Tangent line to the parametric curve


Uploaded by TheIntegralCALC on 03.07.2012

Transcript:
Hi everyone! Today we’re going to talk about how to calculate
the derivative of a parametric curve and then find the equation of the tangent line at a
particular point t. We’ll calculate three values to complete this problem. First, we’ll
find the x-coordinate at the given point t. Second, we’ll find the y-coordinate at the
given point t. Then finally, we’ll calculate the derivative of the parametric curve and
use it to find the slope m at the given point t.
Let’s get started.
And you’ll remember that a parametric curve can’t be defined by one function of y in
terms of x so in order to define it we have to have two equations, one for y and one for
x, they are both in terms of a third parameter t. And in this particular problem, we’ve
been given our two equations, one for y and one for x, and asked to find the derivative
of this parametric curve at the point t equals pie over two. So you remember from our work
with derivatives before that in order to find the derivative of a function at a point we
just need to find the equation of the tangent line at that point. So we can model the equation
of the tangent line with the point slope form of the equation for line which is y minus
y-one equals m, for slope, times x minus x-one or x sub one. And with this equation, we’re
going to need to find three things, we need to find y sub one and x sub one, which will
be the values that our x and y equations return to us when we plug in the point t equals pie
over two, and we need m, which we’ll need to find by taking the derivative of these
parametric equations. First, let’s find x sub one and y sub one
by plugging the point t equals pie over two into our equations for x and y. Before we
do that, let’s go ahead and simplify our x and y equations. We can simplify them somewhat
by distributing the three across this value here. So, we’ll get three t minus three
sine t for x and for y we’ll get three minus three cosine of t. That will make it a little
easier for us to plug t equals pie over two to these equations. So let’s say that if
we plug in t equals pie over two to our x equation, we’ll get three times pie over
two minus three sine of pie over two. Obviously, that will give us three pie over two out in
front, minus sine of pie over two which will be one, three times one is just three so we’ll
get minus three. And we can simplify this somewhat by finding a common denominator so
we’ll get three pie over two minus six over two which will give us three times pie minus
two all over two. And since this is the value we found when we plugged in pie over two to
our x equation, this will be equal to x sub one. Now we need to find y sub one and we’ll
do that by plugging pie over two in for t into our y equation. So we’re going to plug
that in here and we’ll get y equals three minus three cosine of pie over two. When we
evaluate cosine of pie over two we get zero so we’ll get three minus three times zero
or three minus zero which of course is just three. And we know now that three is equal
to y sub one. So now that we have these two points, we can go ahead and plug them in to
our points slope form of the equation of the tangent line. So we’ll get y minus three
is equal to m times x minus three times pie minus two all over two.
And now, all we need to find is m. So, in order to find m we need to take the derivative
of our parametric curve. We have two equations for our parametric curve so we’re going
to need a formula to take the derivative of these two equations at one time. And the way
that we’re going to write out our formula is this, the derivative of dy over dx or the
derivative of our function will be equal to the derivative of our equation for y with
respect to t divided by the derivative of our equation for x with to t, so dy over dx
is equal to dy/dt over dx/dt. So you’ll notice that we’ll need to calculate the
derivative of y with respect to t and plug that in to our numerator and then calculate
the derivative of x with respect to t and plug that in to our denominator. So let’s
go ahead and do that. We’ll take the derivative of our equation for y and again we’re looking
now at this equation that we simplified. This is the equation for y and we want to take
the derivative of that with respect to t. So the derivative of three, that first term,
will just be zero, and then the derivative of negative three cosine of t will be positive
three sine t, it’s actually negative three sine t but because we have that negative already
there it turns into a positive. And now that we have the derivative of our y function we
can go ahead and take the derivative of x. For x, we’ll be looking at this equation
here that we simplified, three t minus three sine of t, and we’ll be taking the derivative
of that with respect to t. So, the derivative of three t with respect t to is just three
and then the derivative of negative three sine t is negative three cosine of t. When
we simplify this, we’ll see that we get three sine of t and then in our denominator
we can factor out a three and we’ll get one minus cosine of t. And you can see now
that we’ll get our threes to cancel and our simplified derivative equation will be
sine of t all over one minus cosine of t. Now, if we had just been asked to find generically
the derivative of this parametric curve, this here would be our answer, sine of t divided
by one minus cosine of t. But because we’ve been asked to find the derivative at a particular
point, we’ll want to plug in pie over two for t so that we can get something we can
plug in for m into our equation for the tangent line. So, we’ll go ahead and plug in pie
over two to our derivative function and we’ll get sine of pie over two divided by one minus
cosine of pie over two. We know that sine of pie over two is equal to one so we’ll
get one here in the numerator. We know that cosine of pie over two is equal to zero, one
minus zero just gives us one. So we have one over one and we can see that our final answer
here will be one, that’s what we’ll go ahead and plug in for m.
So plugging one in for m we see that we get y minus three equals, we’ll just go ahead
and leave out the one because it’s redundant and we’ll have x minus three times pie minus
two all over two. And, we can leave our equation in this form but because we’re so close
to slope intercept form why don’t we just go ahead and move that three over to the right
hand side and we’ll finish our equation here as y equals x minus three times pie minus
two all over two and then we’ll have plus three. I want to go ahead and say we’ll
say plus six over two, right, and then we can combine these terms because we have a
common denominator. Let’s go ahead and call this three pie minus six, so, this will turn
into y equals x minus three pie minus six plus six all over two. And, you can see that
the sixes will cancel here and we’ll be left with y equals x minus three pie over
two and that’s it, that’s our final answer. This is the equation of the tangent line at
the point t equals pie over two or the derivative at t equals pie… pie over two of the parametric
equation that’s modeled by these two original equations here for x and y.
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