Find a tangent plane

Question

I am asked to find a tangent plane of $f(x,y) = e^{x\ln y}$ at the point (2,1).

When I ask wolfram alpha this, I am given the line $z=2y-1$.

I don't intuatively understand this, shouldn't there be a tangent plane and not a line at this point?

Answer 1

It's correct. The plane is perpendicular to the y,z axial plane. That is, for all values of $x$, $z=2y-1$

Ref: Wolfram Alpha's 3D Plot z=2y-1

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Let $(x_0,y_0)$ be any point of a surface function $z=f(x,y)$. Then the surface has a nonvertical tangent plane at $(x_0,y_0)$ with equation:

$$z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$$

In this case $f(x,y) = y^x$, so $f_x(x,y) = y^x \ln y, f_y(x,y)=xy^{x-1}$ unless $x=0$

Specifically where $x_0=2, y_0=1$ we have $z= 1^2+1^2\ln 1(x-2)+2\cdot 1^{2-1}(y-1) \\= 2y-1$

This is a plane perpendicular to the y,z plane. The vector parameterisation is: $$(x,y,z) = (s, t, 2t-1)$$

Answer 2

In $\mathbb{R}^2$ $$\nabla f_{|_{P_0}}\cdot(P-P_0)=0$$

$$(\ln y\cdot e^{x\ln y}, \dfrac{1}{y}\cdot e^{x\ln y} )_{|_{(2, 1)}}\cdot((x, y)-(2, 1))=0$$

$$(0, 1)\cdot(x-2, y-1)=0$$

The tangent line in $(2, 1)$ is $y=1$