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UnLoving NoviceWhat are pooples?
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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.
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That's a term when people sculpt there s--- into people.
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UnLoving NoviceHaha X'D
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F(x,y,k)= 2px+qy^2+k(x^3+y^3-1)
Fx= 2p +3kx^2=0
Fy= 2qy+ +3ky^2=0
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)^3*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
if1+(p/q)^3 -
nice copy paste. 10/10
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XD.
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UnLoving NoviceWhat does the fox say?
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It says show me the Carmax
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UnLoving Novice"But what if I don't want to see the car fax?"
"Then uhh....ringsingdingdingdingdingsing"
"But what if I don't like that song?"
"Then uhh....show me the car fax." -
XD
It also might say "aw man!" -
UnLoving NoviceLol XD
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XD
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UnLoving NoviceI love how most of our convos go like this:
"XD"
"XD
"XD
"XD"
"XD"
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