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Mar 21, 2017
01:22 AM

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Mar 21, 2017
01:22 AM

How to draw 3D plot for my mathcad program

Dear all,

I want to draw 3D plot for my mathcad program by considering,

x-axis ---> x

y-axis ---> f(x)

z-axis ---> v

I attached my code below.

Solved! Go to Solution.

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Mar 21, 2017
03:11 AM

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Mar 21, 2017
03:11 AM

Maybe you are looking for something like this:

I have scaled the function by 10^-40 because Mathcad refused to turn the ploz and watch it at different angles without.

f(x,v() is shown in z-direction.

5 REPLIES 5

Mar 21, 2017
02:29 AM

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Mar 21, 2017
02:29 AM

Hi vetri veeran,

to do what you want, we need a function of x and y, ie, v (x, y) = v(x, f (x)), which, unfortunately it is unknown to us.

Mar 21, 2017
03:11 AM

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Mar 21, 2017
03:11 AM

Maybe you are looking for something like this:

I have scaled the function by 10^-40 because Mathcad refused to turn the ploz and watch it at different angles without.

f(x,v() is shown in z-direction.

Mar 21, 2017
04:00 AM

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Mar 21, 2017
04:00 AM

Dear Werner Exinger and F.M,

Thank you for your reply.

I need to ask 2 questions to @Werner Exinger,

1. In your mathcad 3D plot, in the function M,

Is y-axis (f(x)) is in log scale similar as in 2D plot (y-axis).

2. In the 2D plot, y-axis i.e., f(x) ranges from 10^-6 to 10^23, But in the 3D plot

it contains values from 0 to 6*10^10. Is it possible to make the same axis

as in y-axis in 2D plot. If not, could you please explain me what is the difference in 3D plot

after scaling when compared with 2D plot ?

Mar 21, 2017
07:22 AM

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Mar 21, 2017
07:22 AM

Unfortunately you can't log scale an axis in a 3D-plot.

All you could do is to modify your equation.

So the drawback is that you don't see the real f values. 200 means 10^200 as log in mathcad is the base-10-log lg.

I have also shown the one curve you had drawn in your 2D-plot for comparison

In the 3D-plot we have much higher values because f gets rather quick huge when we increase v.

You can set axis limits in the plot format menu

Mar 21, 2017
11:13 AM

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Mar 21, 2017
11:13 AM

Dear Werner Exinger,

Thank you for your kind reply.