Vectors and Calculus

Okay. So it doesn't seem like you're really doing vector calculus. You're just doing parametric equations... though in your terminology, it would technically be called vector calculus. I actually just reread the thread again. I should have mentioned this in my previous post, but the velocity vector is not constant, and neither is the magnitude of the velocity vector... so just ignore this:

The velocity vector is not constant. The magnitude of velocity is constant.

Given that distinction, then yes, tangential acceleration is the derivative of speed, so if acceleration is perpendicular => there is no tangential acceleration => speed is constant.

If this were one-dimensional, then Ruust would be correct, because perpendicularity doesn't exist in one-dimensional space. However, this is on a two-dimensional plane, so one dimensional methods are inconclusive.

62.01H. L'Hospital's rule, improper integrals, sequences and series, convergence tests, power series, Taylor's formula, conic sections, polar coordinates and their applications, parametric equations of curves, vector algebra in the plane and three-dimensional space, derivatives of vector functions, curvature and the unit normal vector, tangential and normal components of acceleration, analytic geometry of three-dimensional space.

Since you learned the unit normal and tangent vectors, I'm going to assume that you also learned that dotting a parametric equation onto the unit tangent vector provides the projection of that parametric equation onto that vector. An easy way to see this is: A•B = |A|*|B|*cos(θ, so A•(B/|B|) = |A|*cos(θ. B/|B| is just the tangent vector for B. cos(θ pretty much adjusts |A| to what it needs to be.

The dotting process is where the a•T part of the line integral comes from. What the line integral is measuring is how much the acceleration vector actually contributes to the path. Suppose there is a particle traveling along a path. To obtain the total contribution of the acceleration vectors to the path, it's logical to sum them up on the path. Therefore:

I = ∫ (a•T) ds

ds is just the scalar length vector, which I hope you've learned before. The scalar length is just sqrt(x'(t)^2+y'(t)^2)*dt (almost the Pythagorean theorem). This is a logical formula, because notice that sqrt(x'(t)^2+y'(t)^2) is just the magnitude of the velocity. Integrating velocity over time gives a displacement. In other words, it produces a path that can be integrated over. So putting this together:

I = ∫ (a•T) ds = ∫ (a•v) dt

Anyways, since you didn't learn the line integral, your answer on the test would simply be:

Since the acceleration vectors are perpendicular to the velocity vectors, they can provide no contributions to the velocity vectors, because the projection of the acceleration vectors onto the velocity vectors, or a•T, is 0. This should have given you nearly full credit (unless your graders are jerks).

Basically with acceleration though, in the book we basically only had that the tangential component was the derivative of speed and the normal component was the square root of the sum of the magnitude of acceleration and the tangential component.

I'm not sure what you're talking about in this section. I'm going to attempt to translate this to symbols:

If:
f(t) = <x(t), y(t)>
f'(t) = <x'(t), y'(t)>

Then:
f''(t) • T = (sqrt(x'(t)^2+y'(t)^2))' = (x'(t)+y'(t)) / sqrt(x'(t)^2+y'(t)^2)

And:
f''(t) • N = sqrt(x''(t)^2+y''(t)^2) + f''(t) • T
= sqrt(x''(t)^2+y''(t)^2) + (sqrt(x'(t)^2+y'(t)^2))'
= sqrt(x''(t)^2+y''(t)^2) + (x'(t)+y'(t)) / sqrt(x'(t)^2+y'(t)^2)

This somehow doesn't seem right. Maybe there is a slim chance that you failed because you had trouble articulating your thoughts.

I left the question blank, and it was worth 20 points. I spent most of my time doing a Taylor series approximation, which I'm still uncertain how to do (Approximate sine of 1 degrees using a = pi/3).

I'm actually not sure how to do this, but I do know a different way to do this:

Y = sin(π/180)

y = sin(0)
dy/dx = cos(x)
dy = cos(x) dx

y + dy = Y
sin(x) + cos(x)*dx = Y
sin(0) + cos(0)*π/180 = Y
π/180 = Y

For this reason, it's a standard procedure to use sin(θ ≈ θ for small θ. I wish I had brought my calculus textbook home, but I didn't. (I'm on Spring Break for a week.)

On an unrelated note, do universities usually factor in grades from transfer credit?

If the credit didn't come from somewhere shady, then yes.