Question

Question #26b83

Answer 1

`(dV)/(dt) = 2sqrt(S/pi) (cm^3)/sec`

Explanation:

The rate of increase of the area of the surface can be expressed as:

`(dS)/(dt) = 8 (cm^2)/sec`

We can find the rate of increase of the volume with the chain rule:

`(dV)/(dt) = (dV)/(dS) * (dS)/(dt)`

We must then express the volume as function of the surface.
As:

`S=4pir^2 => r = sqrt(S/(4pi))`

`V = 4/3pir^3 = (4pi)/3 (sqrt(S/(4pi)))^3 = S^(3/2)/(6sqrtpi)`

`(dV)/(dS) = 1/(6sqrtpi) 3/2 S^(1/2) = 1/4 sqrt(S/pi)`

So, finally:

`(dV)/(dt) = (dV)/(dS) * (dS)/(dt) = 8*1/4 sqrt(S/pi) = 2sqrt(S/pi)`

Answer 2

Rate of Change of volume is `4r \ cm^3s^(-1)`

Explanation:

Let us set up the following variables:

`{(V, "Volume of the sphere",cm^3), (S, "Surface Area of the sphere",cm^2), (t, "time",sec) :}`

The Volume of the sphere is given by the standard formula:

`V=4/3pir^3`

Differentiating wrt `r` we get:

`(dV)/(dr) = 4pir^2`

The Surface Area of the sphere is given by the standard formula:

`S = 4pir^2`

Differentiating wrt `r` we get:

`(dS)/(dr) = 8pir`

We are given that:

`(dS)/dt = 8 \ cm^2s^-1`

By the chain rule:

`(dV)/dt = (dV)/(dr) * (dr)/(dS) * (dS)/(dt)`
`" " = (4pir^2) * (1/(8pir)) * (8)`
`" " = 4r \ \ cm^3s^(-1)`

Answer 3

http://mathoverflow.net/questions/73492/how-misleading-is-it-to-regard-fracdydx-as-a-fraction