Equation of normal line | Derivative applications | Differential Calculus | Khan Academy


So we have the
function f of x is equal to e to the
x over x squared. And what I want to
do in this video is find the equation,
not of the tangent line, but the equation of the normal
line, when x is equal to 1. So we care about the
equation of the normal line. So I encourage you to pause this
video and try this on your own. And if you need a little bit
of a hint, the hint I will give you is, is that the
slope of a normal line is going to be the
negative reciprocal of the slope of
the tangent line. If you imagine a
curve like this, and we want to find a
tangent line at a point, it’s going to look
something like this. So the tangent line is
going to look like this. A normal line is perpendicular
to the tangent line. This is the tangent line. The normal line is going to
be perpendicular to that. It’s going to go just like that. And if this has a
slope of m, then this has a slope of the
negative reciprocal of m. So negative 1/m. So with that as a
little bit of a hint, I encourage you to find the
equation of the normal line to this curve, when x equals 1. So let’s find the slope
of the tangent line. And then we take the
negative reciprocal, we can find the slope
of the normal line. So to find the slope
of the tangent line, we just take the derivative here
and evaluate it at x equals 1. So f prime of x,
and actually, let me rewrite this a little bit. So f of x is equal to e to the
x times x to the negative 2. I like to rewrite it this
way, because I always forget the whole
quotient rule thing. I like the power
rule a lot more. And this allows me to
use the power rule. I’m sorry, not the power
rule, the product rule. So this allows me
to do the product rule instead of
the quotient rule. So the derivative
of this, f prime of x, is going to be the
derivative of e to the x. Which is just e to the x times
x to the negative 2, plus e to the x times the derivative
of x to the negative 2. Which is negative 2x to
the negative 3 power. I just used the power
rule right over here. So if I want to evaluate
when x is equal to 1, this is going to
be equal to– let me do that in that
yellow color like. I like switching colors. This is going to be
equal to, let’s see, this is going to be
e to the first power. Which is just e times
1 to the negative 2, which is just 1 plus e to the
first power, which is just e, times negative 2. 1 to the negative 3 is just 1. So e times negative 2. So let me write it this way. So minus 2e. And e minus 2e is just going
to be equal to negative e. So this right over here, this is
the slope of the tangent line. And so if we want the
slope of the normal, we just take the
negative reciprocal. So the negative reciprocal
of this is going to be, well the reciprocal
is 1 over negative e, but we want the
negative of that. So it’s going to be 1/e. This is going to be the
slope of the normal line. And then if we, and
our goal isn’t just to the slope of
the normal line, we want the equation
of the normal line. And we know the
equation of a line can be represented
as y is equal to mx plus b, where m is the slope. So we can say it’s going to be
y is equal to 1/e– remember, we’re doing the normal
line here– times x plus b. And to solve for b, we
just have to recognize that we know a point
that this goes through. This goes through
the point x equals 1. And when x equals 1, what is y? Well, y is e to the 1st
over 1, which is just e. So this goes to the
point 1 comma e. So we know that when x is
equal to 1, y is equal to e. And now we can just solve for b. So we get e is
equal to 1/e plus b. Or we could just subtract
1 over e from both sides, and we would get b is
equal to e minus 1/e. And we could obviously right
this as e squared minus 1/e if we want to
write it like that. But could just leave
it just like this. So the equation of
the normal line– so we deserve our drum
roll right over here– is going to be y is equal
to 1/e times x, plus b. And b, plus b, is all of this. So plus e minus 1/e. So that right there is our
equation of the normal line.

13 thoughts on “Equation of normal line | Derivative applications | Differential Calculus | Khan Academy

  1. i have question. for the eq.
    x^2+y^2=(2X^2+2y^2-x)^2
    @ point (1/2,0)? the y'=-1/0. what is the tangent and normal line? pls. reply sir. thanks☺.

  2. can someone tell me why do we need to find the normal and tangent lines? why do we need them, as in the purpose. Thank you.

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