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Finding the distance from origin to a tangent line

  1. Sep 22, 2015 #1
    1. The problem statement, all variables and given/known data

    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)

    2. Relevant equations

    Distance= ##\frac{\left |a_{0}+b_{0}+c \right |}{\sqrt{a^{2}+b^{2}}}##

    3. 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.
     
  2. jcsd
  3. Sep 22, 2015 #2

    Mark44

    Staff: Mentor

    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).
     
  4. Sep 22, 2015 #3
    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?
     
  5. Sep 22, 2015 #4
    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?
     
  6. Sep 22, 2015 #5

    Mark44

    Staff: Mentor

    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.
     
  7. Sep 22, 2015 #6

    Mark44

    Staff: Mentor

    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.
     
  8. Sep 22, 2015 #7
    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?
     
  9. Sep 22, 2015 #8

    Mark44

    Staff: Mentor

    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)##.
     
  10. Sep 22, 2015 #9
    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?
     
  11. Sep 22, 2015 #10

    Mark44

    Staff: Mentor

    No. These are the coordinates of the point of tangency. They are parameters, values that aren't known, but don't change.
    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.
     
  12. Sep 22, 2015 #11
    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?
     
    Last edited: Sep 22, 2015
  13. Sep 22, 2015 #12

    Mark44

    Staff: Mentor

    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).
     
  14. Sep 22, 2015 #13
    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?
     
  15. Sep 23, 2015 #14

    Mark44

    Staff: Mentor

    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
     
  16. Sep 23, 2015 #15
    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.
     
  17. Sep 23, 2015 #16

    Mark44

    Staff: Mentor

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