Question

Could you please explain why we get slope of tangent to parabola = 2(y/x). References in the reply below.

Answer 1

Section 39, Lecture 16, 37:11.
http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/unit-2-applications-of-differentiation/part-c-mean-value-theorem-antiderivatives-and-differential-equations/session-39-introduction-to-differential-equations/

Problem: Find curves perpendicular to parabolas.
We derive the following O.D.E
\[\frac{dy}{dx}=\frac{-1}{2(y/x)}\]

How did we get 2(y/x)?

On unrelated note, is y=x^4 considered a parabola?

Answer 2

In session 40, clip 1, Prof. Jerison shows how differential equations can be used to solve the problem of finding the general equation of a curve with the property that the tangent at each point on the curve is twice as steep as a ray from the origin to that point. The slope of a line from the origin to a point on the curve is y/x (rise over run), so the slope of the tangent has to be 2(y/x) by the condition set in the problem. That's where we get 2(y/x). And of course a line perpendicular to that one has a slope that is the negative reciprocal of 2(y/x).

The graph of y=x^4 looks similar to a parabola, but it isn't a parabola. Many calculus courses include a unit on analytic geometry where you would get that information but this course is so crammed with other stuff that we don't get into the formal definitions of parabola and other curves. A parabola is the set of points equidistant from a fixed point called the focus and a line called the directrix. The graph of y=x^4 doesn't satisfy this condition.

Answer 3

Perhaps a few more words of explanation are in order. The formula used here for the slope of the tangent to a parabola, 2y/x, was derived from the conditions provided by the problem. We're aware, of course, that y=x^2 is a parabola and that the slope of a tangent to that parabola is simply 2x. But 2y/x is the same as 2x because y=x^2, which gives us, by substitution, 2x^2/x = 2x.

Answer 4

Referring to your first post first paragraph, 'In session 40, clip 1, Prof. Jerison shows ....". I don't quite understand. And then, as I am writing this reply, it dawned on me. I understand now that slope of rays are (y/x), and since parabola tangents are always twice of that, they are 2(y/x). But still, a few things are amiss.
1) if 2(y/x) is equivalent to 2x, substituting 2x in the ODE gives a different answer
\[\frac{dy}{dx}=\frac{-1}{2x}\]
\[y=-\frac{1}{2}ln(x)\]

2) What I find it hard to get my head around is that slopes are usually expressed in terms of x or for fixed points, numerical values. For example, for the curve y=f(x) and for a particular point p on the curve, slope will be a numerical value. For any point x, it is expressed as another function f'(x). Here, 2(y/x) is expressed in both y and x. This is a bit confusing.

3) We are lucky that parabolas has this property where its tangents are twice that of rays from origin. What about other curves which does not? say, how do we find curves perpendicular to sin(x)?

Answer 5

4) addendum to point (2). dy/dx can be defined implicitly, which will then express slope in terms of both x and y. But the "y" here comes from original equation. Now, we are deriving a new ODE which represents a different curve. Doesn't putting y there mix it up with the y from ODE?

Answer 6

Perhaps Creeksider shouldn't have made the added remark, which may have only served to add confusion. Let's step back for a moment to recall that we're talking about two different problems. The second one happens to use results from the first one, but it's a different problem.

The first problem asks us to find the class of equations for which the slope of the curve at any point is double the slope of a ray from the origin to that point. For any point on the curve, the y value will be whatever number is assigned by this unknown function to the x value, so we can express the coordinates of the point as (x, y). That means a ray from the origin has a slope equal to y/x. We want a curve that has twice as much slope, so the slope of the curve has to be 2y/x.

You're right in saying that it's unusual to have slope expressed in terms of x and y rather than only in x. We're forced to do that here, however, because of how the problem was stated -- and the point of working through this exercise is to show how a problem of this kind can be solved using separation of variables.

When I made my added remark about substitution, I didn't mean to suggest that we could use substitution to solve the problem. I merely wanted to point out that the solution of the first part of the problem makes sense in terms of what we already know about the derivative of a function that produces a parabola. We have to solve the problem as shown by Prof. Jerison, but after we solve it and learn that the solution is a parabola, we can see that it makes sense because when y=x^2 we have dy/dx = 2x = 2y/x. But this would not have anything to do with the second part of the problem, where we're trying to find functions that produce graphs perpendicular to parabolas. Those functions are completely different, so we get nonsense if we substitute x^2 for y.

To find curves perpendicular to y=sin x, we start by observing that the slope of that curve is dy/dx = cos x. A curve perpendicular to this would have a slope that is the negative reciprocal, or -1/(cos x) which is -sec x, so we would need an integral for that expression. We know the integral of sec^2 x is tan x, but at this point in the course we don't have the tools to attack the problem of integrating -sec x.

Answer 7

You are right about stepping back. I am treading in the region of over-analyzing this issue. Things get a bit clearer after I took a break and come back at it again.

So, things that make sense so far are
1) problem 1 is stated in such a way that we are forced to use 2(y/x) [checked]
2) problem 2 used result from problem 1 [checked]
3) I now get what you mean about 2x. When we verify the result, we get slope of parabola (2x). [checked]

Let me throw out all other things I have said. Forget them.

But still one thing remain. Suppose we do problem 2 without knowledge of problem 1. And let's simplify the problem by stating we just need to find curve perpendicular to one specific parabola y=x^2. The way I would handle this will be by noting that tangent is 2x.
Then, ode will become dy/dx= -1/2x which again fall back to my previous dilemma. It gives solution y = -(1/2) ln (x). It seems like we must somehow add y there to get correct answer.

Answer 8

Btw, I am willing to close this question as the original question has already been answered. But I will appreciate if creeksider could continue to discuss upon this thread. Thanks.