A bucket that weighs 3 lb and a rope of negligible weight are used to draw water from a well that is 90 ft deep. The bucket is filled with 36 lb of water and is pulled up at a rate of 2.5 ft/s but water leaks out of a hole in the bucket at a rate of .25 lb/s. Find the work done in pulling the bucket to the top of the well?
A. Show how to approximate the required work by a Riemann sum (let x be the height in feet above the bottom of the well. Enter xi∗as xi)
B. Express the Integral
C. Evaluate the integral

a) Lim(0-inf)  Work = (36 - 0.1*xi )*dx

b) Work = integral( (36 - 0.1*xi ) ).dx

c) Work = 2835 lb-ft

Step-by-step explanation:

Given:

- The weight of the bucket W = 3 lb

- The depth of the well d = 90 ft

- Rate of pull = 2.5 ft/s

- water flow out at a rate of = 0.25 lb/s

Find:

A. Show how to approximate the required work by a Riemann sum (let x be the height in feet above the bottom of the well. Enter xi∗as xi)

B. Express the Integral

C. Evaluate the integral

Solution:

A.

- At time t the bucket is xi = 2.5*t ft above its original depth of 90 ft but now it hold only (36 - 0.25*t) lb of water at an instantaneous time t.

- In terms of distance the bucket holds:

                           ( 36 - 0.25*(xi/2.5)) = (36 - 0.1*xi )

- Moving this constant amount of water through distance dx, we have:

                            Work = (36 - 0.1*xi )*dx

B.

The integral for the work done is:

                           Work = integral( (36 - 0.1*xi ) ).dx

Where the limits are 0 < x < 90.

C.

- Evaluate the integral as follows:

                           Work = (36xi - 0.05*xi^2 )

- Evaluate limits:

                           Work = (36*90 - 0.05*90^2 )  

                            Work = 2835 lb-ft

the correct answer is 828.07

if x decreases by 4 then y increases by 10

step-by-step explanation: