Plot y^2 = x^2 + 1 (The expression to the left of the equals sign is not a valid target for an assignment.)
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%Clear memory
clear;
%Number of points
N = 10000;
%Preallocate memory
x = zeros(1,N);
y1 = zeros(1,N);
x = -5 + (5+5)*rand(1,N);
y1^2 = x^2 + 1;
plot(x,y1), grid on;
I get the following error:
The expression to the left of the equals sign is not a valid target for an assignment.
How can I plot this first on x,y axis, then on x^2 and y^2?
Thanks in advance.
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採用された回答
MA
2014 年 11 月 1 日
clear all
clc;
N = 10000;
x = zeros(1,N);
y1 = zeros(1,N);
y2 = zeros(1,N);
x = -5 + (5+5)*rand(1,N);
for i=1:N
y1(i)=(x(i)^2 + 1)^(1/2);
y2(i)=-(x(i)^2 + 1)^(1/2);
end
plot(x,y1)
hold on
plot(x,y2)
grid on
good luck
1 件のコメント
David Young
2014 年 11 月 1 日
Seems a shame to give an answer that uses a loop when it would be more natural and simpler to use the vector form.
その他の回答 (4 件)
David Young
2014 年 11 月 1 日
The error message says it all: you can't assign something to y1^2. Remember that in a programming language, the "=" operator isn't like "=" in algebra. What it does is to assign the value of the expression to its right to the variable (or, in MATLAB, to the matrix elements) given by the expression on its left.
So you need a variable on the left here. How about taking the square root?
y1 = sqrt(x.^2 + 1);
That might not get you all the way (because the square root has more than one solution) but it's a start.
Two things to note. You need .^, not ^. And the preallocation step isn't necessary; in fact it just wastes time and adds complexity.
0 件のコメント
Roger Stafford
2014 年 11 月 1 日
編集済み: Roger Stafford
2014 年 11 月 1 日
You can generate this hyperbola plot using hyperbolic functions as parameters:
t = linspace(-4,4,500);
x = sinh(t);
y = cosh(t);
plot(x,y,'y-')
axis equal
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