In Closing The rules for adding and subtracting integers apply to all rational numbers. The sum of two rational numbers (e.g., −1 + 4.3) can be found on the number line by placing the tail of an arrow at −1 and locating the head of the arrow 4.3 units to the right to arrive at the sum, which is 3.3. To model the difference of two rational numbers on a number line (e.g., −5.7 − 3), first rewrite the difference as a sum, −5.7 +(−3), and then follow the steps for locating a sum. Place a single arrow with its tail at −5.7 and the head of the arrow 3 units to the left to arrive at −8.7.

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.