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        riddles >> medium >> Really Hard Precalculus Project
        
(Message started by: warriors837 on May 19^th, 2009, 8:03am)

Title: Really Hard Precalculus Project
Post by warriors837 on May 19^th, 2009, 8:03am

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.

Title: Re: Really Hard Precalculus Project
Post by towr on May 19^th, 2009, 8:40am

You can solve it by drawing a few triangle in the diagram.

And if I managed not to screw it up, you can get:
[hide]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.[/hide]

Title: Re: Really Hard Precalculus Project
Post by Immanuel_Bonfils on May 19^th, 2009, 1:29pm

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 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif+W sin http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif= b , and

blocking the horizontal corridor, (L-X) sin http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif+W cos http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif= a .

Eliminating X we get L = b sec http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif+ a csc http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif- 2 W csc(2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif) .

Title: Re: Really Hard Precalculus Project
Post by Grimbal on May 20^th, 2009, 1:31pm

on 05/19/09 at 08:40:20, towr wrote:

[hide]L=a/sin(t) + b/cos(t) - w tan(t) - w/tan(t)[/hide]

[hide]... = a/sin(t) + b/cos(t) - w/(sin(t)cos(t))[/hide]

Title: Re: Really Hard Precalculus Project
Post by towr on May 21^st, 2009, 12:47am

on 05/20/09 at 13:31:32, Grimbal wrote:

[hide]... = a/sin(t) + b/cos(t) - w/(sin(t)cos(t))[/hide]

And that equal's Immanuel's result, yeah.

Title: Re: Really Hard Precalculus Project
Post by Grimbal on May 21^st, 2009, 12:59am

Uh, right. I am not familiar with sec and csc. I never use them.

Title: Re: Really Hard Precalculus Project
Post by towr on May 21^st, 2009, 1:53am

on 05/21/09 at 00:59:21, Grimbal wrote:

Uh, right. I am not familiar with sec and csc. I never use them.

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.

Title: Re: Really Hard Precalculus Project
Post by rmsgrey on May 21^st, 2009, 5:58am

on 05/21/09 at 01:53:50, 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.

... or you could have simplified it to (1/2)sin(2t) and had the added bonus of being correct ;)

Title: Re: Really Hard Precalculus Project
Post by towr on May 21^st, 2009, 8:25am

on 05/21/09 at 05:58:22, rmsgrey wrote:

... or you could have simplified it to (1/2)sin(2t) and had the added bonus of being correct ;)

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.

Title: Re: Really Hard Precalculus Project
Post by Grimbal on May 21^st, 2009, 9:47am

I personally dislike the introduction of sin(2t) which gives the impression there is an angle of 2·theta somewhere in the diagram.

Title: Re: Really Hard Precalculus Project
Post by towr on May 21^st, 2009, 10:23am

on 05/21/09 at 09:47:34, 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.

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?

Title: Re: Really Hard Precalculus Project
Post by Grimbal on May 21^st, 2009, 2:22pm

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. :P

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.

Title: Re: Really Hard Precalculus Project
Post by warriors837 on May 22^nd, 2009, 9:03am

Here is the complete work, just look at the attachment...

Title: Re: Really Hard Precalculus Project
Post by towr on May 22^nd, 2009, 2:56pm

on 05/22/09 at 09:03:21, warriors837 wrote:

Here is the complete work, just look at the attachment...

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 :P)