# Definite Integral Work

• chukie
In summary, the bucket takes 30 seconds to get to the top of the well. The weight of the bucket is 32.5 pounds and the water remaining in the bucket is 1.4 pounds.

#### chukie

A bucket that weighs 3 lb and a rope of negligible weight are used to draw water from a
well that's 60 feet deep. Suppose the bucket starts with 37 lb of water and is pulled up by a rope at 2 ft/sec, while water leaks out of the bucket at a rate of 1/4 lb/sec.

A. How long does it take for the bucket to get to the top of the well? Write an equation that
expresses the total weight of the bucket as a function of time, as the time varies from 0 until the time the buckets gets to the top.

B. Recall that work equals force times distance. Calculate the work done in lifting the bucket
to the top of the well, keeping in mind that here force is equal to weight.

For A:I did 60ft/2ft/s=30 sec so it takes 30 secs for the bucket ot get to top of well
equation: Weight=40+ integral sign from 0 to 30 (-1/4)dx =32.5 lb

For B: i did total work= integral sign from 0 to 60 (32.5x)dx

i don't know if I am doing this right. I've searched the internet and found a question similar to mine (question #6):
http://count.ucsc.edu/~bauerle/images/19BW07HWK/Homework 6.4 Math 19B Winter 2007, Bauerle.pdf"

and I've used the same steps that i did for this question but i got the wrong answer.

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chukie said:
For A:I did 60ft/2ft/s=30 sec so it takes 30 secs for the bucket ot get to top of well
equation: Weight=40+ integral sign from 0 to 30 (-1/4)dx =32.5 lb
There isn't any integral involved here yet. You have to express your answer in terms of t, time. The weight would be wt of bucket + water remaining. Find an expression for the amount of water remaining at some given time t. No integration is required.

For B: i did total work= integral sign from 0 to 60 (32.5x)dx
One way you could do this would be to change the variables of your integration. Work done is given by $$\int^b_a \vec{F} \cdot d \vec{r} = \int^{b(t)}_{a(t)} \vec{F}(t) \cdot \frac{d\vec{r}}{dt} dt$$.

In this case since the problem is strictly one dimensional you can treat the vector functions as scalar functions. You have to do a change of variables from r to t. Firstly express r in terms of t. r is the distance from the bottom of the well to the top. Then you can continue from there.

Defennder said:
There isn't any integral involved here yet. You have to express your answer in terms of t, time. The weight would be wt of bucket + water remaining. Find an expression for the amount of water remaining at some given time t. No integration is required.

One way you could do this would be to change the variables of your integration. Work done is given by $$\int^b_a \vec{F} \cdot d \vec{r} = \int^{b(t)}_{a(t)} \vec{F}(t) \cdot \frac{d\vec{r}}{dt} dt$$.

In this case since the problem is strictly one dimensional you can treat the vector functions as scalar functions. You have to do a change of variables from r to t. Firstly express r in terms of t. r is the distance from the bottom of the well to the top. Then you can continue from there.

so for A:the equation would be W(t)=40-(1/4)t

for B: i dun quite understand wut u mean.do u mean i do the integral from 0 to 30? if that's the case i don't know wut the equation that I am trying to integrate is

Do you know the difference between "dun" and "don't" or "wut" and "what"?

chukie said:
so for A:the equation would be W(t)=40-(1/4)t

for B: i dun quite understand wut u mean.do u mean i do the integral from 0 to 30? if that's the case i don't know wut the equation that I am trying to integrate is
You have to first do a change of variables from r to t, as I said earlier. The integral for work done in terms of r (displacement) goes from 0 to 60. But if you evaluate in terms of t, it goes from 0 to 30. Replace the limits of r with that of t and the corresponding integrand also changes.

The reason why the change of variables is made is because the earlier part of the question asked you express the weight of the bucket + water as a function of time, which makes its easier to evaluate the work done in t instead or r. You could of course evaluate it in terms of r as well, but you now have to find an expression for the total load weight as a function of r.

okay thanks. i finally got the answer =)

## 1. What is a definite integral?

A definite integral is a mathematical concept that represents the area under a curve on a graph. It is denoted by ∫, and has a lower and upper limit that define the range of integration.

## 2. How is the definite integral calculated?

The definite integral is calculated using the fundamental theorem of calculus, which involves finding the antiderivative of the function being integrated and evaluating it at the upper and lower limits of integration.

## 3. What is the purpose of using a definite integral?

The definite integral is commonly used to find the area under a curve, but it also has many other applications such as calculating volume, determining the average value of a function, and solving various physics and engineering problems.

## 4. How is the definite integral related to the derivative?

The definite integral and the derivative are inversely related. The derivative of a function is the slope of the tangent line at any given point, while the definite integral represents the accumulation of the derivative over a given interval.

## 5. Are there any alternative methods for calculating a definite integral?

Yes, there are other methods for calculating definite integrals such as Riemann sums, the trapezoidal rule, and Simpson's rule. These methods may be used when the antiderivative of a function is not easily found or when the limits of integration are not constant.