Vectors and Calculus

This was a question on my final, which I totally failed, got a D+ overall and can't move on to the next class in the series T~T. I'm hoping my reasoning is wrong, because this just hit me when I woke up this morning.

So the problem was, if an acceleration vector is perpendicular to its velocity vector, then prove that velocity is constant.

So if acceleration is perpendicular to velocity, then it only has a normal component and its tangential component is 0, and the tangential component is the derivative of the velocity, at least I think it is. So, integrating the tangential component, 0, gives you 0+c, and c is a constant, so velocity would be constant?

If a is perpendicular to v and at + ac = a, then either:
- at is also perpendicular to v (which is nonsense), or
- |at| = 0 (in which case a = ac and points perpendicular from v)

=> at = (v/|v|)*(d|v|/dt) = 0
<=> v = 0 (I don't really know how a vector can equal a number, so yeah) or d|v| = 0
=> d|v| = 0

At least that's what I got from this page: http://en.wikipedia.org/wiki/Tangential_...celeration

Haven't really done this material.

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.

Okay, I got it, thanks guys ^O^

This is pretty upsetting, but, what's done is done.

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.

We didn't learn line integrals, that's in the next class in the series apparently, this is the class I took:

162.01H. 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.

I had to take a week and a half off for a funeral because we had to do it both Western styled and Asian styled, and in that week, they covered parametric and 2D vectors, at least I made it back in time for 3D vectors and I could get something out of that. 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 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 emailed the Math advising people to see if I could move on (Pass 162H with a C or written permission of the Honors Math Dept Chair) or at least, make up the course. I can take 153.01, which is basically the last two thirds of the class, to forgive the grade I got in 162, but I don't think that applies to me since I'm not technically in college yet and they don't recommended it if you get a D or above.

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

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.

Sorry about the confusion, I tried to make the distinction by saying Vectors and Calculus instead of Vector Calculus.

I've only done dots between two vectors, and ds is just like the element of arc length, but for parametric, right?

So what I mean is, basically the book has a = a[SIZE="1"]t[/SIZE]T + a[SIZE="1"]n[/SIZE]N

If we take the square of the magnitude of a, |a|^2 = a[SIZE="1"]t[/SIZE]^2 + a[SIZE="1"]n[/SIZE]^2 and solving for a[SIZE="1"]n[/SIZE]^2 gives a[SIZE="1"]n[/SIZE] = sqrt(|a|^2 - a[SIZE="1"]t[/SIZE]^2)

And then for a[SIZE="1"]t[/SIZE], it just gives a[SIZE="1"]t[/SIZE] = d^2 s/dt^2 = d(|v|)/dt

I meant coefficients of the normal and tangential components.

The first "formula" should be very intuitive. Any vector can be broken down into a number of "components" that can be summed together. In this case, the vector is split up into two components: one tangential to a different function and the other normal to the function.

a = a•T + a•N

This makes sense, since dotting a and T gives the projection of a onto T, and likewise for a and N. The second formula isn't really a formula. It's a trick to find the N vector, which is usually really annoying to find. The original formula for the N vector is N = T'/|T'|. As you can tell from the formula, it takes two steps to solve for it from a. That's why the second formula is used: solving N from a directly is much faster.

I'm not really sure what's meant by your third formula. Maybe your book provides a proof or example?



Edit: Very elegant, Russt.

The example in the book uses position vector R = (cos t + t sin t)i + (sin t - t cos t)j.
Then v = dR/dt = (t cos t)i + (t sin t)j, then of course a = dv/dt = (cos t - t sin t)i + (t cos t + sin t)j.
|v| is then the square root of the sum (t cos t)^2 + (t sin t)^2) = square root of t^2(cos^2 t + sin^2 t) = t, using cos^2 t + sin^2 t = 1.
Then it says that a[SIZE="1"]t[/SIZE] is equal to the derivative of of |v|, which is d(t)/dt = 1

I guess this is because the tangential component is the one that has an effect on velocity so it would be the second derivative of position?

From my book.

T = v/|v|, by definition. Rearranging gives:

v = |v| T.

Differentiating both sides of the above equation with respect to t, and using product rule:

a = v' = |v|' T + |v| T'

Also, N = T'/|T'|, by definition of N. Substituting into the above equation:

a = |v|' T + |v||T'| N

So the tangential component of acceleration is equal to |v|' and the normal component is equal to |v||T'|.

Well, the Taylor series approximation is a pretty simple concept, and I grasped it pretty easily at first, but I guess I lost it.
To create the series, you take the nth derivative at a and divide by n! starting at n=0 with f(a)/0! = f(a), so the series is f(a)+f'(a)(x-a)+(f''(a)(x-a)^2)/2+(f'''(a)(x-a)^3)/6 and so on. But first, it asks for an approximation that's accurate to like 4 decimal places with the least number of terms, so you use the remainder formula, the (n+1)th derivative at c, which since we're dealing with sin and cos would always be less than 1, divided by (n+1)! times (x-a)^(n+1) and that has to equal 10^-4 or less, which I think I figured out would only take two terms (when I did it the wrong way).

I think I've finally figured out what I did wrong. My problem was that I had no idea how to handle x-a and I was under the impression that x-pi/3 = pi/180 when really I should've done pi/180 - pi/3 = -59pi/180.

So then it's just guess and check with the remainder, I mean we didn't learn any guaranteed method to get the accuracy we wanted, but I guess n! might provide some clues.

R_7(x) = |f'''©|(-59pi/180)^8/8!, |f'''©| = |-cos c| < 1, so we have (-59pi/180)^8/8! which is 3.14 x 10^-5, which means the last term must be to the power of 7, which is 8 terms all together

So we need 7 terms of the Taylor series expansion so
f(pi/3) = sqrt(3)/2
f'(pi/3) = 1/2
f''(pi/3) = -sqrt(3)/2
f'''(pi/3) = -1/2

sqrt(3)/2 + (-59pi/180)/2 - (sqrt(3)(-59pi/180)^2)/4 - (-59pi/180)^3/12 + (sqrt(3)(-59pi/180)^4)/48 + (-59pi/180)^5/240 - (sqrt(3)(-59pi/180)^6)/1440 - (-59pi/180)^7/10080

Which, what do ya know, is 0.017427 and sin(pi/180) 0.017452

Too late.

This next part is somewhat unrelated.

Ohio State is also on Spring Break, and I'm unsure whether I should take 153 next quarter or not, with AP tests coming up and all. First, and this is the reason I didn't enjoy 161H (because of AP Calc AB), I've already been taught the material (whether I learned it is questionable, I mean, I feel I have, but, well, D+), and second, even if the D+ is forgiven, it'll still be there on my transcript for the rest of my life.

Which, what do ya know, is 0.017427 and sin(pi/180) 0.017452

Lol. Shoving in π/180 directly into your calculator would give you 0.017453, which is much more accurate. ><" It's good that you figured it out, then.

Ohio State is also on Spring Break, and I'm unsure whether I should take 153 next quarter or not, with AP tests coming up and all. First, and this is the reason I didn't enjoy 161H (because of AP Calc AB), I've already been taught the material (whether I learned it is questionable, I mean, I feel I have, but, well, D+), and second, even if the D+ is forgiven, it'll still be there on my transcript for the rest of my life.

Why are you taking AP tests in college? o_O

It's the opposite, I attend college in addition to being a senior in high school, where I have pretty much a full schedule with 4 APs and an IB. Also I just found out I got a C, prof must've curved grades. Its nothing to be proud of, but it's way better than a D. Unfortunately I've just been informed that I need a min GPA of 2.5 to continue taking classes, but I'm gonna talk to the advisor tomorrow and I might be able to take 263H or 254, now I can finally find out what partial derivatives and multiple integrals are. Yay!

Oh and the prof that teaches the series only let's you use techniques we've covered in class. It was frustrating going from the power rule and such in Calc AB to the limit definition of derivatives in 161H.