BAM! Mr. Tarrou. In this calculus lesson we are going to extend on our new
knowledge about how to find derivatives of parametric functions or functions
defined by parametric equations we just got done talking about how we determined
where parametric equations are increasing intervals where they're
increasing and decreasing and concave up and concave down we're gonna do two
examples in this video of well finding the equation of a tangent line at a
given point and for this first example or given point is negative 5 5 and we're
looking at a curve defined by the parametric equations of X is equal to T
squared minus 6 and Y is equal to T squared minus 4t ok and keeping all of
this in this idea or form of parametric equations we're going to find DX DT and
dy DT and then basically put dy DT over DX DT and get dy DX the derivative of Y
with respect to X so we have 4 DX DT we're going to bring the 2 down reduce
that power by 1 so we're looking at just a simple 2t and for a derivative of Y
with respect to T what we're going to have 2 t minus 4 now combining those two
individual derivatives of X and Y with respect to T and writing out our dy DT
over DX DT giving us dy DX we have 2t minus 4 over 2t
okay and of course these are all even values so we can reduce that and say
that we have t minus 2 over T now we want to find the equation of the tangent
line the line tangent to this curve defined by these parametric functions at
the point negative 5 5 well that's an XY coordinate so that means that that's a
problem if we have a derivative that is in respect to T's so we need to find out
at what value of time or T do we want to find the slope of this tangent line so
what we're going to do is we're going to put this value of negative 5 into the
equation for X we're going to put the value of positive 5 in for Y and solve
for T and see what happens okay now see we have two values of T right here we
have a T which is positive 1 a negative 1 well do I find the derivative dy/dx do
I find that at T equals 1 or T equals negative 1 that's why we have to plug
the value of in this particular example x equals negative 5 into the parametric
equation defining X and then 5 into the parametric equation that's going to give
us analyze our vertical movement our Y value if you will and we're going to use
the T that satisfies both conditions
okay so now we know that when X is equal to negative 5 that correlates with a T
value of positive on a negative 1 and for the Y value to equal positive 5 well
that gives us t values of that happens when T equals 5 and when T equals
negative 1 but of course you know we're talking about the curve
where both of these parametric's you know both of these parametric equations
are working together to create this one term or curve excuse me and that means
that we want to use the T value of negative one that satisfies both the
condition of X B and negative five and five Y being equal to positive five at
that same point in time so we want to find dy DX the derivative when T is
equal to negative one and that's going to come out to just simply plug in
negative one into our derivative we get
well that's going to be negative three over negative 1 which is positive three
so now we have a slope and we have a point of tangency of course we're going
to write our equation of the tangent line in just a regular rectangular form
at y equals well point-slope form y minus y sub 1 equals M times X minus X
sub 1 and our tangent line comes out to be Y minus the given Y value of 5 is
equal to our slope when T is equal to negative 1 which is 3 times y minus 5
oops that would be X minus a negative 5 and that this is a tangent line to the
curve at negative 5 5 defined by the parametric equations of X equals T
squared minus 6 and Y is equal to T squared minus 4t I got one more example
coming up right now and it's going to involve some trig functions Pam example
number two and the last one we're going to find all points if there are any of
horizontal and vertical tangency to the curve which is defined about the
parametric equations X is equal to cosine of theta and Y is equal to 2
times the sine of 2 theta and we're going to be looking for solutions where
theta is between 0 into pi well we're going to talk about
horizontal and vertical tangency points of tangency we need to be looking at the
derivative so again we're going to be looking at DX now and this time with
respect to theta instead of respect to T like the last time so we're looking at D
X D theta of course the coasts derivative of cosine is negative sine so
we're looking at negative sine theta we've got D Y D theta all right so the
derivative of sine is cosine and we're looking at two cosine of two theta and
the derivative of two theta chain rule with respect to theta is
equal to force - all right so setting up our dy/dx we have for cosine of two
theta over negative sine theta in it again it's kind of weird right to say
that we have a derivative with respect to X you know a change of rise over run
when your independent variable doesn't match that that the variable that we're
taking the derivative with respect to but that is our derivative of this curve
well where do you have horizontal tangent lines you have horizontal
tangent lines when the derivative is equal to zero and our derivative is a
rational function so when the numerator is equal to zero the entire slope here
or value of our function is gonna be equal to zero so we're looking at four
for the horizontal tangency points okay let's try this again
my handwriting isn't always neat okay so we're just solving a simple little
equation here we're going to buy both sides by four and we have the cosine of
two theta is equal to zero take the inverse cosine of both sides and we have
two theta is equal to the inverse cosine of zero which course is going to be well
PI over 2 in 3 PI over 2 but hopefully a member from precalculus that when we're
solving trigonometric equations with multiple angles our last step here is to
divide by 2 and we're looking for all possible solutions that are between 0 &
2 pi so we actually need to go 2 times around
that unit circle so that when we divide by 2 we have our well in this case
there'll be 4 angles all within that rotation of 0 to 2 pi so going to keep
going around so we have PI over 2 3 PI over 2 4 5 PI over 2 keep counting
around we have 5 6 7 PI over 2 on that unit circle now we're in divide
everything by 2 and get theta is equal to PI over 4 3 PI over 4 5 PI over 4 and
7 PI over 4 now the last step here is to take these values of you know these
angle measures PI over 4 3 PI over 4 5 PI over 4 and 7 PI over 4 and plug those
back into the original parametric equations we have here so we can find
again those points of tangency I'm just going to finish this last step and then
we'll I'll just reveal those solutions but we're going to give you the the 4 in
this case points of tangency where we have a horizontal tangent line that's
what I'm doing I'm plugging these 4 angle measures into x equals cosine of
theta and Y is equal to 2 sine of 2 theta now for our vertical tangency
points
we're kind of doing the same thing as the first example finding sort of that
and it wasn't asked to find the equations of the tangent lines but their
horizontal vertical lines we you know x equals number and y equals number in our
rectangular format okay so again our derivative is a
rational function of course our rational function is going to be undefined a
slope which is undefined our vertical tangent tangency points is going to be
negative we're going to be looking for negative sine of theta to be equal to
zero so we have the sine of theta equals zero theta is equal to the inverse sine
of zero and sine is equal to zero of course at zero and pi okay that's pretty
much it I'm gonna go ahead and say you know like I'm a normal sign off I'm is
true I am go to your homework and I'm gonna reveal those four points of
tangency for the horizontal and the two points of tangency for children' tangent
lines just so that you have those solutions again I miss true go to your
homework
you