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Simple math help

Blogs > Cloud
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Cloud
Profile Blog Joined November 2004
Sexico5880 Posts
Last Edited: 2009-12-04 00:45:43
December 04 2009 00:33 GMT
#1
Hey people. It's the first time I'm asking this (whoops ok second one ), but it's been 2 days since I've been trying this problem which should be easy as hell for me (it's just simple calculus) and it's driving me crazy.

If someone could shed some light on it, it would be lovely.

"A fence is 27/8 meters tall, and there is a house 8 meters away from it, find the shortest possible stair that goes from the ground, over the fence and touches the house."

Here, I made a drawing of it:

+ Show Spoiler +
[image loading]


This is simply about finding a function (for the stair length (L)) and then diferentiating it and finding its minimum point (by equaling the derivative to 0 and solving the equation, etc.)

The problem is that I have tried to postulate the function in half a dozen ways now and the derivative (equaled to 0) is never an equation below grade 4, and that just seems too complicated.

I have tried using Θ, like cosΘ = (x+8)/L, then cosΘ = x/sqrt(x²+8²). And then just substituting.

Or like sinΘ= 27/(8x) and then squaring both the cosine and sine and summing them, etc.

I've tried using H instead of Θ too and and well as I have said, I haven't found an equation below 4th grade.

So if anyone could help it would be greatly appreciated.

By the way, the answer is supposed to be 125/8 meters if that helps.

BlueLaguna on West, msg for game.
Ame
Profile Joined October 2009
United States246 Posts
Last Edited: 2009-12-04 00:49:51
December 04 2009 00:46 GMT
#2
Apologies if you've tried this and I'm just bad at reading formulas, but:

Have you tried breaking it up into a sum of 2 triangles that have the same angles? One with base x, another with base 8 meters.

*edit* Synapse>> Yoink.
synapse
Profile Blog Joined January 2009
China13814 Posts
Last Edited: 2009-12-04 01:20:46
December 04 2009 00:46 GMT
#3
You would split the triangle you drew in your picture into 2 similar right triangles and a rectangle. You can put both of the hypotenuses of the two right triangles in terms of x, and adding the two equations gives you L = f(x)... It is a pretty faggy looking function, but I'm pretty sure that's how you solve to the problem. Differentiate and solve for 0.

(haha a few seconds late ~)

EDIT: actually, nvm...

apparently the differentiated function I wrote would look something like
[image loading]


anyway, using this apparently very complicated method, your derivative looks like:
http://www.wolframalpha.com/input/?i=d/dx(sqrt(x^2+(27/8)^2)+sqrt(64+((8/x)*27/8)^2))

solving for x when L' = 0
http://www.wolframalpha.com/input/?i=d/dx(sqrt(x^2+(27/8)^2)+sqrt(64+((8/x)*27/8)^2)) = 0

x=9/2 fits with your answer of L=125/8
:)
Zortch
Profile Blog Joined January 2008
Canada635 Posts
Last Edited: 2009-12-04 01:18:11
December 04 2009 00:55 GMT
#4
So here is my thought, though it didn't work out for me.

L^2=(x+8)^2 + H^2

tan(theta)=27/8x from the little triangle
tan(theta)=H/(x+8) from the big triangle

So we can equate these:

27/8x=H/(x+8)
Solving for H:
H=27(x+8)/8x

Then we can sub into our equation for L:
L^2=(x+8)^2+ (27(x+8)/8x)^2

Then take the derivative w/r to x and gogogo.

it didn't work out for me, but I'll give it another shot. I don't see my error yet.

EDIT: need L^2 heh, i'll try it now
not working...whyyy?
Ah so i used that program the guy above me used...coool

http://www.wolframalpha.com/input/?i=d/dx(sqrt(x^2+16x+64+(729/64)+(729/(4x))+(729/(x^2))))

I get a root in the derivative at 9/2
Evaluating at my function gives me 125/8.

I imagine that finding that root by hand would be tough.
Respect is everything. ~ARchon
Cloud
Profile Blog Joined November 2004
Sexico5880 Posts
December 04 2009 01:15 GMT
#5
On December 04 2009 09:46 synapse wrote:
apparently the differentiated function I wrote would look something like
[image loading]


Yeah that ends up being something like a 10th grade equation

On December 04 2009 09:55 Zortch wrote:
So here is my thought, though it didn't work out for me.

L^2=(x+8)^2 + H^2

tan(theta)=27/8x from the little triangle
tan(theta)=H/(x+8) from the big triangle

So we can equate these:

27/8x=H/(x+8)
Solving for H:
H=27(x+8)/8x

Then we can sub into our equation for L:
L^2=(x+8)^2+ (27(x+8)/8x)^2

Then take the derivative w/r to x and gogogo.

it didn't work out for me, but I'll give it another shot. I don't see my error yet.

EDIT: need L^2 heh, i'll try it now


I hadn't tried using both theta and H before, but your function seems rather big as well ~.~

Maybe there's no hope but to solve a quartic equation. Which sounds ridiculous for such a simple exercise.
BlueLaguna on West, msg for game.
incnone
Profile Joined July 2009
17 Posts
Last Edited: 2009-12-04 01:19:04
December 04 2009 01:18 GMT
#6
Zortch's way works well. It's easiest to minimize L^2 (instead of taking square roots and minimizing L) -- that way there aren't any square roots to deal with when you differentiate. You ought find

d(L^2)/dx = 2(x+8)(1 - 27^2/(8x^3));

setting this equal to zero makes solving for x easy -- no quartic equation, just x^3 = 27^2/8.
tredmasta
Profile Blog Joined June 2008
China152 Posts
Last Edited: 2009-12-04 01:19:33
December 04 2009 01:18 GMT
#7
I had to use Mathematica, but I was able to get the correct answer.

Basically you split the area under the ladder into two triangles and a rectangle. One of the triangles has base x and height 27/8, while the other has base 8 and unknown height H. Since the 2 triangles are similar, we do 27/(8x) = H/8 to get that H = 27/x.

Now we take our two triangles and write L in terms of x:

L = sqrt(x^2 + (27/8)^2) + sqrt(8^2 + (27/x)^2)

I took the derivative of this using Mathematica and it gave me a root at x=9/2.

Plug this back into the equation for L to get that L=125/8.

Hope this helps, I still have no idea how to do it by hand.

Edit: Seems like people above me got the same thing
<3 한승연 김태연!!!
Cloud
Profile Blog Joined November 2004
Sexico5880 Posts
Last Edited: 2009-12-04 01:26:38
December 04 2009 01:21 GMT
#8
Oh thanks guys, that's pretty cool. I thought of minimizing the square but that just seemed weird to me.
BlueLaguna on West, msg for game.
Fwmeh
Profile Joined April 2008
1286 Posts
Last Edited: 2009-12-04 01:55:57
December 04 2009 01:47 GMT
#9
I think it would be easiest to solve for the angle instead.

We have 27/(8*tan(v))=x and (x+8)/cos(v)=y, together we get (after some rearrangement) 8/cos(v) + 27/(8*sin(v)) = y, if we derive we get y' = - 8sin(v)/cos^2(v) + 27cos(v)/(8*sin^2(v), for extreme value we let y' be 0 and get 64sin^3(v) = 27cos^3(v) => 4sin(v) = 3cos(v) => arctan(3/4) = v.

Edit. Just to finish my calculations, we have

27/(8*0.75) = x = 4.5,

8 + 4.5 = 12.5

12.5/cos(arctan(3/4)) = 12.5/0.8 = 15.625 = 125/8

so that is the analytical solution.
A parser for things is a function from strings to lists of pairs of things and strings
Cloud
Profile Blog Joined November 2004
Sexico5880 Posts
December 04 2009 03:13 GMT
#10
Pretty cool solution as well Fwmeh thanks.
BlueLaguna on West, msg for game.
Pawsom
Profile Blog Joined February 2009
United States928 Posts
Last Edited: 2009-12-04 03:53:11
December 04 2009 03:53 GMT
#11
On December 04 2009 10:21 Cloud wrote:
Oh thanks guys, that's pretty cool. I thought of minimizing the square but that just seemed weird to me.

[image loading]
skyglow1
Profile Blog Joined April 2005
New Zealand3962 Posts
Last Edited: 2009-12-09 04:49:33
December 09 2009 04:47 GMT
#12
Sorry for bumping this thread guys but I wanted to just give it some more closure (actually...to give myself closure ).

I was reading this thread and thought the approach of minimising the square was a smart one. It intuitively makes sense that if the original function produces only positive values, then the squared function should have the minimum point at the same x value.

What bugged me was how to prove this mathematically for a general positive function. I decided to play around with it a bit on paper and scribble some things because I like doing little math problems. I came up with this 'proof' (it's probably not very proper but I only have high school level math under my belt so I'm noob ) and it surprised me how simple it was actually:

Let y=f(x) and let f(x) > 0
dy/dx = f'(x)
To find minimum point, let dy/dx = 0 which means f'(x) = 0
--> solve and get x = a where "a" is the x value of the minimum point on the original curve

Now y^2 = f(x) x f(x)
d(y^2)/dx = f'(x)f(x) + f(x)f'(x) (product rule) = 2f'(x)f(x)
Let d(y^2)/dx = 0 which means 2f'(x)f(x) = 0
--> either:

1. f(x) = 0 --> no real solutions because f(x) > 0 or
2. f'(x) = 0 --> x = a which is the same solution we get by differentiating the original function. Woo!

Probably noone cares but now I can sleep easy tonight
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