Welcome to a video that will show
how to determine the curvature of a curve
defined by a vector value function.
The curvature at a point measures how sharply
the curve bends or how quickly it changes direction.
And just for reference, the curvature of a line
would be zero since a line does not bend
and the curvature of a circle would be equal to
one over r or the reciprocal of the radius.
So we think of a small circle.
It turns or changes direction very quickly
compared to a very large circle.
So the reciprocal of a small radius
would be very large and the reciprocal
of a large radius would be very small.
More formally, the curvature measures
how quickly the direction of the unit tangent vector
changes with respect to a the change in arc length.
So the curvature is equal to the magnitude
of dT ds expressed in either way here.
Notice this requires that the unit tangent vector function
be defined in terms of s or arc length.
Before we take a look at the formula
that we're going to use in this video,
let's take a look at curvature in action.
What we're seeing here in this video
are the unit tangent vector, unit normal vector,
and this blue circle represents the radius
which would be used to determine the curvature.
Remember the curvature would be one divided by
the given radius of the circle.
So if we approach the bend in this curve,
we expect the curvature to increase
or the radius of the circle to decrease as we see here.
Again, remember the curvature would be one over r
where r is the radius of this blue circle.
Again as we approach this next bend or curve,
the radius decreases, increasing the curvature and so on.
Let's go and take a look at one in 3D.
Here the unit tangent vector is the red vector.
The unit normal vector is the blue vector,
and this light blue circle would be used
to determine the radius where the curvature
would be one over r.
Again, as we approach the bend,
you'll notice that the radius of the circle decreases,
and as the curve straightens out, the radius increases.
Let's go and take a look at the formula
that we're going to use to determine
the curvature of a curve.
So again here we're measuring the change
in the unit tangent vector with respects to arc length
and then determining the magnitude of that.
Using the chain rule, we could perform substitution
to come up with dT dt divided by ds dt.
Remember dT dt would be the derivative
of the unit tangent vector value function.
Now ds dt would be the velocity vector value function.
And remember that v of t is equal r prime of t.
That's where this comes from.
There's also another formula that we can use
to determine curvature as we see down here below.
This alternative formula is helpful,
because remember that sometimes determining the derivative
of the unit tangent vector value function can be very messy
and if that's the case, you might want to
use this alternative formula.
We'll do one using this formula
and we'll do another example using the second formula.
Let's determine the curvature of r of t
equals two cosine t comma two sine of t.
In order to use this formula here,
we first need to find the unit tangent vector value function
or T of t.
So derivative of two cosine t is negative two sine t.
And the derivative of two sine t would be two cosine t.
We go ahead and divide this by the magnitude.
So the square root of four sine squared t
plus four cosine squared t.
Let's go ahead and simplify this.
If we factor this radicand we'd have
four times sine squared plus cosine squared.
Well sign squared plus cosine squared is equal to one
and the square root of four would be two.
So we'd have two here.
We can also factor out a two
from this vector value function.
We'd have two times negative sine t cosine t.
This simplifies nicely so T of t
is equal to an x component of negative sine t
and a y component of cosine t.
Let's go and find the derivative
of this vector value function.
The derivative of negative sine t
is going to be negative cosine t
and the derivative of cosine t is negative sine t.
Okay, we're almost there.
Looking back to the curvature formula,
we need to determine the magnitude of T prime of t
and then we also need to determine the magnitude
of our prime of T which is here.
So the magnitude of T prime of t
is equal to the square root of
this would be cosine squared t
plus this would be sine squared t.
Well that's going to be equal to
the square root of one which is equal to one.
And now we need the magnitude of our prime of T
which is going to be similar.
We have the square root of,
well negative two sine t squared
is going to be four sine squared t
and two cosine t squared is going to be
four cosine squared t.
Well if we factored out the four,
we'd have four times sine squared plus cosine squared.
That's going to give us one.
So we have the square root of four times one.
The square root of four which is equal to two.
So the curvature is equal to one divided by two or one half.
And going back up to our vector value function
where the x component is two cosine t
and y is equal to two sine t,
if we converted that to a rectangular equation,
it's going to be x squared plus y squared equals four.
So we have a circle with a radius of two.
Remember if the circle has a radius two,
the curvature would be the reciprocal or one over two.
And this just confirms what we said before
that the curvature of a circle
is the reciprocal of the radius.
Let's go and take a look at another example.
So before we decide on a formula,
we're going to go ahead and try to determine
the unit tangent vector value function
and then see how difficult it is
to determine the derivative of that.
So T of t would be equal to
the derivative of r of t divided by its magnitude,
so we'll have one for the x component.
Derivative of t squared divided by two would be t,
and the derivative of t cubed
divided by t would be t squared.
Divided by the magnitude which would be
the square root of one squared or one
plus t squared plus t squared squared or t to the fourth.
Now looking at this vector value function,
it would be pretty messy to determine the derivative
of this vector value function.
So in this problem, let's go ahead
and use the alternative formula for curvature,
given by this formula here.
So to help us out here, remember this was
r prime of t divided by the magnitude of r prime of t.
And the reason I mention that is
we're going to need r prime of t
cross with r double prime of t.
Let's go ahead and write that over here.
R prime of t has components one, t, and t squared,
and r double prime of t would be zero, one, and two t.
So to use this second formula,
what we're going to have to do
is cross r prime of t and r double prime of t
and then find the magnitude and then divide that by
the magnitude of r prime of t raised to the third power.
Remember this cross product would be
the vector formed by the three by three determinate
where the first row is i, j, k,
the second row would be one, t, t squared,
the third row would be zero, one, and two t.
So this first two by two determinate will be
formed by eliminating row one and column one
so we'll have t, t squared, one, and two t.
Here we're going to eliminate row one, column two
so we'll have one, t squared, zero, two t.
And here we'll eliminate row one, column three.
We'll have one, t, zero, one.
So the result would be two t squared
minus one t squared, that's t squared.
Here we're going to have two t minus zero,
but there's a minus sign here.
It's going to be negative two t for the y component
and here we're going to have one minus zero
so the z component will be one.
Okay, now that we have the cross product,
we need to find the magnitude of this
and that's going to be our numerator.
And then we need to find the magnitude of our prime of t
and raise it to the third power
for our denominator of the curvature.
So let's go ahead and take that information
onto the next slide to finish this up.
Okay, so here's our formula.
Here's the cross product that we just found.
Here's our prime of t.
Let's go ahead and find the magnitudes now.
So t squared to the second will be t to the fourth.
Negative two t squared will be positive four t squared
plus one squared or plus one.
And then for the magnitude of our prime of t,
we're going to have one squared or one
plus t to the second plus t to the second squared,
that'll be t to the fourth.
To finish this up, we know we want the curvature
when t is equal to one.
So our numerator's going to be
the square root of this when t is equal to one
so we'll have one to the fourth
plus four times one plus one.
That's going to be the square root of six.
Remember our denominator here is cubed,
so the square root of one plus one squared
plus one the fourth, that'll be three.
It's going to be the square root of six.
Now the square root of three to the third
is going to be three square root three.
Again we can simplify this.
Remember this is the square root
of two times the square root of three.
All over three square root three, this simplifies out.
So our curvature is going to be equal to
square root of two over three.
And it's probably good to get a decimal
approximation for that.
It's approximately .47
And that's going to have to do it for this video.
I hope you found these two examples helpful.