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Mathcad 14: graphing functions

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Mathcad 14: graphing functions

Hello All,

I'm practicing graphing functions but there seem to be only two worksheets, 'Graphing and Differentiation', 'Solving numberically' with the latter not having much to practice on.

Are there more worksheets, examples so that I could see how to graph more comples trigonometric functions, e.g. f(t) = t^5 ln(3t + 4) , or f(t) = 2/sqrt(3) cos(2t) + 2sin(2t), etc, etc. within various ranges of π ?

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Re: Mathcad 14: graphing functions

Fixed in eight key strokes.

View solution in original post

12 REPLIES 12
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Re: Mathcad 14: graphing functions

Rosanella di-Costanzo wrote:

Hello All,

I'm practicing graphing functions but there seem to be only two worksheets, 'Graphing and Differentiation', 'Solving numberically' with the latter not having much to practice on.

Are there more worksheets, examples so that I could see how to graph more comples trigonometric functions, e.g. f(t) = t^5 ln(3t + 4) , or f(t) = 2/sqrt(3) cos(2t) + 2sin(2t), etc, etc. within various ranges of π ?

See if the attached helps.

Have a go yourself, and if your attempt doesn't work, upload your worksheet so that we can see exactly what you might be doing wrong.

Alan

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Re: Mathcad 14: graphing functions

Thank you Alan for the worksheet with examples

The problem seems to arise when I set limits, e.g. t := -п, п/2..2п (How do I attach a file on this forum?)

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Re: Mathcad 14: graphing functions

To attach a file, click on "Use advanced editor" in the upper right corner of the edit window, an attach files /browse will appear at the bottom of the window.

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Re: Mathcad 14: graphing functions

Here I'm setting ranges from -Pi to +Pi

Fred Kohlhepp wrote:

To attach a file, click on "Use advanced editor" in the upper right corner of the edit window, an attach files /browse will appear at the bottom of the window.

Do! I didn't see that, sorry

Here is what I mean by setting a range for -π < t < π (see attached file)

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Re: Mathcad 14: graphing functions

Rosanella di-Costanzo wrote:

Here is what I mean by setting a range for -π < t < π (see attached file)

The problem lies in your range definition. A range variable definition takes the forms firstNumber,secondNumber...lastNumber. Consequently, -π,0...π expands to just three elements -π, 0 and π. ... that was weird, my Firefox keyboard suddenly started inputting in Russian and then crashed when I tried to update the message.

Stuart

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Re: Mathcad 14: graphing functions

Fixed in eight key strokes.

View solution in original post

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Re: Mathcad 14: graphing functions

Many thanks to all. I now see where I was going wrong.

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Re: Mathcad 14: graphing functions

Fred Kohlhepp wrote:

Fixed in eight key strokes.

Hah! fixed in 6 key strokes (7 if you include ctl-d as two separate keys ... oh, all right, and 3 mouse clicks).

We should remember the wise words of Master Kenobi: "Use the Features, Luke." and use the auto-ranging capability of the 2D plot component. ctl-d to "turn off" the range variable ("t" must be undefined for auto-ranging to work), then type -π into the left-hand x-axis place-holder and π into the right-hand one.

Stuart

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Re: Mathcad 14: graphing functions

Hmmm, loking at your worksheet I wondered, why you didn't need to vectorize g(t) in contrary to f(t).

I played around a little bit and found something I don't understand - maybe someone can shed some light on it.

Let v be a vector consisting of values greater than 0, f(t):=t*ln(t) and g(t):=t*ln(t)+t.

g(v) results in a vector (without vectorizing) and f(v) results in a simple number (which seems to represent the sum of the function results of every single element in vector v.

My only (for this post, at least) question is: WHY????? What makes that "+t" in g(t) the function g(t) so different with respect to f(t)? (OK, thats a second question). Simply don't unserstand that behaviour.

Worksheet in MC11-format to demonstrate attached.

WR

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Re: Mathcad 14: graphing functions

winfrod rager wrote:

Hmmm, loking at your worksheet I wondered, why you didn't need to vectorize g(t) in contrary to f(t).

I played around a little bit and found something I don't understand - maybe someone can shed some light on it.

Let v be a vector consisting of values greater than 0, f(t):=t*ln(t) and g(t):=t*ln(t)+t.

g(v) results in a vector (without vectorizing) and f(v) results in a simple number (which seems to represent the sum of the function results of every single element in vector v.

My only (for this post, at least) question is: WHY????? What makes that "+t" in g(t) the function g(t) so different with respect to f(t)? (OK, thats a second question). Simply don't unserstand that behaviour.

Worksheet in MC11-format to demonstrate attached.

WR

because t*ln(t) is a vector product (ie, t dot ln(t)), which is f. g adds the vector t to t.ln(t) .... try f(t)+t

Stuart

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Re: Mathcad 14: graphing functions

Yes, of course! Shame on me.

Now it all makes sense and explains, too, why function t*t is different from t^2.

Thanks alot, Stuart!

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Re: Mathcad 14: graphing functions

No worries.

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