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  1. #1
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    Last 3 (Calculus)

    Three more and we're done...
    As always, help much appreciated.
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  3. #2
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    Re: Last 3 (Calculus)

    I don't understand (1) and (2).

    (3): see screenshot below.
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  4. #3
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    Re: Last 3 (Calculus)

    I don't understand it either. <img src=/S/confused.gif border=0 alt=confused width=15 height=20>

    But here is some additional information from my daughter:
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    Re: Last 3 (Calculus)

    My work is attached.
    Q1 is Even since it is symmetrical about the X axis. [f(x) = f(-x)]

    Other than the plot one could go through the rigamarole of showing that:

    |x| - (x)2 +1 = |-x| + (-x)^2 + 1 ....

    Q2 I would say is also even but it must be thought as a function of Y not X (it is not a function of X)
    3x = |y|
    becomes
    x = |y| / 3
    f(y) = |y| / 3

    It is thus even since
    f(y) = f(-y)
    |y| / 3 = |-y| / 3

    Steve
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    Re: Last 3 (Calculus)

    Hans / Steve:

    Thank you once again. None of this makes much sense to me <img src=/S/igiveup.gif border=0 alt=igiveup width=31 height=23>, but I'm sure there's plenty here for my daughter (Tricky-ette) <img src=/S/thankyou.gif border=0 alt=thankyou width=40 height=15> to work with. She needs to set up her own lounge account. Or find herself a "math board"! <img src=/S/smile.gif border=0 alt=smile width=15 height=15>

  7. #6
    WS Lounge VIP sdckapr's Avatar
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    Re: Last 3 (Calculus)

    An even function is just one that is symmetric about the y-axis (y= x^2 is a simple example). Whether the X is positive or negative the y-value is the same.

    An odd function is one like y= x^3. They are not symmetrical about the y-Axis, but essentially are identical if rotated 180 degrees.

    Most functions are neither...

    Steve

  8. #7
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    Re: Last 3 (Calculus)

    > Q2 I would say is also even but it must be thought as a function of Y not X (it is not a function of X)

    I'm guessing that's the reason for the question specifying that it's a relation, not necessarily (and in this case, simply "not") a function. It is therefore, as you show, an even relation... or simply "even".

    Alan

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    Re: Last 3 (Calculus)

    Having done my share of physics teaching, I think I can see where this is all heading. I'm guessing that this mixture of material relating to absolute values and odd/ even functions/ relations is all geared towards illustrating the difference between distance and displacement. Hopefully the attached will help to clarify what's going on.

    I've plotted your Q3 equation for velocity, together with a very similar function. The shaded area between the curve and the horizontal axis is the displacement over the 15 second interval. This can be evaluated using the integral that Hans provided. It also happens to be equal to the distance covered, which is what the question asks.

    Now consider the very similar function below. Here we have areas both above and below the horizontal axis. Hans's integral, applied to this function, would again give the displacement over the 15 second interval. But it would NOT give the total distance covered. Why? Because the areas below the axis are negative displacements i.e. they represent motion back towards the starting point, as indicated by a negative velocity. To get the total distance travelled, one must sum the absolute values of all areas between the curve and the horizontal axis. This will mean finding the values at which the velocity curve crosses the axis and solving the integral between each pair of points. Then summing the absolute values of all the integrals involved.

    Alan
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    Re: Last 3 (Calculus)

    I should really add here, that if the function is easily visualized graphically, like the ones in the examples, and if it is only needed to calculate the distance (not the displacement) then a simpler method may be evident. For instance:

    f(x) = 3 + 7cos(x)
    = {7 + 7cos(x)} - 4

    The first two terms constitute a function entirely on or above the x-axis, so the absolute values of all areas involved will be given by a single integral. The third term gives a very simple negative integral, whose absolute value is easily calculated.

    Alan

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    Re: Last 3 (Calculus)

    In regards to Q3, tricky-ette was curious why calculus was needed to find a solution. Though her "back-of-envelope' math proved to be incorrect, she was certain that she could solve the problem using basic trig. So, she concluded that the teacher was looking for something more than just the answer.

    I've printed your additional comments and examples [and placed it by her books] and I know she will be enlightened and fascinated; as for me, there's no hope. I'm just a big fan of money-math!! <img src=/S/money.gif border=0 alt=money width=17 height=15>

    <img src=/S/thankyou.gif border=0 alt=thankyou width=40 height=15> <img src=/S/cheers.gif border=0 alt=cheers width=30 height=16>

  12. #11
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    Re: Last 3 (Calculus)

    > I'm just a big fan of money-math!!

    OK, you show me how to solve all my money problems and I'll show you how to solve all your physics problems!
    Good deal IMO <img src=/S/grin.gif border=0 alt=grin width=15 height=15>

    Alan

  13. #12
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    Re: Last 3 (Calculus)

    The reason her simple solution failed is that a d=v*t formula only works for a constant v, or an appropriate average v, correctly calculated over the particular interval. The latter needs to be an average speed, not velocity, since we're wanting distance not displacement. It's simplest to use the method shown.

    Alan

    These links might help in explaining the concepts:
    Distance and Displacement
    Speed and Velocity

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