Main content
Course: Differential Calculus > Unit 2
Lesson 11: Quotient ruleNormal to y=𝑒ˣ/x²
Sal finds the equation of the line normal to the curve y=eˣ/x² at the point (1,e). Created by Sal Khan.
Want to join the conversation?
Sal,
I understand the process on how to find the Normal Line, what I can't seem to find anywhere is WHY do I need to find it, what does this tell me?
Tangent is instantaneous speed...so what is the normal line?
Anne Zimzores(22 votes)
Knowing the normal at a point on a curve allows for 3D graphics like in games and movies. It also might be useful in physics on curved surfaces to determine the normal force something like a curved hill is exerting to counteract the force of gravity pulling you down.
The derivative/tangent line is like the slope of a hill or mountain at a certain point, the normal line is like someone sticking a flag down at that point perpendicular to the ground and seeing which way the flag is pointing.(35 votes)
- Should there be parenthesis around the e-1/e in the final equation?(12 votes)
The parentheses are not needed here since the order of operations tells you which expressions to evaluate first. For this example, multiplication and division are first; addition and subtraction would be next. The full order of operations rule is to evaluate in this order: Parentheses, Exponents, Multiplication, Division, Addition, Subtraction. Some people find it helpful to remember order of operations using the term PEMDAS, which is the first letter of the order of operations above. So P stands for Parentheses, E for Exponents, etc.
If we follow the order of operations for the equation, we see that the equation evaluates the same way regarless of whether or not there are parentheses. For instance, let's imagine that x = 2 and let's evaluate e to 2.71. We would then have: y = (1/2.71)*2 + 2.71 - 1/2.71. We evaluate multiplication first: y = 2/2.71 + 2.71 - 1/2.71. Then we evaluate division: y = .738 + 2.71 - .369. Then addition: y = 3.448 - .369. Finally subtraction: y = 3.079. If there were parentheses around the expression (e - 1/e) then that section would be evaluated first: 2.71 - 1/2.71. Division would be first: 2.71 - .369. Then subtraction: 2.341. Then we would evaluate the rest of the expression beginning with multiplication, etc. But the answer would be the same.
That said, parentheses might help clear up how the final equation maps onto the equation of a line. So that m lines up with 1/e; x with x; and (e - 1/e) with b. But again, the parentheses are not necessary.(16 votes)
at2:58why is (e-2e) and not (e+2e) if you did (-2(1^-3)) would that be 2 because 1^-3 would be -1 and -2 times -1 would make a +2(15 votes)- Sal skipped some steps in his evaluation, but he is correct. You have 1^-3 = -1, which is where the error is in your reasoning. 1^-3 is the same as 1/1^3. 1/1^3 becomes 1/1 = 1 (not -1). Review the negative exponent videos if that seems tricky: http://www.khanacademy.org/math/arithmetic/exponents-radicals/negative-exponents-tutorial/v/negative-exponents
Here is more detail: e^x(-2x^-3) evaluates to -2e, not to +2e. First, let's substitute 1 for x. That yields: e^1(-2*1^-3), which we could rewrite as: e^1 * (-2/(1^3)). Now let's evaluate each step: e^1 = e and 1^3 = 1, so we have e * (-2/1). -2/1 evaluates to -2, so we have e * -2, which evaluates to -2e.(7 votes)
- So the derivative of e^x is e^x ? looking at the product rule(11 votes)
- One of the most important facts in all of calculus is that the derivative of e^x is e^x. You can't use the product rule or the power rule to get this result because x is in the exponent. We get the result through a longer chain of reasoning.(6 votes)
Maybe i missed the section about "normal line," but i wanted to know ...
1. what is normal line exactly. and 2. what does normal line tell us about the function?(8 votes)
The normal line is defined as:-1/f'(x)
It doesn't tell us much about the function. It is useful for other applications especially graphics where the normal is used to evaluate angles of reflection for 3D rendering.(8 votes)
- Why is the normal line perpendicular to the tangent line? Why would you ever need the normal line and where does it come from?(5 votes)
In this context, "normal" is a synonym of perpendicular or orthogonal. So, by definition of what a normal line is, it must be perpendicular.
Normal lines are quite useful in architecture, engineering and oftentimes in the natural sciences. Thus, we need robust math dealing with normal lines.(8 votes)
Hi, I, really, want to know that why the normal line or the perpendicular line on tangent line slope is defined as thenegative of a reciprocal slope of tangent line? I know why by plotting numbers on y-x-axis coordinate, but it doesn't convince me. And also, why do we need to know about a normal line?(5 votes)- I am not sure that that is how it is actually defined, though it is a correct statement. A better visualization of why might be to remember your geometry and draw a picture: Consider the triangles (0,0), (1,0), (1,m) and (0,0), (−m,0), (−m,1). Elementary geometry reveals immediately that both triangles are congruent, hence n=−1/m (remember, translation and rotation preserve angles!).
Among the uses of the normal line:
1) Suppose you have a point p=(x_0, y_0, z_0) on some plane, and a normal to the plane n=<a,b,c>, then the equation of the plane is a(x-x_0) + b(y-y_0) + c(z-z_0) = 0, Now you can tell if a given point is on the plane or not.
2) If you have a surface S, then the gradient vector at some point p on S is always orthogonal (or normal) to S at that point. Gradients are really important in later calculus (check out the book, "div, grad, curl and all that").
Everything that you get exposed to in math is setting you up for something further, it is all a building block to a greater structure, and these structures in turn become building blocks for even more concepts.
Here is an advanced video by Sal using the normal vector in 2 dimensions in preparation for understanding the concept of divergence of a vector field:
https://www.khanacademy.org/math/multivariable-calculus/line_integrals_topic/2d_divergence_theorem/v/constructing-a-unit-normal-vector-to-a-curve(3 votes)
- why b is same for both normal and tangent line ?(2 votes)
- The letter b is used to indicate the y-intercept. We would use the same letter in the general format of the equation of a line regardless of whether we're talking about the tangent line or the normal line. The actual value of the y-intercept would be different for these two lines (except in a problem where we take a tangent for a point where the curve intersects the y-axis), so although we might refer to b in the equations for both lines, we replace b with different values when we solve for the equations of the lines.(9 votes)
What if the tangent line's slope is 0? Then the equation and the slope of the normal line won't be defined?(2 votes)- If the slope is 0, it means the line is horizontal. It will take the form y=c where c is some number.
The perpendicular will have an undefined slope. But it can still be written as an equation because the line will be perfectly vertical. It will take the form x=c where c is some number.(3 votes)
At3:25, shouldn't (-e)^-1 be equal to -(1/e) instead of 1/e?(1 vote)
f'(x) = d/dx(e^x/x^2)
f'(x) = (d/dx(e^x)•x^2 - e^x•d/dx(x^2))/(x^2)^2
how do u do from the first to the second step?(1 vote)
2:58Video transcript
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.