- #1

Cyrus

- 3,150

- 16

x=f(t) and y=g(t), and that y can be written in terms of t, in other words, y=h(x)

then the chain rule gives us, dy/dx = (dy/dt)/(dx/dt).

Thats fine. The same argument holds for polar coordinates where t is replaced by theta. Looking at polar coordinates for a second, an example he has is r=1+sin(theta). Here, I don't see how it is possible to write y=h(x) directly. Its not possible to write y as a function of x in all cases. I thought this method should work ONLY when that condition is met. However, from vector calculus, I already know that this is just a special 2d case, where you can write x(t)i+y(t)j+0k. And the derivative w.r.t time or theta will give you the tangent. So the slope is just going to be (dy/dt)/(dx/dt) IN ALL CASES. Here it is not necessary to use the chain rule or the condition y=h(x). Am I wrong? ( I usually am so it wouldent surprise me) But my gripe is that he's finding the tanget to a polar curve when one of the conditions in his proof, that y=h(x) is not met. Also, I thought that the reason why we used polar coordinates was becuase we can graph functions in a polar system that we cannot in a standard cartesian coordinate system, so it would seem rather odd to have the condition y=h(x) in the proof in the first place, since that's typically something we can't do, and so we have to use polar.