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