Answer:
Decreasing rate of thickness when x=3 in is [tex]\dfrac{dx}{dt}= -7.28\times 10^{-3} in/min[/tex]
Explanation:
Lets assume that thickness of ice over the spherical iron ball =x
So radius diameter of Sphere=5+x in
  Inner radius=5 in
So volume V=[tex]\dfrac{4}{3}\pi [(5+x )^3-5^3][/tex]
     V=[tex]\dfrac{4}{3}\pi [x^3+75x^2+15x][/tex]
Now given that ice melts rate=[tex] 15in^3/min[/tex]
[tex]\dfrac{dV}{dt}= -15in^3/min[/tex]
[tex]\dfrac{dV}{dt}=\dfrac{4}{3}\pi [3x^2+150x+15]\dfrac{dx}{dt}[/tex]
When x=3 in [tex]\dfrac{dV}{dt}= -15in^3/min[/tex]
[tex]-15=\dfrac{4}{3}\pi [3\times 3^2+150\times 3+15]\dfrac{dx}{dt}[/tex]
[tex]\dfrac{dx}{dt}= -7.28\times 10^{-3} in/min[/tex]
So decreasing rate of thickness when x=3 in is [tex]\dfrac{dx}{dt}= -7.28\times 10^{-3} in/min[/tex]
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