Answer
\[\frac{dz}{dt}=-18\]
Work Step by Step
$x^2+y^2+z^2=9$ ___(1)
Differentiate (1) with respect to $t$
$2x\large\frac{dx}{dt}$+$2y$ $\large\frac{dy}{dt}$+$2z$ $\large\frac{dz}{dt}=0$ ___(2)
Using given data $\large\frac{dx}{dt}=5\;,\;\frac{dy}{dt}=4\;,\;$ $\small(x,y,z)=(2,2,1)$
(2) becomes
$2(2)(5)+2(2)(4)+2(1)\Large\frac{dz}{dt}$=0
$-36=2\Large\frac{dz}{dt}$
$\Large\frac{dz}{dt}$=$-18$
Hence, $\Large\frac{dz}{dt}$=$-18$.