SOLUTION: Hi, I have got a question that is to do with calculus but i couldn't find a section covering it above so I put in alegaebra equations, sorry if that was a problem.
Ok the quest
Question 336003: Hi, I have got a question that is to do with calculus but i couldn't find a section covering it above so I put in alegaebra equations, sorry if that was a problem.
Ok the question I have that i need help with is "Using an algebraic method, determine the equation of the catenary that you believe best models your necklace". Baiscally I had to make earlier a parabola by tracing a chain and then find an equation to match that parabola which i had difficulty with. Now later on, on a much harder question it was required for the parabola to be shitfed upwards on the y-axis and the equation to be calculated.
I have obtained three coordinates that I tried to use but to no avail, they are
Turning Point (0, 10) and (30,27) and (54,79). If a negative x-vaule would make it easier then those three a coordinate that could also be used is- (-54, 79). Any help would be greatly appreciated. I was attempting to use the formula y=ax^2+bx+c, as this formula is what is needed to be solved, that a-value, b-value and c-value.
Any help would be greatly appreciated.
Thank you
Keiran Found 2 solutions by stanbon, Edwin McCravy:Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website! "Using an algebraic method, determine the equation of the catenary that you believe best models your necklace". Baiscally I had to make earlier a parabola by tracing a chain and then find an equation to match that parabola which i had difficulty with. Now later on, on a much harder question it was required for the parabola to be shitfed upwards on the y-axis and the equation to be calculated.
I have obtained three coordinates that I tried to use but to no avail, they are
Turning Point (0, 10) and (30,27) and (54,79). If a negative x-vaule would make it easier then those three a coordinate that could also be used is- (-54, 79). Any help would be greatly appreciated. I was attempting to use the formula y=ax^2+bx+c, as this formula is what is needed to be solved, that a-value, b-value and c-value.
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Using:::::: ax^2 + bx + c = y
(0,10)::::: 0 + 0 + c = 10
(30,27):::: a*30^2+b*30+c = 27
(54,79):::: a*54^2+b*45+c = 79
---------------------------------
Using Matrix:
0::::0::::1::::10
30^2:30:::1::::27
54^2:54:::1::::79
-------------------------
a = 0.0296
b = -0.3222
c = 10
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Note: By the way, a catenary is not a parabola.
A catenary has its own equation form. Check
Google to see more about catenaries.
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Cheers,
Stan H. Answer by Edwin McCravy(20055) (Show Source): You can put this solution on YOUR website!
Substitute (0, 10), (30,27) and (54,79) into
and get
So now the equation is:
Substitute the second point (54,79):
Substitute the third point (30,27):
Solve this system of equations:
Solve that by Cramer's rule:
So the equation
is now
Edwin