dawoodr 0 Posted November 4, 2013 (edited) Hello! Well, It's another math question that I want to know the answer of. This is not homework, I just want to know how to solve this. Unfortunately I have no idea what to do so if you can solve this one for me I would really appreciate it because I just need to look at your equations to understand how to solve this, would be good if you wrote a few words aswell, as an explanation. Here it goes: Put together two parabola sections to a roller coaster with a steady shift in P. What are the equations of the parabolas. Just a tip; They are asking for the equations of the curves on the image, There are two curves that meet in the coordinate P Regards! Edited November 4, 2013 by dawoodr 0 Share this post Link to post Share on other sites

studiot 1897 Posted November 4, 2013 (edited) with a steady shift in P. 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. Edited November 4, 2013 by studiot 0 Share this post Link to post Share on other sites

imatfaal 2480 Posted November 4, 2013 I think you have to be allowed to assume that the two curves are mirror images (about the line y=10) of each other. I think in that case you can get two equations with two unknowns - otherwise you have too little information 0 Share this post Link to post Share on other sites

mathematic 93 Posted November 4, 2013 Alternate so I think you have to be allowed to assume that the two curves are mirror images (about the line y=10) of each other. I think in that case you can get two equations with two unknowns - otherwise you have too little information Alternate possibiliity. Left piece: f(0) = 25, f(10) = 10, f'(10) = h. Right piece, g(10) = 10, g'(10) = h, min g(x) = 0. If you need an additional condition, f'(0) = 0. 0 Share this post Link to post Share on other sites

dawoodr 0 Posted November 4, 2013 (edited) with a steady shift in P. With this I mean that they both share the coordinate P. The thing is that I am not a native english speaker, I got it as my third language actually, so I am sorry if I wrote the problem incomprehensible. Edited November 4, 2013 by dawoodr 0 Share this post Link to post Share on other sites

studiot 1897 Posted November 5, 2013 (edited) So you have two parabolas 1) y = ax^{2 }+ b 2) y' = a'x^{2 }+ 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 = -15/100x^{2 }+ 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. Edited November 5, 2013 by studiot 0 Share this post Link to post Share on other sites

studiot 1897 Posted November 5, 2013 (edited) OK it is more complicated than I first thought. So the general parabola is y = ax^{2} + 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 a_{1}, b_{1} and c_{1} [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 P_{1 }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 P_{2} = [math]y = 0.15{x^2} - 6x + 55[/math] Edited November 5, 2013 by studiot 1 Share this post Link to post Share on other sites

imatfaal 2480 Posted November 5, 2013 done it as well and my answer seems different will post in a moment after I have written up eq 1 must be of the form ax^2 + c = y ; if there were to be a bx term then it would not have turning point on the y axis ax^2 + c = y ; when x = 0 y = 25 c= 25 ax^2 +25 = y; when x=10 y=10 100a+25 = 10 a= -15/100 eq 1 --> -.15x^2 +25=y dy/dx = -.3x ; at x=10 dy/dx = -3 eq2 Ax^2 +Bx+C = y what do we know x=10 y=10 (1) dy/dx(10) =-3 (2) dy/dx = 0 when y =0 (3) (1) 100A+10B+C=10 (2) dy/dx = 2Ax+B 20A+B=-3 (3) dy/dx = 2Ax+B = 0 when y=0 x= -B/2A when y=0 sub in value for x A(-B/2A)^2 +B(-B/2A)+C=0 simplify -B^2/4A+C = 0 three equation in three unknowns solve A = 9/40 B= -15/2 C= 125/2 therefore eq1 --> [latex]-.15x^2+25=y[/latex] eq 2 --> [latex] 9x^2/40 - 15x/2 +125/2=y[/latex] they fit the known parts (not the symmetrical part that I introduced) TP1 at (0,25) EQ1 and EQ2 go through (10,10) and have same slope at (10,10) EQ2 touches x-axis (ie has two equal roots) 0 Share this post Link to post Share on other sites

studiot 1897 Posted November 5, 2013 (edited) 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. Edited November 5, 2013 by studiot 0 Share this post Link to post Share on other sites

imatfaal 2480 Posted November 5, 2013 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 of 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. Not a typo 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" dy/dx = -0.3x so when x = 10 dy/dx= -3 0 Share this post Link to post Share on other sites

studiot 1897 Posted November 5, 2013 dy/dx = -0.3x so when x = 10 dy/dx= -3 Ah. Thanks for the clarification. 0 Share this post Link to post Share on other sites

dawoodr 0 Posted November 5, 2013 (edited) Thanks alot studiot! The right answers was written like this so I think you're right [math]f(x) = 25-15x^2, 0equal<x<10equal[/math] [math]g(x) = 9/40 * (x-50/3)^2, x>10equal[/math] Edited November 5, 2013 by dawoodr 0 Share this post Link to post Share on other sites

studiot 1897 Posted November 5, 2013 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. 0 Share this post Link to post Share on other sites

imatfaal 2480 Posted November 5, 2013 dawoodr - I took the liberty of opening your post and copying the latex out and reformating below - hope that's ok. And that looks more like my solution but seems to be missing a decimal point from the x^2 coefficient of f(x) [latex]f(x) = 25-15x^2[/latex] 0<x<=10 [latex]g(x) = 9/40 * (x-50/3)^2[/latex] x>=10 and the latex implementation seems to go from bad to worse - really not sure what is wrong with it, i cannot get the greater or equals to work so left as plain text. I think the key assumption was mathematics that the line "with a steady shift in p" meant that the slope of f must match the slope of g at (10,10). i am a bit ashamed that studiot worked out a lovely solution with symmetric curves and I now think that was a bum steer from me 0 Share this post Link to post Share on other sites

studiot 1897 Posted November 5, 2013 (edited) i am a bit ashamed that studiot worked out a lovely solution with symmetric curves and I now think that was a bum steer from me 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. Edited November 5, 2013 by studiot 0 Share this post Link to post Share on other sites

overtone 200 Posted November 5, 2013 (edited) The assumption of symmetry around the point 10,10 in the two parabolas contradicts the problem as pictured. The one drops 15 units from the max, the other drops only 10 units to the min. Nevertheless the information is adequate - the key is that a smooth joining requires equality of slope, a common sense observation not adequately specified. So I'm seconding imtfaal's posting. edit in - already noted, crossed, never mind edit - clarification of the already duplicative - going away now - nice problem here. Edited November 5, 2013 by overtone 0 Share this post Link to post Share on other sites

studiot 1897 Posted November 5, 2013 the other drops only 10 units in 10 run. 20 is not the repeated root in Imatfaal's solution, y is 2.5 at this x. 0 Share this post Link to post Share on other sites

imatfaal 2480 Posted November 6, 2013 20 is not the repeated root in Imatfaal's solution, y is 2.5 at this x. yep - the roots at minimum are both (50/3,0) What curve would an engineer use? Off the top of my head I would say catenary as the natural curve that a heavy rope/chain/construction will form. 0 Share this post Link to post Share on other sites

studiot 1897 Posted November 6, 2013 What curve would an engineer use? Off the top of my head I would say catenary as the natural curve that a heavy rope/chain/construction will form. 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. 0 Share this post Link to post Share on other sites

imatfaal 2480 Posted December 1, 2015 ! Moderator Note New Question by Mike B split to new thread. 0 Share this post Link to post Share on other sites