Question

Find the minimum distance from the point (1,1,14) to the paraboloid given by the equation ?=?2+?2.

Answer 1

Let (x,y,z).be the points in the given curve

Then distance to the point (1,1,14) from x,y,z is given by

Let d​​​​​​2​ be the f

​​​

Now we need to find partial derivative of each function F and g with respect to x,y,z and equate as follows

Fx= kGx

Fy=kGy

Fz=kGz

Let's find Fx,Fy,Fz and Gx,Gy,Gz

Fx = 2(x-1) ,Fy=2(y-1) ,Fz=2(z-1)

Similarly

Gx= 2x , Gy=2y ,Gz= -1

So

2(x-1) = k*2x. , 2(y-1)= k*2y , 2(z-14)=k*-1

Express x,y,z in terms of

2x-2=2kx

2x-2kx=2

2x(1-k)=2

x=-1/k-1

Similarly

Y=-1/k-1

2(z-14)=-k

2z-28=k

2z=k+28

Z= (k+28)/2

Put this x,y,z in the equation of parabloid

G(x) we get

Multiply with 2(1-k)2 we get

By re arranging above equation we wil get a cubic polynomial in k

We need to root of this cubic polynomial

On solving we get k has three values

k1=-27.99 , k2=1.369 k3= 0.626

by putting value of k1,k2,k3 in eq of x,y,z we get three points as shown

X=-1/k-1 ,y=-1/k-1 ,z = (k+28)/2

(0.0345,0.0345,0.005) , (-2.71,-2.71,14.6845) and (2.67,2.67,14.313)

By examin

Ans:

It's clear that closest point is (2.67,2.67,14.313)