Vectors and Calculus

One of the first things you should have learned is that line integrals sum the individual components of the gradient of a potential along a path. The formula for this is:

I = ∫ (a•T) ds, where:

a = <x''(t), y''(t)>
T = <x'(t), y'(t)> / sqrt(x'(t)^2 + y'(t)^2)
ds = sqrt(x'(t)^2 + y'(t)^2)*dt

Notice that T*ds just simplifies to v*dt. Therefore, we get:

I = ∫ (a•v) dt

I = ∫ <x''(t), y''(t)> • <x'(t), y'(t)> dt

Now I'm assuming you did the following, based on your first post:

I = ∫ <x''(t), 0> • <0, y'(t)> dt = ∫ 0 dt = C1

I = ∫ <0, y''(t)> • <x'(t), 0> dt = ∫ 0 dt = C2

Therefore, you concluded that since C1 and C2 are constant, they represent the velocity, which is also a constant. Let's consider the definition of the line integral. The line integral sums all vectors along a line. Thus, the result of the line integral isn't the velocity; it's the sum of the acceleration vectors.

Pursuing this further, the vectors being summed are completely independent of the line over which the vectors are summed. In other words, ∫ x''(t) dt is not x'(t) in a line integral. (The only time this is ever true, I = 1/2 x'(t)^2 + 1/2 y'(t)^2 + C1*x''(t) + C2*y''(t). However, this doesn't apply here.) In the general case:

I = ∫ <x''(t), y''(t)> • <x'(t), y'(t)> dt

I = ∫ (x''(t)*x'(t) + y''(t)*y'(t)) dt

Notice that the above is actually just:

I = x''(t)*∫ x'(t) dt + y''(t)*∫ y'(t) dt

So if the force is perpendicular to the velocity, one of the two terms gets killed:

I = x''(t)*∫ 0 dt + 0*∫ y'(t) dt

I = C1*x''(t)

Or:

I = 0*∫ x'(t) dt + y''(t)*∫ 0 dt

I = C2*y''(t)

This is where the constant of integration should come from. It has nothing to do with the velocity. In fact, the constant of integration just represents the scalar magnitude of the line.

Good luck next quarter. I'm also taking vector calculus next quarter, but for the first time.