again you do not need to solve the quadratic equation. Just consider the discriminant. My apologies I said the discrinimant was the square root bit, it is only the expression under the root sign (q2 - 4pr ) and does not include the sign itself. (q2 - 4pr ) > 0 q2 > 4pr with p = 3, q = 2a and r = 1 4a2 > 4.3.1 > 12 a2 > 3 For there to be a solution of the quadratic. So there is no solution when a2 < 3 [math] - \sqrt 3 [/math][math] < a < [/math][math]\sqrt 3 [/math] The last line in your post 7 is not quite correct since a can be positive or negative, but both have the same value of a2. What I have shown above is the correct deduction from the statement that a2 < some number

You obviously realize that the derivative is a quadratic (in x) The formulae for the two roots of a quadratic equation are (I have used p, q, and r for the constants to avoid confusion with your equation) [math]p{x^2} + qx + r = 0[/math] [math]x = \frac{{ - q + \sqrt {{q^2} - 4pr} }}{{2p}}[/math] [math]x = \frac{{ - q - \sqrt {{q^2} - 4pr} }}{{2p}}[/math] However you do not want these. What you need is the square root part [math]{\sqrt {{q^2} - 4pr} }[/math] This is called the discriminant. Since you cannot take a square root of a negative number (q2 - 4pr ) must be greater than or equal to zero q2 > 4pr Or in the range where q2 < 4pr The quadratic has no solution. Since the quadratic has no solution in this range. there are no extreme points in this range. Can you substitute values from your equation for p. q and r to calculate this range now? It is an easy couple of lines that leads directly to the correct solution.

OK so you have an amateur interest in maths. That's good. The derivative is [math]\frac{{dy}}{{dx}} = 3{x^2} + 2ax + 1[/math] How about telling us if you understand this and know what its relationship to extreme points is.. I can't help you if you don't help me understand what you already know. This is a use of calculus question.

Hello dawoodr, I see you are a new member. The forum has a special homework section where you should have posted this homework question, along with your attempt at solution. Since the hint suggests the derivative, what is the derivative and in particular what sort of polynomial is it? What is the condition that the derivative must satisfy at an extreme point?