In this section we want to find the tangent lines to the
parametric equations given by,
To do this let’s first recall how to find the tangent line
to at . Here the tangent line is given by,
Now, notice that if we could figure out how to get the
derivative from the parametric equations we could simply
reuse this formula since we will be able to use the parametric equations to
find the x and y coordinates of the point.
So, just for a second let’s suppose that we were able to
eliminate the parameter from the parametric form and write the parametric
equations in the form . Now, plug the parametric equations in for x and y. Yes, it seems silly to
eliminate the parameter, then immediately put it back in, but it’s what we need
to do in order to get our hands on the derivative. Doing this gives,
Now, differentiate with respect to t and notice that we’ll need to use the Chain Rule on the right
Let’s do another change in notation. We need to be careful with our derivatives
here. Derivatives of the lower case
function are with respect to t while
derivatives of upper case functions are with respect to x. So, to make sure that we
keep this straight let’s rewrite things as follows.
At this point we should remind ourselves just what we are
after. We needed a formula for or that is in terms of the parametric
formulas. Notice however that we can get
that from the above equation.
Notice as well that this will be a function of t and not x.
As an aside, notice that we could also get the following
formula with a similar derivation if we needed to,
Derivative for Parametric Equations
Why would we want to do this? Well, recall that in the arc length section of the Applications of Integral
section we actually needed this derivative on occasion.
So, let’s find a tangent line.
Example 1 Find
the tangent line(s) to the parametric curve given by
Note that there is apparently the potential for more than
one tangent line here! We will look
into this more after we’re done with the example.
The first thing that we should do is find the derivative
so we can get the slope of the tangent line.
At this point we’ve got a small problem. The derivative is in terms of t and all we’ve got is an x-y coordinate pair. The next step then is to determine the value(s)
of t which will give this
point. We find these by plugging the x and y values into the parametric equations and solving for t.
Any value of t
which appears in both lists will give the point. So, since there are two values of t that give the point we will in fact
get two tangent lines. That’s
definitely not something that happened back in Calculus I and we’re going to
need to look into this a little more.
However, before we do that let’s actually get the tangent lines.
t = 2
Since we already know the x and y-coordinates of
the point all that we need to do is find the slope of the tangent line.
The tangent line (at t
t = 2
Again, all we need is the slope.
The tangent line (at t
= 2) is then,
Now, let’s take a look at just how we could possibly get two
tangents lines at a point. This was
definitely not possible back in Calculus I where we first ran across tangent
A quick graph of the parametric curve will explain what is
going on here.
So, the parametric curve crosses itself! That explains how there can be more than one
tangent line. There is one tangent line
for each instance that the curve goes through the point.
The next topic that we need to discuss in this section is
that of horizontal and vertical tangents.
We can easily identify where these will occur (or at least the t’s that will give them) by looking at
the derivative formula.
Horizontal tangents will occur where the derivative is zero
and that means that we’ll get horizontal tangent at values of t for which we have,
for Parametric Equations
Vertical tangents will occur where the derivative is not
defined and so we’ll get vertical tangents at values of t for which we have,
Vertical Tangent for
Let’s take a quick look at an example of this.
Example 2 Determine
the x-y coordinates of the points
where the following parametric equations will have horizontal or vertical
We’ll first need the derivatives of the parametric
We’ll have horizontal tangents where,
Now, this is the value of t which gives the horizontal tangents and we were asked to find
the x-y coordinates of the
point. To get these we just need to
plug t into the parametric
equations. Therefore, the only
horizontal tangent will occur at the point (0,-9).
In this case we need to solve,
The two vertical tangents will occur at the points (2,-6)
For the sake of completeness and at least partial
verification here is the sketch of the parametric curve.
The final topic that we need to discuss in this section
really isn’t related to tangent lines, but does fit in nicely with the
derivation of the derivative that we needed to get the slope of the tangent
Before moving into the new topic let’s first remind
ourselves of the formula for the first derivative and in the process rewrite it
Written in this way we can see that the formula actually
tells us how to differentiate a function y
(as a function of t) with respect to x (when x is also a function of t)
when we are using parametric equations.
Now let’s move onto the final topic of this section. We would also like to know how to get the
second derivative of y with respect
Getting a formula for this is fairly simple if we remember
the rewritten formula for the first derivative above.
Second Derivative for
Let’s work a quick example.
Example 3 Find
the second derivative for the following set of parametric equations.
This is the set of parametric equations that we used in
the first example and so we already have the following computations
We will first need the following,
The second derivative is then,
So, why would we want the second derivative? Well, recall from your Calculus I class that
with the second derivative we can determine where a curve is concave up and
concave down. We could do the same thing
with parametric equations if we wanted to.