Question

An inverted pyramid is being filled with water at a constant rate of 70 cubic centimeters per second. The pyramid, at the top, has the shape of a square with sides of length 8 cm, and the height is 15 cm. Find the rate at which the water level is rising when the water level is 4 cm.

Answer 1

`(dh)/(dt)=1.45cmsec^(-1)`

Explanation:

Rate of Water Fill
`R=(dv)/(dt)=70cm^3`

Length
`l=8cm`

Height
`H=15cm`

Water level
`L_w= 4cm`

Generally the equation for relationship b/w h and a is mathematically given by

Since by the properties of similar triangles

`k=(h)/(1/2)`

Let

`h=15cm \\\\a=8cm`

`k=(h)/(1/2a)`

`k=(15)/(4)`

Therefore

`(h)/(1/2a)=(15)/(4)`

`a=(8h)/(15)`

Generally the equation for volume of Pyramid is mathematically given by

`V=(1)/(3)ah^2h`

Subsitute a

`V=(1)/(3)((8h)/(15))h^2h`

Therefore

`(dv)/(dt)=(((1)/(3)((8h)/(15))h^2h))/(dt)`

`(dv)/(dt)=(64)/(255)(h^2(dh)/(dt))`

Since

`(dv)/(dt)=70cm^3s^(-1)`

Therefore

`70cm^3s^(-1)=(64)/(255)(h^2(dh)/(dt))`

`(dh)/(dt)=(70)/(169)*(225)/(64)`

`(dh)/(dt)=1.45cmsec^(-1)`