Differentiation- finding curvature

[jonnburton]

I have been trying to follow the working in my book on this topic, but there is one step which I can't understand. Can anyone help me unpick what is going on here?

The curvature \(\displaystyle \frac{d\theta}{ds} = \frac{1}{R}\)

\(\displaystyle \frac{dy}{dx}=tan\theta\) and \(\displaystyle \frac{dx}{ds} = cos\theta\)

\(\displaystyle \frac{dy}{dx}=tan\theta\). Need to differentiate with respect to s.

\(\displaystyle \frac{d}{ds}(\frac{dy}{dx}) = \frac{d}{ds}tan\theta\)

But this next step is what I can't understand:

\(\displaystyle \frac{d}{dx}(\frac{dy}{dx}) * \frac{dx}{ds} = \frac{d}{d\theta}(tan\theta)*\frac{d\theta}{ds}\)



I can't see how the left hand side is equivalent to the right hand side here.


On the left hand side we have got: the derivative of \(\displaystyle tan\theta\) with respect to x, multiplied by \(\displaystyle \frac{dx}{ds}\)

On the right hand side we have got the derivative with respect to \(\displaystyle \theta\) of \(\displaystyle tan\theta\), multiplied by \(\displaystyle \frac{d\theta}{ds}\)

I know that this is almost certainly the application of the chain rule, however I can't see how it is working here.

 

[DrPhil]

I have been trying to follow the working in my book on this topic, but there is one step which I can't understand. Can anyone help me unpick what is going on here?

The curvature \(\displaystyle \frac{d\theta}{ds} = \frac{1}{R}\)

\(\displaystyle \frac{dy}{dx}=tan\theta\) and \(\displaystyle \frac{dx}{ds} = cos\theta\)

\(\displaystyle \frac{dy}{dx}=tan\theta\). Need to differentiate with respect to s.

\(\displaystyle \frac{d}{ds}(\frac{dy}{dx}) = \frac{d}{ds}tan\theta\)

But this next step is what I can't understand:

\(\displaystyle \frac{d}{dx}(\frac{dy}{dx}) * \frac{dx}{ds} = \frac{d}{d\theta}(tan\theta)*\frac{d\theta}{ds}\)



I can't see how the left hand side is equivalent to the right hand side here.


On the left hand side we have got: the derivative of \(\displaystyle tan\theta\) with respect to x, multiplied by \(\displaystyle \frac{dx}{ds}\)

On the right hand side we have got the derivative with respect to \(\displaystyle \theta\) of \(\displaystyle tan\theta\), multiplied by \(\displaystyle \frac{d\theta}{ds}\)

I know that this is almost certainly the application of the chain rule, however I can't see how it is working here.

Yes, it is the Chain Rule.
Let \(\displaystyle g(x) = \dfrac{dy}{dx}\)

Then \(\displaystyle \displaystyle \dfrac{dg}{ds} = \left( \dfrac{dg}{dx} \right)\left(\dfrac{dx}{ds} \right)\)

 

[jonnburton]

Yes, it is the Chain Rule.
Let \(\displaystyle g(x) = \dfrac{dy}{dx}\)

Then \(\displaystyle \displaystyle \dfrac{dg}{ds} = \left( \dfrac{dg}{dx} \right)\left(\dfrac{dx}{ds} \right)\)

Thank you DrPhil. I need to give this some more thought; perhaps the reason it seems confusing is the fact that the initial function is of the form \(\displaystyle \frac{dy}{dx}\)

 

[DrPhil]

Thank you DrPhil. I need to give this some more thought; perhaps the reason it seems confusing is the fact that the initial function is of the form \(\displaystyle \frac{dy}{dx}\)

Yes, I thought that was the trouble - but dy/dx is "just another function of x." By considering it to be g(x), that confusion should be cleared up. A second derivative is

\(\displaystyle \displaystyle \dfrac{d}{dx}\left[ \dfrac{d}{dx}\left( f(x) \right)\right] = \dfrac{d^2}{dx^2}f(x)\)