Answer:
[tex]\frac{dr}{dt} = 0.106 in/min[/tex]
[tex]\frac{dA}{dt} = 8 inch/min[/tex]
Explanation:
Rate of melting of ball is given as
[tex]\frac{dV}{dt} = 12 \frac{in^3}{min}[/tex]
now we know that
[tex]V = \frac{4}{3}\pi r^3[/tex]
now we have
[tex]\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}[/tex]
so we have
[tex]\frac{dr}{dt} = \frac{\frac{dV}{dt}}{4\pi r^2}[/tex]
Diameter = 6 inch
Radius = 3 inch
[tex]\frac{dr}{dt} = \frac{12}{4\pi(3^2)}[/tex]
[tex]\frac{dr}{dt} = 0.106 in/min[/tex]
Now rate of change in area is given as
[tex]A = 4\pi r^2[/tex]
so we have
[tex]\frac{dA}{dt} = 8\pi r\frac{dr}{dt}[/tex]
[tex]\frac{dA}{dt} = 8\pi(3)(0.106)[/tex]
[tex]\frac{dA}{dt} = 8 inch/min[/tex]
