CHRISTINE BREINER: Welcome back to recitation.
In this video, I'd like us to do the following two problems,
both related to the same position vector.
So we're starting off with a position vector defined as r
of t is equal to 1 minus 2t squared i plus t squared j
plus negative 2 plus 2t squared k.
So that's our position vector, and I'd like us to do the
following two things.
And you'll notice this problem is pretty much just a
computational problem.
We're going to make sure that we know what these things I'm
about to talk about are, how you define them, and how you
get from the position vector to each of these things.
So we want to compute the velocity, the speed, the
acceleration, and find the unit tangent
vector for r of t.
And then, the second part, we want to compute the arc length
of the trajectory from t equals 0 to t equals 2.
So I'll give you a moment to do that problem.
Why don't you pause the video, work on the problem.
When you're ready to check your work, bring the video
back up and I'll show you how I do it.
OK, welcome back.
Well, hopefully, you felt comfortable with answering
these questions.
So now I will answer them and you can compare your
answers with mine.
So let me start off with part a.
Part a, the first thing we're going to
do is find the velocity.
So velocity is really-- all we need to do is take the
derivative of the position vector with respect to t.
So I'm just going to take r prime of t.
And now I'm going to write it in the shorthand notation that
you've seen with the brackets to denote that it's not a
point, but it's a vector.
So this is what you've seen to denote a vector
rather than a point.
So the derivative with respect to t of the first component is
just negative 4t.
The derivative with respect to t of the second component is
just 2t, because we had t squared, so when we take its
derivative, we just get 2t.
And the third component was negative two plus 2t squared,
so when I take its derivative, I get a 4t, so that is
actually v of t, OK?
And then the next thing I asked you to do is determine
the speed, and the speed, of course, is just the length of
the velocity vector, right?
So we just need to find the length of v. Now, to do that,
to remind ourselves what we do for that, we actually take the
inner product of v with itself, the dot product of v
with itself, and then we take the square root of that.
So let's look at what the dot product will be.
Let me find the squared thing first, and then I will take
the square root.
So v dotted with v, the first component I'm going to have
negative 4t quantity squared, so that's
going to be 16t squared.
And then the second component is going to be 2t quantity
squared, so I'm going to have plus 4t squared.
And the third component is going to be another-- it's
going to be 4t quantity squared, so I
get another 16t squared.
So when I add those up, I believe I get 36t squared?
Yes, good.
And so then, I just have to take the square root of both
sides to get what the speed actually is instead of the
square of the speed.
So I get 6t, OK?
So that's the velocity; that's the speed.
Now I need to find the acceleration and I need to
find the unit tangent vector.
OK, so let me see.
I will come over here.
Let me step off here and I will find the acceleration and
the unit tangent vector.
So the acceleration, if you remember, the acceleration is
actually just the derivative of the velocity
with respect to t.
So the acceleration is going to be the derivative of
negative 4t is just negative 4.
The derivative of 2t is just 2, and the derivative of 4t is
just 4, all with respect to t.
OK.
So the acceleration vector is equal to negative 4 comma 2
comma 4, so you see this actually has constant
acceleration.
So at any point, your acceleration is always this
value, so it's not surprising that our velocity is
increasing, and actually, it's increasing-- you'll notice,
each of these components is constant.
The velocity, each of the components is linear, and if
we went back, we look at the position vector, each of those
components is quadratic.
And this is exactly what you expect from just your
understanding of derivatives in single-variable calculus.
If you start off with a constant and you find an
antiderivative, it's going to be linear, and you find
another antiderivative, you're going to have a quadratic, so
we shouldn't be surprised by any of that.
Now we have one more thing to do with Part a, and that is to
find the unit tangent vector.
And that's fairly easy, because all we have to do is,
if you notice, we have the velocity vector and we have
its length.
And so to find the unit tangent vector, all we have to
do is take the velocity and divide it by its length, and
that will normalize it.
That means that its length will be one at that point,
because you're taking a vector, dividing by its
length, so the length of the new vector will have to be
length one.
So let me write that down.
And actually, I guess the point to remember here is that
the velocity vector is tangent to the path you're carving
out, to the trajectory.
OK, so this is a vector.
This is a scalar.
So I'm going to take 1 over 6t, and I'm going to multiply
it by negative 4t, 2t, 4t, and this gives me, when I do my
division, looks like it gives me a negative 2/3,-- right--
1/3, 2/3.
So that is the unit tangent vector, OK?
OK, and now, we have one more point we want to make, and
that is having to do with the arc length of the trajectory.
That was the second part of this problem, was to find the
arc length of the trajectory from t equals 0 to t equals 2.
So let me just draw another line here.
And what we want to do there then is-- really what we want
to do is we want to integrate the speed, right?
We want to integrate the speed from 0 to 2.
So this--
let me come over here--
this absolute v, you might have also seen it written as
ds dt, right?
And so we want to integrate this in dt--
in t-- sorry-- from 0 to 2.
And so we come over here.
We want to integrate from 0 to 2, 6t dt.
That's fairly easy.
That's going to be 6t squared over 2, evaluated from 0 to 2.
And so when I write that down, I'm going to get 6 times 4
divided by 2, 24, divided by 2, I just get 12, and the
other term is 0.
So the arc length is--
of the trajectory from 0 to 2 is just 12 units.
So this really was a purely computational type of problem.
All we were doing, if you come back over here and you recall
what we were trying to do, is we started off
with a position vector.
We just did a lot of computation.
We found the velocity, the speed, the acceleration, the
unit tangent vector, and then we just wanted to find the arc
length of the trajectory.
So this is all very computational, but just to
make sure we understood what all the terms meant and how
they were related to one another.
So I'll stop there.