Найти косинус угла между градиентами скалярных полей u=6x^2+y^2+3z^2, v=1/(6x+y+3z) в точке Мo (-5;3;-2)
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mikael2Найти косинус угла между градиентами скалярных полей u=6x^2+y^2+3z^2, v=1/(6x+y+3z) в точке Мo (-5;3;-2) u'x=12x=-5*12=-60 u'y=2y=2*3=6 u'z=6z=-6*2=-12 grad u = (-60;6;-12)
v=(6x+y+3z)⁻¹ v'x=-6/(6x+y+3z)²=-6/(-5*6+3-5)²=-6/1024 v'y=-1/(-5*6+3-5)²=-1/1024 v'z=-3/1024 grad v = (-6/1024;-1/1024;-3/1024)
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u'x=12x=-5*12=-60 u'y=2y=2*3=6 u'z=6z=-6*2=-12
grad u = (-60;6;-12)
v=(6x+y+3z)⁻¹ v'x=-6/(6x+y+3z)²=-6/(-5*6+3-5)²=-6/1024
v'y=-1/(-5*6+3-5)²=-1/1024 v'z=-3/1024
grad v = (-6/1024;-1/1024;-3/1024)
скалярное произведение (grad u,grad v)=60*6/1024-6/1024+36/1024=
=1/1024(360-6+36)=390/1024=195/512
|grad u|=√(3600+36+144)= √3780=2√945
|grad v|=1/1024√(36+1+9)=√46/1024
cosa=(grad u,grad v)/|grad u|*|grad v|
195/512 195*1024 195
cosa = ------------------------- = ------------- ≈ 0.93
512*2√945*√46 √945*√46