Equations of Tangents and Normals using Differentiation Video Lecture | Mathematics (Maths) Class 12 - JEE

hi now in this video what I want to do
is show you how we go about finding the
equation of a tangent to a curve and
also how we go about finding the
equation of a normal to a curve and to
demonstrate this I've got this curve
here y equals some function of X and if
we take a point on the curve let's say
we take this point here and this point
has coordinates say x1 y1 now a tangent
to the curve at this point here is a
line that just touches the curve at that
point something like this so this is our
tangent and we're going to be wanting to
find the equation of the tangent and
also mention normals what is a normal
well a normal at this point is a line
that is perpendicular to the tangent at
that point so we're looking say at
something like this and obviously these
lines they carry on okay outwards but so
I've just shown part of these lines so
this line then is a normal we'll just
write that in there now when it comes to
working out the equations of these lines
we should be familiar with the fact that
any line has the form Y minus y1 equals
the gradients M multiplied by X minus x1
where x1 y1 represent a point on the
line and it will be this point here and
as I say M is the gradient now to get
the gradient for the tangent what we
need to do is just differentiate the
curve in other words we'll just put here
that M is equal to dy by DX or if you're
working with function notation it
is f dash of X now if we want the
gradient though at this particular point
we've got to let X equal X 1 and it will
give us that particular gradient then
now when it comes on to the equation of
the normal the normal is going to have
the form again y minus y1 equals the
gradient I'll just leave that blank for
a moment though multiplied by X minus x1
now what is the gradient well since
these two lines are perpendicular to one
another there's a relationship between
the gradients we should be familiar with
the fact that if you've got a gradient
say M the perpendicular gradient is
minus 1 over m ok the negative
reciprocal of this so that the product
of these two gradients thus if you
multiplied them together in other words
comes to negative 1 now what I want to
do is run through a numerical example
and I've got a question here where we've
got to find the equation of the tangent
and normal to the curve y equals x
squared plus 1 at the point where x
equals 3 so when I'm doing a question
like this it's always good idea to draw
a sketch of your curve and it just so
happens that the curve y equals x
squared plus 1 would be a parabola
passing through one on the y axis and
it's going to look something like this
ok so what I need to do first of all is
we'll differentiate the curve to get the
gradient at any point so we'll just
start off with the fact that we're given
y equals x squared plus 1 and if we
differentiate this with respect to X in
other words find dy by DX
then differentiating x squared gives us
2x and the constant goes to zero so this
gives us the gradient of the tangent to
the curve at any point X well I want to
work out what this point is we know that
the x coordinate x1 is going to be 3 so
I need to get the y coordinate the
corresponding y coordinate so what I'll
say is when x equals 3 we know that the
y coordinate will be 3 squared plus 1
if we substitute it into there and that
comes to 10 so we've got this point x1y1
now it's coordinates are 3 10
what we need is M the gradient of the
tangent in order to get the equation of
the tangent so what we can do is we'll
work out what the value of d-y-by-dx is
when X is 3 so which is put here and dy
by DX equals well it's 2 times X so it's
going to be 2 times 3 which is clearly 6
a positive gradient which is what we
would expect here so therefore we've got
the gradient of the tangent let's just
put it down here as a reminder gradient
of tangent okay will be equal to 6 and
it means that therefore the gradient of
the normal here if we apply the
perpendicular gradient rule that is the
product the gradient should come to
negative 1 all we had to do was just do
the negative reciprocal of this so in
other words minus 1 over 6 notice how
they multiplied together to give you
negative 1 so I've got everything I need
now to get the equation of the tangent
and the equation of the normal with the
equation of the
tangent I'll put an intro here rather
than just writing it down just gives the
reader some idea hopefully of what we're
doing so we'll just say the equation of
the tangent and I'll also put at the
particular point which is 3/10 and we'll
just say what it is and using this
format here it's going to be Y minus y1
which is 10 okay equals M the gradient
the gradients of the tangent is 6
multiplied by X minus x1 which is 3 and
you could leave it like that quite often
you get in a question which just says
find an equation for the tangent and
that would be sufficient you might want
to put it in the form y equals MX plus C
if you do then we just carry on and we
can expand the bracket and add 10 to
both sides so we get 6x we'd have minus
18 and then if I add the 10 to both
sides it's going to come out at minus 8
so that would be the equation of the
tangent if I want to put in the form y
equals MX plus C and it's always good to
look at it like this because again you
can just check to see whether it looks
sensible from my sketch here this isn't
quite sensible because okay we've got a
positive grade in here but the y
intercept would be at minus 8 so it
would be a lot steeper coming down and
intercepting the y axis here at negative
8
well let's go on and work out the
equation of the normal so we just put a
subtitle here again equation of the
normal
and again I'll just say that what
coordinates we're doing that at and
that's at 3:10 and what is it going to
be well again using this form here it's
going to be y minus y1 so it's still 10
equals the gradient which is now minus
1/6 multiplied by X minus x1 which is
the 3 now again I could leave it like
this if it said write down an equation
of the normal but maybe I want to write
it in the form ax plus B y plus C equals
0 where a B and C are integers if I'm
asked to do that it's always a good idea
to just multiply by your fraction here 6
in this example so what we'll get is 6y
and then minus 60 equals and if we
multiply this by 6 we're just going to
be left with minus 1 times the bracket
so that's going to give me minus X plus
3 now if I add X to both sides and
subtract 3 from both sides I'm going to
end up with X then plus the 6y and then
we've got minus 60 minus another 3 which
is going to be minus 63 and that equals
0 and that's in the form ax plus bu y
plus C equals 0 which sometimes you
asked to do so hope that's given you
some idea then on how we can go about
working out equations of tangents then
and normals to curves
you