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Second Derivatives?

  1. May 31, 2003 #1


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    Ok..Im new to calculus and I'm looking for a simple explanation here...

    Lets say u have a function f(x)..the derivative of that function at point x is the slope of the tangent line, right? What would the second derivative be? How is that represented on a graph?
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  3. Jun 1, 2003 #2


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    The attachment (once it's shown... I think it needs to be approved before it is accessible) shows a function, its derivative and its second derivative.

    The bottomline:
    At each x, the function has a "height".
    Its derivative says if at each x the function is going up or down (for instance, if the function is going down, the derivative is negative).

    The second derivative tells you if the first is going up or down. Maybe you can gain some intuition with the interest on a bank account: say you have $1000 saved. If you get $1 per week due to interests:
    1. you have a positive slope.
    2. such slope is constant. You get $1 more every week. No more, no less.
    3. i.e., the "slope of the slope" is zero.

    One way you could have a second derivative different from zero is if your bank were in trouble and reducing the interest every week, so that instead of $1 every week, you got 95 cents the first week, then 90, the 85, etc.

    1. Your still have savings (f(t)>0),
    2. They are still increasing every week (f'(t)>0), but
    3. The weekly increase is getting smaller as time passes (f''(t)<0).

    Of course, it would also be possible that the weekly increase (f'(t)) was getting bigger (not only you are getting more money every week, but the addition is every time bigger than the previous one). A single value of a function does not imply that its derivatives will be positive or negative.

    Now, if you think about it, a positive second derivative means that the function is increasing more rapidly as time (say) passes. In a graph, this means that the graph is "curving up", not just increasing.

    (I'll coment on the attachment once it becomes visible).

    Attached Files:

    • c1.gif
      File size:
      6.6 KB
  4. Jun 1, 2003 #3
    Another common example about second derivatives is the acceleration.
    if x is the position
    v=dx/dt is the velocity
    a=d2x/dt2 is the acceleration
    Now we know that the acceleration is the rate of change of velocity, if velocity is increasing, "a" is +ve, if it's decreasing (e.g. car breaking) "a" becomes -ve.
  5. Jun 1, 2003 #4


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    As for how it looks on a graph, that is probably best demonstrated by looking at the curves with constant second derivative.

    The set of all curves f(x) with constant second derivative s is:

    f(x) = (s / 2) x2 + A x + B

    where A and B can be anything.

    A small section of a curve where the second derivative is positive will look like a small piece of an upwards opening parabola. Where the second derivative is negative it will look like a downwards opening parabola, and where the second derivative is zero it will look like a line.
  6. Jun 1, 2003 #5


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    Also the second derivative measures "curvature".

    If the second derivative is positive, the graph is curving "upward" (like y= x2). If the second derivative is negative, the graph is curving downward (like y= -x2).

    The larger (in absolute value) the second derivative is, the more the graph curves.
  7. Jun 1, 2003 #6


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    Ok, the attachment is visible now.

    If you look at the first plot (the gaussian), you can see that it has a positive slope before x=0 (i.e., it "goes up"), and negative for x>0.

    The first derivative agrees with this. For x<0, the derivative has positive values, and it has its maximum value at about x=-0.7, where the slope for f is the highest.

    The second derivative is positive in the places where f(x) curves upward (from -[oo] to -0.7), then it goes negative (-0.7 to 0.7) where the function curves downwards, and goes back to positive after 0.7.

    Notice also how the biggest absolute value for the second derivative is at x=0, which is where the original function is most curved.
  8. Jun 1, 2003 #7


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    Alright, thanks, i kinda get it more now...yeah, thinking about the velocity/acceleration thing helps too
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