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Junk: What are the hardest mathematic question you have ever seen?
#1
Greetings. I am about to start with the mathematic again, so I would like to see if someone could actually come up with a question which I can't answer.

Only mathematic questions are allowed here.

Also, please explain the question as good as possible, cause then it would be easier for me to understand.

Be cool Cool
- Mark Topper
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#2
2y - 5
Because I didn't know the value of y.
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#3
(08-11-2012, 09:25 PM)Oqud Wrote: 2y - 5
Because I didn't know the value of y.

Well, it still have to be possible to figure out. As the y could be everything is it the sum is not set. Cool
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#4
Decent problem (Calc 3)

1) Find the flux of the vector field F(x,y,z) = (e^-y) i - (y) j + (x sinz) k across σ wit outward orientation where σ is the portion of the elliptic cylinder r(u,v) = (2cos v) i + (sin v) j + (u) k with 0 ≤ u ≤ 5, 0 ≤ v ≤ 2pi.

2) Find the work done by the force field F(x, y, z) = (x + y) i + (xy) j - (z^2) k on a particle that moves along line segments from (0,0,0) to (1,3,1) to (2,-1,4).

Answer:
Code:
∫∫s F dS ... F = e^(-y) i - y j + x sin(z) k
s: the elliptic cylinder r(u,v) = 2 cos(v) i + sin(v) j + u k
... 0 ≤ u ≤ 5, 0 ≤ v ≤ 2π

∫∫s F dS = ∫∫∫v ∇·F dV
∇·F = x cos(z) - 1

= ∫∫∫v x cos(z) - 1 dx dy dz
v: x² + 4y² ≤ 4 ; 0 ≤ z ≤ 5

let u = x , v = 2y , z = z
∂(u,v)/∂(x,y) = 2

= 1/2 ∫∫∫v u cos(z) - 1 du dv dz
v: u² + v² ≤ 4 ; 0 ≤ z ≤ 5

let u = r cos(θ) ; v = r sin(θ) ; z = z
∂(u,v)/∂(r,θ) = r

= 1/2 ∫∫∫ (r cos(θ) cos(z) - 1) r dr dθ dz
{(r,θ,z) | 0 ≤ r ≤ 2 ; 0 ≤ θ ≤ 2π ; 0 ≤ z ≤ 5}

= 1/3 ∫∫ (4 cos(θ) cos(z) - 3) dθ dz
{(θ,z) | 0 ≤ θ ≤ 2π ; 0 ≤ z ≤ 5}

= -2π ∫ dz
{(z) | 0 ≤ z ≤ 5}

= -10π

Also, this one isn't too bad without the divergence theorem
because obviously there isn't any flux across the ends.

∫∫s F dS

= ∫∫s F(u,v)·<∂r/∂u x ∂r/∂v > du dv

∂r/∂u x ∂r/∂v = <-cos(v), 2 sin(v), 0>
F(u,v) = <e^(-sin(v)),-sin(v),2 cos(v) sin(u)>

= ∫∫ -cos(v) e^(-sin(v)) - 2 sin²(v) du dv
{(u,v) | 0 ≤ u ≤ 5 ; 0 ≤ v ≤ 2π}

= 5 ∫ -cos(v) e^(-sin(v)) - 2 sin²(v) dv
{(v) | 0 ≤ v ≤ 2π}

= -10π
Answer (1): -10π


2. ∫c F·dr .... F = (x + y) i + (xy) j - (z^2) k ; c: from (0,0,0) to (1,3,1) to (2,-1,4)

→ (0,0,0) to (1,3,1)
x = t ; y = 3t ; z = t ; 0 ≤ t ≤ 1
dx = 1 dt; dy = 3 dt; dz = 1 dt

∫c F·dr = ∫ (x + y) dx + xy dy - z^2 dz
= ∫ (t + 3t) + 9t² - t² dt [0,1]
= 14/3

→ (1,3,1) to (2,-1,4)
x = t + 1 ; y = 3 - 4t ; z = 1 + 3t
dx = 1 dt; dy = -4 dt; dz = 3 dt

∫c F·dr = ∫ (x + y) dx + xy dy - z^2 dz
= ∫ (t + 1 + 3 - 4t) - 4(t + 1)(3 - 4t) - 3(1 + 3t)^2 dt [0,1]
= ∫ -11t^2 - 17t - 11 dt [0,1]
= -139/6

14/3 - 139/6 = -37/2

Answer (2): -37/2
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#5
This thread has been junked for a reason, leave it!
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