Solve this

[Chemical]

Yah, I don't know if you got rid of the homework forum or what?
But I require some assistance with this algebraic equation:

The problem, as a whole, is that I have two circles and I need to find the x-intercepts. x² + (y-50)² = 50² and x² + y² = r² (radius unknown).

Help!

[Tavrobel]

Yes, they did remove Study Hall.

So, to find the x intercepts of the equation, we should set y=0. This is because an x intercept is when the graph of an equation crosses the x axis, which means it does not go up or down in the vertical (y) direction.

Given: x² + y² = r²

Problem Statement: x² + (y - 50)² = 50²

Substitution for y = 0:
x² + (0 - 50)² = 50²

Simplify:
x² + (-50)² = 50²
x² + 2500 = 2500
x² = 2500 - 2500 = 0

x² = 0
x = +/- 0

This graph only touches the x axis at one point: (0,0)

In case you were wondering, the radius of your circle is 50. In addition, your graph is that of a circle that has been shifted up. Two y intercepts, one at (0,0) and another at (0,100) exist.

[rubah]

one of the ph.D students gave me this problem today xD

Suppose that you have a perfectly round world with one airport and many planes. With a full tank of gas, a plane can fly halfway around the world. however, you don't want to crash the plane! You can, though, refuel in midair from another airplane. Each plane holds the same amount of fuel and gets the same efficiency. How many planes do you need (minimum) to nurse one plane all the way around the globe? You aren't allowed to crash any of the helper planes.

[blackmage_nuke]

x² + (y-50)² = 50² equation 1
x² + y² = r² equation 2

x² = 50²-(y-50)²
sub into equation 2
50² - (y - 50)² + y² = r²
expand the bracket
50² - (y² - 100y + 2500) + y² = r²
100y = r²
y = r²/100
sub into 2
x² + r²/100 = r²
x² = r² - r²/100
x² = 99r²/100

x = +-√(99r²/100)
x = +-(√(99)r/10)

so the circles intersect at
(√(99)r/10 , r²/100) and (-√(99)r/10 , r²/100)

edit: oops my bad didnt read the question (i thought it asked where the circles intersect)

@ rubah's question

Assuming you can refuel the moment all you're fuel runs out I'd say 3 (and possibly 2), also ignorng any rotation of the world...

two planes set off at the same time
quater way around the world one refuels the other and flies back to the airport
when the second plane reaches the point halfway around the world a third plane is sent from the airport in the opposite direction
when the second plane reaches 3/4 around the world the third plane refuels the second plane and they both fly back to the airport together

alternatively if we can refuel at the airport then you can refuel the first plane and send it off in the oppsoite direction instead of the third plane, meaning you will only need 2 planes

horribly wrong forgot that the first refueling would take away the first plane's fuel.

[Chemical]

Thanks Tav uhmmmm how about this though:

These are two different circles. One has a fixed radius of 50, the other I have to determine based on some other criteria that isn't relevant at this point. I should have been more clear that I need to find the x-values of the points where the two circles intercept! From what I understand, I have to substitute my solved y value into the other equation, giving me...

x² + (sqrt(-x²+2500)+50)² = r²

and now solve for x... Ew.

[rubah]

if you fly a helper plane 1/4 the way down, it can still only give the other plane 1/4 a tank, because the other plane hasn't used more than that yet.

[Tavrobel]

one of the ph.D students gave me this problem today xD

Suppose that you have a perfectly round world with one airport and many planes. With a full tank of gas, a plane can fly halfway around the world. however, you don't want to crash the plane! You can, though, refuel in midair from another airplane. Each plane holds the same amount of fuel and gets the same efficiency. How many planes do you need (minimum) to nurse one plane all the way around the globe? You aren't allowed to crash any of the helper planes.

I'm not sure, but I think that this is impossible. If we assume uniform properties for all of the planes, then for every plane we send out, we would need to send more planes to be able to get that refueling plane further up to where the original plane was. For every plane you add, you bring back the average point that each plane could travel.

[Bunny]

News Alert:

Math (http://forums.eyesonff.com/2884124-post8.html)still (http://forums.eyesonff.com/2842284-post26.html) sucks (http://forums.eyesonff.com/2847986-post82.html)!

[Rye]

I moved this to the help forum, since it's the closest we've got to the old study hall forum. Hope you don't mind. ;)

[Chemical]

I moved this to the help forum, since it's the closest we've got to the old study hall forum. Hope you don't mind. ;)

I might since now I'm less likely to get people looking at it... who the hell goes to the Help forum anymore?

[Tavrobel]

These are two different circles. One has a fixed radius of 50, the other I have to determine based on some other criteria that isn't relevant at this point. I should have been more clear that I need to find the x-values of the points where the two circles intercept! From what I understand, I have to substitute my solved y value into the other equation, giving me...

x² + (sqrt(-x²+2500)+50)² = r²

and now solve for x... Ew.

Given: x² + ((-x + 2500)^1/2 + 50)² = r²

Solving without r, we would have to do this problem symbolically. However, since I am a lazy ass, I will assume that r = 50.

FOIL:
x² + (-x² + 2500) + 2500 + 2*(-x + 2500)^1/2 = 2500

Simplify:
x² - x² + 5000 + 2*(-x + 2500)^1/2 = 2500
2*(-x + 2500)^1/2 = -2500
(-x + 2500)^1/2 = -1250

Square both sides:
-x + 2500 = (-1250)²
x = 2500 - (1250)²

... wait what, let me come back to this.

[blackmage_nuke]

OK i say 6 (possibly 4) assuming you can refuel the moment all you're fuel runs out
3 planes set off at the same time

at 1/8 of the distance around the world plane 3 fuels 1 and 2 to maximum
so the fuel in each plane is now
plane 1: full
plane 2: full
plane 3: 1/4 (1/4 to get there, 1/4 to plane 1 and 1/4 to plane 2 leaves 1/4 remaining)
plane 3 has enough fuel to fly back to the airport

at 1/4 distance around the world plane 2 refuels plane 1
plane 1: full
plane 2: 1/2 (1/4 to get there from the previous refuel, 1/4 to plane 1)
thats enough for plane 2 fly back to base

when plane 1 is halfway across the world three planes (4,5 and 6) set out from the airport in the oppsoite direction
when plane 4, 5 and 6 reach 7/8 distance around the world (from the original direction) 4 refuels 5 and 6 so
4: 1/4
5: full
6: full
4 flies back to base

at 3/4 around the world (from the original direction) 5 and 6 (which now have 3/4 fuel) will meet 1 (which now has 0 fuel)
5 and 6 both give plane 1, 1/4 of their fuel each.
1: 1/2
5: 1/2
6: 1/2

This is enough fuel to get all 3 planes back to the airport

alternatively if refueling is allowed at the airport then planes 2 and 3 can be used in place of 5 and 6 meaning only 4 planes will be needed.

also:
if you fly a helper plane 1/4 the way down, it can still only give the other plane 1/4 a tank, because the other plane hasn't used more than that yet.
wouldnt a quater of the way around be half a tank since halfway around is a full tank? Not that it matters since my first answer wouldve caused the first and third plane to crash, but it's all for the greater good!

also

Thanks Tav uhmmmm how about this though:

These are two different circles. One has a fixed radius of 50, the other I have to determine based on some other criteria that isn't relevant at this point. I should have been more clear that I need to find the x-values of the points where the two circles intercept! From what I understand, I have to substitute my solved y value into the other equation, giving me...

x² + (sqrt(-x²+2500)+50)² = r²

and now solve for x... Ew.
then my first solution is what you want
x² + (y-50)² = 50² equation 1
x² + y² = r² equation 2

x² = 50²-(y-50)²
sub into equation 2
50² - (y - 50)² + y² = r²
expand the bracket
50² - (y² - 100y + 2500) + y² = r²
100y = r²
y = r²/100
sub into 2
x² + r²/100 = r²
x² = r² - r²/100
x² = 99r²/100

x = +-√(99r²/100)
x = +-(√(99)r/10)

so the circles intersect at
(√(99)r/10 , r²/100) and (-√(99)r/10 , r²/100)

The circles intercept at
x=√(99)r/10
and
x=-√(99)r/10

My bad (what trav says below)

[rubah]

well, I meant a quarter of a tank. It would be an eighth xD. I just had a weird way of counting, as you'll read below.
Also, I think it's counting planes full of fuel, not literal planes. but that's pretty good either way


I have it down to 5 planes.

The main plane and the first helper plane take off, and the 1st refuels 1/3 of the main plane's tank at the 1/3 mark. First helper plane makes it back safely.
The second plane also leaves with those two, and at the 4/9 fuel mark, he gives 1/9 fuel to the main plane, and makes it back safely.

The main plane now has an extra 4/9 tank of fuel to come back home on.

Okay, so the main plane comes back 4/9 of his tank, at the -5/9 mark. The third helper plane meets her, and gives 2/9 of their tank. Plane 3 and the main plane can each make it to -3/9.

The fourth plane meets them at -3/9 and gives them each 1.5/9 of fuel. The fourth plane makes it back safely, and the main plane and helper are each at -1.5/9

The fifth plane flies to 1.5/9 and gives everyone all the fuel ever.

This can be optimized further, I'm sure, but I lack the brainpower to do it atm

[Tavrobel]

My last post, shut up.

Okay, so we are given two equations:
x² + y² = r²
x² + (y - 50)² = 50²

We are trying to find the values where the x values will intercept. Therefore, we are finding the values for which x will be the same, as it will fulfill both equations. So, first, solve both equations, isolating x.

x² = r² - y²
x² = 50² - (y - 50)²

Since x² in both equations are the same value, we can set them equal to each other.
r² - y² = 50² - (y - 50)²

FOIL the right side part of the equation in parentheses:
r² - y² = 2500 - (y² - 100*y + 2500)

Simplify:
r² - y² = 0 - y² + 100*y
r² = 100*y
y = r²/100

Substitute our newest finding into the second equation:
x² + (r²/100 - 50)² = 50²

FOIL the former (y - 50)² term:
x² + r⁴/100² + 2500 - 50*r²/100 - 50*r²/100 = 2500

Simplify again:
x² + r⁴/100² -100*r²/100 = 2500 - 2500
x² + r⁴/100² - r² = 0
x² = r² - r⁴/100²

Factor r² out from the right side:
x² = r²(1 - r²/100²)

Your x values will depend on the radius of the undetermined circle.
x = +/- (r²(1 - r²/100²))^1/2

NOTE: Something raised to the half in the exponent is the same as a square root symbol.

NOTE: The solution is in a symbolic format. Notice that the radius of the undetermined circle must also be a value less than 100.

MORE NOTE: Substituting in a new y value into the first equation will yield the same result.

y = r²/100
y² = r⁴/100²

x² + y² = r²
x² + r⁴/100² = r²
x² = r² - r⁴/100²
Algebra!



50² - (y - 50)² + y² = r²
expand the bracket
50² - (y² - 100y + 2500) + y² = r²
100y = r²
y = r²/100
sub into 2
x² + r²/100 = r²
x² = r² - r²/100

Algebraic error, you forgot to square the new y term when you plugged it into the equation. Notice that if you intend on substituting y = r²/100 into x² + y² = r², you have to square the y term.

y² = r⁴/10000

[blackmage_nuke]

Ah if we dont count the main plain my solution also has 5 planes

[Chemical]

thanks to all

[rubah]

I asked the guy today, and he said it was 2 helper planes :p obviously we have a long ways yet to go! unless there was some sort of thing about refueling. I should ask him to break it down.

[Manny]

I think my brain just collapsed into itself and became antimatter.

[Tavrobel]

I asked the guy today, and he said it was 2 helper planes :p obviously we have a long ways yet to go! unless there was some sort of thing about refueling. I should ask him to break it down.

Three planes depart:
(x, t) = 0, 0
P₁ = 1
P₂ = 1
P₃ = 1

(x, t) = 1/8, 1
P₁ = 3/4
P₂ = 3/4
P₃ = 3/4

(x, t) = 1/8, 2
P₁ = 1
P₂ = 1
P₃ = 1/4

(x, t) = 1/4, 3
P₁ = 3/4
P₂ = 3/4
P₃ = 0 (port)

(x, t) = 1/4, 4
P₁ = 1
P₂ = 1/2
P₃ = 1 (port)

(x, t) = 3/4, 5
P₁ = 0
P₂ = 1 (port)
P₃ = 1 (port)

(x, t) = 3/4, 6
P₁ = 0
P₂ = 2/3
P₃ = 2/3
P₂ and P₃ are traveling in the negative direction, traveling 1/6 of the way, consuming 1/3 of the tank.

(x, t) = 3/4, 7
P₁ = 0
P₂ = 1 (x = -1/6)
P₃ = 1/3

P₂ travels from -1/6 to -1/4, which is a distance of 1/12, consuming 1/6th of the tank.

(x, t) = 3/4, 8
P₁ = 0
P₂ = 5/6 (x = -1/4)
P₃ = 0 (port)

Move 1/3 of the tank.

(x, t) = 3/4, 9
P₁ = 1/3
P₂ = 1/2
P₃ = 1 (port)

P₃ must travel out. P₁ and P₂ move another 1/6th forward, consuming 1/3 of the tank. 3/4 + 1/6 = 11/12.

(x, t) = 11/12, 10
P₁ = 0
P₂ = 1/6
P₃ = 5/6 (x = -1/12)

(x, t) = 11/12, 11
P₁ = 1/6
P₂ = 1/6
P₃ = 4/6

(x, t) = 1, 12
P₁ = 0
P₂ = 0
P₃ = 1/2

So basically, all three planes went together at first. Plane 3 gave its fuel to the first two, which caused it to head back, and the other two to travel on. The first two went further, and the second plane gave its fuel to the first plane. Plane 2 heads back, while plane 3 refuels. Plane 1 spends its gas, while plane 2 and 3 are refueled.

Plane 2 and 3 go a distance, until plane 3 gives its fuel to plane 2. Plane 3 heads back, while plane 2 makes its way to the first plane. Plane 2 refuels plane 1 once again, while plane 3 is refueled. Plane 3 heads out to refuel both plane 1 and plane 2 and head to port from the negative x direction. All three planes land down.