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parabolic doorway

A doorway is to be parabolic in shape and 2 metres high. At the height of 1 metre above ground level the width of the opening is 1.6 metres. How wide is the door at floor level.

I worked this out to be (8 * Sqrt(2)) / 5 or 2.2627 metres, but I don't think I did it the most effecient way.

You see, At Y = 1 we are told that there difference between the two x intcpts is 1.6metres and at y =2 the distance between them is 0 (because it is a turning point)

So I made up the equation Y = ax(x-b) and substitued Y = 1. Factorised it and once I got to the factorisation and I worked out the distance between the two x intctps was equal to 1.6.. THen I solved this for A and still had a B in the equation, so I did the same for Y=2. Then Once I had to equations in A=.... equations in terms of B I solved them simulataneously to find B, put B back into an equation to find A and found the equation to beY = 25/16x(X-(8*Sqr(2))/5))

Anyway, the x int was easy from there.

I think there must be an easier way.

I also realise that the parabola was supposed to be negative, for some reason A turned out to be 25/16 instead of -(25/16)

I am not sure why this is.

Could someone please tell me the proper way to do this problem because I think there has to be an easier way.

BYe
0
blackdan
• 2
1 Solution

Commented:
blackdan
Defining the doorway to be symetric about the y axis and the x axis to be floor level then the height of the doorway H(x) is since the doorway max height is 2 metres given by

H(x)=2-k*x*x

where k is a constant, (since the doorway is symetric about the y axis).  At x=+/- 0.8  H(x)=1,  hence

1=2-k*(0.8)^2

giving k=1/(0.8)^2 so

H(x)=2-(x/0.8)^2

at floor level H(x)=0 so x satisfies

0=2-(x/0.8)^2

ie (x/0.8)^2=2

which means the parabola cuts the x axis at x=-0.8*sqrt(2) and x=+0.8*sqrt(2) the distance between these 2 points being 1.6*sqrt(2). Hence

width at floor level = 1.6*sqrt(2) =2.2627