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if-U-x-3-y-3-6-sinz-11-then-u-z-1-0-pi-is-




Question Number 78894 by gopikrishnan last updated on 21/Jan/20
if U=x^3 +y^3 +6 sinz+11 then u_z (1,0,pi) is
$${if}\:{U}={x}^{\mathrm{3}} +{y}^{\mathrm{3}} +\mathrm{6}\:{sinz}+\mathrm{11}\:{then}\:{u}_{{z}} \left(\mathrm{1},\mathrm{0},{pi}\right)\:{is} \\ $$
Commented by mathmax by abdo last updated on 22/Jan/20
U(x,y,z) =x^3  +y^3  +6sinz +11 ⇒U(1,0,π)=1+0 +0 +11 =12
$${U}\left({x},{y},{z}\right)\:={x}^{\mathrm{3}} \:+{y}^{\mathrm{3}} \:+\mathrm{6}{sinz}\:+\mathrm{11}\:\Rightarrow{U}\left(\mathrm{1},\mathrm{0},\pi\right)=\mathrm{1}+\mathrm{0}\:+\mathrm{0}\:+\mathrm{11}\:=\mathrm{12} \\ $$

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