1) If a continuously varying force F(x) moves an object on the xy-plane through arc (whose derivative is continous) given by g(x) from x=a to x=b (on the interval [a, b]). Derive a formula for the work done by the force in moving the objects from x=a to x=b along the arc.
2) If the force is F(x) = 2x^-3 and the arc is g(x) = ln x, find the work done if an object is moved from x= 1 to x= 10


Cc. Mrshape, Darivie04, MathsEconomics, and others.

Martinez39:
1) If a continuously varying force F(x) moves an object on the xy-plane through arc (whose derivative is continous) given by g(x) from x=a to x=b (on the interval [a, b]). Derive a formula for the work done by the force in moving the objects from x=a to x=b along the arc.
2) If the force is F(x) = 2x^-3 and the arc is g(x) = ln x, find the work done if an object is moved from x= 1 to x= 10


Cc. Mrshape, Darivie04, MathsEconomics, and others.

Hmmmm.
I will attempt later sir

Very simple question, how much will you pay me to solve it?

So no one is up to the task?

Martinez39:
So no one is up to the task?

Should I post the answer?

Martinez39:
Should I post the answer?

I have solved it!

frankdoz:
I have solved it!

Post your answer.

Martinez39:
1) If a continuously varying force F(x) moves an object on the xy-plane through arc (whose derivative is continous) given by g(x) from x=a to x=b (on the interval [a, b]). Derive a formula for the work done by the force in moving the objects from x=a to x=b along the arc.
2) If the force is F(x) = 2x^-3 and the arc is g(x) = ln x, find the work done if an object is moved from x= 1 to x= 10


Cc. Mrshape, Darivie04, MathsEconomics, and others.

Really cool problem, wish I had seen it earlier. I might drop a solution later today.

OK, check my answer!

Darivie04:

Really cool problem, wish I had seen it earlier. I might drop a solution later today.

Okay then. I will drop my solution tomorrow morning. I await your solution.

frankdoz:
OK, check my answer!

Wrong! Try again.

Martinez39:
Wrong! Try again.

then drop your answer, stop wasting time!

Are you sure F(x) is not equal to 2x^3?

frankdoz:
then drop your answer, stop wasting time!

I wanted to drop the solution this afternoon but Darivie04 said he didn't see the question until today and he would be dropping his solution later today. For his sake, I will withhold my answer until 8:00pm today. Though I wanted to wait till tomorrow but I will unfailingly release my answer by 8:00pm since you are eager to see it. But if Darivie04 drops a solution now (wrong or correct), I will release my answer soon after his.

Darivie04:
Are you sure F(x) is not equal to 2x^3?

F(x) = 2x^-3

.

Darivie04:
2. Work Done = ∞

Wrong. Let me proceed to drop my solution.

Martinez39:
Wrong. Let me proceed to drop my solution.

Wait abeg

2.Work Done = 278.3314206 Joules

MY SOLUTION

— QUESTION ONE ANSWER
Partition the interval [a, b] in n subinterval of width Δx_i. Δx_i is the width of the i -th subinterval. The work done in the i -th subinterval is approximated by

F(c_i) [ Δx_i² + Δy_i² ]^½ —— (1) (force times distance) where c_i is any value of x in the i -th subinterval and Δy_i is the total change in y in the i -th subinterval)

By rewriting, (1) takes the form

F(c_i)[ 1 + (Δy_i/Δx_i)² ]^½ Δx_i

From the Mean Value theorem we know that since g(x) is continous on the interval [a, b] and differentiable on the interval (a, b), then on each subinterval, there exists an x value, say d_i, for which the derivative is equal to the average rate over the subinterval ie.
g'(d_i) = Δy_i/Δx_i (for the i -th subinterval)
So we can write

F(c_i) [ 1 + (g'(d_i))² ]^½ Δx_i

By summing the approximations on each subinterval, we can approximate the work done W by the force F(x) on moving an object on the arc g(x) by

W ≈ Σ F(c_i)[ 1 + (g'(d_i))² ]^½ Δx_i

By taking limit of this Riemann sum as the norm of the sum (||Δ||) approaches zero, we get the exact work done. The means that the work done is

The integral of F(x) [ 1 + (g'(x))² ]^½ from x = a to x = b

Cc. Darivie04, frankoz

Darivie04:
2.Work Done = 278.3314206 Joules

Have you seen this @Martinez39?

Martinez39:
MY SOLUTION

— QUESTION ONE ANSWER
Partition the interval [a, b] in n subinterval of width Δx_i. Δx_i is the width of the i -th subinterval. The work done in the i -th subinterval is approximated by


By rewriting, (1) takes the form

From the Mean Value theorem we know that since g(x) is continous on the interval [a, b] and differentiable on the interval (a, b), then on each subinterval, there exists an x value, say d_i, for which the derivative is equal to the average rate over the subinterval ie.
g'(d_i) = Δy_i/Δx_i (for the i -th subinterval)
So we can write


By summing the approximations on each subinterval, we can approximate the work done W by the force F(x) on moving an object on the arc g(x) by


By taking limit of the sum as the norm of the sum (||Δ||) approaches zero, we get the exact work done. The means that the work done is



Cc. Darivie04, frankoz

I was able to get the question up to this part. It was that force function in no.2 that confused me for awhile.

Darivie04:

Have you seen this @Martinez39?

I have seen it but it is incorrect. Evaluate the definite integral I gave and compare the answers.

Martinez39:
I have seen it but it is incorrect. Evaluate the definite integral I gave and compare the answers.

I have seen my error in the second solution I posted. I also saw that our formulas are actually different so I'll just write down my full solution and put it here.

Darivie04:

I have seen my error in the second solution I posted. I also saw that our formulas are actually different so I'll just write down my full solution and put it here.

Did your answer accord with the definite integral I gave?

Martinez39:
Did your answer accord with the definite integral I gave?

We got different integrals for number 1.

My solution is based on the fact that the work done is the area beneath the force distance curve.

My x_f is gotten from the formula of an arc length.

If I'm making a mistake somewhere I'll be glad to see it.

Darivie04:
My solution is based on the fact that the work done is the area beneath the force distance curve.

Not correct.

Martinez39:
Not correct.

Why is it not correct?

Darivie04:

Why is it not correct?

Your integral is an improper integral because F(x) = 2x^-3 has an infinite discontinuity at x = 0. Moreover, your integral diverges. I guess that was why you got Work Done = ∞.

Martinez39:
Your integral is an improper integral because F(x) = 2x^-3 has an infinite discontinuity at x = 0. Moreover your integral diverges.

But if you use that same force function for a simpler problem say raising an object 20m or any other simple Work related problem. You would have to use a similar integral(with the only difference being the upper bound) and you would still encounter the same problems.

E.g
A variable force F(x)=2x^-3 acts on an object and displaces the object by 20m. Calculate the work done.

Darivie04:

But if you use that same force function for a simpler problem say raising an object 20m or any other simple Work related problem. You would have to use that same integral and you would still encounter the same problems.

True. The reason is because the force function has an infinite discontinuity at x = 0. Any definite integral of the force function with x = 0 as the lower limit of integration is an improper integral that diverges.

That is why I used x = 1 as the lower limit of integration.