Finding slope of a curve at some specific points

Question

Sein Lim 2023-4-17

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https://ww2.mathworks.cn/matlabcentral/answers/1948218-finding-slope-of-a-curve-at-some-specific-points

编辑： Image Analyst 2024-12-3

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?

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Answer 1

Star Strider 2023-4-17

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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.

.

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Star Strider 2023-4-17

在 MATLAB Online 中打开

Data.xlsx

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

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.

.

Star Strider 2024-12-2

在 MATLAB Online 中打开

curveData.mat

That code is not appropriate for your data or what you want to do.

Try this —

load curveData.mat;

x = curveData(1,:);

y = curveData(2,:);

x = x(:);

y = y(:);

% xy = [x(:) y(:)]

figure

plot(x, y, DisplayName='All Data')

hold on

plot(x(10:100), y(10:100), '.r', DisplayName='Selected Data')

plot(x(42), y(42), 'pg', DisplayName='Maximum |Slope|')

hold off

grid

legend(Location='best')

ixv = 10:100;

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

% [maxslope,idx] = max(abs(dydx))

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

yval = y(ixv);

itcpt = yval - dydx .* xval; % Calculate Intercept From Slope & x-Values

Result = table(xval, yval, dydx, itcpt, VariableNames={'X','Y','Slope','Intercept'})

Result = 91x4 table

X Y Slope Intercept _______ ___ ________ _________ -4.7653 1.8 -0.53713 -0.75961 -5.1377 2 -0.5553 -0.85297 -5.4856 2.2 -0.5939 -1.0579 -5.8112 2.4 -0.63436 -1.2864 -6.1162 2.6 -0.67632 -1.5365 -6.4026 2.8 -0.7192 -1.8047 -6.6724 3 -0.76225 -2.086 -6.9274 3.2 -0.80512 -2.3774 -7.1692 3.4 -0.84928 -2.6887 -7.3984 3.6 -0.89659 -3.0333 -7.6153 3.8 -0.94762 -3.4164 -7.8205 4 -1.0027 -3.8418 -8.0142 4.2 -1.0623 -4.3135 -8.197 4.4 -1.1268 -4.8361 -8.3692 4.6 -1.1966 -5.4145 -8.5313 4.8 -1.2723 -6.0542

figure

plot(x, y)

hold on

for k = 1:numel(ixv)

idxrng = (ixv(k)-5) : (ixv(k)+5);

xrng = x(idxrng);

yrng = [xrng ones(size(xrng))] * [dydx((k)); itcpt((k))];

xpt = x(ixv(k));

ypt = [xpt 1] * [dydx(k); itcpt(k)];

hpp = plot(xpt, ypt, 'p');

hpp.MarkerFaceColor = hpp.Color;

plot(xrng, yrng, Color=hpp.Color)

end

ylim([-3 21])

title('All Data')

figure

plot(x, y)

hold on

for k = 1:numel(ixv)

idxrng = (ixv(k)-5) : (ixv(k)+5);

xrng = x(idxrng);

yrng = [xrng ones(size(xrng))] * [dydx((k)); itcpt((k))];

xpt = x(ixv(k));

ypt = [xpt 1] * [dydx(k); itcpt(k)];

hpp = plot(xpt, ypt, 'p');

hpp.MarkerFaceColor = hpp.Color;

plot(xrng, yrng, Color=hpp.Color)

end

ylim([9 11])

title('Detail')

% return

%

% 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

% hold off

% grid

% xlabel('X')

% ylabel('Y')

.

Prasanna Routray 2024-12-3

编辑：Image Analyst 2024-12-3

在 MATLAB Online 中打开

curveData.mat

You are a star as always.

Here is what I tried and it works.

load curveData.mat;

x = curveData(1,:);

y = curveData(2,:);

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

idx=450;

xv = x(idx);% Choose A Random Point

yv = interp1(x(idx:idx+1), y(idx:idx+1), xv); % Interpolate 'y' Vector

dydxv = interp1(x(idx:idx+1), dydx(idx:idx+1), xv); % Interpolate Derivative Vector

angleSlope = atand(dydxv);

if angleSlope<0

angleSlope = angleSlope + 180;

end

figure

plot(x, y)

hold on

plot(xv, yv, '*r')

hold off

text(xv, yv, sprintf(' \\leftarrow Slope at (%.3f, %.3f) = %.3f',xv, yv, angleSlope), 'Horiz','left', 'Vert','middle')

axis([-50 50 -20 220]); % Ensure equal scaling on both axes

Star Strider 2024-12-3

@Prasanna Routray —

My pleasure!

A Vote would be appreciated!

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Answer 2

Image Analyst 2024-12-3

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Wouldn't the average velocity just be the total displacement divided by the total time? I don't see why you need to compute the slope at a bunch of points along the curve and then average them together. That seems like the hard way of doing it.

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Finding slope of a curve at some specific points

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