Hi Jan,

most likely Wolfram|Alpha used this:

ArcLength[{\[Theta], Sin[\[Theta]]}, {\[Theta], 0, 3 Pi/2}] // N

which evaluates to 5.7303. That does actually make sense if you ask about the length of the curve, which should be larger than 3. Here is a simple example which substitutes the Sine curve by three straight lines (or two depending on your philosophical point of view):

Show[ParametricPlot[{\[Theta], Sin[\[Theta]]}, {\[Theta], 0, 3 Pi/2}],
  Graphics[Line[{{0, 0}, {Pi/2, 1}, {Pi, 0}, {3 Pi/2, -1}}]]]

Clearly, the black line is shorter than the sine curve. And the total length of the black lines is:

ArcLength[Line[{{0, 0}, {Pi/2, 1}, {Pi, 0}, {3 Pi/2, -1}}]] // N

which evaluates to 5.58629 and is a lower bound to the length of the sine curve. You are doing something quite different. You look at the vertical motion only. In a two dimensional plot this corresponds to three stretches of lines each of length 1. So you are doing this:

Total[ArcLength /@ (Line /@ Partition[Table[{0, Sin[\[Theta]]}, {\[Theta], 0, 3 Pi/2, Pi/2}], 2, 1])]

which gives 3.

Cheers, M.

PS: In general if you want to know what Wolfram|Alpha does this is a useful approach: Type in an equal sign at the beginning of the input line in Mathematica and then use the same input you used for Wolfram|Alpha:

This evaluates to

which shows what Wolfram|Alpha does. Note that this function is equivalent to my ArcLength function.

Hi,

You left unanswered the second part of my question about the pod with the graphics. I'm not asking what the code is to generate it (I can do that, albeit not at the click of the fingers) but how do you get it from any WolframAlpha query (if that functionality is available, it would be very nice)

Like this?

WolframAlpha["get arclength of sin of x from 0 to 3Pi/2", {{"Plot", 1}, "Content"}]

My question was more general than for this example with the length of an arc (and my MMA code was for something else). I don't know how general your answer is as I notice that WolframAlpha does not use the ArcLength function which is rather new (and I suspect most new functions can be obtained using functions available in older MMA versions)

I am not sure I understand. I think that the answer is quite general in the sense that it gives you what Wolfram|Alpha uses. The first solution (using ArcLength) was just what I intuitively used because it is less typing than the automatic solution that Wolfram|Alpha creates.

Cheers,

M.

But how can you ask WolframAlpha to return the code that created the whole graphics?

I don't think that is possible.

Cheers,

M.