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#1 2006-11-27 02:22:04

Toast
Real Member
Registered: 2006-10-08
Posts: 1,321

Cuboid inside parabola help please

A cuboid needs to be moved through the arch of the equation

.

Consider the rectangular cross-section of the cuboid where the top corners of the rectangle touch the arch where the top corners of the rectangle touch the arch and the bottom of the rectangle is on the x axis. Let 'a' metres be the distance from the edge of the arch at the base (the origin) to the nearest edge of the rectangle.
The units on the axes are in metres.

i)Find the height and width of the rectangle when of the rectangle when a = 0.5
I got h=2.5m and w=2m.
ii)Find the height of the rectangle when its width is 2.5 metres.
I got h=1.375m
iii) Find, using an algebraic method, the width of the rectangle when its height is 4 metres.
I got dizzy.

I attached the image at the bottom.
Thanks.

Last edited by Toast (2006-11-27 02:23:22)

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#2 2006-11-27 05:00:12

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Cuboid inside parabola help please

An important thing for us to recognise here is that the curve is symmetric about x = 1.5. This means that the rectangle will always be centred there, which makes it much easier to work out its width.

I agree with your answer for (i). Because one side of the rectangle is at 0.5, the other side must be at 2.5 and so the width is 2m. The height can be found by substituting 0.5 for x in the equation.

y = -2*0.5(0.5-3) = -1*-2.5 = 2.5m.

I agree with your answer for (ii) as well. If the rectangle's width is 2.5, then half of its width is 1.25, and as it's centred at 1.5 then it must have an edge at 0.25.

Substituting this into the equation gets y = -2*0.25(0.25-3) = -0.5*-2.75 = 1.375m.

For (iii), you just need to use the equation the other way and solve it as a quadratic.

4 = -2x(x-3)
-2x² +6x -4 = 0
x² -3x +2 = 0
(x-1)(x-2) = 0
x = 1 or 2.
So the rectangle starts at x=1 and ends at x=2, meaning its width is 1m.


Why did the vector cross the road?
It wanted to be normal.

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