A bucket is being lifted from the bottom of a 50-foot deep well; its weight (including the water), B, in pounds at a height h feet above the water is given by the function B(h). When the bucket leaves the water, the bucket and water together weigh B(0)=20 pounds, and when the bucket reaches the top of the well, B(50)=12 pounds. Assume that the bucket loses water at a constant rate (as a function of height, h) throughout its journey from the bottom to the top of the well. a. Find a formula for B(h). b. Compute the value of the product B(5). Δh, where Δh=2 feet. Include units on your answer. Explain why this product represents the approximate work it took to move the bucket of water from h=5 to h=7. c. Is the value in (b) an over- or under-estimate of the actual amount of work it took to move the bucket from h=5 to h=7? Why? d. Compute the value of the product B(22).Δh, where Δh =0.25 feet. Include units on your answer. What is the meaning of the value you found? e. More generally, what does the quantity Wslice= B(h)Δh measure for a given value of h and a small positive value of Δh? f. Evaluate the definite integral ∫ B(h) dh. What is the meaning of the value you find? Why?

a) B(h) = -4*h/25 + 20

b)  B(5).dh = 38.4 lb-ft

c) Over-estimate

d)  B(22).dh = 4.12 lb-ft

f)  W = 800 lb-ft

Step-by-step explanation:

Given:

- B(0) = 20 lbs

- B(50) = 12 lbs

- Constant rate of decrease of water.

Find:

Find a formula for  B(h)

Solution:

- We will set up a linear expression for Weight B with respect to height h:

                              B(h) = m*h + C

Where,   m: rate of change of weight with height and C is the initial Weight.

                              m = (12-20) / 50 = -4/25

                              C = 20 lb

Hence,

                              B(h) = -4*h/25 + 20

Find:

Compute the value of the product  B(5).dh. Where, dh = 2ft. Explain why this product represents the approximate work it took to move the bucket of water from h = 5 to h = 7

Solution:

- We will evaluate B(5):

                              B(h) = -4*h/25 + 20

                              B(5) = -4*5/25 + 20 = 19.2

- Hence,  

                     B(5).dh = 19.2*2 = 38.4 lb-ft     Work done to raise bucket

Since the bucket's weight actually did not stay constant, this number is slightly off, but since the bucket's weight did not change by too much over the time period, it will not be off by much.

Find:

Is the value in (b) an over- or under-estimate of the actual amount of work it took to move the bucket from h=5 to h=7.

Solution:

The answer above is an over-estimate of the work it would have taken. The answer above would correspond to the bucket's weight remaining constant after h = 5, but since the bucket continued to get lighter after that point, it would have taken less work than our calculation showed.

Find:

Compute the value of the product  B(22).dh. Where, dh = 0.25ft. Meaning of the value?

Solution:

- We will evaluate B(22):

                              B(h) = -4*h/25 + 20

                              B(22) = -22*5/25 + 20 = 15.6

- Hence,  

                   B(22).dh = 15.6*0.25 = 4.12 lb-ft     Work done to raise bucket

This represents the approximate amount of work it would have taken to raise the bucket   .25  feet further once the bucket had been raised 22 feet.

Find:

More generally, what does the quantity W_slice = B(h).dh measure for a given value of  h  and a small positive value of dh?

Solution:

- It measures the amount of work it would take to raise the bucket an additional   Δ h  feet once you've raised it h  feet already.

Find:

Evaluate the definite integral of W from 0 to 50, What is the meaning off it. And why?

Solution:

                                     W = integral (B(h)).dh

                                     W = integral (-4*h/25 + 20)

                                     W = (-2*h^2/25 + 20*h)  

                                     W = (-2*50^2/25 + 20*50)

                                      W = 800 lb-ft

This represents the amount of work, in foot-pounds, that it would take to raise the bucket from h = 0  to h = 50. This calculation provides the right answer because, in order to find the amount of work, we'd want to take very thin slices of the form W_slice = B(h).dh sum them, and then take the limit as the slices get thinner  Of course, this would be a Riemann sum, and in the limit dh > 0. It would turn into an integral.

the answer to the expression is 83