Question

For time 0≤t≤10 , water is flowing into a small tub at a rate given by the function F defined by F(t)=arctan(π/2−t/10) . For time 5≤t≤10 , water is leaking from the tub at a rate given by the function L defined by L(t)=0.03(20t−t^2−75) . Both F(t) and L(t) are measured in cubic feet per minute, and t is measured in minutes. The volume of water in the tub, in cubic feet, at time t minutes is given by W(t) .
The tub is in the shape of a rectangular box that is 0.5 foot wide, 4 feet long, and 3 feet deep. What is the rate of change of the depth of the water in the tub at time t=6 ?

Answer 1

Answer:

  0.250 ft/min

Step-by-step explanation:

The surface area of the water in the tub is 0.5 ft by 4 ft, or 2 square feet.

At t=6, the rate of change of volume is ...

  V'(6) = F(6) -L(6)

  V'(6) = arctan(π/2 -6/10) - 0.03(20(6)-6^2 -75) = 0.77058 -0.27000

  V'(6) = 0.50058

The rate of change of depth is related to the rate of change of volume by ...

  V = Bh . . . . where B is the base area of the prism (2 ft^2)

  h = V/B

  h' = V'/B

  h'(6) = V'(6)/(2 ft^2) = (0.50058 ft^3/min)/(2 ft^2)

  h'(6) = 0.25029 ft/min

The rate of change of depth is about 0.250 ft/min at time t=6.