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How to compute the gradient at each point of an arbitrary symbolic curve defined by "n" points?

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Hello,
I have a vector of "n" points describing a spline. I would like to compute the gradient at each of the "n" points.
Each point in this vector has 3 components, each of which is a symbolic expression in multiple variables.
The reason I am computing the gradients is to compare 2 curves (one of which I know each point in a 3-D space) with this. My criterion for comparison is that the cross product of gradients at each point must be zero. I am able to compute the gradients of the known curve using the matlab function "gradient". However, when i use the same function to compute for the symbolic points, I get the error
**************************************
Error using sym/gradient (line 21)
First argument must be scalar.
***************************************
Any help on how I may proceed is apprecciated.
For ref:
known_points=[3Xn] % format double
dx_1=gradient(known_points(1,1:end));
dy_1=gradient(known_points(2,1:end));
dz_1=gradient(known_points(3,1:end));
% I am able to compute the above expressions.
unknown_points=[3Xn] %format sym
dx_2=gradient(unknown_points(1,1:end));
dy_2=gradient(unknown_points(2,1:end));
dz_2=gradient(unknown_points(3,1:end));
% I get the highlighted error when i compute the above expressions.
% alternately i have also tried dx_2=gradient(unknown_points(1,1:end), symvar(unknown_points)); to no avial.

採用された回答

VBBV
VBBV 2022 年 2 月 13 日
編集済み: VBBV 2022 年 2 月 13 日
gradient(subs(unknown_points,x,value));
If x is the symbolic variables in array unknown_points then use subs to replace the variable with your value. Then compute the gradient as above.
  2 件のコメント
Suhas Raghavendra Kulkarni
Suhas Raghavendra Kulkarni 2022 年 2 月 13 日
Thank you for your answer. If I understand correctly, you are asking me to replce the desired values in place of the variable.
Unfortunately this is not possible as the symbolic function is actually a very long non-linear function in multiple variables. So I do not have the desired value of all the variable to replace in the expression.
Please correct me if my understanding of your answer is wrong.
VBBV
VBBV 2022 年 2 月 13 日
Ok. In that case use diff without subs function instead of gradient with subs

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