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Implicit differensiation

  1. Oct 13, 2006 #1
    I cant get the problem. can anyone help me please.

    -Find equations for two lines thorugh the origin that are tangent to the curve x^2 - 4x +y^2 + 3 = 0.

    I found dy/dx=(-x+2)/y and put thta into the point slope equation, and then filled in (0,0) for the point, but couldnt get an equation.


    any help appretiated.
     
  2. jcsd
  3. Oct 13, 2006 #2

    arildno

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    "I found dy/dx=(-x+2)/y and put thta into the point slope equation, and then filled in (0,0) for the point, but couldnt get an equation"

    What point??
    Please clarify to us, and not the least to yourself, what sort of points you're after, and what relation the origin has to those points!
     
    Last edited: Oct 13, 2006
  4. Oct 13, 2006 #3
    I am looking for a tangent line that goes through the equation. the question i gave is all that is in the question. Only poing i think we know is that th eline goes through 0,0

    exact question

    Find equations for two lines thorugh the origin that are tangent to the curve x^2 - 4x +y^2 + 3 = 0.
     
  5. Oct 13, 2006 #4

    radou

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    Hint: x^2 - 4x + y^2 + 3 = 0 => (x - 2)^2 + (y - 0)^2 = 1.
     
  6. Oct 13, 2006 #5

    HallsofIvy

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    First, a tangent line does not go through an 'equation', it goes intersects a graph- you mean the tangent line goes through the graph of the relation x^2 - 4x +y^2 + 3 = 0 (that's a circle with center at (2, 0) and radius 1). You seem to be assuming that the lines must be tangent to the curve at the point (0,0) which can't be true: (0, 0) does not satisfy the equation. You are seeking two lines that pass through (0,0) and are tangent to the circle at some point on the circle.

    Call the point of tangencey (x0,y0). Then any line through that point and (0, 0) has equation [itex]y= \frac{y_0}{x_0}x[/itex].
    The slope is [itex]\frac{y_0}{x_0}[/itex]. For what (x0, y0) is the derivative of x2 - 4x +y2 + 3 = 0 equal to [itex]\frac{y_0}{x_0}[/itex]?
     
  7. Oct 13, 2006 #6
    its a circle. how does putting the equation like that help me find the tangent line


    my freind "guessed" a point where the circle looks like it starts to curve away from teh origin, and then used that to solve for the tangent line, and he seems to have gotten the right answer, but we cant figure out how to get x= 1.5 allgebraically.
     
  8. Oct 13, 2006 #7
    Random Tip: When doing implicit differentiation, it helps to replace "y" by "y(x)" so that it is clear that y is a function of x and not just another constant. Most beginners make the mistake of thinking of y as a constant and erroneously conclude that dy/dx = 0 (instead of using the chain rule).
     
  9. Oct 13, 2006 #8

    arildno

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    Continue along the track shown you by HallsofIvy.
    1. Assuming that [itex](x_{0},y_{0})[/itex] is a point of tangency on your curve, what is the slope at [itex](x_{0},y_{0})[/itex]?

    2. Do you have a second expression for the very same slope, given the information that the tangent line at [itex](x_{0},y_{0})[/itex] goes through the origin?

    3. Set the two expressions from 1. and 2. together as an equation!

    4. What is the system of equations you need to solve for [itex](x_{0},y_{0})[/itex]?

    5. Solve that system of equations!
     
  10. Oct 13, 2006 #9
    o yes. thankyou. our teacher made that point in class as well. I use a differnt symbol, but they have the same purpose.

    i'm sorry but i still dont understand how to solve this equation. All of these hints are helping me get a little closer...but i still dont get it. Like, i have no clue where you guys are getting r from... (radius...?)

    thanks for the help thus far.
     
  11. Oct 13, 2006 #10

    arildno

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    Perhaps you could start out with my list, then?
     
  12. Oct 13, 2006 #11

    arildno

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    Doing 1. and 2. for you:
    1. You have already done this, sort of:
    [tex]\frac{dy}{dx}\mid_{(x_{0},y_{0})=\frac{2-x_{0}}{y_{0}}[/tex]
    2. HallsofIvy did this one for you: The slope of a line going through the origin AND [itex](x_{0},y_{0})[/itex] is:
    [tex]\frac{y_{0}}{x_{0}}[/itex]

    Can you carry on now?
     
  13. Oct 13, 2006 #12
    no, sorry, we've never used the constant r- what is it. is it the radius of the circle- correlation coeficiant??
     
    Last edited: Oct 13, 2006
  14. Oct 14, 2006 #13

    HallsofIvy

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    Now you've confused me! I've looked back through all the responses and find no reference to a "constant r"! Did someone edit or delete it?

    Any way, the point is that the circle given, [itex]x^2- 4x+ y^2+ 3= 0[/itex], which is equivalent to [itex](x- 2)^2+ y^2= 1[/itex] is a circle with center at (2, 0) and radius 1. It doesn't pass through (0,0) so you want to find two lines that do go through (0,0) and are tangent to the circle at two other points.

    You could do this geometrically: The line segement from (0,0) to (2, 0), the line segment from (2, 0) to the point of tangency, and the line segment from (0,0) to the point of tangency form a right triangle with one leg of length 1 (the radius of the circle) and hypotenuse of length 2 (from (0,0) to (2, 0)) and so the length of the other leg is [itex]\sqrt{3}[/itex]. Any point (x,y) on the circle must satisfy [itex](x-2)^2+ y^2= 1[/itex] and if the line from (0,0) to (x,y) has length [itex]\sqrt{3}[/itex] then
    also [itex]x^2+ y^2= 3[/itex]. Solve those two equations for the points of tangency.

    But, using calculus, do what I suggested before. The line from (0,0) to (x, y) has slope [itex]\frac{y}{x}[/itex] and that must be the y' at the point on the circle. Using implicit differentiation, 2x- 4+ 2yy'= 0 so [itex]y'= \frac{4-2x}{2y}= \frac{2-x}{y}[/itex]. Solve [itex]\frac{2- x}{y}= \frac{y}{x}[/itex] along with [itex]x^2- 4x+ y^2+ 3= 0[/itex] for x and y. Those two methods should give the same point.
     
  15. Oct 14, 2006 #14

    arildno

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    Have you even bothered to read any replies you've been given?
    Is independent thinking so difficult for you that you are unable to understand clear ideas given to you as hints?
     
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