Many extrema problems involve minimizing/maximizing the distance between objects

Question

Many extrema problems involve minimizing/maximizing the distance between objects which comes down to working with the distance between points which involves square roots. It is simpler to deal with the square of the distance but then the question is, does the square of the distance function have max/min at the same points as the distance function? Show that if f : Rn -> R is always positive (like the distance between objects that are not touching) then gradient(f ^2) = 0 if and only if gradient(f) = 0.

Explanation / Answer

Let,

1.

gradient(f) = 0

Then (df/dx)y,z = 0

(df/dy)x,z = 0

(df/dz)x,y = 0

Now gradient(f2) = (df2/dx)y,z i + (df2/dy)x,z j + (df2/dz)x,y k

= 2f (df/dx)y,z i + 2f(df/dy)x,z j + 2f(df/dz)x,y k
= 2f [gradient(f) ] = 2f*(0) = 0

Thus, gradient(f2) = 0

2.

Now let gradient(f2) = 0

From above we know,

gradient(f2) = 2f[gradient(f) ] = 0

2f[gradient(f) ] =0

gradient(f) = 0 [if f is not equal to 0]

But for f = 0, gradient(f) =0

Hence prooved that then gradient(f ^2) = 0 if and only if gradient(f) = 0.

Related Questions

Navigate

• Browse (All)
• Browse M
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.