A ball has a radius that increases at a rate of 1.5 cm per second

A snowball has a radius that increases in length at a rate of 1.5 cm per second. 
The rate of change of the surface area and volume of the sphere when its radius reaches 40 cm is . . .

The Surface Area : 4πr^2 Volume : 4/3 πr^3 Laju : 1,5cm/(detik )→dr/dt 
The Surface Area : dL/dr = 8πr 
dL/dt = dL/dr × dr/dt=8πr×1,5=12πr saat r = 40→L = 12×π×40 = 480πcm^2/detik 
Volume : dV/dr = 4πr^2 
dV/dt = dV/dr×dr/dt = 4πr^2×1,5 = 6πr^2 
When r = 40→V = 6π(40)^2 = 6π(1.600) = 9.600πcm^3/detik.
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