can you try solving the last two parts

[jjones-6]

i am supposed to

generate  two graphs which i did on vm.x1 and the second of vm vs.x2 and comment on the trends observed

2.develop a 3rd order polynomial fit of vm vs. x2 and use this fit to predict values of vm at the x2 values of 0.06,0.07,0.08,0.09

3.find the value of the derivative of the 3rd polynomial fit function with respect to x2 (use the values from the original x2 data).

[ValeryOchkov]

[jjones-6]

this is great, you did real good, can you complete the question by taking the derivative of the result you obtained please

3. find the value of the derivative of the 3rd polynomial fit function with respect to x2.

4. aslo can you expand the graph of x2. vs vm so that the values of x2 are between 0.01 to 0.09 so that i can predict the values vm using the polynomial fit please

can you attach the mathcad file so that i can manipulate the data. i am using mathcad prime 3.1

Thank man.

[LucMeekes]

"i am using mathcad prime 3.1"

Does not match with the photograph you show in the other thread, nor with the attachment to your first post in this thread. You're using Mathcad 15, not Prime.

Luc

[jjones-6]

yes the file is in math cad15  because I am using the online server through my school but my computer at home has math prime 3.1, mathcad 15 is fine as long as you can you can complte the whole question by finding  the derivative of the 3rd polynomial fit function with respect to x2. and expand the values of x2 after the plot and then send attach the file. You did real got I wanna use the file to see the good work you did. thanks

[LucMeekes]

Here's the whole stuff, that Fred also did, see below, but from a slightly different angle:

Now to fit a 3rd-order polynome:

Interpolate for the given values:

The derivative of f(x) is:

with the coefficients for x and x^2 you can easily see that it hardly depends on x, a good approximation is -111.1.

part 3 of the exercise: what 3rd polynomial fit function?  Only one polynomial fit function was requested (see f(x) above).

But anyway if you apply f(x) to any of the values of x2 (or x1) the result is (very close to) -111.1.

Luc

[Fred_Kohlhepp · Accepted solution]

here