Webassign HW 03 6.4 - Answers/Explanations

Remember in the Webassign section there are other answer sets just not explanations!

For this problem, remember the basic formula for work is the integral of the force equation. This problem is 2000 is the force equation and 1152 feet is the upper bound with 0 being the lower bound.

For this problem, problem gives you the initial bounds and force equation. Again, this simply typing in the inputs. REMEMBER you dx! To solve the integral, take out the 1000 and factor the remaining function. Next using the law of sums break the integral up and solve each separately adding the results together at the end. Or use Symbolab.

For this one, remember the Spring equations. First work equals the integral of the spring constant and X. Also, force equals the spring constant times X (displacement). To begin, the person stretches the spring 11 cm at a force of 2200 dynes (dynes is a measure of force). Then he stretches it 11 more cm. First, the initial 11cm is the bottom bound because it becomes new "natural position" this is because the natural position is for all intensive purposes where you start. And, as such, the top bound is wherever you finish in this case 22cm.

The spring constant is the first thing you need to find before trying to find work. You do this by dividing the force that is took to pull the spring to X displacement by that displacement (AKA the second equation I provided). Then you plug the spring constant, in this case 200, into the integral KX with K being the spring constant. Don't forget dx.

For this problem, you need to remember for any tank problem the work is equal to the integral of (62.5 density of water) (L - X)A(X). L = the distance to where you are pumping and A(X) is the area for the tank when looking at the y-plane. So the integral works out to be 62.5pi integral from 10 to 0 of (10-X) (21)^2.

For the second part the bounds stay the same because the volume of water in the tank doesn't change but L does. So the integral has to be taken as (12-X)(21)^2.

For this problem, you need to find a A(x) as it changes. So use similar triangles to find a the radius at any point along y in this case 1/3. So then applying the same equation as the question before with 6 being your top bound and 3 being your lower bound ( they didn't pump all the water out). So you get 62.5 pi integral (6-x)(x/3)^2.

So this is basically a stupid problem. But here's how it goes... To begin GMm is constant so ignore it for now. That leave the force equation as f(x)= -1/x^2 so work is equal to the integral of force with the bounds beginning this case 4000 and 5200 (or 4000+1200). 

The second part just replace the top bound with infinity. Remember anything 1/infinity equals 0 so basically the top bound doesn't exist.

For this problem, the bounds are given as the interval. Next you need to find the which equation is larger in this case f(x). So then you subtract g(x) from f(x) this is because g(x) undercut the area of f(x) and you are only interested in the area in between the two. Then you can just plug in squaring each and multiplying by pi. Remember that the base equation for a volume of a revolved solid the integral of pi (function)^2.

For this problem, its the same as the top. Just for the bounds you need to set them equal to each other.