Question

Find the point of the graph y=e^(-x) where the normal line to the curve passes through the origin. They told me to either use newtons method or the calculator but I would prefer if I could learn how to use newtons method for this.

Answer 1

(1) Have you considered sketching the graph of y=e^(-x) and that of the tangent and normal lines to the graph such that the normal line passes through the origin?
(2) How would you find the slope of the tangent line to the graph?
(3) How would you find the slope of the normal line to the graph, with the understanding that this normal line passes through the origin?
(4) How would you identify the point of tangency of the tangent line to the graph? The normal line will go through this point, as well as through the origin.
(5) How would you use the basic definition of the slope of a line to find an algebraic expression for the slope of the normal line?
(6) Once you have these two expressions for the slope of the normal line, equate them to each other. (Why would you want to do that?)
(7) For reference: I've found that the point in question is (0.43630, e^(-0.43630)).
Good luck!

Answer 2

Using Newton's Method to derive this result is the only efficient way of doing it, so yes, you're on the right track.

Answer 3

Ok I know how to find the equation of a line tangent to the graph, and I have sketched the graph. The problem I am having is how to use newtons method to find which tangent line crosses the origin.

Answer 4

OK. In that case, please answer the following:
(1) what mathematical expression represents the slope of the normal line to the curve obtained by using the derivative?
(2) What math expression represents the slope of the normal line obtained by using the basic formula for slope of a straight line?
(3) Take the results of (1) and (2) and set them equal to each other.

I'll be glad to help as much as I can, but have very limited time right now. Good luck.

Answer 5

Excuse the correction, but you're trying to determine the point on the graph where the NORMAL line passes through the origin as well as through that point on the graph.

Answer 6

ah yes I see it, ok so we are trying to find the normal line. so essentially by using newtons method, we have to find the roots? or the zeros right?

Answer 7

Yes, Newton's Method is for finding the roots/zeros of a function. First, we have to derive a suitable function. That's why I've asked you those several questions 3-4 minutes ago. Care to try answering them?

Answer 8

(1) Have you considered sketching the graph of y=e^(-x) and that of the tangent and normal lines to the graph such that the normal line passes through the origin?
Graphed it but cant draw it here
(2) How would you find the slope of the tangent line to the graph?
Solve the derivative of the function then depending on the point you would convert it to the slope formula
(3) How would you find the slope of the normal line to the graph, with the understanding that this normal line passes through the origin?
The slope of the normal line is perpendicular to the tangent line
(4) How would you identify the point of tangency of the tangent line to the graph? The normal line will go through this point, as well as through the origin.
Cant answer this
(5) How would you use the basic definition of the slope of a line to find an algebraic expression for the slope of the normal line?
y-n=k(x-h)
(6) Once you have these two expressions for the slope of the normal line, equate them to each other. (Why would you want to do that?)
Not sure

Answer 9

I don't know how to quote your message (as you have done), so will use numbers to identify the sentence I'm discussing.
(2) the slope of the tangent line is simply the derivative of y = e^(-1).
(3) the tangent line and normal line are perpendicular to one another (yes). But my question was: How do you find an algebraic expression for the slope of the NORMAL line, if you already have an expression for the slope of the tangent line?
(4) we know that y = e^(-x), so the point of tangency is simply (x, e^(-x)).
(5) what you have typed here is the EQUATION of a straight line with known slope. Please try again: going back to the very first definition of "slope of a straight line" in algebra, derive an expression for the slope of the line connecting (0,) to (x, e^(-x)).
(6) You do want to equate these two different expressions for slope because that will give you the function whose roots/zeroes you'll find using Newton's Method.

Answer 10

3) y−f(x1)=−1f′(x1)(x−x1).
5) do you mean e^(-x)-0/x-0?
6) Im really not understanding this question

Answer 11

(3) here you're writing the EQUATION of a straight line, whereas I was looking for the derivative of y = e^(-x) (which represents the slope of the tangent line to that curve).
(5) Yes, indeed, I do mean that. Pls be sure to use parentheses to ensure that the calculations are carried out in the proper order: (e^(-x) - 0) / (x - 0). Please simplify this expression, and then label it as "slope of the normal line to the curve at (x,e^(-x))",
(6) We'll discuss this soon. Have you used Newton's Method before? If not, would you be willing to look it up in your textbook and to go through some of the examples there?

I am a teacher myself and now have a student with me, so it may be some time before I can write you again. Hope this helps with your planning.

Answer 12

3) oh ok, the derivative is simple -e^(-x)
6) No, this is my first time using Newton's Method. This is actually the first time hearing about newton's method as I haven't learned it in class yet. I was supposed to answer the questions using the calculator as my teacher wanted us to first learn how to use our calculators but I was interested in how newtons method worked. Thanks for the help though.

Answer 13

How far along are you in solving this problem? Newton's Method is a beautiful tool, but it does take some time to learn it through practice.
As for solving this problem using a calculator: I did come up with a formula that I evaluated again and again on my TI-84 until I ended up with the x value of 0.4263.