FAO: maths people, simultaneous eqns

[smaddy]

im jus tryin to find out the equation of a parabola, havin to work backwardsi have y = -ax^2 - bx - ci kno that c = 0so the first equation is 17.5 = 52.5^2a +52.5b + 0the second equation im using is dy/dx = -2ax - b = 0cos differentiating gives max and mins and at max and mins dy/dx = 0my questions ISIS THIS ALLOWED, can you solve for a and b using the differential equation and the oringinalsafeactually im havin second thoughts, in the second one can i use the same value for x, of 52.5? the coordinates for the maximum = (52.5, 17.5)

[L]

3.142

[Brukfoot]

everything there makes sense to me except the word "parabola", I think I've done it before but can't recall

[Scorpzz]

is that you in the pic

[smaddy]

everything there makes sense to me except the word "parabola", I think I've done it before but can't recall

are u sure, i duno if i believe u if u dunno what a parabola is loolexcept mine is the other way roundso the whole equation is negative

[Brukfoot]

ohhh seen, yeah of course I know what that is, it's just I haven't used that word in ages

[NORA BATTY]

but why should we take you seriously? when i post maths questions even the moderators post rubbish or l ock it. why should i help you

[smaddy]

but why should we take you seriously? when i post maths questions even the moderators post rubbish or l ock it. why should i help you

i dont care if u help mei kno there are people that come in ur threads and are correct, im talkin to themi dont go in ur topics, and im not tryin to test anyone here i jus want helpu can leave nowsafe

[NORA BATTY]

your question doesnt really make sense. what is it you want to do and what do you have to start with? also are you doing A level or are you just mucking around?

[Brukfoot]

can you say exactly what the question asks for and what it gives you,sorry it's just when i do maths i have to start from the beginning when helping people, i can't just jump into where they stopped

[Skola]

im jus tryin to find out the equation of a parabola, havin to work backwardsi have y = -ax^2 - bx - ci kno that c = 0so the first equation is 17.5 = 52.5^2a +52.5b + 0the second equation im using is dy/dx = -2ax - b = 0cos differentiating gives max and mins and at max and mins dy/dx = 0my questions ISIS THIS ALLOWED, can you solve for a and b using the differential equation and the oringinalsafeactually im havin second thoughts, in the second one can i use the same value for x, of 52.5? the coordinates for the maximum = (52.5, 17.5)

i think you have to explain your problem in a more structured fashionbecause people might not be able to follow ur trail of thought which makes sense to youi cant beleive that even tho im good at maths, and its one of the foundational courses of my degree, i cant explain how you work it out!

[smaddy]

its a tied arch bridge, which im designingit is 105m longthe maximum height of the bridge is 17.5mi kno that the shape is a parabolabut in the write up i need to state the equation that we use, or we will lose marksthe rest is as in the first posttell me if u need to kno anything else, but i think thats iti am getting a = 1/306 and that dont look righti dont kno how to run it on excel, and i dont have graph paper to draw it out

[smaddy]

tied arch bridge length 105 mmax height 17.5thereforethe point of maximum = (52.5, 17.5) where x= 52.5 and y=17.5parabola equation/quadratic equation: y= -ax^2 - bx - c 17.5 = -(52.5^2)a -52.5b - 0second equation, differentiating dy/dx = -2ax - b = 0 max is when x = 52.5therefore dy/dx = (-2)(-52.5)(a) - b = 0am i right or wrong?what is ur answer?

[DRC]

I wish I could help but I can't. Coming in here has made me realise how happy I am I no longer study maths Swear eks does that stuff in his sleep

[smaddy]

I wish I could help but I can't. Coming in here has made me realise how happy I am I no longer study maths Swear eks does that stuff in his sleep

yea but im not goin down to his househe is never on msnand he wont get me over the fone

[Jodie]

I wish I could help but I can't. Coming in here has made me realise how happy I am I no longer study maths

Co Signed:D

[Brukfoot]

is the x axis (where y=0) the top of the bridge or the bottom?

the point of maximum = (52.5, 17.5) where x= 52.5 and y=17.5

there seems to be something wrong with this assumption, as you said it is an upside-down parabola

[Skola]

im sorry i cant help lol

[smaddy]

is the x axis (where y=0) the top of the bridge or the bottom?

(0,0) is the left hand side of the bridgex = 52.5 y=17.5 is the top centre coordinates

[smaddy]

omygosh ive jus had a revelationon the right hand sidey = 0x = 105

[Skola]

omygosh ive jus had a revelationon the right hand sidey = 0x = 105

lol propsi always used to start by drawing the problem and filling in everything i knew

[Brukfoot]

lol, aite now you've had that revelation i'm guessing you don't need helpi was still confused on what you were saying but the picture made me understand, i could still do the equation if you want?

[NORA BATTY]

if the arch height is 17.5 and the roots of the equation is x=52.5 and the parabola is symmetrical about x=0 then you are nearly right.solve a(52.5)^2-17.5=0then put the value of a you get into -ax^2+17.5=0 and you're done right?

[smaddy]

YEAAAlook its a lot simpler now, but the answer im gettin looks wrongso here is the 2 new equations-17.5 = (52.5^2)a + 52.5b (1)0 = (105^2)a + 105b (2)thank you all very very much

[smaddy]

if the arch height is 17.5 and the roots of the equation is x=52.5 and the parabola is symmetrical about x=0 then you are nearly right.solve a(52.5)^2-17.5=0then put the value of a you get into -ax^2+17.5=0 and you're done right?

no cos where is b?