# Double differential / differential squared

• Feb 3rd 2010, 07:46 AM
wglmb
Double differential / differential squared
If $\displaystyle r$ and $\displaystyle \theta$ are funtions of $\displaystyle t$, does $\displaystyle \frac{d^2 r}{dt^2}\bigg / \left ( \frac{d\theta}{dt}\right ) ^2$ equal $\displaystyle \frac{d^2 r}{d\theta ^2}$?
• Feb 3rd 2010, 08:11 AM
HallsofIvy
Quote:

Originally Posted by wglmb
If $\displaystyle r$ and $\displaystyle \theta$ are funtions of $\displaystyle t$, does $\displaystyle \frac{d^2 r}{dt^2}\bigg / \left ( \frac{d\theta}{dt}\right ) ^2$ equal $\displaystyle \frac{d^2 r}{d\theta ^2}$?

No. While the first derivative can be "treated like" a fraction, the second derivative cannot.

For example, take $\displaystyle r= t^2$ and $\displaystyle \theta= t^4$. Then $\displaystyle \frac{dr}{dt}= 2t$, $\displaystyle \frac{d\theta}{dt}= 4t^3$, $\displaystyle \frac{d^2r}{dt^2}= 2$ and $\displaystyle \frac{d^2\theta}{dt^2}= 12t^2$.

Now it is easy to see that $\displaystyle \theta= (t^2)^2= r^2$ so $\displaystyle \frac{d\theta}{dr}= 2r= 2t^2= \frac{4t^3}{2t}$$\displaystyle = \frac{\frac{d\theta}{dt}}{\frac{dr}{dt}}$.

But $\displaystyle \frac{d^2\theta}{dr^2}= 2$ while $\displaystyle \frac{\frac{d^2\theta}{dt^2}}{\frac{d^2r}{dt^2}}= \frac{12t^2}{2}= 6t^2$.
• Feb 3rd 2010, 08:20 AM
wglmb
Thanks a lot :)
• Feb 3rd 2010, 03:51 PM
Jester
Quote:

Originally Posted by wglmb
If $\displaystyle r$ and $\displaystyle \theta$ are funtions of $\displaystyle t$, does $\displaystyle \frac{d^2 r}{dt^2}\bigg / \left ( \frac{d\theta}{dt}\right ) ^2$ equal $\displaystyle \frac{d^2 r}{d\theta ^2}$?

To find $\displaystyle \frac{d^2r}{d \theta^2}$ you would need to calculate $\displaystyle \frac{\displaystyle \frac{d}{dt}\left(\displaystyle \displaystyle \frac{dr}{dt}\bigg /\displaystyle \frac{d \theta}{dt}\right)}{\displaystyle \frac{d \theta}{dt}}$.