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CHRISTINE BREINER: Welcome
back to recitation.

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In this video, I'd like
us to do the following two

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problems, both related to
the same position vector.

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So we're starting off with
a position vector defined

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as r of t is equal to 1 minus
2 t squared i plus t squared

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j plus negative 2
plus 2 t squared k.

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So that's our position
vector, and I'd

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like us to do the
following two things.

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And you'll notice this
problem is pretty much

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just a computational problem.

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We're going to make
sure that we know

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what these things I'm
about to talk about are,

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how you define them, and how
you get from the position vector

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to each of these things.

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So we want to compute
the velocity, the speed,

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the acceleration, and find the
unit tangent vector for r of t.

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And then, the
second part, we want

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to compute the arc length of the
trajectory from t equals 0 to t

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equals 2.

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So I'll give you a moment
to do that problem.

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Why don't you pause the
video, work on the problem.

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When you're ready to check your
work, bring the video back up

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and I'll show you how I do it.

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OK, welcome back.

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Well, hopefully, you felt
comfortable with answering

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these questions.

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So now I will answer them and
you can compare your answers

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with mine.

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So let me start off with part a.

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Part a, the first
thing we're going to do

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is find the velocity.

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So velocity is
really-- all we need

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to do is take the derivative
of the position vector

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with respect to t.

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So I'm just going to
take r prime of t.

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And now I'm going to write
it in the shorthand notation

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that you've seen, with the
brackets to denote that it's

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not a point, but it's a vector.

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So this is what
you've seen to denote

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a vector rather than a point.

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So the derivative with respect
to t of the first component

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is just negative 4t.

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The derivative with respect
to t of the second component

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is just 2t, because
we had t squared,

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so when we take its
derivative, we just get 2t.

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And the third component was
negative 2 plus 2 t squared,

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so when I take its
derivative, I get a 4t,

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so that is actually v of t, OK?

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And then the next
thing I asked you

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to do is determine the speed,
and the speed, of course,

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is just the length of the
velocity vector, right?

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So we just need to find the
length of v. Now, to do that,

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to remind ourselves
what we do for that,

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we actually take the
inner product of v

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with itself, the dot
product of v with itself,

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and then we take the
square root of that.

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So let's look at what
the dot product will be.

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Let me find the
squared thing first,

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and then I will take
the square root.

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So v dotted with v,
the first component

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I'm going to have negative
4t quantity squared,

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so that's going to
be 16 t squared.

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And then the second
component is going

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to be 2t quantity
squared, so I'm

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going to have plus 4 t squared.

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And the third component
is going to be

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another-- it's going to
be 4t quantity squared,

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so I get another 16 t squared.

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So when I add those up, I
believe I get 36 t squared?

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Yes, good.

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And so then, I just have to take
the square root of both sides

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to get what the speed
actually is instead

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of the square of the speed.

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So I get 6t, OK?

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So that's the velocity;
that's the speed.

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Now I need to find
the acceleration

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and I need to find the
unit tangent vector.

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OK, so let me see.

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I will come over here.

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Let me step off here and I
will find the acceleration

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and the unit tangent vector.

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So the acceleration,
if you remember,

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the acceleration is
actually just the derivative

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of the velocity
with respect to t.

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So the acceleration
is going to be

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the derivative of negative
4t is just negative 4.

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The derivative of 2t is just
2, and the derivative of 4t

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is just 4, all
with respect to t.

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OK.

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So the acceleration vector is
equal to negative 4 comma 2

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comma 4, so you
see this actually

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has constant acceleration.

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So at any point,
your acceleration

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is always this value,
so it's not surprising

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that our velocity is
increasing, and actually, it's

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increasing-- you'll notice,
each of these components

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is constant.

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The velocity, each of
the components is linear,

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and if we went back, we
look at the position vector,

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each of those
components is quadratic.

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And this is exactly
what you expect

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from just your
understanding of derivatives

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in single-variable calculus.

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If you start off with a constant
and you find an antiderivative,

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it's going to be linear, and
you find another antiderivative,

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you're going to
have a quadratic,

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so we shouldn't be
surprised by any of that.

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Now we have one more
thing to do with Part a,

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and that is to find the
unit tangent vector.

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And that's fairly easy,
because all we have to do

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is-- if you notice, we
have the velocity vector

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and we have its length.

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And so to find the
unit tangent vector,

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all we have to do
is take the velocity

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and divide it by its length,
and that will normalize it.

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That means that its length
will be one at that point,

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because you're taking a
vector, dividing by its length,

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so the length of the new vector
will have to be length one.

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So let me write that down.

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And actually, I guess the
point to remember here

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is that the velocity vector
is tangent to the path you're

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carving out, to the trajectory.

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OK, so this is a vector.

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This is a scalar.

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So I'm going to take
1 over 6t, and I'm

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going to multiply it
by negative 4t, 2t, 4t,

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and this gives me,
when I do my division,

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looks like it gives me a
negative 2/3, right, 1/3, 2/3.

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So that is the unit
tangent vector, OK?

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OK, and now, we have one
more point we want to make,

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and that is having
to do with the arc

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length of the trajectory.

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That was the second
part of this problem,

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was to find the arc length
of the trajectory from t

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equals 0 to t equals 2.

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So let me just draw
another line here.

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And what we want to do
there then is-- really

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what we want to do is we want
to integrate the speed, right?

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We want to integrate
the speed from 0 to 2.

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So this-- let me come over
here-- this absolute v,

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you might have also seen
it written as ds/dt, right?

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And so we want to
integrate this in dt--

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in t, sorry-- from 0 to 2.

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And so we come over here.

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We want to integrate
from 0 to 2, 6t dt.

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That's fairly easy.

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That's going to be 6 t squared
over 2, evaluated from 0 to 2.

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And so when I write
that down, I'm

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going to get 6 times 4
divided by 2, 24 divided by 2,

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I just get 12, and
the other term is 0.

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So the arc length
is-- of the trajectory

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from 0 to 2 is just 12 units.

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So this really was a purely
computational type of problem.

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All we were doing, if
you come back over here

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and you recall what
we were trying to do,

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is we started off with
a position vector.

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We just did a lot
of computation.

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We found the velocity, the
speed, the acceleration,

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the unit tangent
vector, and then

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we just wanted to find the
arc length of the trajectory.

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So this is all
very computational,

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but just to make
sure we understood

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what all the terms
meant and how they

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were related to one another.

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So I'll stop there.