MHB Challenge problem Find k if x=k is tangent to the curve y=x+√(2).e^[(x+y)/√(2)]

[Olinguito]

If the line $x=k$ is tangent to the curve
$$\large y\:=\:x+\sqrt2\,e^{\frac{x+y}{\sqrt2}}$$
what is the value of $k$?

 

[I like Serena]

If the line $x=k$ is tangent to the curve
$$\large y\:=\:x+\sqrt2\,e^{\frac{x+y}{\sqrt2}}$$
what is the value of $k$?

Hi Olinguito!

Here is my attempt.

Spoiler

Take the total derivative:
$$dy\:=\:dx+\sqrt2\,e^{\frac{x+y}{\sqrt2}}\cdot \frac{dx+dy}{\sqrt2}$$
Substitute $dx=0$ to find where we have a vertical tangent:
$$ dy=\sqrt2\,e^{\frac{x+y}{\sqrt2}}\cdot \frac{dy}{\sqrt2}=dy\cdot e^{\frac{x+y}{\sqrt2}} \quad\Rightarrow\quad
y=-x$$
Substitute in the original equation:
$$-x\:=\:x+\sqrt2\,e^{\frac{x+(-x)}{\sqrt2}} \quad\Rightarrow\quad x=-\frac 12\sqrt 2$$
Thus:
$$k=-\frac 12\sqrt 2$$

 

[Olinguito]

Great work! (Clapping) I have a different solution that does not involve calculus – again I’ll wait and see if anyone else wants to have a go.

 

[Olinguito]

My solution.

Spoiler

A rotation of 45° clockwise about the origin sends the point $(x,y)$ to the point $(X,Y)$, where
$$\begin{pmatrix}X \\ Y\end{pmatrix}\ =\ \begin{pmatrix}\frac1{\sqrt2} & \frac1{\sqrt2} \\ -\frac1{\sqrt2} & \frac1{\sqrt2}\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}\ =\ \begin{pmatrix}\frac{x+y}{\sqrt2} \\ \frac{-x+y}{\sqrt2}\end{pmatrix}.$$
Thus the rotation takes the curve
$$\frac{-x+y}{\sqrt2}\ =\ e^{\frac{x+y}{\sqrt2}}$$
to the curve
$$Y\ =\ e^X.$$
The 45° tangent to the latter curve is the line $Y=X+1$. Rotating this 45° counterclockwise about the origin gives
$$\frac{-x+y}{\sqrt2}\ =\ \frac{x+y}{\sqrt2}\,+\,1$$
$\implies\ x\ =\ -\dfrac1{\sqrt2}$.

 

[solakis1]

If the line $x=k$ is tangent to the curve
$$\large y\:=\:x+\sqrt2\,e^{\frac{x+y}{\sqrt2}}$$
what is the value of $k$?

Spoiler

Differentiate both sides w.r.t. y and we get:$$\frac{dy}{dy}= \frac{dx}{dy}+\sqrt2\,e^{\frac{x+y}{\sqrt2}}\frac{1}{\sqrt 2}(\frac{dx}{dy}+\frac{dy}{dy})$$

And now since x=k is per/lar to the x-axis and tangent to the curve we must have : dx/dy =0 and the above formula becomes:$$1=\,e^{\frac{x+y}{\sqrt2}}\Longrightarrow \frac{x+y}{\sqrt 2}=0\Longrightarrow x=-y$$ e.t.c,e.t.c