Differentiate and store results in matrix

I have two matrices. The first matrix is Density and it 311x1. The second matrix is Temperature and it is 311x21. The column in density matrix corresponds to each distinct column in temperature matrix.

So if I want to plot the first line I could just do:

plot(Density, Temperature(:,1))

and second line would just be:

plot(Density, Temperature(:,2))

How can I create a new matrix that takes the derivative of each temperature line (isotherm) with respect to density, and put this in a new matrix, such as diffTemperature?

Answer 1

[Star Strider]

Try this:

Density = (0:310)';                                         % Create Data
Temperature = sin((0:310)'*(0:20)*2*pi/20);                 % Create Data

dTdD = gradient(Temperature, mean(diff(Density)));          % Equal Spacing
dTdD = gradient(Temperature) ./ gradient(Density);          % Unequal Spacing

figure
plot(Density, Temperature(:,5))
hold on
plot(Density, dTdD(:,5))
hold off
grid
title('Example')
legend('Temperature', 'Derivative')
xlim([0 30])

[Benjamin]

What do you mean between equal vs. unequal spacing? Also, the result seems off. The derivative is much different than it should be. If I replace gradient with diff, why is the produced derivative so much different?

[Star Strider]

The gradient (link) function calculates the numerical derivative of a vector or matrix. The output is the same size as the argument vector (or matrix). For a matrix, you must request two outputs, the first is the horizontal gradient along the rows, and the second is the vertical gradient down the columns.

If you want to calculate it as:

dTdD = gradient(Temperature(:,1))

that is not an issue.

It requires that the vector elements are equally spaced with respect to time (or other independent variable). You can get around this by dividing it by the gradient of the independent variable.

The gradient function is more accurate than diff with respect to taking the derivative, because of the central difference method it uses to estimate the derivative.

[Benjamin]

Thanks for your help, I just am hoping to get another follow-up here. I think I am plotting the derivatives incorrectly because gradient and diff are looking like night and day compared to each other.

If I have x data and y data in a matrix. So I have density and pressure, let' say. Density = x, pressure = y. If I want to differentiate pressure wrt to density, would I use:

dPdD = gradient(Pressure(:,1))

or would I use:

dPdD = gradient(Pressure(:,1)) / gradient(Density)

Basically, I have x data and y data and I just want to plot the first and second derivatives.

[Star Strider]

Use:

dPdD = gradient(Pressure(:,1)) / gradient(Density)

To calculate the second derivative, use the gradient function on ‘dPdD’.

[Benjamin]

I want to calculate the derivative on all columns of data, not just the 1st though. Pressure is 310x21. Each column I need to take derivative of. Currently, the equation you provided creates a 310x310 matrix with all zeros except the last column

[Star Strider]

Try this:

[~,dTdD] = gradient(Temperature, mean(diff(Density)));          % Equal Spacing

or:

[~,dTdD] = gradient(Temperature);
dTdD = bsxfun(@rdivide, dTdD , gradient(Density));              % Unequal Spacing

Those should work.

You can use the ‘Unequal Spacing’ version for both. It is slightly less efficient if you have equal spacing, because it requires two gradient calls, and one bsxfun call.

[Benjamin]

Wow, hey that works! What would it be to take 2nd derivative?

[~,ddPddD] = gradient(dPdD, mean(diff(density)));

[Star Strider]

Great!

Yes. That will work for equally-spaced data.

EDIT — An alternative could also be the del2 (link) function on ‘Temperature’ (or any other dependent variable), not the derivative, (although you may need to multiply the result by 4). The syntax is essentially the same.