rate of change in a melting cube of ice

[mares09]

Water is being collected from a block of ice with a square base. The water is produced because the ice is melting in such a way that each edge of the block is decreasing at 2 inches per hour, while the height of the block is decreasing at 3 inches per hour. What is the rate of flow of water into the collecting pan when the base has edge lenght 20 inches and the height of the block is 15 inches?

(make the asumption that water and ice have the same density)

Is a cube and we are trying to figure out at what rate is the volume declining (flow of water into pan).

They say that the rate that the base is decreasing is 2inch/hour so dx/dt = 2 inch / hour (because the base is obviously X)

Also the height is declining at 3 inch / hour so dy/ dt = 3 inch / hour (The height is Y)

The equation for the volume of a cube is V = Base x Height x Width

In this cube there is a square base so Base = Width. In terms of X and Y: V = (X ^ 2 ) times Y (Height)

V = (X ^ 2) ( Y ) this is the equation then we do implicit differentiation: Dv / Dt = 2X (dx/dt) times (dy/dt)

Dv/ Dt = 2 (20) (2) (3)

Dv / Dt = 240 inches / hour.

I don’t know if this is completely right because I have nowhere to plug in the Y = 15 after I do the differentiation. Any help would be appreciated.

 

[skeeter]

x = base length and width, y = height

V = x²y

take the derivative of Volume w/r to time. you have to treat x and y as seperate functions of time and implicitly differentiate ...

dV/dt = x²(dy/dt) + 2xy(dx/dt)

now plug in your given values to determine dV/dt