Some show me steps to solution

Step-by-step explanation:

a) The rate of change in volume is equal to the volume flow rate going in minus the volume flow rate going out.

W'(t) = F(t) − L(t)

At time t = 3, the slope of the line tangent to W(t) is W'(3).

W'(3) = F(3) − L(3)

W'(3) = arctan(π/2−3/10) − 0.03(20(3)−3²−75)

W'(3) ≈ 1.624

Using point-slope form, the linear approximation is:

y − 2.5 = 1.624 (x − 3)

y − 2.5 = 1.624 x − 4.872

y = 1.624 x − 2.372

The approximate volume at t = 3.5 is:

y = 1.624 (3.5) − 2.372

y = 3.312

b) W'(t) = arctan(π/2−t/10) − 0.03(20t−t²−75)

W"(t) = (-1/10) / (1 + (π/2−t/10)²) − 0.03(20−2t)

W"(8) = (-1/10) / (1 + (π/2−8/10)²) − 0.03(20−2(8))

W"(8) = -0.183

At t = 8 minutes, the flowrate is slowing down at a rate of 0.183 ft³/min².

c) If the rate of change of the volume (W'(t)) changes from positive to negative, then there must be a point where W'(t) = 0.

0 = arctan(π/2−t/10) − 0.03(20t−t²−75)

Solving with a calculator, t = 8.149 or t = 14.627.  Since we're only considering 5 < t < 10, a possible point is t = 8.149.

Evaluate W'(t) before and after t = 8.149:

W'(8) = 0.027

W'(9) = -0.129

So yes, there is a time where the rate of change of the volume changes from positive to negative.

d) The tub is rectangular, so the volume of water is equal to the area of the base times the depth of the water:

W(t) = (0.5) (4) h

W(t) = 2h

Taking derivative:

W'(t) = 2 dh/dt

Evaluate at t = 6:

arctan(π/2−6/10) − 0.03(20(6)−6²−75) = 2 dh/dt

0.501 = 2 dh/dt

dh/dt = 0.250

The depth of the water is increasing at 0.250 ft/min.