I continue to get this error and can’t figure out how to fix it, Unable to perform assignment because the size of the left side is 1-by-4 and the size of the right side is 1-b
% Given temperature-depth data
z = [0, -2.3, -4.6, -6.9, -9.2, -11.5, -13.8, -16.1]; % Depth in meters
T = [22.8, 22.7, 22.5, 20.6, 13.9, 11.7, 11.2, 11.1]; % Temperature in Celsius
% Step 1: Plot the data points
figure;
plot(T, z, ‘r*’); % Plotting temperature vs depth with red stars
xlabel(‘Temperature (°C)’);
ylabel(‘Depth (m)’);
title(‘Temperature vs Depth’);
set(gca, ‘YDir’,’reverse’); % Inverting the y-axis
grid on;
% Step 2: Sort the data since the depth must be in increasing order
[z_sorted, sortIndex] = sort(z);
T_sorted = T(sortIndex);
% Step 3: Create a cubic spline interpolation
cs = spline(z_sorted, T_sorted);
% Step 4: Evaluate the spline and its derivatives on a finer grid
z_fine = linspace(min(z_sorted), max(z_sorted), 500);
T_spline = ppval(cs, z_fine);
% Step 5: Find the first and second derivatives of the spline
[breaks,coefs,l,k,d] = unmkpp(cs); % Extracts the pieces of the cubic spline
dcoefs = coefs; % Derivative coefficients
% Each row of dcoefs will be the coefficients of the polynomial of a piece
for j = 1:l
dcoefs(j,:) = polyder(dcoefs(j,:));
end
% Make a pp-form of derivative
csd1 = mkpp(breaks,dcoefs(:,1:k-1));
% First derivative evaluation
T_spline_deriv = ppval(csd1, z_fine);
% Find the second derivative
for j = 1:l
dcoefs(j,:) = polyder(dcoefs(j,:));
end
% Make a pp-form of second derivative
csd2 = mkpp(breaks,dcoefs(:,1:k-2));
% Second derivative evaluation
T_spline_second_deriv = ppval(csd2, z_fine);
% Step 6: Locate the thermocline by finding the depth where the second derivative
% changes sign, and the first derivative is a maximum
inflection_points = find(diff(sign(T_spline_second_deriv)) ~= 0) + 1;
[~, max_gradient_index] = max(abs(T_spline_deriv(inflection_points)));
thermocline_depth = z_fine(inflection_points(max_gradient_index));
thermocline_temperature = T_spline(inflection_points(max_gradient_index));
% Display the thermocline depth and temperature
fprintf(‘The thermocline is located at a depth of %.2f m with a temperature of %.2f°C.n’, …
thermocline_depth, thermocline_temperature);% Given temperature-depth data
z = [0, -2.3, -4.6, -6.9, -9.2, -11.5, -13.8, -16.1]; % Depth in meters
T = [22.8, 22.7, 22.5, 20.6, 13.9, 11.7, 11.2, 11.1]; % Temperature in Celsius
% Step 1: Plot the data points
figure;
plot(T, z, ‘r*’); % Plotting temperature vs depth with red stars
xlabel(‘Temperature (°C)’);
ylabel(‘Depth (m)’);
title(‘Temperature vs Depth’);
set(gca, ‘YDir’,’reverse’); % Inverting the y-axis
grid on;
% Step 2: Sort the data since the depth must be in increasing order
[z_sorted, sortIndex] = sort(z);
T_sorted = T(sortIndex);
% Step 3: Create a cubic spline interpolation
cs = spline(z_sorted, T_sorted);
% Step 4: Evaluate the spline and its derivatives on a finer grid
z_fine = linspace(min(z_sorted), max(z_sorted), 500);
T_spline = ppval(cs, z_fine);
% Step 5: Find the first and second derivatives of the spline
[breaks,coefs,l,k,d] = unmkpp(cs); % Extracts the pieces of the cubic spline
dcoefs = coefs; % Derivative coefficients
% Each row of dcoefs will be the coefficients of the polynomial of a piece
for j = 1:l
dcoefs(j,:) = polyder(dcoefs(j,:));
end
% Make a pp-form of derivative
csd1 = mkpp(breaks,dcoefs(:,1:k-1));
% First derivative evaluation
T_spline_deriv = ppval(csd1, z_fine);
% Find the second derivative
for j = 1:l
dcoefs(j,:) = polyder(dcoefs(j,:));
end
% Make a pp-form of second derivative
csd2 = mkpp(breaks,dcoefs(:,1:k-2));
% Second derivative evaluation
T_spline_second_deriv = ppval(csd2, z_fine);
% Step 6: Locate the thermocline by finding the depth where the second derivative
% changes sign, and the first derivative is a maximum
inflection_points = find(diff(sign(T_spline_second_deriv)) ~= 0) + 1;
[~, max_gradient_index] = max(abs(T_spline_deriv(inflection_points)));
thermocline_depth = z_fine(inflection_points(max_gradient_index));
thermocline_temperature = T_spline(inflection_points(max_gradient_index));
% Display the thermocline depth and temperature
fprintf(‘The thermocline is located at a depth of %.2f m with a temperature of %.2f°C.n’, …
thermocline_depth, thermocline_temperature); % Given temperature-depth data
z = [0, -2.3, -4.6, -6.9, -9.2, -11.5, -13.8, -16.1]; % Depth in meters
T = [22.8, 22.7, 22.5, 20.6, 13.9, 11.7, 11.2, 11.1]; % Temperature in Celsius
% Step 1: Plot the data points
figure;
plot(T, z, ‘r*’); % Plotting temperature vs depth with red stars
xlabel(‘Temperature (°C)’);
ylabel(‘Depth (m)’);
title(‘Temperature vs Depth’);
set(gca, ‘YDir’,’reverse’); % Inverting the y-axis
grid on;
% Step 2: Sort the data since the depth must be in increasing order
[z_sorted, sortIndex] = sort(z);
T_sorted = T(sortIndex);
% Step 3: Create a cubic spline interpolation
cs = spline(z_sorted, T_sorted);
% Step 4: Evaluate the spline and its derivatives on a finer grid
z_fine = linspace(min(z_sorted), max(z_sorted), 500);
T_spline = ppval(cs, z_fine);
% Step 5: Find the first and second derivatives of the spline
[breaks,coefs,l,k,d] = unmkpp(cs); % Extracts the pieces of the cubic spline
dcoefs = coefs; % Derivative coefficients
% Each row of dcoefs will be the coefficients of the polynomial of a piece
for j = 1:l
dcoefs(j,:) = polyder(dcoefs(j,:));
end
% Make a pp-form of derivative
csd1 = mkpp(breaks,dcoefs(:,1:k-1));
% First derivative evaluation
T_spline_deriv = ppval(csd1, z_fine);
% Find the second derivative
for j = 1:l
dcoefs(j,:) = polyder(dcoefs(j,:));
end
% Make a pp-form of second derivative
csd2 = mkpp(breaks,dcoefs(:,1:k-2));
% Second derivative evaluation
T_spline_second_deriv = ppval(csd2, z_fine);
% Step 6: Locate the thermocline by finding the depth where the second derivative
% changes sign, and the first derivative is a maximum
inflection_points = find(diff(sign(T_spline_second_deriv)) ~= 0) + 1;
[~, max_gradient_index] = max(abs(T_spline_deriv(inflection_points)));
thermocline_depth = z_fine(inflection_points(max_gradient_index));
thermocline_temperature = T_spline(inflection_points(max_gradient_index));
% Display the thermocline depth and temperature
fprintf(‘The thermocline is located at a depth of %.2f m with a temperature of %.2f°C.n’, …
thermocline_depth, thermocline_temperature); matrices MATLAB Answers — New Questions
