Sea ice in the Arctic ocean increases and decreases annually. Declines have even

Question

Sea ice in the Arctic ocean increases and decreases annually. Declines have even made it possible for shipping to occur through the "Northwest Passage" in recent summer seasons. The following data, shown in Table 8.10, were obtained from the NASA Earth Observatory website and represent the maximum and minimum sea ice extent, measured in September (minimum) and March (maximum) of the following year. The first line in the table is the average for the previous 20 years. (a) Plot both the September values and March values vs. the year, and add a line representing the average ice extent for 1979-2000. (b) Use the 1979-2000. (c) Use the find command to determine which years were lower than the average February values for 1979-2000. (d) Since the extent of Arctic sea ice has continually decreased from the previous 20 year average, a more interesting calculation would be to find the average sea ice extent over the years from 1999 to 2012. Repeat parts (b) and (c) using this new average. find command to determine which years exceeded the average September values for

Explanation / Answer

Please find the MATLAB code corresponding to the above problem. Please note that:

1) Description to the solution is provided in the comments in between the code.

2) Scipts corresponding to the plots have been commented, which can be uncommneted and added with the suitable title as per the user requirement.

3) Part (c) refers the data corresponding to the month of february, which is absent from the given text. Hence, data corresponding to the month of march has been assumed.

4) In case of any further clarification, please comment.

Hope this helps! If it works, please thumbs up!

Please find the code below!!!

==============================================

clear all;
clc;

% ================ Initialization with given data =================
sept_avg=7.0;
march_avg=15.7;
sept=[6.2 6.3 6.8 6.0 6.2 6.1 5.6 5.9 4.3 4.7 5.4 4.9 4.6 3.6];
march=[15.3 15.6 15.4 15.5 15.1 14.7 14.4 14.7 15.2 15.2 15.1 14.6 15.2 15.0];
year=[1999,2000,2001,2002,2003,2004,2005,2006,2007,2008,2009,2010,2011,2012];
years=size(years);
avg_line(1,1:14)=sept_avg;
avg_line(2,1:14)=march_avg;

%================================================
%================ Part (a) ======================
% Uncomment below to view the respective plot
%================================================

%======= September Plot =========
% plot(year,avg_line(1,:))
% hold on
% plot(year,sept);
% axis([1999 2012 0 10]);
%=================================

%========== March Plot ===========
% plot(year,avg_line(2,:))
% hold on
% plot(year,march);
% axis([1999 2012 0 20]);
%=================================

%========================================================
%================ Part (b) and (c) ======================
% Finding the column indices of the values for which the values are greater
% than the average value for the 1979 - 2000
[r_1,c_1]=find(sept>sept_avg);      % Calculation of September part(b)
display('The required years for the month of september are:');
if size(c_1')~=0
for i=1:size(c_1')
disp(year(c_1(i)));
end
else
    display('No such year found');
end
% NOTE: As the values are always below the average values of 1979-2000 for september,
% the resulting indices will be empty. Hence, the answer will be 'no such year exists'

[r_2,c_2]=find(march<march_avg);    % Calculation for March part(c)
display('The required years for the month of march are:');
if size(c_2')~=0
for i=1:size(c_2')
disp(year(c_2(i)));
end
else
    display('No such year found');
end
% In this case as every year lier below the average from 1990-2000, all the
% years will be displayed

%================================================
%================ Part (d) ======================
%================================================

new_sept_avg=mean(sept);     % Calculation of average from 1999-2012 for september
new_march_avg=mean(march);   % Calculation of average from 1999-2012 for march
[r_1,c_1]=find(sept>new_sept_avg);      % Calculation of September
[r_2,c_2]=find(march<new_march_avg);    % Calculation for March

display('The required years for the month of september are:');
if size(c_1')~=0
for i=1:size(c_1')
disp(year(c_1(i)));
end
else
    display('No such year found');
end
display('The required years for the month of march are:');
if size(c_2')~=0
for i=1:size(c_2')
disp(year(c_2(i)));
end
else
    display('No such year found');
end

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