Answer:
The rate of change of volume is [tex]127.3cm^3/s[/tex].
Step-by-step explanation:
The surface area [tex]A[/tex] of the sphere of radius [tex]R[/tex] is given by
[tex](1).\: \: A = 4\pi R^2[/tex].
and the volume [tex]V[/tex] of the same is
[tex](2).\: \: V = \dfrac{4}{3}\pi R^3.[/tex]
Now we are told that there rate of change of surface area is [tex]30cm^2/s[/tex] or
[tex]\dfrac{dA}{dt} = \dfrac{d}{dt} (4\pi R^2) = 30cm^2/s[/tex]
which means
[tex](3).\: \: \dfrac{dR^2}{dt} = \dfrac{30cm^2/s}{4\pi}[/tex]
From the product rule we see that
[tex]\dfrac{dR^2}{dt} = R\dfrac{dR}{dt}+R\dfrac{dR}{dt} = 2R\dfrac{dR}{dt}[/tex]
which we substitute into equation (3) to get:
[tex](4). \: \; \dfrac{dR}{dt} = \dfrac{30}{8\pi R}[/tex]
Now, we find the rate of change of volume by taking the time derivative of equation (2) as follows:
[tex]$(5). \:\: \frac{dV}{dt} = \dfrac{4}{3}\pi\frac{dR^3}{dt} $[/tex]
From the product rule
[tex]$\frac{dR^3}{dt} = R\frac{dR^2}{dt} + R^2\frac{dR}{dt} $[/tex]
and since
[tex]\dfrac{dR^2}{dt} =2R\dfrac{dR}{dt}[/tex],
we have
[tex]$\frac{dR^3}{dt} = R(2R\frac{dR}{dt} )+ R^2\frac{dR}{dt} $[/tex]
[tex]$\frac{dR^3}{dt} =3 R^2\frac{dR}{dt} $[/tex]
putting this into equation (5) we get:
[tex]$(6). \:\: \frac{dV}{dt} =4\pi R^2\frac{dR}{dt} $[/tex]
from equation (4) we substitute the value of [tex]\dfrac{dR}{dt}[/tex] to get:
[tex]$ \frac{dV}{dt} =4\pi R^2 \dfrac{30}{8\pi R} =15R $[/tex]
[tex]$(7). \:\: \frac{dV}{dt} =15R $[/tex]
Now, when the area is [tex]288\pi cm^2[/tex], the radius is
[tex]4\pi R^2= 288\pi[/tex]
[tex]R = 6\sqrt{2}[/tex],
putting that into equation (7) we get
[tex]$(7). \:\: \frac{dV}{dt} =15(6\sqrt{2} ) $[/tex]
[tex]$ \boxed{ \frac{dV}{dt} =127.3\: cm^3/s} $[/tex]
Thus, the rate of change of volume is [tex]127.3\: cm^3/s.[/tex]
