Respuesta :
Using implicit differentiation, it is found that the water level is rising a rate of 0.001 ft/sec.
The volume of a pool of length l, width w and height h is given by:
[tex]V = lwh[/tex]
Applying implicit differentiation, the rate of change of the volume is given by:
[tex]\frac{dV}{dt} = wh\frac{dl}{dt} + lh\frac{dw}{dt} + lw\frac{dh}{dt}[/tex]
Neither the width nor the length changes, hence [tex]\frac{dl}{dt} = \frac{dw}{dt} = 0[/tex], and:
[tex]\frac{dV}{dt} = lw\frac{dh}{dt}[/tex]
We are given that:
[tex]\frac{dV}{dt} = 0.3, l = 20, w = 40[/tex], hence:
[tex]\frac{dV}{dt} = lw\frac{dh}{dt}[/tex]
[tex]0.8 = 20(40)\frac{dh}{dt}[/tex]
[tex]\frac{dh}{dt} = \frac{0.8}{800}[/tex]
[tex]\frac{dh}{dt} = 0.001 [/tex]
Water level is rising at rate of 0.001 ft/sec.
A similar problem is given at https://brainly.com/question/9543179
