Question

A trough is 12 ft long and its ends have the shape of isosceles triangles that are 3 ft across at the top and have a height of 1 ft. If the trough is being filled with water at a rate of 14 ft3/min, how fast is the water level rising when the water is 4 inches deep

Answer 1

`(dh)/(dt)=0.5ft`

Explanation:

From the question we are told that:

Length
`l=12`

Top length
`l_t=3ft`

Height
`h=1ft`

Rate
`R=14 ft3/min`

Water rise
`w=4`

Generally the equation for Velocity is mathematically given by

`V=frac{1}{2}wh'(l)\\\\V=frac{1}{2}wh'(12)`

`V=18h'^2`

Therefore

`R=18(2h)((dh)/(dt))`

Where

`h=(3)/(4)`

Therefore

`(dh)/(dt)=(R)/(18(2h))`

`(dh)/(dt)=(14)/(18(2.3/4))`

`(dh)/(dt)=0.5ft`