Welcome to a video
that will show how to determine the parametric equations
of a tangent line to a curve
that is defined by a vector valued function.
Let's start off by reviewing the parametric equations
of a line in R three.
A line that is parallel to the vector v
that has components a, b, c,
that passes through the point
x sub one, y sub one, z sub one,
is given by these three parametric equations.
So, what I want you to notice here
is that the vector parallel to the line
with components a, b, c,
give us the direction numbers
for the line given here as the coefficients of t.
And then the point on the line, as we see here,
would be the point of tangency,
having coordinates x sub one, y sub one, and z sub one.
So, in order to determine the parametric equations
of a tangent line to a curve,
we'll have to determine two things.
We'll first determine the point of tangency,
x one, y one, z one,
and then we'll determine a unit tangent vector,
or at least a tangent vector,
and the components of that tangent vector
will give us a, b, and c.
And again, one more thing to review.
In order to find a unit tangent vector,
we need to find the derivative
of the given vector valued function
and then divide by the magnitude.
Again, we don't have to find the unit tangent vector.
A tangent vector will do,
but it is common to determine the components
of the unit tangent vector instead of just a tangent vector.
Let's go and look at an example.
When t is equal to two, we want to determine a set
of parametric equations of the tangent line
to the given plane curve.
Let's first determine the point of tangency
when t is equal to two.
And that would be the terminal point of the vector produced
by this vector valued function when t is equal to two.
So, the x component would be two squared,
and the y component would be three times two.
So, the vector would be four, comma, six.
Which means the point of tangency would be four, six.
We now need to determine the direction numbers
for this tangent line,
and that'll come from the component form
of the unit tangent vector.
So, let's find the unit tangent vector valued function
and then evaluate it when t is equal to two.
So, the derivative of r would be two t, comma, three,
and the magnitude of this would be the square root of...
two t squared
plus three squared.
So, we would have two t, comma, three,
all over the square root of four t squared
plus nine.
Again, this will give us the unit tangent vector.
We could just use our prime of t to find a tangent vector
that's not a unit tangent vector.
So, it's going to evaluate this at t equals two.
So we'd have two times two, that's four.
The y component would be three, and we'll divide
by the square root of four times two squared,
that's going to be 16 plus nine, square of 25 would be five.
So, the unit tangent vector at t equals two
would have an x component of 4/5 and a y component of 3/5.
So now, using the point of tangency
and the components of the unit tangent vector,
we can determine the parametric equations
of the tangent line.
We would have x of t
is equal to four
plus 4/5 times t,
and y of t
would be equal to six
plus 3/5 t.
Now remember, parametric equations are not unique.
What I mean by that is,
notice that we use the direction numbers 4/5 and 3/5
because these came from the unit tangent vector.
While the vector with components four, three
would be a tangent vector, not just a unit tangent vector.
So, we could also have said, x of t is equal to four
plus four t,
and y of t
could equal six
plus three t.
These equations would give the same tangent line.
Let's go and take a look at that line.
In blue we see the graph of the plane curve,
and in red we see the tangent line.
What we're going to see now is the unit tangent vector
approach the value t equals two.
It's this purple vector here with magnitude one,
and notice that when t is equal to two,
that unit tangent vector is right on the tangent line,
verifying that the components of this unit tangent vector
would give us the direction numbers for that line.
Let's go and take a look at one more example.
Here, when t is equal to pi over four,
we want to determine the set of parametric equations
of the tangent line
defined by the given vector valued function.
Let's begin by finding the point of tangency
by determining the vector produced
when t is equal to pi over four.
Here's the 45, 45, mini right triangle
that we'll use to evaluate these trig functions.
The cosine of pi over four
is going to be one over square root two.
Sine of pi over four would be the same.
And the z component is pi over four.
So, the x, y, and z components of this vector
give us the point of tangency.
Of course, we could rationalize these,
but I'm going to go ahead and leave it in this form.
Now, let's go and determine
the unit tangent vector valued function.
So, we're going to find the derivative of r
and then divide by its magnitude.
So, the derivative of cosine would be negative sine t.
The derivative of sine t would be cosine t.
The derivative of t would be one.
We divide by the square root of...
Negative sine t square would be positive sine squared t.
Plus cosine t square would be cosine square t.
And then one squared would be one.
Remember that sine squared plus cosine squared
is equal to one, so we can simplify this.
It would be negative sine t,
cosine t,
one,
and the magnitude would be the square root of one plus one
or the square root of two.
We now need to evaluate this at t equals pi over four.
So, instead of having this denominator here,
I'm going to go ahead and write it as one over square root two
times the components of this vector,
so negative sine, pi over four would be negative one
over square root two.
Cosine pi over four is positive one over square root two,
and then the z component stays one.
Now, let's go ahead and perform
the scalar multiplication here.
So, we're going to have negative one over two or negative 1/2.
For the y component we'll have positive 1/2.
And the z component will be positive one
over square root two.
Now that we have the point of tangency
and the components of the unit tangent vector,
these will give us our direction numbers.
We can now write the parametric equations
for the tangent line.
X of t would be equal to one over square root two
minus 1/2 t.
Y of t would be equal to one over square root two
plus 1/2 t.
And z of t
would be equal to pi over four
plus one over square root two times t.
And as mentioned before,
these parametric equations are not unique.
Let's go and take a look at this graphically as well.
So, what we're seeing here in blue is the space curve.
In red we have the tangent line,
and again notice that the unit tangent vector here in purple
coincides with the line, verifying that the components
of this vector do give us the directional numbers
of the red line.
Okay, hope you found these examples helpful.
Thank you for watching.