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Equations of lines through a point and tangent to a function?

  1. Feb 16, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the equations of all the lines through the origin that are tangent to the curve y = (some complicated cubic function)

    I looked up the question in google and found a much simpler example, y = x^2 passing through (1,-1). However, I don't even get what's going on with this example even though its much more basic.



    2. Relevant equations



    3. The attempt at a solution


    I know the equations of a straight line are derived simply by slope = m = rise/run

    And i'm given a point (1, -1). All I need is a slope.

    The function i'm given is f(x) = x^2

    I can take the derivative of f(x) to get f'(x) = 2x = Mtangent

    So now I have a slope for the point (1, -1)

    So I can create the equation 2x = y+1/x-1

    So assuming i'm on the correct path up until this point, this is where I begin to get confused. If i'm given a point (1, -1) and a slope of say, m = 2, I can get slope intercept form of y = 8x - 9.

    However, the actual Mtangent slope I have is 2x and that gives me the equation
    [2x^2 - 2x -1 = y] which seems to be an equation of a curve.

    But I think my goal is to find the equation of a line, not a curve. Where is my confusion? What does [2x^2 -2x - 1 = y] represent? And how would I actually get the equation of a line?
     
  2. jcsd
  3. Feb 16, 2012 #2
    What ##f'(x)## represents is the slope of the curve represented by the function ##f## at the point ##(x, f(x))##. That is, for ##f(x)=x^3,## we have ##f'(x)=3x^2## and given a point ##(2, 8)## we can say that the slope of the graph of ##f## at this point is the value of the function's derivative ##f'(2)=12##.
     
  4. Feb 16, 2012 #3
    But you chose a point, (2,8), that happens to exist on the graph x^3?

    In my question, the point (1, -1), does not exist on the graph x^2

    I'm not asking what a derivative represents, I think what i'm asking about is the equation created from taking the derivative and setting the derivative equal to a point that does not exist on the graph.

    I am trying to figure out how to create an equation that connects a point on a coordinate system to a point that is tangent to the graph f(x) = x^2

    EDIT: Or maybe i'm misunderstanding your response, i'm not sure.
     
    Last edited: Feb 16, 2012
  5. Feb 16, 2012 #4
    Oh, my bad. Just write (-1) - f(x) = f'(x) (1 - x), so -1 - x2 = 2x(1 - x), and solving gives you x2 - 2x - 1 = 0. Both roots should lead you to valid solutions in this case.

    Edit: This equation is from the point-slope formula and from knowing that m = f'(x), so you're basically solving for an x that produces a line with slope 2x which intercepts (x, x2) and (1, -1).
     
    Last edited: Feb 16, 2012
  6. Feb 16, 2012 #5

    Okay, that seems exactly as I already did, the difference being that I chose points (1,-1) and a random point (x,y), just as the slope is written rise/run = y-y/x-x. And you chose points (1,-1) and (x,x^2).

    Does that mean that my [2x^2 -2x - 1 = y] and your [x^2 - 2x - 1 = 0] are equivelent somehow? When I graph both equations, they are different, but I see that if I substitute x^2 for y and bring it to the other side, i'll get the same equation that you produced.


    Also, as per my original confusion, I thought point slope formula was a formula that represents a line. Are you saying the point slope formula represents parabolas also?

    EDIT: okay, I just solved for both roots and got 1+√2 and 1-√2. Those are both points on the x axis, but what do they mean? Are they x intercepts to the lines I am trying to create an equation for?

    EDIT: okay, looking at the website where I got this sample problem is starting help me figure out the last parts. However it ends up showing me the coordinates of the points on the equation fx = x^2 where the line is tangent. But my goal is not to find two points on the graph, by goal is to find an equation of the line. Would I just use those coordinates to come up with the equation?
     
    Last edited: Feb 17, 2012
  7. Feb 17, 2012 #6
    Okay, so I have this equation: y-y=m(x-x), I have always seen it when a textbook is talking about straight lines. However, now it is being used to get y-(-1)=(2x)(x-1). Where this clearly seems not to be representing a straight line. First of all, a slope is a constant while the m = slope variable here is not a constant because it contains an x.

    Okay so i think i get it a little bit, am I supposed to go from a general derivative to an exact slope? And the only way to do that is by finding some x value that will make that 2x turn into some constant slope?
     
  8. Feb 17, 2012 #7
    Yeah, you're almost there.

    Letting x = 1+√2 gives you an equation for a line

    y - (3 + 2√2) = (2 + 2√2)(x - (1+√2)).

    Plugging in (1, -1) for (x, y) shows that this line contains (1, -1). This line crosses through (1+√2, f(1+√2)) and has slope f'(1+√2), so it is tangent to the graph f at that point.

    y - y1 = m(x - x1) is indeed the equation for a line, provided m, x1, y1 are constants. To find the correct constants, you can solve for them as I've done above.
     
  9. Feb 17, 2012 #8

    HallsofIvy

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    Any line through (1, 1) is of the form y= m(x- 1)- 1. If that is tangent to y= f(x) at some point [itex](x_0, y_0)[/itex] (it is a very bad idea to use just "x" both for the variable and a specific value), then you must have [itex]m(x_0- 1)+ 1= f(x_0)[/itex] (so that it touches the graph) and [itex]m= f'(x_0)[/itex] (so that it is tangent there). That gives you two equations to solve for m and [itex]x_0[/itex].

    In your [itex]f(x)= x^2[/itex] example that would be [itex]m(x_0- 1)+ 1= x_0^2[/itex] and [itex]m= 2x_0[/itex]. Replace m in the first equation by [itex]2x_0[/itex] to get [itex]2x_0^2- 2x_0+ 1= x_0^2[/itex], a quadratic equation for [itex]x_0[/itex].

    (In fact, that equation is just [itex](x_0- 1)^2= 0[/itex] which has the single root [itex]x_0= 1[/itex]. That's because the line is tangent and does not cross through the graph.)
     
  10. Feb 18, 2012 #9
    Thanks for adding clarity to this derivative problem. I was able to solve (and understand) the original example, which was to find the equations of all the lines that pass through the origin and are tangent to the curve f(x) = x^3 + 9x^2 - 16x

    This felt like an interesting problem to me and I was wondering if this concept of finding the equation of a line that passes through a point and is tangent to a function has a direct application in physics? For instance, are there any acceleration problems that involve the mathematics of this particular problem?
     
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