Need help with Line Integrals

[Aranfan]

I need to do the following for webwork:

Find the work done by the force field F(x,y,z) = 4xi + 4yj + 3k on the particle that moves along the helix r(t) = cos(t)i + sin(t)j + 5tk where t is between 0 and π

When I set up and do the problem I get: 75*2*π2
Which the computer says is wrong. I can't figure out where I messed up.

I set it up as:
S02π -4cos(t)sin(t) + 2sin(t)cos(t) + 75t dt


Where did I mess up?

[Fingolfin_Noldor]

Why are you integrating from 0 to 2 pi?

And I think you got your dot product wrong. Check your working.

[Aranfan]

Gah! Yeah, I transcribed that wrong, here's how I actually set it up:

S02π -4cos(t)sin(t) + 4sin(t)cos(t) + 75t dt

As to integrating from 0 to 2π, those are the endpoints of t. Where else am I supposed to integrate from?

[Kuroneko]

Use the fundamental theorem of line integrals with F = ∇(2x²+2y²+3z). If you want to do it the long way, dW = F·dr = F·v dt, where v is the velocity vector. If you set up the dot product correctly, the x and y contributions should cancel.
Edit: The problem explicitly defined the path from 0 to pi.

[Kuroneko]

How did you get 75 under that integral?

[Aranfan]

I messed that up too. It was from 0 to 2π

Edit:

How did you get 75 under that integral?

F(r(t)) = < 4cos(t), 4sin(t), 15t>
r'(t) = < -sin(t), cos(t), 5>

F(r(t)) * r'(t) = -4sin(t)cos(t) + 4sin(t)cos(t) + 75t



I really need help with line integrals.

[Surlethe]

Let's try to make this intuitive from the perspective of work. If you have a force pushing in or against the direction of an object's motion, the work done will be Fd, where F is the force and d is the displacement (adjust the sign of F to indicate direction). If the object is constrained to move along a straight path, then work done will be , where d is the displacement vector and F is the work vector.

Now, for a smooth curve , locally at r(t) it is straight and well-approximated by the differential dr = r'(t)dt. Moreover, at r(t) the vector field F is nearly constant and well-approximated by F(r(t)). Therefore, we can approximate the work done in a very small interval of time (t,t+dt) by . Slapping on integral signs and globalizing, we have
.

How does this apply to your problem? Well, Spoiler

and
,
so
.

The cheap way Kuroneko suggests is because in this case, so Stokes' Theorem applies.
Spoiler

Stokes' Theorem, applied to line integrals, says that for a conservative vector field (i.e., the gradient of some scalar function) the integral is merely the scalar function evaluated between the endpoints. It's the line integral form of the fundamental theorem of calculus. To wit:

If you want the vector field to have a z-component of 15t instead of 3t, adjust all of the above calculations accordingly.

[Darth Mordius]

Quibble: Stoke's theorem is the one with the line integral around a closed loop being the surface integral of the curl. I think the applicable theorem in this case is just called the fundamental theorem for line integrals.

[Aranfan]

Oh. I get it now. Duh. Thanks Surlethe. I was messing up F(r(t)).

[Surlethe]

Quibble: Stoke's theorem is the one with the line integral around a closed loop being the surface integral of the curl. I think the applicable theorem in this case is just called the fundamental theorem for line integrals.

In my understanding, when it's presented in higher generality, Stokes' theorem says that the integral of a derivative on a region is the integral of the function around the boundary. In this case, the region is the curve, the derivative is the field, the function is the potential, and the boundary is the endpoints of the curve.

[Darth Mordius]

Quibble: Stoke's theorem is the one with the line integral around a closed loop being the surface integral of the curl. I think the applicable theorem in this case is just called the fundamental theorem for line integrals.

In my understanding, when it's presented in higher generality, Stokes' theorem says that the integral of a derivative on a region is the integral of the function around the boundary. In this case, the region is the curve, the derivative is the field, the function is the potential, and the boundary is the endpoints of the curve.

Quite possibly; I don't speak math, I just use it.

[Kuroneko]

A physicist might also observe that the projection of the path to the xy plane is clearly circular, and hence the xy velocity components must be orthogonal to the (radial) xy components of the field. Hence, only the z components can contribute to work. But as some point it's better to just get some practice with the general method.

In differential geometry, Stokes' theorem is the general result that the integral of an n-form over a manifold with boundary being equal to the integral of the corresponding (n+1)-form over the boundary. A smooth function is by definition a 0-form and its gradient is the differential (or rather, its dual): df = (∂f/∂xi) dxi = f,i dxi. In R³, the curl of a vector field corresponds to a certain 1-form (giving the Stokes' theorem from elementary calculus), and the divergence to a 2-form (giving Gauss' theorem).

[starslayer]

In my understanding, when it's presented in higher generality, Stokes' theorem says that the integral of a derivative on a region is the integral of the function around the boundary. In this case, the region is the curve, the derivative is the field, the function is the potential, and the boundary is the endpoints of the curve.

You have it backwards. The integral of the derivative over the boundary equals the integral of the function over the region.

[Kuroneko]

No, I'm the one who mistyped n+1 instead of n-1. Surlethe is correct. IntA dF = Int∂A F. The most trivial example is in the reals, with the manifold being an interval I = [a,b], in which case we get the fundamental theorem of calculus
Int[a,b] dF = Int{a,b} F = F(b) - F(a),
with the two-point boundary {a,b}.

[Surlethe]

Cheap answer: the ∂ goes from the integrand to the region of integration.

[starslayer]

Oh, son of a bitch. I misread my calculus book. Yes, of course you're right. You'd think I would remember this stuff...