Yes.kent davidge said:so this has only made solving the problem harder than it must be, right?
If you do this, only one of your polar coordinates will be a free parameter as the other will be fixed. For example, consider the line ##x = 1##. On this line, ##r^2 = 1 + y^2## which means that, if you fix ##\theta##, then ##r^2(1-\sin^2\theta) = 1## and so ##r = 1/\cos\theta##. The relation between ##y## and ##\theta## is therefore ##y = \tan\theta##, which is just a (non-linear) change of parametrisation of your line.kent davidge said:If I have a physical problem, say, a particle which is constrained to move in the ##y## direction, which means that its ##x## coordinate remains fixed, does it make sense to write ##y## in terms of polar coordinates? That is, ##y = r \sin\theta##. Since now I have two parameters ##r,\theta## varying, so this has only made solving the problem harder than it must be, right?
Polar coordinates are a system of representing points in a plane using a distance from the origin and an angle from a fixed reference direction.
In 1-dimensional problems, polar coordinates are used to represent points along a single axis, typically the x-axis. This allows for a more efficient and intuitive way of solving problems involving circular or rotational motion.
One advantage of using polar coordinates in 1-dimensional problems is that it simplifies the equations and makes them easier to solve. It also allows for a more visual representation of the problem, making it easier to understand and analyze.
Yes, there are some limitations to using polar coordinates in 1-dimensional problems. It is not suitable for all types of problems and may not be as accurate as other coordinate systems in certain situations.
To convert from polar coordinates to Cartesian coordinates, you can use the formulas x = r * cos(theta) and y = r * sin(theta). To convert from Cartesian coordinates to polar coordinates, you can use the formulas r = sqrt(x^2 + y^2) and theta = arctan(y/x).