Derivatives of Functions in Parametric Form Video Lecture | Mathematics (Maths) Class 12 - JEE

so what we have here is X being defined
in terms of T and y being de taught
defined in terms of T and then if you
were to plot over all of the T values
you get a pretty cool plot just like
this so you know you try to equals zero
figure out what x and y are T is equal
to 1 figure out what x and y are and all
the other T's and then you get this
pretty cool looking graph but our the
goal in this video isn't just to
appreciate the coolness of graphs or
curves defined by parametric equations
but we actually want to do some calculus
and in particular we want to find the
derivative we want to find the
derivative of Y with respect to X the
derivative of Y with respect to X when T
when T is equal to negative one-third
and if you are inspired I encourage you
to pause and and try to solve this and
I'm about to do it with you if you
incase you you already data or you just
want me to alright so the key is is well
how do you find the derivative with
respect to X derivative of Y with
respect to X when they're both defined
in terms of T and the key realization is
the derivative of Y with respect to X
with respect to X is going to be equal
to is going to be equal to the
derivative of Y with respect to T over
the derivative of X with respect to T if
you were to view these differentials as
numbers well this would actually work
out that work out mathematically now it
gets a little bit non rigorous when you
start to do that but if you thought
about that it's an easy way of thinking
about why this actually might make sense
the derivative of something versus
something else is equal to the
derivative of Y with respect to T over X
with respect to T all right so how does
that help us well we can figure out the
derivative of X with respect to T and
derivative of Y with respect to T the
derivative of X with respect to T is
just going to be equal to let's see the
derivative of the outside with respect
to the inside it's going to be 2 sine
whoops
derivative of sine is cosine 2 cosine of
1 plus 3t times the derivative of the
inside with respect to T so that's going
the derivative of 1 is just zero
derivative of 3t with respect to T is 3
so times 3 that's the derivative of X
with respect to D I just use a chain
rule here derivative of the outside to
sine of something with respect to the
inside so derivative of this outside to
sine of something with respect to 1 plus
3t is that right over there and the
derivative of the inside with respect to
T is just our 3 now the derivative of Y
with respect to T is a little bit more
straightforward derivative of Y with
respect to T we just apply the power
rule here 3 times 2 is 6 T to the 3
minus 1 power 6 T squared so this is
going to be equal to 6t squared 6 T
squared over well we have the 2 times
the 3 so we have 6 times cosine of 1
plus 3t and then our 6 is canceled out
and we are left with we are left with T
squared over cosine of 1 plus 3t and if
we care when T is equal to negative 1/3
when T is equal to negative 1/3 this is
going to be equal to well this is going
to be equal to negative 1/3 squared
negative 1/3 squared over over over the
cosine of 1 plus 3 times negative 1/3 is
negative 1 so it's 1 plus negative 1 so
it's the cosine of 0 and the cosine of 0
is just going to be 1 so this is going
to be equal to positive positive 1/9 now
let's see if we can visualize what's
going on here
so let me draw a little table here so so
I'm going to plot I'm going to think
about t + x + y so t + x + y so when T
is equal to negative 1/3 well our X is
going to be this is going to be sine of
0 so our X is going to
zero and our Y is going to be what to
over or negative to over 27 so we're
talking about we are talking about the
point 0 comma negative 2 over 27 so that
is that point right over there so that's
the point where we're trying to find the
slope of the tangent line and it's
telling us that that slope is 1/9 that
slope is 1/9 so if we move
I guess one way to think about it is if
we move for one two three four and a
half we're going to move up half so if I
wanted to draw the tangent line right
there
it would look something like something
like that something something something
like that and see if we go one two three
four and a half is well yeah just like
that is pretty close so that's what we
just figured out we figured out that the
slope of the tangent line right at that
point is 1/9 so it's not only neat to
look at but I guess somewhat useful