Find the points on the cone z^2 = x^2 + y^2 that are closest to the point (4, 2,

Question

Find the points on the cone z^2 = x^2 + y^2 that are closest to the point (4, 2, 0). Find the local maximum, local minimum values and saddle point(s) of the function f(x, y) = y^2 - 2y cos x, 1 lessthanorequalto x lessthanorequalto 7 the boundary of a lamina consists of the semicircles y = Squareroot l - x^2 and y = Squareroot 4 - x^2 along with the portions of the x-axis that joins them. Find the center of mass of the lamina, if the density at any point is inversely proportional to the distance from the origin.

Explanation / Answer

1) point on cone z2=x2+y2 is (x,y,z)

distance from cone to point (4,2,0) is ,d=[(x-4)2+(y-2)2+z2]

point is closest when distance is minimum

distance is minimum when f(x,y,z)=(x-4)2+(y-2)2+z2 is minimum

f(x,y)=(x-4)2+(y-2)2+(x2+y2)

fx=2(x-4)+2x,fy=2(y-2) +2y

fx=4x-8,fy=4y-4

for critical points fx =0 , fy =0

4x-8=0,4y-4=0

x=2 , y=1

fxx=4,fyy=4,fxy=0, D=fxxfyy-(fxy)2=16>0

fxx>0, so critical point is minimum

z2=x2+y2

z2=22+12

z=-5,5

so points on cone closest to the point (4,2,0) are (2,1,-5),(2,1,5)

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