Find the curvature and radius of curvature
of the vector function at t equals five pi over six.
The graph of r of t is shown below in blue,
the graph has an ellipse in the xy plane.
Let's review curvature and radius of curvature.
The curvature at a point measures how sharply
the curve bends or how quickly it changes direction.
The curvature of a line is zero,
the curvature of a circle is equal to
the reciprocal of the radius.
More formally, the curvature measures how quickly
the direction of the unit tangent vector changes,
with respect to a change in arc length s.
When we have a space curve in three dimensions,
we normally use this formula here
to determine the curvature.
In our case, because we have a plane curve,
we'll be using a simplified formula,
which we'll talk about in just a moment,
let's first look at an animation.
Here we have a function graphed in black,
in blue we will see the circle of curvature
as we animate a point along the curve.
If we stop here for a moment, remember,
the radius of the circle of curvature
and the curvature of the function
are reciprocals of one another,
so right here, the circle is fairly large,
which means your radius is fairly large,
resulting in a small curvature.
But as we approach this bend in the curve,
the circle of curvature gets smaller
and the curvature of the function gets larger.
So right here, the radius of the circle of curvature
is small, resulting in a much larger curvature.
Let's animate this point in the circle of curvature
along the curve.
Again, the curvature is equal to the reciprocal
of the radius of the blue circle,
which is the circle of curvature.
Going back to our example,
because we have a two dimensional vector function,
the z component is equal to zero,
so if we applied this formula for the curvature,
with a z component of zero,
the curvature formula simplifies
to the formula we see here,
so for this example, we'll be using the simplified
formula for the curvature,
would again only applies when we have a plane curve,
not when we have a space curve.
So for the given vector function x of t
is equal to seven cosine t,
and y of t is equal to three sine t.
Now we need to find the first and second derivatives
of x of t and y of t.
So x prime of t is equal to negative seven sine t,
x double prime of t is equal to negative seven cosine t,
and y prime of t is equal to three cosine t,
and y double prime of t is equal to negative three sine t,
and now we can determine the curvature as a function of t.
The curvature K is equal to,
in the numerator we have the absolute value
of x prime of t times y double prime of t,
which is negative seven sine t
times negative three sine t
minus x double prime of t times y prime of t,
which is negative seven cosine t
times three cosine t.
So we have the absolute value
of the difference of these two products.
In the denominator we have the quantity
x prime of t squared plus y prime of t squared
raised to the 3/2 power,
which gives us the square of negative seven sine t
plus the square of three cosine t,
and all this is raised to the 3/2 power,
and now let's begin simplifying.
In the numerator we have 21 sine squared t
minus negative 21 cosine squared t,
which is equivalent of plus 21 cosine squared t.
We have the absolute value of this sum.
In the denominator we have the quantity 49 sine squared t
plus nine cosine squared t,
and this is all raised to the 3/2 power.
Let's continue on the next slide.
Looking at the numerator, we have a common factor of 21.
We factor out the 21, we have 21 times the quantity
sine squared t plus cosine squared t.
You should recognize that sine squared t
plus cosine squared t is equal to one,
and therefore the numerator simplifies
to the absolute value of 21, which is 21,
and the denominator remains the same,
we have 49 sine squared t
plus nine times cosine squared t raised to the 3/2 power.
And now that we have the curvature as a function of t,
we need to determine the value at t equals five pi over six.
So the curvature at t equals five pi over six
is equal to 21 divided by the quantity
49 times sine squared five pi over six,
plus nine times cosine squared, five pi over six,
and all of this is raised to the 3/2 power.
We now need to find sine five of pi over six
and cosine five pi over six,
which we can find on the unit circle,
we'll using reference triangles.
Five pi over six would be
this angle here in standard position,
where the reference angle is
pi over six radian at 30 degrees,
and therefore we can label the opposite side one,
the hypotenuse two, and the long leg square root of three,
but because we're in the second quadrant
where x is negative,
this would be negative square root three,
so using the reference angle Theta,
we can determine sine five pi over six,
and cosine five pi over six.
Sine is equal to the ratio
of the opposite side of hypotenuse, which is 1/2,
cosine is equal to the ratio
of the adjacent side to the hypotenuse,
which is negative square root three over two.
Which means the curvature at five pi over six
is equal to 21 divided by the quantity
49 times 1/2 squared
plus nine times the square of
negative square root three divided by two,
and all this is raised to the 3/2 power.
Simplifying, we have 21
divided by the quantity, 1/2 squared is 1/4,
so we have 49/4 plus,
the square of negative square of three over two is 3/4,
nine times 3/4 is 27/4,
all this is raised to the 3/2 power,
and 49/4 plus 27/4
is equal to 76/4, which is equal to 19,
so the denominator simplifies to 19 raised to the 3/2 power.
So the curvature at five pi over six
is equal to 21 divided by 19 raised to the 3/2 power,
which is approximately equal to 0.25356.
Now if we go back to our question,
we're also asked to determine the radius of curvature,
and remember, the curvature is equal to the reciprocal
of a radius of the circle of curvature,
which means radius is equal to one over curvature.
So the radius of curvature, which I'll call r,
at five pi over six is equal to one over the curvature,
or the reciprocal of the curvature,
which is 19 to the 3/2 divided by 21,
which is approximately 3.94377.
Now if we go back to the first graph for a moment,
the circle of curvature for the ellipse
at t equals five pi over six is shown here,
and the radius of this circle,
which again is the circle of curvature,
is approximately 3.94,
and the curvature of the ellipse
at t equals five pi over six is approximately 0.25.
I hope you found this helpful.