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) .

(a) At time t=3 , there are 2.5 cubic feet of water in the tub. Write an equation for the locally linear approximation of W at t=3 , and use it to approximate the volume of water in the tub at time t=3.5 .

(b) Find W′′(8) . Using correct units, interpret the meaning of W′′(8) in the context of the problem.

(c) Is there a time t , for 5
(d) 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 ?

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.

where are the options?

step-by-step explanation: