Graph Derivative in 3D View (or rotate function)

BeksOmega shared this question 2 years ago
Answered

Hello!

I am attempting to demonstrate what a derivative is visually. I have a function, and then then I graph all of the slopes of the function as a plane in the 3d view. The x and y define the two points, and the z is the slope between those points.

Here's an example of the sin function:

d3a09bb903e227fe576fe84ce9ad4c77

The result of this is that if you take all of the limit of the z vaules along the diagonal you get the derivative of the function. This is because along the diagonal x = y, so the z value is essentially the slope at a point.

Hopefully that explains sort of the idea behind it.

What I would like to be able to do is draw a line showing all of those z values along the diagonal. The problem is that since they're actually undefined, this gets tricky :/

What I've Tried

1) Use a plane intersection. The plane is vertical and goes through (1, 1). This would be correct, but doesn't work because all of the points along the diagonal are actually undefined.

2) Graph the derivative and then rotate it. I rotated the function 90 degrees around the x and then 45 degrees around the z. This doesn't quite line up :/ I'm not sure if there is a different set of rotations that would work.

3) g(x, y) = If(x==y, f'(x)) -- Geogebra doesn"t seem to know how to interpret this. It allows me to enter the command, but nothing is displayed.


Does anyone know how I could draw the derivative of the function in my 3D view so that it lines up with the plane of the function?

Any help is very much appreciated! Thank you for your time!

--Beka

Best Answer
photo

Ad 3) Maybe try to use command Curve:

Curve(t, t, f'(t), t, -3, 3)

Comments (2)

photo
3

Ad 3) Maybe try to use command Curve:

Curve(t, t, f'(t), t, -3, 3)

photo
1

Yay that works! Thank you so much :D

photo
© 2023 International GeoGebra Institute