Question

How would you apply the second derivative test to find the points on the surface x^2 = 4(y-2)^2 +(z-2)^2 closest to the point (0,0,2)

Answer 1

The problem is equivalent to finding (x,y,z) that minimizes d, where \(d=\sqrt{x^2+y^2+(z-2)^2}\), or alternatively, you can minimize \(d^2=x^2+y^2+(z-2)^2\).
Substitute \(x^2 = 4(y-2)^2 +(z-2)^2\) into d^2 since the point must lie on the surface and you are left with a function of y and z that you can use the Second Derivative Test on.

Answer 2

how would I apply the second derivative test

Answer 3

Correction: Instead of eliminating x, z should be eliminated, i.e. replace \((z-2)^2\) with \(x^2-4(y-2)^2\), so that the function we are minimizing is \(f(x,y)=2x^2+y^2-4(y-2)^2\).
Then begin finding the critical points of f.
Afterwards, classify them as minimum, maximum, or saddle point. The one you want among the critical points is the minimum.

Answer 4

To classify the critical point (a,b), let
\[D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-(f_{xy}(a,b))^2\]
1. If D>0 and f_xx>0, then (a,b) is a minimum.
2. If D>0 and f_xx<0, then (a,b) is a maximum.
3. If D<0, then (a,b) is a saddle point.
4. If D=0, then no information is gained on (a,b).