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subinator
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This is a problem that has been bugging me all day. While working with the well-known dydx = rdrdθ, where r is a function of θ I divided both sides of the equation by dxdθ to get dy/dθ = r(dr/dx)
For the left side, I use y = rsinθ and derive with respect to θ to get dy/dθ = sinθdr/dθ + rcosθ. For the left side, I use r^2 = y^2 + x^2, and derive both sides dx, to get
2r(dr/dx)=2y(dy/dx) + 2x, which simplifies to r (dr/dx)=y(dy/dx) + x.
I then put both of these equalities in the equation to get
sinθ(dr/dθ) + rcosθ = y(dy/dx) + x. Knowing that x = rcosθ, and that y/r = sinθ
I subtract x then divide y from both sides to get
(1/r)(dr/dθ) = (dy/dx).
However, this contradicts the proof that
dy/dx =((dr/dθ)sinθ+rcosθ)/((dr/dθ)cosθ-rsinθ).
I just want to know what went wrong. (I have checked mathematically, these two functions will NOT give the same results at most points.)
For the left side, I use y = rsinθ and derive with respect to θ to get dy/dθ = sinθdr/dθ + rcosθ. For the left side, I use r^2 = y^2 + x^2, and derive both sides dx, to get
2r(dr/dx)=2y(dy/dx) + 2x, which simplifies to r (dr/dx)=y(dy/dx) + x.
I then put both of these equalities in the equation to get
sinθ(dr/dθ) + rcosθ = y(dy/dx) + x. Knowing that x = rcosθ, and that y/r = sinθ
I subtract x then divide y from both sides to get
(1/r)(dr/dθ) = (dy/dx).
However, this contradicts the proof that
dy/dx =((dr/dθ)sinθ+rcosθ)/((dr/dθ)cosθ-rsinθ).
I just want to know what went wrong. (I have checked mathematically, these two functions will NOT give the same results at most points.)
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