Slope of Level Curve b at Point B: How to Determine It?

  • Context: MHB 
  • Thread starter Thread starter Yankel
  • Start date Start date
  • Tags Tags
    Functions Implicit
Click For Summary

Discussion Overview

The discussion revolves around determining the slope of a level curve of the function \(f(x,y)=x^{\alpha }y^{\beta }\) at a specific point, given that the slope of another level curve at a different point is known. The conversation includes exploration of implicit differentiation, the relationship between the slopes of different level curves, and the implications of constants in the function.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants clarify that level curves represent where \(f(x,y)\) is constant and suggest setting \(a=x^{\alpha}y^{\beta}\) to find the slope.
  • One participant derives the expression for the slope of the level curve using implicit differentiation, resulting in \(\frac{\partial y}{\partial x}=-\frac{\alpha x^{\alpha -1}}{\beta y^{\beta -1}}\).
  • Another participant questions how to find the coordinates of point A, given the presence of two unknown constants, and proposes a potential coordinate form.
  • There is a discussion about the implications of the known slope of -3 at point A and how it relates to the slope at point B.
  • Some participants express confusion about the relationship between the slopes of the level curves and the constants \(\alpha\) and \(\beta\), particularly whether the slope at point B must also be -3.
  • One participant corrects a previous differentiation error and notes that the slope expression is independent of \(x\), leading to further questions about the consistency of the slope across different points.
  • Another participant emphasizes that the slope of level curve \(b\) can only be determined with knowledge of the ratio of \(y\) to \(x\), indicating that the slopes are parallel along lines through the origin.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the relationship between the slopes of the level curves and the constants involved. There is no consensus on whether the slope at point B must equal -3, with some suggesting it could be 3 based on the book's answer, while others maintain it should be negative.

Contextual Notes

Participants note the complexity introduced by the constants \(\alpha\) and \(\beta\) and the implications of using implicit differentiation. The discussion highlights the need for additional information to resolve the relationship between the slopes definitively.

Yankel
Messages
390
Reaction score
0
Hello all,

In the attached photo, I have two level curves of the function:

\[f(x,y)=x^{\alpha }y^{\beta }\]

where alpha and beta are constants. In addition, I have the line

\[y=2x\]

It is known that the slope of the level curve a at the point A is -3. I need to find the slope of the level curve b at the point B.

Any ideas or hints ? Thanks ! :confused:

View attachment 2350
 

Attachments

  • Capture.PNG
    Capture.PNG
    1.7 KB · Views: 109
Physics news on Phys.org
Level curves are where $f(x,y)$ is a constant, right? So, for curve $a$, why not set $a=x^{\alpha}y^{\beta}$. Could you find $y'$?
 
yes... it is

\[\frac{\partial y}{\partial x}=-\frac{\alpha x^{\alpha -1}}{\beta y^{\beta -1}}\]

how do I proceed from here ?
 
Last edited:
So what could you say about point $A$? Could you find its coordinates, perhaps? You know it's on both the level curve and the line $y=2x$.
 
How can I find the coordinates of A if I have 2 unknown constants ?

I think that it should be A(a/2,a), but I am not sure, and don't know how to proceed even if it is true... :confused:
 
The final answer by the way should be -3, I really don't know how to get there.
(Sadface)
 
Yankel said:
How can I find the coordinates of A if I have 2 unknown constants ?

I think that it should be A(a/2,a), but I am not sure, and don't know how to proceed even if it is true... :confused:

For point $A$ you have the condition:
$$-3=-\frac{\alpha x^{\alpha-1}}{\beta (2x)^{\beta-1}}. $$
The target variable is $m$, where
$$-m=-\frac{\alpha x^{\alpha-1}}{\beta (2x)^{\beta-1}},$$
where the $x$'s here are not necessarily the same as the $x$'s above. So here, you see we've used the fact that $y=2x$. We have not used the fact that we're on level curves. That may or may not be helpful, as it introduces another unknown. Can you try to work out a ratio, where something cancels? Try simplifying the expression we've got.
 
this is a complicated one, isn't it ? (Whew)

Level curves will add another unknown in addition to alpha and beta, which are already a problem.

I tried simplifying your expression, there is not much I can do with it.

\[3=\frac{\alpha }{\beta }\cdot \frac{1}{2^{\beta -1}}\cdot x^{\alpha -\beta }\]

But if the x's are not the same x's (slightly confusing), how can it help me when I replace 3 for m ? Logic say that somehow I am supposed to use 3, otherwise I wouldn't be given it.
 
Yankel said:
yes... it is

\[\frac{\partial y}{\partial x}=-\frac{\alpha x^{\alpha -1}}{\beta y^{\beta -1}}\]

how do I proceed from here ?
You seem to have differentiated this wrongly. The level curve is $x^\alpha y^\beta = a$. If you use implicit differentiation on that equation, you get $(\alpha x^{\alpha-1})y^\beta + x^\alpha \bigl(\beta y^{\beta-1}\frac{dy}{dx}\bigr) = 0.$ Cancel $x^{\alpha-1}y^{\beta-1}$ and solve for $\frac{dy}{dx}$ to get $\dfrac{dy}{dx} = \dfrac{\alpha y}{\beta x}$. At each point on the line $y=2x$, that has the same value $\dfrac{2\alpha}{\beta}.$
 
  • #10
Yes, you are right ! (Blush) I did it all wrong.

I have calculated the derivative again, and I got it similar to yours, but with a minus before the expression (dy/dx = -(Fx/Fy)).

Still, I can't see how to progress from here. I mean, even if I do replace y for 2x, and by that x also goes, I still have alpha and beta. What I still miss here is how to use the information about -3 to find what I want.

The expression I got after replacing y=2x is independent of x. So if for level curve a, the slope is -3, shouldn't it be also -3 for b ? I don't get why it is 3 (unless the book answer is incorrect).
 
Last edited:
  • #11
Yankel said:
Yes, you are right ! (Blush) I did it all wrong.

I have calculated the derivative again, and I got it similar to yours, but with a minus before the expression (dy/dx = -(Fx/Fy)).

Still, I can't see how to progress from here. I mean, even if I do replace y for 2x, and by that x also goes, I still have alpha and beta. What I still miss here is how to use the information about -3 to find what I want.

The expression I got after replacing y=2x is independent of x. So if for level curve a, the slope is -3, shouldn't it be also -3 for b ? I don't get why it is 3 (unless the book answer is incorrect).
You're right, I left out a minus sign. It should be $\dfrac{dy}{dx} =- \dfrac{\alpha y}{\beta x}$. You said in comment #6 that the answer should be $-3$, so presumably that is what the book says. The slope of the level curve at $B$ is clearly negative, as you can see from the diagram. So the answer could not possibly be $3$.
 
  • #12
This was getting confusing, so I checked again, and the answer in the book is 3 and not -3, I was wrong in answer #6.

I do get what you say, it must be negative...

But just to understand, is it correct that it's -3 because the expression is not dependent on x, and thus at every point on y it will be -3 ?
 
  • #13
We can only determine the slope on the level curve $b$ when we know the ratio of $y$ to $x$, that is to say: which line it lies on.

For BOTH level curves, we have:

$\dfrac{dy}{dx} = -\dfrac{\alpha y}{\beta x}$

which depends on just two things: the constant $-\dfrac{\alpha}{\beta}$ and the ratio $\dfrac{y}{x}$.

Put another way, on any line through the origin, the tangent lines through the intersection points are parallel, which the drawing suggests as well.
 

Similar threads

MHB Level Curves and Partial Derivatives
  • · Replies 3 ·
Replies
3
Views
6K
Undergrad Partial derivatives of the function f(x,y)
  • · Replies 6 ·
Replies
6
Views
3K
MHB -apc.3.3.01 Find the equation of the curve that passes through the point (1,2)
  • · Replies 11 ·
Replies
11
Views
3K
Undergrad Gradient Vectors: Perpendicular to Level Curves
  • · Replies 4 ·
Replies
4
Views
2K
MHB Drawing Level Curves: Solving K=0 Problem
  • · Replies 2 ·
Replies
2
Views
2K
High School Confusion on Implicit Differentiation
  • · Replies 16 ·
Replies
16
Views
3K
High School How to Find the Intersection of a Logarithmic Curve and a Tangent Line?
  • · Replies 14 ·
Replies
14
Views
2K
Undergrad How to evaluate the enclosed area of this implicit curve?
  • · Replies 2 ·
Replies
2
Views
3K
MHB How Do I Find a Point on a Curve with a Specific Slope?
  • · Replies 2 ·
Replies
2
Views
1K
High School Tangents to a curve as functions of different variables
  • · Replies 5 ·
Replies
5
Views
2K