Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

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 $$\frac{\dot x}{\dot y} = \frac{3}{4t} = \frac{3}{4} y^{-1/4} = \frac{dx}{dy}.$$

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.