Mathematics-I (3110014)

BE | Semester-1   Winter-2019 | 17-01-2020

Q3) (c)

Find the maximum and minimum distance from the point 1,2,2 to the sphere x2+y2+z2=36.

Let (x,y,z) be any point on the sphere. Its distance D from the point (1,2,2) is,

D=x-12+y-22+z-22

D2=fx,y,z=x-12+y-22+z-22      .........[1]

Let, φx,y,z=x2+y2+z2-36      .........[2]

Now, Auxiliary equation is,

fx,y,z+λ·φx,y,z=0

x-12+y-22+z-22 +λ·x2+y2+z2-36=0      .........[3]

Taking partial differentiation of Eq. 3 with respect to x,

2(x-1)+λ(2x)=0

λ(2x)=-2(x-1)

λ=1-xx=1x-1

Taking partial differentiation of Eq. 3 with respect to y,

2(y-2)+λ(2y)=0

λ(2y)=-2(y-2)

λ=2-yy=2y-1

Taking partial differentiation of Eq. 3 with respect to z,

2(z-2)+λ(2z)=0

λ(2z)=-2(z-2)

λ=2-zz=2z-1

Comparing values of λ,

1x-1=2y-1

1x=2y

y=2x

1x-1=2z-1

1x=2z

z=2x

Substituting the values of y and z in given eq. of sphere,

x2+y2+z2=36

x2+4x2+4x2=36

9x2=36

x2=4

x=±2

Now,

y=2x=2(±2)=±4

z=2x=2(±2)=±4

So, Stationary Point are (2,4,4)   & (-2,-4,-4).

Putting Values for x,y & z to find distance,

D=x-12+y-22+z-22=2-12+4-22+4-22=3

D=x-12+y-22+z-22=-2-12+-4-22+-4-22=9

So, Minimum distance is 3 and maximum distance is 9.