2007-09-21

Today's nonsense
- Luisa's wrong, Spiller right
- Katie's still getting lost on which class she's in
- Dakota is still watchless

Unit vectors

Unit vectors have a magnitude of 1- because it's a unit vector, like a unit circle. Unit vectors point in their corresponding directions. We have the i unit vector which points horizontally and the j unit vector which points in the vertical directions. Then there's also the k unit vector which points in the z direction. Each begins at the origin. Remember, you can move around vectors as much as you want, so unit vectors are with respect to the origin and the origin is as you define it, so you're not necessarily on an x-y plane.

ex 1) You walk 3 km west, then 4 km headed 60 degrees north of east, and then 6 km north 40 degrees west (N-40-degrees-W). 60 degrees north of east is the same thing as 30 degrees east of north, and remember that 90 is the key here. So in this scenario, what is the resultant displacement? So that's: -3i + 0j (not pointing up or down) + then you figure out a triangle via SOHCAHTOA because the "60 degrees north of east" is in the first quadrant so it's going to be (cos60 = adjacent/hypothesis) -> (x=4cos60) -> so that's going to be what you do as !THAT THING! (which is 2) * i. So far: -3i + 0j + 2i + "same but different"- this time you find the y-value of that triangle, and so sin60=y/4, so then y=4sin60,, so you get: -3i + 0j + 2i + 3.464j .... and now what about that final part where you go 6 km north and 40 degrees west? So the angle from the x-axis to the vector is going to be 50 degrees in this scenario. So what's the component form of this vector? -3.857i + 4.596j. So to get your resultant, you add up all of the i's and j's and you get your ultimate displacement vector (call it the 'd' vector). So VectorD = -4.856i + 8.06j
-- But we are not done yet. We are trying to find the resultant displacement, not the resultant displacement vector. So if I want to find how far, we're going to do the magnitude, and so what quadrant is this displacement vector? It's in quadrant 2, so it's to the top-left somewhere, So the magnitude of that vector is going to be the hypotanuse of the right triangle, so if I want to find the magnitude of the vector, I'm going to do a^2 + b^2 = c^2, so the absolute value of VectorD = 9.410 km. So now how do you get the direction? You have to find the angle that you want, the angle of the resultant displacement vector- so it's going to be the arctan of 8.06/4.357 or whatever. So that turns out to be 58.927 degrees. So now we have the resultant magnitude and the resultant direction.

How do we report this answer? 58.927 degrees North of West. Or you can say: West 58.927 degrees North. Or you can say: North 31.7310 degrees West. But we're not going to get into that. Typically you do North/South first, and they said that you started with North and then either go East or West. So in that case we would have to subtract. So our final answer is: 9.410 km 58.927 degrees North of West.

You took the arctan because you were trying to figure out the angle, in order to get the final direction of the resultant direction, so that we may report our answer properly. So whenever you are looking for the angle and you know some sides, then you take the arc{trigfunc}.

Position, Velocity, and Acceleration

The position vector of a particle is a vector drawn from the origin of a reference frame to the xy position of the particle.
VectorR = x(VectorI)+y(VectorJ)
So we can talk about an average velocity vector:
How would you find the average velocity vector? VectorAVG = ( change in VectorR ) / (change in time)

So now we can get now to instantaneous velocity vectors, which is when there's no delta-t and delta-t would be approaching zero?
Instantaneous velocity vector
VectorV = lim (as delta-t -> 0) of ( delta VectorR)/(delta-t) = d(VectorR) / dt
The derivative of the position vector, is the velocity vector. So you would take the derivative of the two components, separately, so next we are going to write:

Average acceleration vector
VectorA_avg = (change in VectorV)/(change in t)
Instantaneous acceleration vector
VectorA = lim of ( delta-VectorV / delta-t ) as delta-t -> 0 == dVectorV / dt

The magnitude of any vector, VectorV = a(VectorI) + b(VectorJ)
So this is just rehashing the idea of using the Pythagorean theorem.
So, the magnitude of that vector is going to be abs(VectorV) = sqrt(a^2 + b^2)
Likewise, for a vector that might be VectorV = a(VectorI) + b(VectorJ) + c(VectorK)
abs(VectorV) = sqrt(a^2 + b^2 + c^2)

So let's ask a couple theoretical questions or two. The magnitude of the displacement of a particle is, what the distance the object traveled? How does displacement compare to distance? Displacement is less than or equal to distance. So there are some simple questions here. So let's take a look at #5 on the homework or on the same pages. It says: a man walks around a circular arc from x=5,y=0 to the final position x=0,y=5, so what's the overall displacement? You just use half the circle. So part of his displacement is the distance, then plus his angle, so you use the pythagorean theorem to find the distance. You know the magnitude is going to be the sqrt(5^2 + 5^2) so since it's displacement you also need to include an angle. So abs(VectorDisplacement) = 5sqrt(2). So now we need an angle. The angle is 45, because it's isoscoles triangle. So it's 45 degrees North of West. Or: North 45 degrees West. Or 45 Degrees N of W.

Quick mention of prime factoring

VectorC = VectorA + VectorB.
VectorC = aVectorI + bVectorJ
VectorA = dVectorI + eVectorJ
VectorB = fVectorI + gVectorJ
One of the coefficients on VectorA or VectorB could have been negative.