# Graphing vectors and partial derivatives

GiantZebra shared this question 2 years ago

I was given a problem involving the function f(x) =sqrt(2x^2-y).

I was asked to find the direction of maximum increase and the maximum increase (and the same for the direction of minimum increase and minimum increase) for f(3,2).

I found the numbers, but I can't figure out how to graph them so that I can picture it. Any ideas?

1

this?

Files: foro.ggb
1

the up attached has an error

Files: foro.ggb
1

Looks like it, but I don't understand your steps. Is it possible to explain that way I can try it out myself?

1

la explicacion está en cualquier libro o apunte sobre el gradiente de una función

puedo explicar los pasos en español pero en resumen las derivadas parciales dan la dirección de máxima pendiente y su modulo es la máxima pendiente así pues el vector es (f_x , f_y , |(f_x,f_y)|) es el vector que deseo graficar desde el punto B; esto se puede hacer con el comando vector()

```The gradient can be interpreted as the "direction and rate of fastest increase". If at a point p, the gradient of a function of several variables is not the zero vector, the direction of the gradient is the direction of fastest increase of the function at p, and its magnitude is the rate of increase in that direction
https://en.wikipedia.org/wi...```