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Homework Help: I'm new here

  1. May 13, 2004 #1
    We are currently interpreting derivatives using the delta process in my analytical geometry class. The problem is: Calculate the coordinates (x,y) of the point at which the graph y= 0.027x^2 + 8.3x -7.5 is at a 35 degree angle from the horizontal. I have figured the derivative to be equal to 0.054x + 0.027 delta x +8.3
    Question #1: Is the derivative correct?
    Question #2: What do I do now?
    Thanks
     
  2. jcsd
  3. May 14, 2004 #2

    HallsofIvy

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    No, that is not the derivative. What you have calculated is the "difference quotient":
    (f(x+ deltax)- f(x))/detax. To find the derivative itself, take the limit as deltax goes to 0. In this particular case, that's very easy: it is exactly the same as setting deltax equal to 0: 0.054x +8.3.

    You should know that the derivative is the same as the "slope of the tangent line" and that the slope is equal to tan(θ) where θ is the angle the line makes with the x-axis (horizontal). In this problem you are told that that is 35 degrees.

    Find the tangent of 35 degrees, set 0.054x +8.3 equal to that, and solve for x.
     
  4. May 14, 2004 #3

    arildno

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    I'd like to add to HallsofIvy's comment a link to concepts you are probably familiar with from analytical geometry:

    When your teacher introduced you to "the derivative", I would think that he/she first talked about secants/secant lines and then about tangents/tangent lines. (?)

    In particular, you may have learnt that the slope of the tangent line at some point of a curve may be found as the limit of the slopes of secant lines associated with that point when the distance between the two points (on the curve) defining a secant line goes to zero.

    To make maths out of this:
    1.
    Let the two points on the curve be:
    [tex]P_{1}=(x,y(x)), P_{2}=(x+\bigtriangleup{x},y(x+\bigtriangleup{x}))[/tex]
    2.
    Since you've got two points, [tex]P_{1},P_{2}[/tex], you can evidently draw a straight line between them!!
    This straight line is called the secant line S with respect to the points [tex]P_{1},P_{2}[/tex].

    3.
    Now, I would think that you know that a straight line L in the plane usually can be represented as a function Y(X)=Ax+B, where Y(X) is the vertical coordinate Y at a point at L, while X is the horizontal coordinate of the same point at L.
    A and B are constants for L (equal values for all choices of X!); B is called the Y-intercept (lies on the Y-axis, X=0), while A is called the slope of L.
    4.
    Going back to our secant line S, how can I find its slope?
    We know 2 things about S:
    a) If X=x, then Y(X)=y(x) (i.e, we're at the point [tex]P_{1}[/tex])
    b) If [tex] X=x+\bigtriangleup{x}[/tex], then [tex]Y(X)=y(x+\bigtriangleup{x})[/tex] (i.e, we're at the point [tex]P_{2}[/tex])

    Going back to the general equation for a line L, we must therefore have for S:
    [tex]A*x+B=y(x), A*(x+\bigtriangleup{x})+B=y(x+\bigtriangleup{x})[/tex]
    Solving these equations for A and B, we find:
    [tex]A=\frac{y(x+\bigtriangleup{x})-y(x)}{\bigtriangleup{x}}[/tex]
    [tex]B=y(x)-A*x[/tex]

    Hence, we may represent the Y-coordinate of a point on S, [tex]S_{Y}[/tex], as a function of the horizontal coordinate, X, like this:
    [tex]S_{Y}(X)= \frac{y(x+\bigtriangleup{x})-y(x)}{\bigtriangleup{x}}X+y(x)-A*x[/tex]

    This is the way in which the secant line S can be represented in the usual manner of a line L.

    5.
    We are interested in the slope of S, [tex]\frac{y(x+\bigtriangleup{x})-y(x)}{\bigtriangleup{x}}[/tex]
    This is called the quotient of differences, as HallsofIvy says.
    6.
    In order to find the slope of the tangent line at point [tex]P_{1}[/tex], we evaluate the slope expression from S as we let the difference between the values of horizontal coordinates of [tex]P_{1},P_{2}[/tex] shrink to zero.
    (That difference is [tex]\bigtriangleup{x}[/tex]).

    Geometrically, this limiting process has the interpration that we evaluate the slopes of different secant lines which have [tex]P_{1}[/tex] in common, but where each secant line's [tex]P_{2}[/tex] is chosen to be progressively closer to [tex]P_{1}[/tex].
    The tangent line's slope is found when [tex]P_{2}[/tex] becomes [tex]P_{1}[/tex].

    The derivative of y at x, [tex]\frac{dy}{dx}[/tex], is the name of the slope of the tangent line at [tex]P_{1}[/tex].
     
    Last edited: May 14, 2004
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