Question

A spherical iron ball 10 in. in diameter is coated with a layer of ice of uniform thickness. If the ice melts at a rate of 15 in cubed Over min, how fast is the thickness of the ice decreasing when it is 3 in​ thick?

Answer 1

Decreasing rate of thickness when x=3 in is
`(dx)/(dt)= -7.28* 10^(-3) in/min`

Step-by-step 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=
`(4)/(3)\pi [(5+x )^3-5^3]`

V=
`(4)/(3)\pi [x^3+75x^2+15x]`

Now given that ice melts rate=
`15in^3/min`

`(dV)/(dt)= -15in^3/min`

`(dV)/(dt)=(4)/(3)\pi [3x^2+150x+15](dx)/(dt)`

When x=3 in
`(dV)/(dt)= -15in^3/min`

`-15=(4)/(3)\pi [3* 3^2+150* 3+15](dx)/(dt)`

`(dx)/(dt)= -7.28* 10^(-3) in/min`

So decreasing rate of thickness when x=3 in is
`(dx)/(dt)= -7.28* 10^(-3) in/min`