Error question :The expression to the left of the equals sign is not a valid target for an assignment

Hi, i'm trying to run this sintax:

t = [0,2*pi];

r^2 = 17^2*cos(2t)+sqrt(6^4-17^4*sin(2t)^2)

polar(t,r)

and when ii run it, it says: The expression to the left of the equals sign is not a valid target for an assignment and i don't know how to make it work, i know that is a simple sintax, but i am a beginner :)!

Thanks!

[mohammad azsad]

yes it is wrong you can write >> t=[0:(2*pi/100):2*pi];r=sqrt((17.^2)*cos(2*t)+sqrt(((6.^4)-(17.^4)).*((sin(2*t)).^2))) polar(t,r)

Answer 1

[Harry]

Try this:

% Define a vector of time values
dt = 0.01;
t = 0:dt:2*pi;

r = sqrt(17^2*cos(2*t)+sqrt(6^4-17^4*sin(2*t).^2));
polar(t,r);

This is what I changed:

1) To define t on the interval [0,2*pi], you must create a vector of numbers (for example [0,0.01,0.02,...]).

2) The error you saw happened because you had "r^2" on the left hand side of an equation. In fact, you want to assign a value to "r", so just take the square root of both sides.

3) In order to calculate sin(2t)^2, you must use the ".^2" operator, since sin(2*t) is a vector and you want to raise every element to the power 2.

[Mihai]

It works now ,thanks, but if you see in my sintax, r is r^2 this is how it looks like : r^2=b^2*cos(2t)+sqrt(a^4-b^4*sin(2t)^2)

[Mihai]

Ok, i understand what have you done, thank you very much an have a good day!

[Harry]

Great. Please accept my answer if you are happy with it.