Calculus, all content (2017 edition)
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?
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.(33 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:
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 the
negative 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:
- 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)
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.