Solving Implicit Function: Tangent Point & Level Curve Equation

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Discussion Overview

The discussion revolves around finding the tangent point and level curve equation for the function \(f(x,y)=x+4y^{2}\) that tangents the curve \(y=\frac{8}{x}\) in the first quadrant. Participants are exploring the use of implicit functions in this context.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant seeks assistance in determining the tangent point and level curve equation, expressing confusion about the problem.
  • Another participant explains that a level curve of \(f(x,y)\) is defined by the equation \(x+4y^2=k\) and suggests differentiating implicitly to find the derivative \(y'\).
  • A participant provides their calculations for the implicit derivative as \(\frac{dy}{dx}=-\frac{1}{8y}\) and the immediate derivative as \(y'=-\frac{8}{x^{2}}\), leading to the values \(x=8\), \(y=1\), and \(k=12\).
  • Another participant confirms their understanding and asks if their calculations are correct.

Areas of Agreement / Disagreement

There is no consensus on the correctness of the calculations or the approach taken, as participants are still in the process of verifying each other's work and understanding.

Contextual Notes

The discussion includes assumptions about the derivatives and the conditions for tangency, which remain unresolved. The dependency on the definitions of the functions and the level curve is also noted.

Who May Find This Useful

Participants interested in implicit functions, calculus, and the analysis of curves may find this discussion relevant.

Yankel
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Hello all,

I need some help with this one, I do not have a clue how to even begin.

the level curve of

\[f(x,y)=x+4y^{2}\]

tangents the function

\[y=\frac{8}{x}\]

in a point at the first quarter. What is the tangent point, what is the equation of the level curve ?

This question need to involve implicit functions. I don't get it... :confused:

thanks !
 
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Yankel said:
Hello all,

I need some help with this one, I do not have a clue how to even begin.

the level curve of

\[f(x,y)=x+4y^{2}\]

tangents the function

\[y=\frac{8}{x}\]

in a point at the first quarter. What is the tangent point, what is the equation of the level curve ?

This question need to involve implicit functions. I don't get it... :confused:

thanks !
A level curve of $f(x,y)$ is the set of points at which $f(x,y)$ takes a constant value $k$ say. So start with the equation $x+4y^2=k$, and differentiate it implicitly to get an expression for $y'.$ Next, differentiate $y=8/x$ to get another expression for $y'$. If the two curves are tangent to each other, then they must have the same value for $y'$ at that point. So put the two expressions for $y'$ equal to each other and you will get an equation for the point $(x,y)$. Use that together with the equation $y=8/x$ to find $x$ and $y$. Finally, use the equation $x+4y^2=k$ to find $k$.
 
Thank you !

Did I do it correctly ?

the implicit derivative is:

\[\frac{dy}{dx}=-\frac{1}{8y}\]

The immediate derivative is:

\[y'=-\frac{8}{x^{2}}\]

leading to x=8, y=1 and k=12 ?

Thanks for your explanation, very helpful.
 

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