Math Question

[ProxyAmenRa]

It was orientation week at university today so I decided to celebrate by giving my first year engineering students a math problem. To further celebrate all things related to engineering and studying engineering, I am going to give you kiddos the same problem. After a few days I will upload a hint if you none of your have solved it.

Preface:

Imagine you have circle with a line indicating the diameter drawn horizontally through it. You have another line parallel to the first, with a length of the circle's radius, intersecting the circle's circumference. There is an area 'A' between the second line and the circle's circumference.

Question:

Put forward a method for a calculating the area given that you know the circle's diameter.

Advice:

Start off by drawing the circle and two lines.

 

[QuickTwist]

I have not had the education to solve this question but will enjoy working on it non the less.

Question, Do you need to use a sign wave to solve the problem?

 

[ProxyAmenRa]

I have not had the education to solve this question but will enjoy working on it non the less.

Question, Do you need to use a sign wave to solve the problem?

I used a cosine function once in my solution.

cos(a)=adjacent/hypotenuse

 

[QuickTwist]

I know nothing about the properties on a cosine function so I will have to solve it a different way?

 

[ProxyAmenRa]

I know nothing about the properties on a cosine function so I will have to solve it a different way?

If you understand

tan(a)=opp/adj
sin(a)=opp/hyp

you should understand

cos(a)=adj/hyp

where opp is the side of a right angle triangle opposite to the angle a, adj is the side of a right angle triangle adjacent to the angle a and the hypotenuse of triangle is its longest side.

 

[QuickTwist]

I might have a solution:

[Edit] 2pi *r^2 / 2 *.707

 

[Ex-User (9086)]

(2*pi*r^2-(3*sqrt(3)*r^2))/12~~(6.283*r^2-5.196*r^2)/12=(1.086*r^2)/12

circular segment area`/`equilateral triangle

 

[Coolydudey]

It was orientation week at university today so I decided to celebrate by giving my first year engineering students a math problem. To further celebrate all things related to engineering and studying engineering, I am going to give you kiddos the same problem. After a few days I will upload a hint if you none of your have solved it.

Preface:

Imagine you have circle with a line indicating the diameter drawn horizontally through it. You have another line parallel to the first, with a length of the circle's radius, intersecting the circle's circumference. There is an area 'A' between the second line and the circle's circumference.

Question:

Put forward a method for a calculating the area given that you know the circle's diameter.

Advice:

Start off by drawing the circle and two lines.

So there are two lines, one through the center length 2r (r the radius), another parallel to it which is a chord of length r?

Without using trig (why not?) we have by similar triangles that the distance between the two lines is r/2. Call the two points where the chord intersects the circle A,B and the centre O. Call the two points where the diameter meets the circle C,D (C closest to B, clockwise ordering). OAB is equilateral, so the arc AB is 1/3 of the circle (120 degrees).
The area of the shape is 2*the area of the sector OAD + area of triangle OAB. now the sector OAD sees an arc which is a 1/12 of the circle (30 degrees), so has area 1/12*pi*r^2. The equilateral triangle is side r, so (by Pythagoras, avoiding trig) has area sqrt(3)/2*r^2.
The answer is hence (1/6*pi+sqrt(3)/4)*r^2.

Isn't this a bit basic? Chord non-parallel but length r still clearly gives the same result, should have used that.

 

[QuickTwist]

.707=angle at 45 degrees.

 

[QuickTwist]

Very well could be:

2pi * r^2 / 2 * .637

.637= average of half sign wave

 

[Ex-User (9086)]

Okay, it's r^2/2*(1/3*pi-sina)~~r^2/2*0.1811(6)=0.09058(3)*r^2

which is a more elegant way to put my previous calculation

where r is radius, pi is 3.1415, sina is 0.866

it's more like math trolling, not a serious question

 

[Coolydudey]

an interesting functional equation for the forum I just came up with:
What is the most general thing* we can say about the functions f,g (what they are or what their relationship is) when they obey
f(x)g(y)=f(y)g(x)
f(x)+g(y)=f(y)+g(x)
for all x,y?

You find very few that are just above easy, and this is one (most are either easy or quite hard).

*By most general thing I mean a non-trivial rule they must obey but to which no refinements can be made.

 

[QuickTwist]

Its not a troll thread because there is a right answer which I'm sure PAR will tell us eventually whether or not we are correct.

f(x)g(y)=f(y)g(x) are inverse of each other
In the case of f(x)+g(y)=f(y)+g(x), x and y are the same number and the functions are the same.

Mystery solved.

 

[BigApplePi]

It was orientation week at university today so I decided to celebrate by giving my first year engineering students a math problem. To further celebrate all things related to engineering and studying engineering, I am going to give you kiddos the same problem. After a few days I will upload a hint if you none of your have solved it.

Preface:

Imagine you have circle with a line indicating the diameter drawn horizontally through it. You have another line parallel to the first, with a length of the circle's radius, intersecting the circle's circumference. There is an area 'A' between the second line and the circle's circumference.

Question:

Put forward a method for a calculating the area given that you know the circle's diameter.

Advice:

Start off by drawing the circle and two lines.

I didn't check Cooleydooley but if we are talking the same construction, I get:

Area A = r² [pi/6 - (square root of 3)/4] where r = 1/2 diameter or Area A = about 3 percent of the area of the circle.

 

[QuickTwist]

What angle is the line that is the length of r on a sign wave? I ask because I went about solving the problem in a different (not conventional) way.

 

[Coolydudey]

Its not a troll thread because there is a right answer which I'm sure PAR will tell us eventually whether or not we are correct.

f(x)g(y)=f(y)g(x) are inverse of each other
In the case of f(x)+g(y)=f(y)+g(x), x and y are the same number and the functions are the same.

Mystery solved.

No. Just wrong. Also, they satisfy both at the same time

I didn't check Cooleydooley but if we are talking the same construction, I get:

Area A = r² [pi/6 - (square root of 3)/4] where r = 1/2 diameter or Area A = about 3 percent of the area of the circle.

Yes. 3% seems way too small doesn't it?

 

[QuickTwist]

No. Just wrong. Also, they satisfy both at the same time

You're clearly looking too far into it. If f(x) * g(y) = f(y) * g(y) then either they are/have the same function with the same values for x and y (which I ruled out because that would be far too simplistic and not worth mentioning) or x function/y function= y function/x function. Essentially f o g = g o f. The latter set of functions is very self explanatory because it is addition. Also if they satisfy both at the same time you would have to write it as f(x)g(y)=f(y)g(x)=f(x)+g(y)=f(y)+g(x), in which case x=2 and y=2, simplified down to its basic level.

This shit is elementary.

 

[ProxyAmenRa]

Isn't this a bit basic?

Yes. Yes it is.

it's more like math trolling, not a serious question

First week, first year engineering students seem not to be able to do it.

 

[QuickTwist]

It has been at least 8 years since I saw anything like that question. My memory is shit as it is and I don't use that kind of math very often.

Anyways if the problem hasn't been solved yet this is the answer:

((pi*r^2)/2)-(1/6*pi*r^2)

 

[Ada]

Yes. Yes it is.



First week, first year engineering students seem not to be able to do it.

They couldn't do it? Or were you met with a field of silence when you asked for the answer?

Approach:
Area of segments + Area of equilateral triangle
(1/3*pi*r^2) + (r/(2*tan(30))*r/2)

QuickTwist: Your answer's implying that the area's 1/6 of the total?

 

[ProxyAmenRa]

They couldn't do it? Or were you met with a field of silence when you asked for the answer?

I wondered around as they were supposed to be attempting the problem. Some gave an attempt. Yes, when I asked for the answer I was met by ~150 blank expressions. :P

Approach:
Area of segments + Area of equilateral triangle
(1/3*pi*r^2) + (r/(2*tan(30))*r/2)

QuickTwist: Your answer's implying that the area's 1/6 of the total?

Everyone here looks like they're right. BAP has the right idea.

I will scan and upload my on working for everyone to scrutinize tomorrow.

 

[Coolydudey]

You're clearly looking too far into it. If f(x) * g(y) = f(y) * g(y) then either they are/have the same function with the same values for x and y (which I ruled out because that would be far too simplistic and not worth mentioning) or x function/y function= y function/x function. Essentially f o g = g o f. The latter set of functions is very self explanatory because it is addition. Also if they satisfy both at the same time you would have to write it as f(x)g(y)=f(y)g(x)=f(x)+g(y)=f(y)+g(x), in which case x=2 and y=2, simplified down to its basic level.

This shit is elementary.

wrong. as have been your answers to the OP problem btw

 

[QuickTwist]

Maybe you could explain what was wrong?

 

[crippli]

I learned these things once, but it is all a blur now. I do believe this will provide the answer. (didn't read the replies).

Area= (Area half sector (60/360)pi x r2 - Area triangle ((r/2)x sq(r2-(r/2)2))/2 ) x 2 (full area)

 

[Coolydudey]

Maybe you could explain what was wrong?

Eventually. Want to give others a chance too.

 

[BigApplePi]

I saw this problem as having two parts:
1. Interpreting the question so the picture could be drawn.
2. Performing the calculation.

 

[Coolydudey]

Right well to anyone interested:
Clearly the more strict condition is the second. Rearranging, and setting y=0, we have f(x)-g(x)=f(0)-g(0)=c, a constant.
so for all x, we have f(x)=g(x)+c. Substituting into the first:
f(x)g(y)=f(y)g(x) -> g(x)g(y)+cg(y)=g(x)g(y)+cg(x) -> c(g(y)-g(x))=0 for all x,y. So either g is constant or c=0.
Hence f=g or f and g are two arbitrary constant functions.

 

[Trebuchet]

I saw this problem as having two parts:
1. Interpreting the question so the picture could be drawn.
2. Performing the calculation.

Yeah, actually I first calculated the area between the two parallel lines, then reread the question and had to subtract.

Careful reading: your friend in math. (Or maths, for those who spell it that way.)

 

[Base groove]

I'll take a stab at it.

If the chord length is equivalent to the radius (c=r) then it will subtend an equilateral triangle (theta @ = 60 degrees or pi/3).

The rest is straightforward.

A1 = (1/2)r^2 ( @ - sin@)
A2 = (1/2)pi*r^2

A2-A1 and simplify/execute.

 

[ProxyAmenRa]

I learned these things once, but it is all a blur now. I do believe this will provide the answer. (didn't read the replies).

Area= (Area half sector (60/360)pi x r2 - Area triangle ((r/2)x sq(r2-(r/2)2))/2 ) x 2 (full area)

This is correct. ^_^

Everyone is this thread had the right idea. XD

---

Well, I look forward to another year of techning engineering students programming, math and construction engineering.

(I cry a little inside. It is too painful.)

 

[Ada]

Haha, it's their first week, they're probably still assessing how scary you are.

 

[QuickTwist]

They are still highschoolers give them some time to get the kinks out. Its been a long summer for them.

 

[crippli]

Thanks for the brain tease.

Yeah, actually I first calculated the area between the two parallel lines, then reread the question and had to subtract.

Me too. Until I read the Q again. Reading the question three times would probably be the best. I doubt a math teacher would subtract much from the score though, even if one did the area between the circles. As it's the same equation twice.

 

[envirodude]

A = pi r² / 2 - (area of equilateral triangle with sides r + pi r² (120/360) )