Guest boysetsfire Posted April 1, 2004 Posted April 1, 2004 I had this question on a test today and just want to know if i was on the right path untill i got too frustrated and gave up. Find the closest points to the origin on the surface x^2-yz=5. This is what i did d=sqrt( (x-0)^2 + ( y-0 )^2 + (z-0 )^2 ) then i took the f(x,y) = x^2 + y^2 +( (x^2-5)/y )^2 took partial derivatives but then i couldnt solve for either x or y. -Marc
wolfson Posted April 1, 2004 Posted April 1, 2004 D^2 = x^2 + y^2 + z^2 = y^2 + y^z + z^2 + 5 Now note that if z is non-zero then the expression (y^2 + y^z + z^2) is always greater than zero, Therefore the smallest value of D^2 (and hence D) will be when z = y = 0. Thus finding the partial derivitives of this expression w.r.t. to y and z and setting them equal to 0. Threrfore 5,0,0. Oh and note also that (-5,0,0) is also a point.
wolfson Posted April 2, 2004 Posted April 2, 2004 Oops... mistake The points should be (sqroot5,0,0) and (-sqroot5,0,0).
bloodhound Posted April 13, 2004 Posted April 13, 2004 for general questions like this, u can use the method of lagrange multipliers. basically if u are given a function f and u have to minimise or maximise the value of f given a condition that another function g=0 then if u form another function F(x,y,z,lamda)=f(x,y,z)-lamda*g(x,y,z) and then u find the partial derivatives , Fx,Fy,Fz,Flamnda, and solve Fx=Fy=Fz=Flamnda=0 , then those values of (x,y,z) will minimise the value of f(x,y,z) with the condition that g(x,y,z)=0 In this case we are minimising (x^2+y^2+z^2)^(1/2) which is the same as minimising(x^2+y^2+z^2) set that = f(x,y,z) . also we are given the condition that g(x,y,z)=x^2-yz-5=0 now take F(x,y,z,lamda)=f(x,y,z)-lamda*g(x,y,z) =x^2+y^2+z^2 - lamda*(x^2-yz-5) so now find all the partial derivatives. equate them to 0 and solve them. that is ur values which will minimise ur function This method can be applied to function of as many variables as u like with as many conditions eg. minimise f(r,s,t,v) given g(r,s,t,v)=0 and h(r,s,t,v)=0 u create F(r,s,t,v,lambda, mu)=f(r,s,t,v)-lambda*g(r,s,t,v) - mu*h(r,s,t,v) u find the partial derivatives of F w.r.t r,s,t,v, lambda and mu and equate them to 0 and solve them. Those valuesof (r,s,t,v) minimises or maximises the function f(r,s,t,v) and so it can be extended to functions of as many variables as u like 1
bloodhound Posted April 13, 2004 Posted April 13, 2004 ill just do the given question for an example we have F(x,y,z,lamda)=x^2+y^2+z^2 - lamda*(x^2-yz-5) finding the four partial derivatives and equating them to 0 we get 4 simultaneous equations. 2x-2lambda*x=0 which gives us lambda = 1 partial dev wrt y gives us 2y+lambda*z=0 since lambda = 1 we get 2y+z=0 partial dev wrt z gives us 2z+lambda*y=0 since lambda = 1 we get 2z+y=0 therefore from the last two equations , the only solution is y=0,z=0 Now, Partial Dev wrt lambda gives us 5-x^2=0 which gives us x=plus.minus(sqrt(5)) Therefore (x,y,z)=(+-sqrt(5),0,0) minimises the distance from the origin given =0 But if we look at x^2+y^2+z^2 we see that its nonnegative and not bounded above there fore the points(+-sqrt(5),0,0) minimises the distance
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