Solving Implicit Function: Tangent Point & Level Curve Equation

[Yankel]

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...

thanks !

 

[Opalg]

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...

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 $k$.

 

[Yankel]

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.

 

[Opalg]

Did I do it correctly ?

View attachment 2312 (Rock)

 

[RLBrown]

Attached is a graphic of n=12
View attachment 2317

Click Here