slope at defined co-ordinates

3 vues (au cours des 30 derniers jours)
Gavin Seddon le 15 Fév 2017
Hello I have defined my distance function as
d = (5757*x^3)/100000 - (2227*x^2)/10000 + (1303*x)/2500 + 25/2
I then differentiate this to get diff
(5757*x^3)/100000 - (2227*x^2)/10000 + (1303*x)/2500 + 25/2
ans =
(1303*x)/2500 - (4999*x^2)/100000 + 25/2
how would I calculate the gradient at varying X values?
Thanks. Gav.
0 commentairesAfficher -2 commentaires plus anciensMasquer -2 commentaires plus anciens

Connectez-vous pour commenter.

Réponse acceptée

Star Strider le 15 Fév 2017
The gradient is simply the derivative, since yours is a univariate function. I would create a vectorized anonymous function from it, and use it as I would any other function:
dd_dx = @(x) (1303*x)/2500 - (4999*x.^2)/100000 + 25/2;
then call it as:
x = linspace(0, 10);
figure(1)
plot(x, dd_dx(x))
grid
0 commentairesAfficher -2 commentaires plus anciensMasquer -2 commentaires plus anciens

Connectez-vous pour commenter.

Catégories

En savoir plus sur Assembly dans Help Center et File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by