Math Problems

tbh i would have been disappointed if noone had bothered to answer.

to say somethin relevant: i thought about intergration of the area of triangles with one angle as variable but i have to find a possibility of getting the area that is counted several times in that case. i dunno if that's the way to go

wait, I have an idea
polar coordinates. I could be wrong I am typing this with lack of sleep but I believe the shape could be described by r=1-sin(theta)*cos(theta)/sqrt(2) which you would integrate from 0 to pi/2 which I integrate by squaring then integrating normally then dividing by one half. I could be wrong someone confirm or tell me I am wrong please. so I need to integrate 1-2sin(theta)cos(theta)/sqrt(2)+sin(theta)cos(theta)*sin(theta)cos(theta)/2 which I can turn into 1-sin(2theta)/sqrt(2)+sin(2theta)^(2)/8 which I can turn into 1-sin(2theta)/sqrt(2)+(1-cos(4theta))/16 which integrates into
(theta+cos(2theta)/2sqrt(2)+(theta-sin(4theta)/4)/16)/2. right? I know a lot of trig identity. so we get (17pi/64-1/2sqrt2) or close to .4809 right?
again I could be wrong.

Quote from: iznih on September 06, 2009, 09:12:20 AM

Quote from: iznih on September 05, 2009, 06:25:20 PM

ah,i think i get it. one boundary is a quarter of a circle

someties i'm ashamed of myself 
it's clearly not a circle

That's not come up for me, but yes, the answer probably is gay porn.

wait, I have an idea
polar coordinates. I could be wrong I am typing this with lack of sleep but I believe the shape could be described by r=1-sin(theta)*cos(theta)/sqrt(2) which you would integrate from 0 to pi/2 which I integrate by squaring then integrating normally then dividing by one half. I could be wrong someone confirm or tell me I am wrong please. so I need to integrate 1-2sin(theta)cos(theta)/sqrt(2)+sin(theta)cos(theta)*sin(theta)cos(theta)/2 which I can turn into 1-sin(2theta)/sqrt(2)+sin(2theta)^(2)/8 which I can turn into 1-sin(2theta)/sqrt(2)+(1-cos(4theta))/16 which integrates into
(theta+cos(2theta)/2sqrt(2)+(theta-sin(4theta)/4)/16)/2. right? I know a lot of trig identity. so we get (17pi/64-1/2sqrt2) or close to .4809 right?
again I could be wrong.

anyway to stay on topic does anyone know if I am right or not.


second eddy baby quit quoting him.

2am over here so no guarantees: i found a solution for the upper boundary of the area, i doubt i'm able to intergrate it in finite time, a numerical solution yields an area of 0,2945.  
if that's correct i'll post the derivation of the upper boundary

2am over here so no guarantees: i found a solution for the upper boundary of the area, i doubt i'm able to intergrate it in finite time, a numerical solution yields an area of 0,2945. 
if that's correct i'll post the derivation of the upper boundary

something somewhere went wrong. I am assuming you meant .26945. anyway I would like to see your integral, I would also like if you could take a look at mine and see if I messed up anywhere.

A line segment AB of length 1 has endpoints A, constrained to the nonnegative y axis, and B, constrained to the nonnegative x axis.  Let the region S of the xy plane be defined as follows.  A point P is in S if P lies within a right triangle formed by A, B, and the origin for some A,B satisfying the condition above.

What is the area of S?  What is the length of its boundary?  Find an equation of each smooth segment of the boundary of S in Cartesian coordinates.

The equation for the line segment AB with endpoints A(0, y) and B(x, 0) is:

u(t) - y = (y - 0)/(0 - x) *(t - 0) = -y*t/x

u(t) = y*(1 - t/x)

However, the parameters x and y are not independent since AB has lenght 1:

x² + y² = 1 => y = √(1 - x²), 0 x 1

So, we have a family of lines given by the parameter 0 ≤ x ≤ 1:

u(t; x) = √(1 - x²) * (1 - t/x)

The region S that they are talking about is delimited by the positive x and y axes and the envelope of the family of curves u(t; x). The equation of the envelope is found from the condition:

∂u/∂x = -[x/√(1 - x²)]*(1 - t/x) + √(1 - x²)*t/x² = 0 /. x²*√(1 - x²)

- x²*(x - t) + (1 - x²)*t = 0

-x³ + x²*t + t - x²*t = 0

x³ = t

x = t^1/3

Substitute this value in the equation of the line and we get:

u(t; t^1/3) = √(1 - t^2/3)*(1 - t/t^1/3) = (1 - t^2/3)^3/2, 0 ≤ t ≤ 1

The area of the region S is given by the integral:

A = Int[u dt, {t, 0, 1}]

A = Int[(1 - t^2/3)^3/2 dt,{t, 0, 1}]

This integral can be reduced to Euler Beta function by the substitution:

z = t^2/3 => t = u^3/2, t = 0 -> u = 0, t = 1 -> u = 1, dt = 3/2 u^1/2 du

A = 3/2 Int[u^1/2 (1 - u)^3/2,{u, 0, 1}]

A = 3/2 B(3/2, 5/2)

Using the relation between the Euler Beta and Euler Gamma function, we get:

A = 3/2 Γ(3/2) Γ(5/2)/ Γ(4) = (3/2) * (1/2) * (3/2) *  (1/2) * [Γ(1/2)]² / 6 = 3*π/32

For the other parts of the problem do it yourself!

somehow i arrived at 3pi/32 
seems i lost a factor of 2 somewhere, my solution itself seems not that bad

You are right. I missed one of the twos.

somehow i arrived at 3pi/32 
seems i lost a factor of 2 somewhere, my solution itself seems not that bad

Winrar.  What was your method?

BTW, this is what the region S looks like:

AHAHAHA UR ALL SUCH GEEEKS!!

IM SOOO GLAD IM NOT A GEEEEEEEK LAIK U PPLZ.

BTW, this is what the region S looks like:

there has got to be a simple polar graph for this

Quote from: Euclid on September 04, 2009, 08:18:55 PM

A line segment AB of length 1 has endpoints A, constrained to the nonnegative y axis, and B, constrained to the nonnegative x axis.  Let the region S of the xy plane be defined as follows.  A point P is in S if P lies within a right triangle formed by A, B, and the origin for some A,B satisfying the condition above.

What is the area of S?  What is the length of its boundary?  Find an equation of each smooth segment of the boundary of S in Cartesian coordinates.

The equation for the line segment AB with endpoints A(0, y) and B(x, 0) is:

u(t) - y = (y - 0)/(0 - x) *(t - 0) = -y*t/x

u(t) = y*(1 - t/x)

However, the parameters x and y are not independent since AB has lenght 1:

x² + y² = 1 => y = √(1 - x²), 0 x 1

So, we have a family of lines given by the parameter 0 ≤ x ≤ 1:

u(t; x) = √(1 - x²) * (1 - t/x)

The region S that they are talking about is delimited by the positive x and y axes and the envelope of the family of curves u(t; x). The equation of the envelope is found from the condition:

∂u/∂x = -[x/√(1 - x²)]*(1 - t/x) + √(1 - x²)*t/x² = 0 /. x²*√(1 - x²)

- x²*(x - t) + (1 - x²)*t = 0

-x³ + x²*t + t - x²*t = 0

x³ = t

x = t^1/3

Substitute this value in the equation of the line and we get:

u(t; t^1/3) = √(1 - t^2/3)*(1 - t/t^1/3) = (1 - t^2/3)^3/2, 0 ≤ t ≤ 1

The area of the region S is given by the integral:

A = Int[u dt, {t, 0, 1}]

A = Int[(1 - t^2/3)^3/2 dt,{t, 0, 1}]

This integral can be reduced to Euler Beta function by the substitution:

z = t^2/3 => t = u^3/2, t = 0 -> u = 0, t = 1 -> u = 1, dt = 3/2 u^1/2 du

A = 3/2 Int[u^1/2 (1 - u)^3/2,{u, 0, 1}]

A = 3/2 B(3/2, 5/2)

Using the relation between the Euler Beta and Euler Gamma function, we get:

A = 3/2 Γ(3/2) Γ(5/2)/ Γ(4) = (3/2) * (1/2) * (3/2) *  (1/2) * [Γ(1/2)]² / 6 = 3*π/32

For the other parts of the problem do it yourself!

okay where can I learn how to do this? any good web sites? I never learned how to use parametric to solve integrals very well.

Quote from: iznih on September 07, 2009, 05:56:16 PM

somehow i arrived at 3pi/32 
seems i lost a factor of 2 somewhere, my solution itself seems not that bad

Winrar.  What was your method?

similar to bridget's method, i used x and the angle formed by the x-axis and AB though. that leads to

y=sin(phi)-xtan(phi)

for a given x  dy/dphi=0 leads to cos(phi)^3=x

i plugged that into the formula for y and was a bit lost cause i had no idea on how to integrate it. i showed it a mate and he somehow solved it  . seems he only wrote the remaining sine as sqrt(1-cos^2) and plugged in cos(phi)^3=x leading to bridget's integral

since no one else is posting I will post a question I enjoyed solving in a past physics class. the problem once you understand it is quite easy to solve.
you have a refrigerator that ways 150 Kg and is 1 meter deep and 2.5 meters tall, the center of mass is in the center and the coefficient of static friction is .4. what is the maximum height you can push on the fridge and it will move forward instead of tipping.

since no one else is posting I will post a question I enjoyed solving in a past physics class. the problem once you understand it is quite easy to solve.
you have a refrigerator that ways 150 Kg and is 1 meter deep and 2.5 meters tall, the center of mass is in the center and the coefficient of static friction is .4. what is the maximum height you can push on the fridge and it will move forward instead of tipping.

1.25 metres?

Edit: That should have been sort of obvious, now that I come to think about it. I did it a far longer way than was necessary.

Quote from: optimisticcynic on September 12, 2009, 12:16:09 AM

since no one else is posting I will post a question I enjoyed solving in a past physics class. the problem once you understand it is quite easy to solve.
you have a refrigerator that ways 150 Kg and is 1 meter deep and 2.5 meters tall, the center of mass is in the center and the coefficient of static friction is .4. what is the maximum height you can push on the fridge and it will move forward instead of tipping.

1.25 metres?

Edit: That should have been sort of obvious, now that I come to think about it. I did it a far longer way than was necessary.

How on earth??   

GEEEK.

At least I can take solace in the fact that knowing this isn't ever going to help anyone outside of school.

In the real world, math genius = pizza boy. So there.

Quote from: optimisticcynic on September 12, 2009, 12:16:09 AM

since no one else is posting I will post a question I enjoyed solving in a past physics class. the problem once you understand it is quite easy to solve.
you have a refrigerator that ways 150 Kg and is 1 meter deep and 2.5 meters tall, the center of mass is in the center and the coefficient of static friction is .4. what is the maximum height you can push on the fridge and it will move forward instead of tipping.

1.25 metres?

Edit: That should have been sort of obvious, now that I come to think about it. I did it a far longer way than was necessary.

Eh you aren't the only one. when we were first asked this in my physics 1 class I was the only one who used under a page of work. for some reason people have trouble with this one. the over complicate it.

Eh you aren't the only one. one we were first asked this in my physics 1 class I was the only one who used under a page of work. for some reason people have trouble with this one. the over complicate it.

What the hell? I did it a needlessly complicated way, and even I had under a page of work. Tell the other people in your physics class that they are fucking retarted.

Quote from: optimisticcynic on September 12, 2009, 08:53:29 AM

Eh you aren't the only one. one we were first asked this in my physics 1 class I was the only one who used under a page of work. for some reason people have trouble with this one. the over complicate it.

What the hell? I did it a needlessly complicated way, and even I had under a page of work. Tell the other people in your physics class that they are fucking retarted.

A. that would be hard since I took the class in high school over 1.5 years ago and
B. we didn't have a teacher. give them a little slack.

Now that I look at it again, my way may not have been that unnecessarily complicated. What way did you do it?

Here is a good physics one dealing with Geodesics

Show that  U^BU^alpha;_B=fU^alpha
where f is a scalar function, parametrizes the same geodesic as:  U^BU^alpha;_B=0

Good Luck

I think the unspoken rule here is to post problems you can solve yourself.

I think the unspoken rule here is to post problems you can solve yourself.

Oh I can solve it. It is a problem that I had for one of my classes

Now that I look at it again, my way may not have been that unnecessarily complicated. What way did you do it?

I made an equation 9.81*150*.4*(force of static friction.)h=150*9.81*.5(torque caused by gravity.) solve for h. most stuff canceled.

Quote from: ﮎingulaЯiτy on September 12, 2009, 04:05:52 PM

I think the unspoken rule here is to post problems you can solve yourself.

Oh I can solve it. It is a problem that I had for one of my classes

Given the content of your previous posts and the nature of the problem, I'm calling BS and saying you stole this from a textbook or a class website.

Quote from: cbarnett97 on September 12, 2009, 04:49:31 PM

Quote from: ﮎingulaЯiτy on September 12, 2009, 04:05:52 PM

I think the unspoken rule here is to post problems you can solve yourself.

Oh I can solve it. It is a problem that I had for one of my classes

Given the content of your previous posts and the nature of the problem, I'm calling BS and saying you stole this from a textbook or a class website.

Your powers of observation are outstanding, especially since i said that it came from one of my classes

Quote from: Euclid on September 12, 2009, 08:05:05 PM

Quote from: cbarnett97 on September 12, 2009, 04:49:31 PM

Quote from: ﮎingulaЯiτy on September 12, 2009, 04:05:52 PM

I think the unspoken rule here is to post problems you can solve yourself.

Oh I can solve it. It is a problem that I had for one of my classes

Given the content of your previous posts and the nature of the problem, I'm calling BS and saying you stole this from a textbook or a class website.

Your powers of observation are outstanding, especially since i said that it came from one of my classes

I'm guessing you stole this problem from the Caltech website, and I'm fairly certain you never went there.

In fact, I just found the problem to which you refer in last academic year's course materials.  How original.