4. 0.52 / 2.08 points details my notes previous answers ask your teacher a bucket that weighs 4 lb and a…

4. 0.52 / 2.08 points details my notes previous answers ask your teacher a bucket that weighs 4 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 42 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 0.25 lb/s. find the work done in pulling the bucket to the top of the well. 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 x*_i as x_i.) lim_(n -> ∞) sum_(i = 1)^n ( )δx express the work as an integral. ∫_( )^( ) ( ) dx evaluate the integral. ft - lb

4. 0.52 / 2.08 points details my notes previous answers ask your teacher a bucket that weighs 4 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 42 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 0.25 lb/s. find the work done in pulling the bucket to the top of the well. 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 x*_i as x_i.) lim_(n -> ∞) sum_(i = 1)^n ( )δx express the work as an integral. ∫_( )^( ) ( ) dx evaluate the integral. ft - lb

Answer

Explanation:

Step1: Determine the force as a function of height

The initial weight of the bucket with water is $4 + 42=46$ lb. The water leaks out at a rate of $0.25$ lb/ft. Let $x$ be the height above the bottom of the well. The weight of the bucket - water system at height $x$ is $F(x)=(4 + 42)-0.25x=46 - 0.25x$. The distance the bucket needs to be lifted from height $x$ to $x + \Delta x$ is $\Delta x$. The work done in lifting the bucket - water system through a small distance $\Delta x$ is $\Delta W\approx F(x)\Delta x=(46 - 0.25x)\Delta x$.

Step2: Set up the Riemann sum and the integral

We want to find the work done in lifting the bucket from $x = 0$ to $x = 90$. The work $W$ is given by the limit of a Riemann sum: $W=\lim_{n\rightarrow\infty}\sum_{i = 1}^{n}(46-0.25x_i)\Delta x$, where $\Delta x=\frac{90 - 0}{n}$ and $x_i=i\Delta x$. This limit is equivalent to the definite integral $W=\int_{0}^{90}(46 - 0.25x)dx$.

Step3: Evaluate the integral

We use the power - rule for integration $\int x^n dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$). [ \begin{align*} \int_{0}^{90}(46 - 0.25x)dx&=\int_{0}^{90}46dx-\int_{0}^{90}0.25xdx\ &=46x\big|{0}^{90}-0.25\times\frac{x^{2}}{2}\big|{0}^{90}\ &=46\times90-0.25\times\frac{90^{2}}{2}\ &=4140-0.25\times\frac{8100}{2}\ &=4140 - 0.25\times4050\ &=4140 - 1012.5\ &=3127.5 \end{align*} ]

Answer:

$3127.5$ ft - lb