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I read that if $x$ and $y$ depend solely on $t$, then:

$$\frac{\mathrm dx}{\mathrm dy}=\frac{\mathrm d{\dot x}}{\mathrm d{\dot y}}$$

It makes sense to me because by the chain rule:

$$\frac{\mathrm d{\dot x}}{\mathrm d{\dot y}}=\frac{\frac{\mathrm dx}{\mathrm dt}}{\frac{\mathrm dy}{\mathrm dt}}=\frac{\frac{\mathrm dx}{\mathrm dy}\frac{\mathrm dy}{\mathrm dt}}{\frac{\mathrm dy}{\mathrm dt}}=\frac{\mathrm dx}{\mathrm dy}$$

However, this seems to be a counter-example.

$$\mathrm x(y)=y^{3/4}=t^{3}$$ $$\mathrm y(t)=t^{4}$$

$$\frac{\mathrm dx}{\mathrm dy}\frac{\mathrm dy}{\mathrm dt}=\frac{\mathrm dx}{\mathrm dt}$$ $$\frac{\mathrm dx}{\mathrm dy}{\mathrm 4t^{3}}={\mathrm 3t^2}$$ $$\frac{\mathrm dx}{\mathrm dy}=\frac{\mathrm 3}{\mathrm 4t}$$

$$\frac{\mathrm d{\dot x}}{\mathrm d{\dot y}}\frac{\mathrm d{\dot y}}{\mathrm dt}=\frac{\mathrm d{\dot x}}{\mathrm dt}$$ $$\frac{\mathrm d{\dot x}}{\mathrm d{\dot y}}{\mathrm 12t^{2}}={\mathrm 6t}$$ $$\frac{\mathrm d{\dot x}}{\mathrm d{\dot y}}=\frac{\mathrm 1}{\mathrm 2t}$$

Where have I gone wrong here?

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1 Answer 1

up vote 2 down vote accepted

Your first displayed equation is wrong. It should read: $$\frac{dx}{dy} = \frac{\dot{x}}{\dot{y}}$$ Note that, in your example, $\dot x = 3t^2$ and $\dot y = 4t^3$, so indeed $$\frac{\dot x}{\dot y} = \frac{3}{4t} = \frac{3}{4} y^{-1/4} = \frac{dx}{dy}.$$

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Thanks for the help on this. I've posted another question regarding the situation that prompted this question here: math.stackexchange.com/questions/311572/… –  Gerber Feb 22 '13 at 22:16

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