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Predicting the Shape of Complicated Functions

  1. Dec 10, 2006 #1
    This isn't a homework question-- it is a general question about curve sketching in multivariable calculus.

    In order to draw graphs in 3D, we use level sets, then-- if the level sets are something "nice" like circles or parallel lines-- we take a cross section when y=0 and then transform the cross section on to the level sets.

    However, it is at the stage of the cross sections where I begin to be confused. We are given a function such as [tex]e^{-x^2}[/tex]. I do not already know what this function looks like.
    What is a quick way to draw this graph (or any other) without going through the whole curve sketching process (learned in single variable calculus)?
    Are there any tricks to drawing graphs quickly that you don't already know?
    Another example is [tex]sin(x^2)[/tex]. (I already know what this looks like-- but how would I figure it out quickly if I didn't?).

    Thank you.
     
  2. jcsd
  3. Dec 10, 2006 #2

    arildno

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    The first one: "Something that goes to zero dreadfully fast to each side, starting at x=0 with the value 1"

    Second: Even function that oscillates faster and faster. The zeroes of the function is given by [itex]x=\sqrt{n\pi}[/itex] where n is some whole number.
     
  4. Dec 10, 2006 #3
    For the first one, you would think that it approaches zero in the same way that [tex]e^{-x}[/tex] does-- i.e. concave up.
    However, in reality, it is concave down. How do you explain this without taking the second derivative?
     
  5. Dec 10, 2006 #4

    arildno

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    Well, you see that in a region around x=0, the curvature of the graph is negative, whereas further out on either side, the curvature changes sign and becomes positive.

    So you are incorrect in believing it is always negative.
     
  6. Dec 10, 2006 #5
    I know that it changes... however, it still doesn't look like [tex]e^{-x}[/tex].
    And, referring to your first descriptions, my question was how you think of those properties-- I didn't want you to just tell me the properties.
     
  7. Dec 10, 2006 #6

    arildno

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    1. I don't know what YOU mean by "look like"
    2. I don't see the point of this silly thread.
     
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