Show that the gradient is perpendicular to a point

[Pallatinus]

Homework Statement

$W = x^2+5y^2$
Show that $\nabla W$ is perpendicular to the level curves of W at $(X_0, 0)$

Homework Equations

$\nabla f(x,y) = <\frac {\partial f} {\partial x} , \frac {\partial f} {\partial y}>$

The Attempt at a Solution

I know that the gradient is always perpendicular to the level curves and the dot product of two vectors is 0 when they are perpendicular to each other.
Then,
$\nabla W(X_0,0) = <\frac {\partial W(X_0,0)} {\partial x} , \frac {\partial W(X_0,0)} {\partial y}>$
$\nabla W(X_0,0) = <2X_0, 0>$
Now I need other vector to show that their dot product is 0, but I don't have sure how to proceed.

 

[Mark44]

Homework Statement

$W = x^2+5y^2$
Show that $\nabla W$ is perpendicular to the level curves of W at $(X_0, 0)$

Homework Equations

$\nabla f(x,y) = <\frac {\partial f} {\partial x} , \frac {\partial f} {\partial y}>$

The Attempt at a Solution

I know that the gradient is always perpendicular to the level curves and the dot product of two vectors is 0 when they are perpendicular to each other.
Then,
$\nabla W(X_0,0) = <\frac {\partial W(X_0,0)} {\partial x} , \frac {\partial W(X_0,0)} {\partial y}>$
$\nabla W(X_0,0) = <2X_0, 0>$
Now I need other vector to show that their dot product is 0, but I don't have sure how to proceed.

Each level curve is an ellipse whose equation is $x^2 + 5y^2 = C$ for some $C \ge 0$. Each ellipse lies in a plane that is parallel to the x-y plane. Can you find the slope of the tangent to any of these curves?

 

[Pallatinus]

Each level curve is an ellipse whose equation is $x^2 + 5y^2 = C$ for some $C \ge 0$. Each ellipse lies in a plane that is parallel to the x-y plane. Can you find the slope of the tangent to any of these curves?

Yes, now I know what should I do:
$\frac {dy} {dx}= \frac {-F_x} {F_y} = \frac {-2x} {10y}$
The slope of the normal line is $\frac {10y} {2x}$ and therefore they are orthogonal at any X and Y.
Is this answer right?
There are another solution that doesn't require the slope?

 

[Mark44]

Yes, now I know what should I do:
$\frac {dy} {dx}= \frac {-F_x} {F_y} = \frac {-2x} {10y}$
The slope of the normal line is $\frac {10y} {2x}$ and therefore they are orthogonal at any X and Y.
Is this answer right?

"therefore they are orthogonal at any X and Y" -- who are "they"? And why do you think "they" are orthogonal?
You haven't shown that the gradient of W is perpendicular to the level curves at (X₀, 0).

There are another solution that doesn't require the slope?

 

[Pallatinus]

"therefore they are orthogonal at any X and Y" -- who are "they"? And why do you think "they" are orthogonal?
You haven't shown that the gradient of W is perpendicular to the level curves at (X₀, 0).

(Sorry, I need to be more specific)
I mean that the tangent line and the line that contains $(X_0,0)$ and have the gradient as its vector , If their slopes are negative reciprocal, aren't the lines orthogonal?

 

[Mark44]

(Sorry, I need to be more specific)
I mean that the tangent line and the line that contains $(X_0,0)$ and have the gradient as its vector , If their slopes are negative reciprocal, aren't the lines orthogonal?

Yes, in most cases, but not in this particular case, because of the point in question.

How did you come up with this:

The slope of the normal line is $\frac {10y} {2x}$

I saw that you calculated the gradient of W, but you didn't show this step.

 

[Pallatinus]

Yes, in most cases, but not in this particular case, because of the point in question.

How did you come up with this:
I saw that you calculated the gradient of W, but you didn't show this step.

I forgot to show it. If the slope (m), is $m = \frac {\Delta x} {\Delta y} = \frac {f_y} {f_x} = \frac {10y} {2x}$
If what I'm doing is wrong because of the point $(X_0,0)$, what should I do?

 

[Mark44]

I forgot to show it. If the slope (m), is $m = \frac {\Delta x} {\Delta y} = \frac {f_y} {f_x} = \frac {10y} {2x}$

m is the slope of the tangent line, right? That would be $m = \frac {\Delta y} {\Delta x} = \frac {10y} {2x}$.
What's the slope of the tangent to any of the level curves at the point (X₀, 0)?

If what I'm doing is wrong because of the point $(X_0,0)$, what should I do?

Include this point in your calculations.

 

[Pallatinus]

m is the slope of the tangent line, right? That would be $m = \frac {\Delta y} {\Delta x} = \frac {10y} {2x}$.
What's the slope of the tangent to any of the level curves at the point (X₀, 0)?

My mistake, I know that the slope is $\frac {\Delta y} {\Delta x}$. And this is the slope of the normal line. the slope of the tangent is $\frac {-2x} {10y}$ as I showed before.
The slope of the tangent would not exist, and the slope of the normal line would be 0.

Include this point in your calculations.

Yes, I have noticed this before, There will not exist a slope since the Y is 0.
This is why I haven't sure how to proceed, since this is the only way that I thought.

 

[Mark44]

My mistake, I know that the slope is $\frac {\Delta y} {\Delta x}$. And this is the slope of the normal line. the slope of the tangent is $\frac {-2x} {10y}$ as I showed before.
The slope of the tangent would not exist, and the slope of the normal line would be 0.
Yes, I have noticed this before, There will not exist a slope since the Y is 0.
This is why I haven't sure how to proceed, since this is the only way that I thought.

Since the slope of the gradient at the point in question is 0, and the slope of the tangent line at the same point is undefined, the two lines are orthogonal.

 

[Pallatinus]

Since the slope of the gradient at the point in question is 0, and the slope of the tangent line at the same point is undefined, the two lines are orthogonal.

Indeed, I was not sure if I can imply that with a undefined slope. But It seen's that I can.
Thank you for the help and patience.