Finding the distance from origin to a tangent line

[TyroneTheDino]

Homework Statement

Find the distance between the origin and the line tangent to $x^\frac{2}{3}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$ at the point P(x,y)

Homework Equations

[/B]
Distance= $\frac{\left |a_{0}+b_{0}+c \right |}{\sqrt{a^{2}+b^{2}}}$

The Attempt at a Solution

To begin I find the derivative of the original equation. I derive implicitly to find the line tangent to the original curve, and think this is where I begin to go wrong.

I get $\frac{dy}{dx}=\frac{\sqrt[3]{y}}{\sqrt[3]{x}}$, but something tells me that if I want to find a linear equation for the tangent to the curve. I don't want to define the slope this way.

$\frac{2}{3}x^{\frac{-1}{3}}+\frac{2}{3}y^{\frac{-1}{3}}\left (\frac{dy}{dx} \right )=0$ Is what I get before I define what dy/dx is. Is it okay if I just plug dy/dx into this equation. That would give me...

$\frac{2}{3}x^{\frac{-1}{3}}+\frac{2}{3}y^{\frac{-1}{3}}\left( \frac{\sqrt[3]{y}}{\sqrt[3]{x}} \right )=0$

I am not sure if this is valid, but either way after finding the derivative. I am not sure how to evaluate it to get the slope to define the tangent line. I know I am suppose to end up with a y=mx+b equation, I'm not sure how to devise it though.

 

[Mark44]

Homework Statement

Find the distance between the origin and the line tangent to $x^\frac{2}{3}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$ at the point P(x,y)

It would be better to call this point P(x₀, y₀). That way you will know that you're working at a specific, but unspecified, point.

Homework Equations

[/B]
Distance= $\frac{\left |a_{0}+b_{0}+c \right |}{\sqrt{a^{2}+b^{2}}}$

The Attempt at a Solution

To begin I find the derivative of the original equation. I derive implicitly to find the line tangent to the original curve, and think this is where I begin to go wrong.

I get $\frac{dy}{dx}=\frac{\sqrt[3]{y}}{\sqrt[3]{x}}$, but something tells me that if I want to find a linear equation for the tangent to the curve. I don't want to define the slope this way.

You have a sign error in dy/dx.

$\frac{2}{3}x^{\frac{-1}{3}}+\frac{2}{3}y^{\frac{-1}{3}}\left (\frac{dy}{dx} \right )=0$ Is what I get before I define what dy/dx is. Is it okay if I just plug dy/dx into this equation.

No, you don't plug dy/dx into this -- you solve this equation for dy/dx, which I think you already did. To get the slope of the curve at P(x₀, y₀), substitute these coordinates into your expression for dy/dx. Then, to get the equation of the tangent line, use the point-slope form of the equation of a line: y - y₀ = m(x - x₀).

That would give me...

$\frac{2}{3}x^{\frac{-1}{3}}+\frac{2}{3}y^{\frac{-1}{3}}\left( \frac{\sqrt[3]{y}}{\sqrt[3]{x}} \right )=0$

I am not sure if this is valid, but either way after finding the derivative. I am not sure how to evaluate it to get the slope to define the tangent line. I know I am suppose to end up with a y=mx+b equation, I'm not sure how to devise it though.

 

[TyroneTheDino]

It would be better to call this point P(x₀, y₀). That way you will know that you're working at a specific, but unspecified, point.
You have a sign error in dy/dx.
No, you don't plug dy/dx into this -- you solve this equation for dy/dx, which I think you already did. To get the slope of the curve at P(x₀, y₀), substitute these coordinates into your expression for dy/dx. Then, to get the equation of the tangent line, use the point-slope form of the equation of a line: y - y₀ = m(x - x₀).

So I did what you said and defined the tangent line as
$y=\left ( -\frac{\sqrt[3]{y_{o}}}{\sqrt[3]{x_{o}}} \right )x+b$.

Does that look correct? And if so then how would I solve for "b" or even apply the distance formula to an equation defined this way?

 

[TyroneTheDino]

Ok using this equation, I went ahead and tried to find the distance to the specific point. Because the beginning is the origin. I just said that y₀=Y-0 and x₀=X-0 making the new equation for the line:

$y=\left ( -\frac{\sqrt[3]{y}}{\sqrt[3]{x}} \right )x$.

And putting it into Ax+By+C form I get:

$y+\left ( \frac{\sqrt[3]{y}}{\sqrt[3]{x}} \right )x=0$

Using the distance formula I find the magnitude of (ax)^2+(by)^2 which I found to be:

$\sqrt{\frac{y^{\frac{2}{3}}}{x^{\frac{2}{3}}}x^2+y^2}$

and then I found the what the magnitude of A^2+B^2 which is:

$\sqrt{1+\frac{y^{\frac{2}{3}}}{x^{\frac{2}{3}}}}$

The distance would be:
$\frac{\sqrt{\frac{y^{\frac{2}{3}}}{x^{\frac{2}{3}}}x^2+y^2}}{ \sqrt{1+\frac{y^{\frac{2}{3}}}{x^{\frac{2}{3}}}}}$

With simplification would this be the correct way to find distance?

 

[Mark44]

So I did what you said and defined the tangent line as
$y=\left ( -\frac{\sqrt[3]{y_{o}}}{\sqrt[3]{x_{o}}} \right )x+b$.

Does that look correct? And if so then how would I solve for "b" or even apply the distance formula to an equation defined this way?

It's not wrong, but it's not what I said. You used the slope-intercept form (y = mx + b), not the point-slope form. You have an expression for the slope and a point on the line, so the point-slope form is the most convenient to use. As for what you did, you could solve for b, since you have the slope and a point on the curve/tangent line.

 

[Mark44]

Ok using this equation, I went ahead and tried to find the distance to the specific point. Because the beginning is the origin. I just said that y₀=Y-0 and x₀=X-0 making the new equation for the line:

I don't see that doing this is advantageous at all. Y - 0? That's just Y, and X - 0 is just X. Are you trying to transform things so that the point of tangency is the new origin? Then the old origin is going to be just as bad.

$y=\left ( -\frac{\sqrt[3]{y}}{\sqrt[3]{x}} \right )x$.

And putting it into Ax+By+C form I get:

$y+\left ( \frac{\sqrt[3]{y}}{\sqrt[3]{x}} \right )x=0$

Using the distance formula I find the magnitude of (ax)^2+(by)^2 which I found to be:

$\sqrt{\frac{y^{\frac{2}{3}}}{x^{\frac{2}{3}}}x^2+y^2}$

and then I found the what the magnitude of A^2+B^2 which is:

$\sqrt{1+\frac{y^{\frac{2}{3}}}{x^{\frac{2}{3}}}}$

The distance would be:
$\frac{\sqrt{\frac{y^{\frac{2}{3}}}{x^{\frac{2}{3}}}x^2+y^2}}{ \sqrt{1+\frac{y^{\frac{2}{3}}}{x^{\frac{2}{3}}}}}$

With simplification would this be the correct way to find distance?

 

[TyroneTheDino]

It's not wrong, but it's not what I said. You used the slope-intercept form (y = mx + b), not the point-slope form. You have an expression for the slope and a point on the line, so the point-slope form is the most convenient to use. As for what you did, you could solve for b, since you have the slope and a point on the curve/tangent line.

If I put my equation in point slope form noting that the initial point is at (0,0) and I'm going to P(x,y) I would get

y-0= $\frac{x^{\frac{1}{3}}}{y^{\frac{1}{3}}}(x-0)$

Which would be the same thing as y=$\frac{x^{\frac{1}{3}}}{y^{\frac{1}{3}}}x$

But then how do you find distance when using point slope form rather than y=mx+b?

 

[Mark44]

If I put my equation in point slope form noting that the initial point is at (0,0) and I'm going to P(x,y) I would get

y-0= $\frac{x^{\frac{1}{3}}}{y^{\frac{1}{3}}}(x-0)$

This is wrong. You're assuming that the tangent line goes through (0, 0). What you know about the tangent line is its slope, $\frac{-y_0^{1/3}}{x_0^{1/3}}$ and a point on the line that I'm calling $P(x_0, y_0)$.

Which would be the same thing as y=$\frac{x^{\frac{1}{3}}}{y^{\frac{1}{3}}}x$

But then how do you find distance when using point slope form rather than y=mx+b?

 

[TyroneTheDino]

This is wrong. You're assuming that the tangent line goes through (0, 0). What you know about the tangent line is its slope, $\frac{-y_0^{1/3}}{x_0^{1/3}}$ and a point on the line that I'm calling $P(x_0, y_0)$.

Ok the point slope equation would be:

$y-y_{0}=-\frac{\sqrt[3]{y}}{\sqrt[3]{x}}\left ( x-x_{0} \right )$,

Am I supposed to solve for Y₀ and X₀? How will this form be applied to finding the distance between the points?

 

[Mark44]

Ok the point slope equation would be:

$y-y_{0}=-\frac{\sqrt[3]{y}}{\sqrt[3]{x}}\left ( x-x_{0} \right )$,

Am I supposed to solve for Y₀ and X₀?

No. These are the coordinates of the point of tangency. They are parameters, values that aren't known, but don't change.

How will this form be applied to finding the distance between the points?

Your goal is to find the distance between the origin and the tangent line, not the distance between two points (which is pretty easy). The equation above is that of the tangent line. You have a formula in post #1 under relevant formulas, but it's wrong. See https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line.

 

[TyroneTheDino]

I still don't think I am understanding.

Using the equation from wikipedia for distance:
$\frac{\left | ax_{0}+by_{0}+c \right |}{\sqrt{a^2+b^2}}$

I can define "a" as the value:
$\frac{\sqrt[3]{y}}{\sqrt[3]{x}}$
and "b" as the value: 1

Because in the slope form equation I have (y-y₀)=$\frac{\sqrt[3]{y}}{\sqrt[3]{x}}$(x-x₀)

and c would be zero in this case.

Or am i still misunderstanding?

 

[Mark44]

I still don't think I am understanding.

Using the equation from wikipedia for distance:
$\frac{\left | ax_{0}+by_{0}+c \right |}{\sqrt{a^2+b^2}}$

I can define "a" as the value:
$\frac{\sqrt[3]{y}}{\sqrt[3]{x}}$
and "b" as the value: 1

OK to both.

Because in the slope form equation I have (y-y₀)=$\frac{\sqrt[3]{y}}{\sqrt[3]{x}}$(x-x₀)

You have the sign of the slope wrong again.
BTW, this is called the point-slope form of the line for reasons that should be obvious.

and c would be zero in this case.

No. Move all of the terms over to one side to get Ax + By + C = 0. The constant C won't be zero.

Or am i still misunderstanding?

The coefficients are a little messy, but the main ideas are pretty simple, and should be things that you mastered back in algebra or precalculus. You seem to be struggling with a very elementary concept, the equation of a line. It might be helpful to step back from the problem and make a list of the things you need to do in this problem. To find the distance from (0, 0) to a particular line, you need the equation of the line. To get the equation of the line you need a point on the line and the slope of the line. To get the slope of the line, you need to find the derivative of the function, and evaluate it at the point P(x₀, y₀).

 

[TyroneTheDino]

OK to both.
You have the sign of the slope wrong again.
BTW, this is called the point-slope form of the line for reasons that should be obvious.
No. Move all of the terms over to one side to get Ax + By + C = 0. The constant C won't be zero.

The coefficients are a little messy, but the main ideas are pretty simple, and should be things that you mastered back in algebra or precalculus. You seem to be struggling with a very elementary concept, the equation of a line. It might be helpful to step back from the problem and make a list of the things you need to do in this problem. To find the distance from (0, 0) to a particular line, you need the equation of the line. To get the equation of the line you need a point on the line and the slope of the line. To get the slope of the line, you need to find the derivative of the function, and evaluate it at the point P(x₀, y₀).

I know it shouldn't be that hard, but I'm having such problems.

So if the constant C is not zero how do I find it?

From what I understand you saying I move all terms to one side to get:
$(y-y_{0})+\frac{\sqrt[3]{y}}{\sqrt[3]{x}}(x-x_{0})+c=0$

So C is:

$-(y-y_{0})-\frac{\sqrt[3]{y}}{\sqrt[3]{x}}(x-x_{0})$

Correct?

 

[Mark44]

I know it shouldn't be that hard, but I'm having such problems.

So if the constant C is not zero how do I find it?

From what I understand you saying I move all terms to one side to get:
$(y-y_{0})+\frac{\sqrt[3]{y}}{\sqrt[3]{x}}(x-x_{0})+c=0$

So C is:

$-(y-y_{0})-\frac{\sqrt[3]{y}}{\sqrt[3]{x}}(x-x_{0})$

Correct?

No.
Change this equation:
$(y-y_{0})+\frac{\sqrt[3]{y}}{\sqrt[3]{x}}(x-x_{0})+c=0$
so it looks like this:
Ax + By + C = 0

 

[TyroneTheDino]

No.
Change this equation:
$(y-y_{0})+\frac{\sqrt[3]{y}}{\sqrt[3]{x}}(x-x_{0})+c=0$
so it looks like this:
Ax + By + C = 0

Ok so
$\frac{\sqrt[3]{y}}{\sqrt[3]{x}}x_{0}+y_{0}+C=0$
$C=-\frac{\sqrt[3]{y}}{\sqrt[3]{x}}x_{0}-y_{0}$
correct?

So the beginning of this equation would be the magnitude of
$\frac{\sqrt[3]{y}}{\sqrt[3]{x}}x_{0}+y_{0}-\frac{\sqrt[3]{y}}{\sqrt[3]{x}}x_{0}-y_{0}$

Over

$\sqrt{\left ( \frac{y^{\frac{2}{3}}}{x^{\frac{2}{3}}}+1 \right )}$

Correct?And the numerator of the fraction is actually supposed to be magnitude of $\frac{\sqrt[3]{y}}{\sqrt[3]{x}}x_{0}+y_{0}-\frac{\sqrt[3]{y}}{\sqrt[3]{x}}x_{0}-y_{0}$

So I will have something that looks like

$\frac{\sqrt{\left (\frac{\sqrt[3]{y}}{\sqrt[3]{x}}x_{0} \right )^{2}+\left (y_{0} \right )^{2}+\left (\frac{\sqrt[3]{y}}{\sqrt[3]{x}}x_{0} \right )^2+\left (y_{0} \right )^{2}}} {\sqrt{\left ( \frac{y^{\frac{2}{3}}}{x^{\frac{2}{3}}}+1 \right )}}$
And then once I have that is it if I plug x into x₀ and y into y₀

Correct?

Sorry for asking so much. I hope I am getting on the right track.

 

[Mark44]

Ok so
$\frac{\sqrt[3]{y}}{\sqrt[3]{x}}x_{0}+y_{0}+C=0$
$C=-\frac{\sqrt[3]{y}}{\sqrt[3]{x}}x_{0}-y_{0}$
correct?

No. The equation should be of the form Ax + By + C = 0. x₀ and y₀ are not variables.

So the beginning of this equation would be the magnitude of
$\frac{\sqrt[3]{y}}{\sqrt[3]{x}}x_{0}+y_{0}-\frac{\sqrt[3]{y}}{\sqrt[3]{x}}x_{0}-y_{0}$

Over

$\sqrt{\left ( \frac{y^{\frac{2}{3}}}{x^{\frac{2}{3}}}+1 \right )}$

Correct?And the numerator of the fraction is actually supposed to be magnitude of $\frac{\sqrt[3]{y}}{\sqrt[3]{x}}x_{0}+y_{0}-\frac{\sqrt[3]{y}}{\sqrt[3]{x}}x_{0}-y_{0}$

So I will have something that looks like

$\frac{\sqrt{\left (\frac{\sqrt[3]{y}}{\sqrt[3]{x}}x_{0} \right )^{2}+\left (y_{0} \right )^{2}+\left (\frac{\sqrt[3]{y}}{\sqrt[3]{x}}x_{0} \right )^2+\left (y_{0} \right )^{2}}} {\sqrt{\left ( \frac{y^{\frac{2}{3}}}{x^{\frac{2}{3}}}+1 \right )}}$
And then once I have that is it if I plug x into x₀ and y into y₀

Correct?

Sorry for asking so much. I hope I am getting on the right track.