Answer
(a) $2m^3/s$ (increasing)
(b) $0$ (no change)
(c) $0$ (no change)
Work Step by Step
Identify the given quantities: $\frac{dx}{dt}=1$ m/sec, $\frac{dy}{dt}=-2$ m/sec, $\frac{dz}{dt}=1$ m/sec, $x=4,y=3,z=2$
(a) Volume: $V=xyz$, $\frac{dV}{dt}=xy\frac{dz}{dt}+xz\frac{dy}{dt}+yz\frac{dx}{dt} =4(3)(1)+4(2)(-2)+3(2)(1)=2m^3/s$ (increasing)
(b) Surface Area: $A=2(xy+xz+yz)$, $\frac{1}{2}\frac{dA}{dt}=x\frac{dy}{dt}+y\frac{dx}{dt}+x\frac{dz}{dt}+z\frac{dx}{dt}+y\frac{dz}{dt}+z\frac{dy}{dt}=4(-2)+3(1)+4(1)+2(1)+3(1)+2(-2)=0$ thus $\frac{dA}{dt}=0$ (no change)
(c) Length of diagonal: $s^2=x^2+y^2+z^2$, $2s\frac{ds}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}+2z\frac{dz}{dt}=2(4(1)+3(-2)+2(1))=0$, thus $\frac{ds}{dt}=0$ (no change)
