Rate of Change of Slice Area

First we find an expression for the distance $d(t)$ of the center of the sphere (which is the origin) from the plane.

$d(t) = | -\sin(\pi / 3 ) (12 - (24/50) t ) |$

Now the area of the cross section ( assuming $d(t) \le 10$ ) is given by

$A(t) = \pi ( 100 - d^2(t) ) = \pi ( 100 - \sin^2(\pi/3) (24/50 t - 12 )^2 )$

Differentiating gives the desired result

$\frac{dA}{dt} (t) = \pi ( - 2 \sin^2(\pi/3) (24/50 t - 12 ) ( 24/50 ) )$

Substituting t = 15, gives

$d(15) = | - \sin(\pi/3) ( 12 - (24/50)(15) ) | = 4.15692 \le 10$

Hence,

$\frac{dA}{dt} (15) = \pi ( - 2 \sin^2(\pi/3) (24/50 (15) - 12 ) ( 24/50 ) ) = \boxed{ 10.8573 }$