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

In summary, the conversation discusses finding the slope of a level curve at a given point. The function $f(x,y)=x^{\alpha}y^{\beta}$ is introduced, along with the line $y=2x$. The slope of the level curve a at point A is given as -3 and the goal is to find the slope of the level curve b at point B. The conversation goes into finding the coordinates of point A and using implicit differentiation to calculate the derivative $\frac{dy}{dx}$. The final answer is determined to be -3, but there is confusion regarding the negative sign in the derivative and the discrepancy between the book's answer and the expected answer.
  • #1
Yankel
395
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:

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  • #2
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 $y'$?
 
  • #3
yes... it is

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

how do I proceed from here ?
 
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  • #4
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 $y=2x$.
 
  • #5
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:
 
  • #6
The final answer by the way should be -3, I really don't know how to get there.
(Sadface)
 
  • #7
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.
 
  • #8
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.
 
  • #9
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).
 
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  • #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.
 

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

What is an implicit function problem?

An implicit function problem is a type of mathematical problem where the relationship between two variables is given in an implicit form, rather than in an explicit form. This means that the relationship between the variables cannot be easily solved for one variable in terms of the other.

How is an implicit function problem different from an explicit function problem?

An explicit function problem is one in which the relationship between the variables is given in an explicit form, meaning that one variable can be solved for in terms of the other. In contrast, an implicit function problem does not have a clear solution for one variable in terms of the other and often requires more complex methods to solve.

What are some common techniques used to solve implicit function problems?

Some common techniques used to solve implicit function problems include implicit differentiation, the implicit function theorem, and the method of Lagrange multipliers. These methods involve manipulating the equations of the problem to isolate one variable and solve for it in terms of the other.

What are some real-world applications of implicit function problems?

Implicit function problems have many real-world applications in fields such as physics, engineering, economics, and computer science. For example, they can be used to model the relationship between variables in a physical system or to optimize a system for maximum efficiency.

How can I improve my skills in solving implicit function problems?

To improve your skills in solving implicit function problems, it is important to have a strong foundation in algebra, calculus, and other related mathematical concepts. Additionally, practicing with a variety of problems, seeking help from a tutor or teacher, and studying different solution techniques can also be beneficial.

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