The relation between the normal and the slope of a cylindrical curve

In summary, the conversation revolves around the formula for the relation between the normal and slope of a curve, which is shown in the picture. The questioner is confused because they do not have the curve equation and are curious how the given equation was derived. The writer assumes the use of the gradient operator to obtain the normal vector on a cylindrical curve, but they do not have the function F and are unsure why the writer used the only phi component. The questioner is seeking help in obtaining the formula based on the curve coordinates.
  • #1
baby_1
159
15
As you can see in this picture:
pAyHm.jpg
This explanation "relation between the normal and the slope of a curve" is formulated here:

$$\frac{1}{\rho} \frac{d\rho }{d\psi }=\tan\left(\frac{\theta+\psi}{2}\right)$$

I got confused because I don't have the curve equation(regarding the slope of the curve and normal vector) and I am curious to know how the above equation is derived.

First I assume that the normal vector on cylindrical curve is going to obtain via gradian operator:
$$\bigtriangledown F=\frac{\partial F }{\partial r}\hat{ar}+\frac{\partial F }{r\partial \phi}\hat{\phi}+\frac{\partial F }{\partial z}\hat{az}$$
but I don't have the F function, and as you can see the above question the writer assume F as $$\rho$$ that I don't understand where it comes from and why the writer used the only phi component.

I will be grateful if you could help me to obtain the formula based on the curve coordinates.
 
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  • #2
@baby_1 what has this to do with textbooks? :wideeyed:
 
  • #3
Thank you Malawi_glenn,
Yes, it seems the administrators changed the group of my question. My question is part of an article instead of a textbook.
 

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