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Topic: Really Hard Precalculus Project (Read 857 times) |
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warriors837
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As the diagram attached suggests, a rectangular beam of length L and width W has become tightly wedged because it would not fit when someone tried to move it around a corner formed by two intersecting corridors, one of width, a, and the other of width b. The beam makes an angle of theta, with one wall, as shown, Express the length of the beam L in terms of the other quantities a, b, W, and theta.
The diagram is an attached file.
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towr
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Re: Really Hard Precalculus Project
« Reply #1 on: May 19th, 2009, 8:40am » |
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You can solve it by drawing a few triangle in the diagram.
And if I managed not to screw it up, you can get:
a/y =sin(t)
b/x =cos(t)
v/w=tan(t)
w/u=tan(t)
L=x+y-u-v
L=a/sin(t) + b/cos(t) - w tan(t) - w/tan(t)
I extended the line of lower side of the beam, u and v are the extra pieces on both ends, and x and y are the length from the corner to the ends of the beam.
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| « Last Edit: May 19th, 2009, 8:42am by towr » |
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Immanuel_Bonfils
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Re: Really Hard Precalculus Project
« Reply #2 on: May 19th, 2009, 1:29pm » |
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If X is the distance from the beams lowest corner to the convex edge of the corridors intersect, we have :
blocking the vertical corridor, X cos  +W sin = b , and
blocking the horizontal corridor, (L-X) sin +W cos = a .
Eliminating X we get L = b sec + a csc - 2 W csc(2) .
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| « Last Edit: May 19th, 2009, 1:30pm by Immanuel_Bonfils » |
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Grimbal
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Re: Really Hard Precalculus Project
« Reply #3 on: May 20th, 2009, 1:31pm » |
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on May 19th, 2009, 8:40am, towr wrote:| L=a/sin(t) + b/cos(t) - w tan(t) - w/tan(t) |
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... = a/sin(t) + b/cos(t) - w/(sin(t)cos(t))
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towr
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Re: Really Hard Precalculus Project
« Reply #4 on: May 21st, 2009, 12:47am » |
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on May 20th, 2009, 1:31pm, Grimbal wrote:
... = a/sin(t) + b/cos(t) - w/(sin(t)cos(t)) |
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And that equal's Immanuel's result, yeah.
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Grimbal
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Re: Really Hard Precalculus Project
« Reply #5 on: May 21st, 2009, 12:59am » |
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Uh, right. I am not familiar with sec and csc. I never use them.
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| « Last Edit: May 21st, 2009, 1:00am by Grimbal » |
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towr
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Re: Really Hard Precalculus Project
« Reply #6 on: May 21st, 2009, 1:53am » |
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on May 21st, 2009, 12:59am, Grimbal wrote:| Uh, right. I am not familiar with sec and csc. I never use them. |
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Yeah, me neither. But you could have simplified sin(t)cos(t) to 2sin(2t) [edit]1/2 sin(2x)[/edit]; that way you use one less trigonometric function.
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| « Last Edit: May 21st, 2009, 8:22am by towr » |
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rmsgrey
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Re: Really Hard Precalculus Project
« Reply #7 on: May 21st, 2009, 5:58am » |
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on May 21st, 2009, 1:53am, towr wrote:
Yeah, me neither. But you could have simplified sin(t)cos(t) to 2sin(2t); that way you use one less trigonometric function.
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... or you could have simplified it to (1/2)sin(2t) and had the added bonus of being correct 
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towr
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Re: Really Hard Precalculus Project
« Reply #8 on: May 21st, 2009, 8:25am » |
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on May 21st, 2009, 5:58am, rmsgrey wrote:
... or you could have simplified it to (1/2)sin(2t) and had the added bonus of being correct |
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Meh. I was perfectly content to leave it as L=a/sin(t) + b/cos(t) - w tan(t) - w/tan(t)
Needless manipulation just breeds errors.
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Grimbal
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Re: Really Hard Precalculus Project
« Reply #9 on: May 21st, 2009, 9:47am » |
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I personally dislike the introduction of sin(2t) which gives the impression there is an angle of 2·theta somewhere in the diagram.
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towr
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Re: Really Hard Precalculus Project
« Reply #10 on: May 21st, 2009, 10:23am » |
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on May 21st, 2009, 9:47am, Grimbal wrote:| I personally dislike the introduction of sin(2t) which gives the impression there is an angle of 2·theta somewhere in the diagram. |
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In that case, isn't it better to just keep the terms separate? Because what does sin(t)cos(t) correspond to in the diagram?
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Grimbal
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Re: Really Hard Precalculus Project
« Reply #11 on: May 21st, 2009, 2:22pm » |
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Actually, it's the third height of a right triangle with hypotenuse 1 and angle t (the other 2 being sin(t) and cos(t). It even gives some nice symmetry to my formula. 
But my point was that sin(2t) looks ugly to me. It doesn't recombine in an obvious way with sin(t) and cos(t). By keeping sin(t) and cos(t), you can factor them out, and get for instance:
L = (a·b - (a-w/cos(t))·(b-w/sin(t)))/w
which you might or might not find useful.
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| « Last Edit: May 21st, 2009, 2:23pm by Grimbal » |
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warriors837
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Here is the complete work, just look at the attachment...
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| « Last Edit: May 22nd, 2009, 9:04am by warriors837 » |
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towr
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Re: Really Hard Precalculus Project
« Reply #13 on: May 22nd, 2009, 2:56pm » |
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on May 22nd, 2009, 9:03am, warriors837 wrote:| Here is the complete work, just look at the attachment... |
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You can simplify things a bit more quickly by using the fact that tan(90 - t) = 1/tan(t).
And it isn't entirely clear from the diagram that x and y include the segments u and v respectively; it now seems they end where the beam ends rather than go up to the wall. (Which admittedly wasn't clear from my description either )
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