# Gradient command producing multiple functions

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Gavin Seddon el 5 de Dic. de 2016
Comentada: Jan el 14 de Feb. de 2017
Hello, when I use the gradient command I am given multiple functions as the answer. I assume these are the gradient at each 'root'. Is it necessary to substitute X at a specific position to generate the gradient at the position? Furthermore, will quiver generate the slope field?
G.
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Gavin Seddon el 14 de Feb. de 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 and sorry for the delay. Gav.
Jan el 14 de Feb. de 2017
@Gavin: Please do not attach a new question as a comment to another. Open a new thread instead.

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Jan el 5 de Dic. de 2016
The functions you obtain from the gradient function are the derivatives with respect to the vector of the used variables. See https://www.mathworks.com/help/symbolic/gradient.html .
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Gavin Seddon el 6 de Dic. de 2016
Thank you, this is the page I used originally. Clearly I am getting the derivative of my initial function. I need the differential tangent to my curve at a specific X position. It is this page that shows the slope/vector field. As mentioned earlier should I pursue evaluating my gradient at a particular x position? Thank you again.
Jan el 9 de Dic. de 2016
It would be easier to answer, if you provide some details. To get the gradient at a specific point, you have to insert the coordinate values into the obtained formula.

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### Más respuestas (1)

Guillaume el 5 de Dic. de 2016
Ah, language. It's so hard to get yourself understood if you don't use the correct terms...
The gradient function does not return functions. Not one, not multiple. It returns numerical matrices / vectors.
It will return as many outputs as you have requested, so if you ask for multiple output, you'll get multiple outputs. If you ask for one, you'll only get one. Each output is the gradient of the input matrix along the corresponding dimension. I have no idea what you mean by 'root'
I also have no idea what this X you want to substitute is. You get the gradient at each point in your original matrix. If you want the gradient at points in between you either interpolate your input, or interpolate your output.
Finally, what is a slope field. quiver just plots arrow in the direction you specify. If it's the gradient of you matrix you want to plot with quiver, just pass the output of gradient to quiver (as per the example in the doc of gradient).
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Jan el 5 de Dic. de 2016
If you call gradient with symbolic expressions, you can obtain functions, see https://www.mathworks.com/help/symbolic/gradient.html .
Guillaume el 5 de Dic. de 2016
Ah... I don't have the symbolic toolbox, so wasn't aware of this gradient overload.

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