Find partials for ##f## and the equation of plane

[toforfiltum]

Homework Statement

Function is $f(x,y)=((x-1)y)^\frac{2}{3}$,$\space\space(a,b)=(1,0)$

a) Calculate $f_{x}(a,b)$ and $f_{y}(a,b)$ at point $(a,b)$ and write the equation for the plane.

Homework Equations

The Attempt at a Solution

So,

$f_{x}(x,y)=\frac{2y^\frac{2}{3}}{3(x-1)^\frac{1}{3}}$

$f_{y}(x,y)=\frac{2(x-1)^\frac{2}{3}}{3y^\frac{1}{3}}$

Therefore, at $(1,0)$, $f_{x}(x,y)=0$ and $f_{y}(x,y)=0$

Equation of plane at $(1,0)$ is $z=0$.

What I'm doing seems right, but there is a nagging feeling that I'm wrong.

So thoughts?

Thanks.

 

[PeroK]

Homework Statement

Function is $f(x,y)=((x-1)y)^\frac{2}{3}$,$\space\space(a,b)=(1,0)$

a) Calculate $f_{x}(a,b)$ and $f_{y}(a,b)$ at point $(a,b)$ and write the equation for the plane.

Homework Equations

The Attempt at a Solution

So,

$f_{x}(x,y)=\frac{2y^\frac{2}{3}}{3(x-1)^3}$

$f_{y}(x,y)=\frac{2(x-1)^\frac{2}{3}}{3y^3}$

Therefore, at $(1,0)$, $f_{x}(x,y)=0$ and $f_{y}(x,y)=0$

Equation of plane at $(1,0)$ is $z=0$.

What I'm doing seems right, but there is a nagging feeling that I'm wrong.

So thoughts?

Thanks.

Your differentiation of $x^{\frac23}$ is wrong.

 

[Ray Vickson]

Homework Statement

Function is $f(x,y)=((x-1)y)^\frac{2}{3}$,$\space\space(a,b)=(1,0)$

a) Calculate $f_{x}(a,b)$ and $f_{y}(a,b)$ at point $(a,b)$ and write the equation for the plane.

Homework Equations

The Attempt at a Solution

So,

$f_{x}(x,y)=\frac{2y^\frac{2}{3}}{3(x-1)^3}$

$f_{y}(x,y)=\frac{2(x-1)^\frac{2}{3}}{3y^3}$

Therefore, at $(1,0)$, $f_{x}(x,y)=0$ and $f_{y}(x,y)=0$

Equation of plane at $(1,0)$ is $z=0$.

What I'm doing seems right, but there is a nagging feeling that I'm wrong.

So thoughts?

Thanks.

Yes, indeed, both of your partial derivatives are incorrect. I don't get $\frac{2}{3} - 1 = 3$, as you seem to have written.

 

[toforfiltum]

@Ray Vickson @PeroK

Oh, so sorry. I've corrected this in the edited OP.

 

[Mark44]

The Attempt at a Solution

So,
$f_{x}(x,y)=\frac{2y^\frac{2}{3}}{3(x-1)^\frac{1}{3}}$

$f_{y}(x,y)=\frac{2(x-1)^\frac{2}{3}}{3y^\frac{1}{3}}$

Therefore, at $(1,0)$, $f_{x}(x,y)=0$ and $f_{y}(x,y)=0$

The corrected partials are now correct, but the values you show at (1, 0) aren't. Notice that in both cases, both the numerator and denominator are zero.

 

[toforfiltum]

The corrected partials are now correct, but the values you show at (1, 0) aren't. Notice that in both cases, both the numerator and denominator are zero.

Is it undefined in both cases because they involve a division by zero?

 

[Mark44]

Is it undefined in both cases because they involve a division by zero?

Yes. Both partials are undefined at (1, 0).

 

[toforfiltum]

Yes. Both partials are undefined at (1, 0).

So that means there is no equation that exists for the tangent plane at that point, right?

 

[Mark44]

So that means there is no equation that exists for the tangent plane at that point, right?

Yes

 

[toforfiltum]

Yes

Thanks!