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Directional derivative and gradient in advanced mathematics
Solution:

Radial unit direction:

(x0,y0,z0)/[√(x0)? +(y0)? +(z0)? ]

Therefore, the radial direction angle is:

cosα=x0/√[(x0)? +(y0)? +(z0)? ]

cosβ=x0/√[(x0)? +(y0)? +(z0)? ]

cosγ=z0/√[(x0)? +(y0)? +(z0)? ]

Function u=(x? /a? )+(y? /b? )+(z? /c? The directional derivative of this radial direction is:

u/? r0

=u'x cosα+u'y cosβ+u'z cosγ

=2(x0)? /a? √[(x0)? +(y0)? +(z0)? ] + 2(y0)? /b? √[(x0)? +(y0)? +(z0)? ] +2(z0)? /c? √[(x0)? +(y0)? +(z0)? ]

=2[(x0)? /a? +(y0)? /b? +(z0)? /c? ]/√[(x0)? +(y0)? +(z0)? ]

=2[b? c? (x0)? +a? c? (y0)? +a? b? (z0)? ]/a? b? c? √[(x0)? +(y0)? +(z0)? ]

The gradient modulus of M0 point is:

|gradu(x0,y0,z0)|

=√[(u'x0)? +(u'y0)? +(u'z0)? ]

=√{[2(x0)? /a? ]? +[2(y0)? /b? ]? +[2(z0)? /c? ]? }

=2√{[b? c? (x0)? +a? c? (y0)? +a? b? (z0)? ]}/a? b? c?

According to the meaning of the question:

u/? R0=|gradu(x0, y0, z0)|, then:

2【b? c? (x0)? +a? c? (y0)? +a? b? (z0)? ]/a? b? c? √[(x0)? +(y0)? +(z0)? ]

=2√{[b? c? (x0)? +a? c? (y0)? +a? b? (z0)? ]}/a? b? c?

Therefore:

a=b=c