Answer:
7/16π
Step-by-step explanation:
The surface area of the snow ball = 4πr^2
Let the surface area= y
Then y= 4πr^2
The derivative can be taken with respect to time as
dy/dt= (8πr)dr/dt...........eqn(1)
We were told that its surface area decreases at a rate of 7 cm2/min.
Then dy/dt=-7
diameter = 8 cm. cm/min, the radius r= 4cm/min
r= 4
Substitute the values of r into eqn(1)
dy/dt=(8π×4)dr/dt
-7=(32π) dr/dr
dr/dt= -7/32π................eqn(**)
But we know that diameter d= 2× radius
d=2r............eqn(*)
We can find the derivatives with respect to t as
dd/dt=2dr/dt
Then substitute en(**) we have
dd/dt=2(7/32π)
dd/dt= -14/32π
Divide both numerator and denominator by 2
dd/dt= -7/16π
Hence, the diameter decreases by the rate of 7/16π
