We're given a vector valued function r of t
and asked to find parametric equations
of a line tangent to r of t at the point
nine comma three comma negative two natural log two.
Let's begin by reviewing the parametric equations
of a line in space.
A line parallel to vector v
with components a comma b comma c
that passes through the point
x sub one, y sub one comma z sub one
is given by the following parametric equations.
So vector v is called the direction vector
where a, b, and c are the directional numbers.
Because we want our line to be tangent to the curve
given by the vector valued function,
we can use a tangent vector for our direction vector
to find our directional numbers a, b, and c.
But before we do this, let's begin by
determining the value of t that would give us
this point of tangency.
We want to determine what value of t
would make four t to the third plus five equal to nine,
which should be the same value of t
that makes t squared plus two t to the fourth
equal to three, which should be the same value of t
that makes negative two natural log two t
equal to negative two natural log two.
So looking at the z component,
notice how it's easy to see t would be equal to one.
But again to emphasize this,
let's go ahead and perform the substitution
and find r of one.
So r of one would be equal to an x component of
four times one to the the third plus five.
The y component would be
one to the second plus two times one to the fourth,
and the z component would be
negative two natural log two times one.
So again notice how this vector
does have the terminal point at the point of tangency,
which should be
nine comma three comma negative two natural log two.
This is important because once we find r prime of t,
we'll have to find r prime of one
to find the tangent vector at that location.
So now let's find r prime of t,
which will give us our tangent vector.
So for r prime of t, the x component would be
the derivative of four to the third plus five.
With respect to t, that would be 12 t squared.
The y component would be the derivative of
t squared plus two t to the fourth.
That would be two t plus eight t to the third.
The z component would be the derivative of
negative two natural log two t.
We could use a log property to write this as a sum,
but let's go ahead and find the derivative
using the chain rule.
So we'd have negative two times one over two t
times the derivative of two t, which would be two.
Notice how this does simplify to just
negative two divided by t.
So again, r prime of t would have
an x component of 12 t squared,
a y component of two t plus eight t to the third,
and a z component of negative two divided by t.
And because we know the point of tangency occurs
when t equals one, to find our tangent vector,
we'll now determine the vector r prime of one.
So r prime of one, again, will be our direction vector
for our tangent line.
So we'd have an x component of 12 times one squared,
a y component of two times one
plus eight times one to the third,
and a z component of negative two divided by one.
So our tangent vector r prime of one
has an x component of 12,
a y component of ten,
and a z component of negative two.
Now we have all the information we need.
Again the point of tangency will give us
x sub one, y sub one, and z sub one,
which we know is nine, three,
and negative two natural log two.
And the components of the tangent vector
give us our directional numbers,
which are a, b, and c.
So a is 12, b is 10, and c is two.
Of course these are not unique.
We could use any scalar multiple of this vector
for our directional numbers.
So the parametric equation of the tangent line
would be x of t equals nine
plus 12 times t, or 12 t.
Y of t would be equal to three
plus 10 times t,
and z of t would be equal to
negative two natural log two
minus two t.
Remember parametric equations of a line are not unique,
but these are probably the most obvious parametric equations
for our tangent line.
Let's go ahead and verify this graphically.
So looking at our graph, this black curve is given by
r of t, our vectored valued function.
That point of tangency is this blue point here,
and the red tangent line is graphed here in red.
Notice how this does verify our answer is correct,
and this purple vector here is the tangent vector
which we used for our directional vector
or for our directional numbers.
I hope you found this helpful.