Finding slope of a curve at some specific points

Hi.

I did the sediment settling experiment and obtained the experimental data. My task is to calculate the average particle velocity. To compute this, I must calculate the average gradient of multiple gradients at different points. Can I know how to do this?

Answer 1

[Star Strider]

Probably the easiest way to calculate the instantaneous slope is:

dydx = gradient(y) ./ gradient(x);

All that is necessary after that is to index into it by using the index of the appropriate ‘x’ values if those are known.

If the gradient is monotonic (either increasing or decreasing), it is straightforward to get the slope at any ‘x’ value using the interp1 function:

slope_at_xval = interp1(x, dydx, xval);

If the slopes are not monotonic, this is still possible, however the code becomes a bit more complicated.

.

[Sein Lim]

If i want to find many gradients at diffeetn points, how to do this?

[Star Strider]

Calculate ‘dydx’ and then use interp1 to return the gradient at whatever you define ‘xval’ to be.

It can be a vector, and if so, ‘slope_at_xval’ will be a vector of slopes at those points.

[Sein Lim]

I apreciate your reply.

[Star Strider]

My pleasure!

If my Answer helped you solve your problem, please Accept it!

To illustrate this by plotting tangent lines at random ‘x’ values —

x = linspace(0, 4500, 1E+4);
y = 85 - [80; 70; 60].*(1-exp(-[3;2;1]*0.001*x));

dydx = gradient(y) ./ gradient(x);

xval = rand(1,3)*max(x)

xval = 1×3

1.0e+03 * 0.7819 1.6721 3.7315

for k = 1:size(y,1)
    dydx_at_xval(k,:) = interp1(x, dydx(k,:), xval);
    y_at_xval(k,:) = interp1(x, y(k,:), xval);
end

figure
plot(x, y)
hold on
for k1 = 1:size(y,1)
    for k2 = 1:numel(xval)
        b = y_at_xval(k1,k2)-dydx_at_xval(k1,k2)*xval(k2);          % Calculate Tangent Line Y-Intercept
        xrng = [-500 500]+xval(k2);
        tangent = xrng*dydx_at_xval(k1,k2) + b;
        plot(xval(k2), y_at_xval(k1,k2), 'ms')
        plot(xrng, tangent, '-r')
    end
end
hold off
grid
xlabel('X')
ylabel('Y')

.

[Sein Lim]

I am sorry, but can u create code base on my data?

[Star Strider]

Sure!

It took a bit to adapt my previous code (using row vectors) to yours (using column vectors). This illustrates the derivatives by plotting tangents to the curves at random points —

% x = linspace(0, 4500, 1E+4);
% y = 85 - [80; 70; 60].*(1-exp(-[3;2;1]*0.001*x));

T1 = readtable('Data.xlsx', 'VariableNamingRule','preserve')

T1 = 146×4 table
    Var1    40g     60g     80g 
    ____    ____    ____    ____

      0     84.8    86.4    86.7
     30       78    81.5      83
     60       69      76      79
     90     60.5      71      76
    120     52.5      66    72.5
    150       44    61.5    69.5
    180       36      56      66
    210       27    51.7      64
    240       20    44.7    61.8
    270       18    43.5    59.8
    300     16.6    40.5    58.1
    330     15.5    38.5    56.3
    360     14.8    36.5    54.7
    390     14.1    34.8      53
    420     13.6    33.2    51.5
    450     13.1    31.6      50

x = T1.Var1;
y = T1{:,2:4};
VN = T1.Properties.VariableNames;

[~,dy] = gradient(y);                                               % Gradient Of 'y' Values
dydx = dy ./ gradient(x);                                           % Derivative

xval = rand(1,3)*min(max(x))                                        % Choose 'x' Values (Can Be Any Appropriate Value)

xval = 1×3

1.0e+03 * 0.8746 4.0382 1.4755

for k = 1:size(y,2)
    dydx_at_xval(:,k) = interp1(x, dydx(:,k), xval);                % Calculate Derivatives AT 'xval'
    y_at_xval(:,k) = interp1(x, y(:,k), xval);                      % Calculate 'y' Values At 'xval'
end
% dydx_at_xval
% y_at_xval

figure
plot(x, y)
hold on
for k1 = 1:size(y,2)
    for k2 = 1:numel(xval)
        b = y_at_xval(k2,k1)-dydx_at_xval(k2,k1)*xval(k2);          % Calculate Tangent Line Y-Intercept
        xrng = [-250 250]+xval(k2);                                 % Tangent Line 'x' Range
        tangent = xrng*dydx_at_xval(k2,k1) + b;                     % 'y' Values For Tangent Line
        plot(xval(k2), y_at_xval(k2,k1), 'ms')                      % Plot Specific '(x,y)' Pair For Tangent Line
        plot(xrng, tangent, '-r')                                   % Plot Tangent Line
    end
end
hold off
grid
xlabel('X')
ylabel('Y')
legend(VN{2:end}, 'Location','best')

I am not certain what you want with respect to the derivatives, however the data appear smooth so there are essentially no problems with noise. This illustrates the derivates as tangents to the curves at various points on the x-axis. If noise were a problem, the data would need to be smoothed first.

.

[Prasanna Routray]

Hi.. this fails for my data. It fails because of data repetation as I want it all along the curve.

Is there a different way to get the slope or angle at each point on the curve and plot?

load curveData.mat;

x = curveData(1,:);

y = curveData(2,:);

dydx = gradient(y) ./ gradient(x);

xval = x(10:100); % I'm trying to get the slope for 90 points 10 to 100.

for k = 1:size(y,1)

dydx_at_xval(k,:) = interp1(x, dydx(k,:), xval);

y_at_xval(k,:) = interp1(x, y(k,:), xval);

end

figure

plot(x, y)

hold on

for k1 = 1:size(y,1)

for k2 = 1:numel(xval)

b = y_at_xval(k1,k2)-dydx_at_xval(k1,k2)*xval(k2); % Calculate Tangent Line Y-Intercept

xrng = [-500 500]+xval(k2);

tangent = xrng*dydx_at_xval(k1,k2) + b;

plot(xval(k2), y_at_xval(k1,k2), 'ms')

plot(xrng, tangent, '-r')

end

end

hold off

grid

xlabel('X')

ylabel('Y')