Carri's busy doing random stuff
- Locked due to inactivity on Aug 4, '16 4:27pm
Thread Topic: Carri's busy doing random stuff
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So I'm going to sit here and work out advanced math problems.
Determine the regions of the pq-plane for which the function f(x,y) = 2px + qy^2 restricted to the curve x^3 + y^3 = 1 has exactly one and exactly three critical points, respectively.This should work. -
Can you help me with figuring out my math homework ?
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I'm just derping in the shadows of the internet. owo
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so y=0 and x=1
2q+3ky=0 so k=-2q/3y=-2p/3x^2
q/y=p/x^2 so y=qx^2/p and (q/p)^3x*x^6+x^3-1=0
call x^3=z
(p/q)^3z^2+z-1=0 if 1+4(p/q)^3>0 there are two roots for x^3 and so for x and added to (1,0) there are three critical points
If 1+4(p/q)^3=0 (p/q)^3=-1/4 and -1/4z^2+z-1=z^2-4z+4=(z-2)^2=0 and z=2 and x=2^1/3
if 1+(p/q)^3 < 0 only one critical point (1,0)
Well that was easy enough. -
And I can't even figure out powers of powers!
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f---ing hell, I did this in word and missed the first three lines.
F(x,y,k)=2px+qy^2+k(x^3+y^3-1)
Fx=2p+2kx^2=0
Fy= 2qy+3ky^2=0
^ Insert that at the beginning or else it won't make sense.
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