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## #26 2006-02-04 17:39:49

irspow
Member
Registered: 2005-11-24
Posts: 457

### Re: A problem!!!!

No John, the arm is 40cm long.  I know this getting to be a long thread, but there is a good illustration posted somewhere above. When the arm is vertical only 10cm is exposed.

I think that my last post raised serious questions about the conditions stated in the problem.  I used equivalent trigonomic functions and found that they would not produce the same values that are stipulated  by the functions originally given.  I think that we are missing something here.

In other words, this type of problem is quite easy to solve normally, but all of the conditions stated in the original problem can not be satisfied.

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## #27 2006-02-04 23:07:34

johny
Member
Registered: 2005-11-19
Posts: 34

### Re: A problem!!!!

I dont want to be pain, but just one last question becuase i too am getting a bit confused.....so i need to simplify the equation given on post no. 18???? And that would give me the area????

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## #28 2006-02-05 03:38:08

irspow
Member
Registered: 2005-11-24
Posts: 457

### Re: A problem!!!!

If you use Ricky's conclusions, then the area would be;

2{[∫y1 - y2 dx] + [∫y3 - y2]}

If y1 = x²/50 + 30,  y2 = x²/50,  y3 = 40 - x/√3 then the above becomes;

2 {[30x]0 to b + [40x - x²/√(12) - x³/150]b to x max}

That would indeed give you the area that you seek, but you would have to have the correct limits of integration.  This determination is where all of the discussion has taken place.

Right now you just have;

-x³/75 -x²/√3 + 80x + b³/75 + b²/√3 - 20b = A

But that is assuming that you can figure out what the proper value of x max and b are to plug into the above formula.  If you read the last several posts there are some questions raised as to what exactly they should be given the original conditions of the problem.  Are you absolutely sure that you did not leave anything out?  Especially check out posts 18 through 24 again to see what I am talking about.

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## #29 2006-02-05 04:34:03

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

### Re: A problem!!!!

irspow, I'm not understanding your objections at all.

The problem gives two curves and two lines, and want you to find the area inbetween all of them.

What's wrong with this?

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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## #30 2006-02-05 04:50:14

irspow
Member
Registered: 2005-11-24
Posts: 457

### Re: A problem!!!!

Look at post #23 and #24 to see where I am coming from.  It is the varying length of the wiper blade that is causing my confusion.

And shouldn't the y3(right boundary function) be y = 40 - x/√3 ?

If I follow all of the rules of finding the intersections of the lines and using them as the limits of integrations I get:

y1 = y3(my y3 that is) at x = 1.63899453, we'll call this b

y3 = y2 at x = 32.55915054, we'll call this c

We will use 0 = a for the origin

We will assume the functions given are correct so that;

y1 = x²/50 + 30

y2 = x²/50

y3 = 40 - x/√3

We will multiply the integrals by two because the two halves are symmetrical.

2 {∫y1 - y2 dx]from a to b + ∫y3 - y2 dx]from b to c}   which gives;

2(∫30 dx]a to b + ∫40 - x/√3 - x²/50 dx]b to c)

2{ [30x]a,b + [40x - x²/√12 - x³/150]b,c}

If you plug in the values of a, b, and c you will get;

A = 1501.301771cm²

That is strictly by the rules.

The problem is with the points b and c. (y1 at b  and  y2 at c)

pb = (1.638994538, 30.05372606)

pc = (32.55915054, 21.20196568)

We all know how to find the distance between these two points, and in this case it equals;

blade length = √[(dx)² + (dy)²]

That measurement contradicts the stipulated conditions, but it was done using the equations given for the endpoints of the wiper blade.

This is what I was talking about all along in a formal way.  The blade should be 20cm long at the end of its swing and 30cm at the origin.  I could not and still can not find where the discrepency lies.

Last edited by irspow (2006-02-05 07:15:07)

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## #31 2006-02-05 05:11:08

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,588

### Re: A problem!!!!

I think that the wiper blade is centered on the end of the arm, so when vertical, then 40 minus 30/2 is exposed of the arm, not 40 minus 30.  Why would you hold the wiper at the bottom?

igloo myrtilles fourmis

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## #32 2006-02-05 07:29:00

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

### Re: A problem!!!!

And shouldn't the y3(right boundary function) be y = 40 - x/:raic3?

Yep.  Big mistake on my part.  Nice catch.

When I find the length of the wiper blade based on my functions (with the √3 change), I get the blade goes from 23.5307 to 30cm.  Not exactly 20-30, but close.  Is the question wrong?  Is the interpretation of it?  Not sure.

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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## #33 2006-02-05 07:38:32

irspow
Member
Registered: 2005-11-24
Posts: 457

### Re: A problem!!!!

Yeah, so it isn't just me.  My numbers say that the blade actually increased in length instead of the opposite, from 30 to >32 cm.  Something is fishy.  Can you at least see why I have been making all of the noise here now?

It all stemmed from how the intersection points were defined.  And the ho-hum, by the way, varying length blade does indeed make the situation more complicated.  Like I have said all along, this problem should have been really simple, but something is not quite right.

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## #34 2006-02-05 07:44:04

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

### Re: A problem!!!!

Well, the way I read it, the functions given by the question are describing the changing length of the blade.  It does change, after all, but it is 3cm off from 20.  Is the 20-30cm change just supposed to be approximate?  If so, then the equations as stated properly represent the area of integration.  If not, then there may be some oversight in assigning the functions or 120° by the creator.

Or the function don't represent the changing lenght of the blade, in which case I believe this question is impossible.

I'd rather go with the first option personally.

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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## #35 2006-02-05 07:48:35

irspow
Member
Registered: 2005-11-24
Posts: 457

### Re: A problem!!!!

I interpreted the functions given like this;

y1 represents the height of the top of the blade

y2 represents the height of the bottom of the blade

Is that what you just said?  haha

I believe either y1 or y2 is incorrect.

edit*

Interestingly, using simple geometry and assuming no blade length variations produces a pretty good approximation.

120° represents 1/3 of a circle

1/3 π(40)² - 1/3 π(10)² = 500π ≈ 1570.796327cm²

Last edited by irspow (2006-02-05 08:21:41)

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## #36 2006-02-05 08:07:42

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

### Re: A problem!!!!

Yes, but why are you saying the changing lenght of the blade complicates the problem?  Those functions describe that change.

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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## #37 2006-02-05 08:31:43

irspow
Member
Registered: 2005-11-24
Posts: 457

### Re: A problem!!!!

Has anyone definitively answered the question?  I tried my best in post #30.  I thought that I followed mathematical convention exactly, yet it does not match all of the information given in the original problem. For all of the talk and views that this thread has gotten, my humble calculation is the only one that offers the area in question?  I would say with the talent in this forum and the lack of a definitive answer would indeed suggest that the changing length of blade has complicated the problem.

In fact, without the blade variation, no integration is even needed.  See my edit in 35.

edit

I see your point Ricky.  If all of the information was correct from the beginning, yes, it would be a rather simple calculation.  The changing length would not make it more difficult if the functions were defined correctly.

Last edited by irspow (2006-02-05 08:38:37)

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## #38 2006-02-06 08:12:14

johny
Member
Registered: 2005-11-19
Posts: 34

### Re: A problem!!!!

Seriously guys thanks alot for giving soo much time to it!!!!! If i find something out then i wil let u know.........and once again thanks

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## #39 2006-02-06 14:43:30

irspow
Member
Registered: 2005-11-24
Posts: 457

### Re: A problem!!!!

johny,  the answer in post #30 is correct given the functions you had and the angle of swing.  What we are still questioning is only why the length of blade does not match what you said originally.

A ≈ 1501.3cm²

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