0
$\begingroup$

I am confused as to how to take the total derivative $\frac{dKE}{dt}$, where $KE$ is the kinetic energy.

I know that $KE = 1/2 *m * \dot{\vec r} \cdot \dot{\vec r}$. From here, if I take derivative of both sides w.r.t. $dt$, how should I apply the chain rule and simplify?

Attempted Solution:

$$\frac{dKE}{dt} = \frac{\partial KE}{\partial t} \frac {dt}{dt}*\frac{\partial KE}{\partial {\vec r}} \frac {d{\vec r}}{dt}*\frac{\partial KE}{\partial \dot {\vec r}} \frac {d\dot {\vec r}}{dt}* ...$$ How do I proceed from here? The answer in most books is simply $m*\dot {\vec r}\cdot \ddot {\vec r}$, which I totally agree with if $KE$ is a function of $\dot {\vec r}$ only. Why should one assume that? Clearly, $\dot {\vec r}$ comes from ${\vec r}$ and $t$.

$\endgroup$
  • $\begingroup$ Related: physics.stackexchange.com/q/9122/2451 $\endgroup$ – Qmechanic Sep 19 at 16:33
  • $\begingroup$ @Qmechanic Thanks. I read the linked question and kind of understand what the answers mean, but when applied to a real example it fails. Check out my first comment, under the answer by Ekrem, for the example case. $\endgroup$ – User 10482 Sep 19 at 18:05
0
$\begingroup$

The equation you wrote is correct. All you need to do is take the partial derivative. I think this is where you got confused. You can check the difference between partial and total derivative. So here is the answer:

$\dfrac{\partial KE}{\partial t} =\dfrac{\partial ( \dfrac{1}{2}\cdot m \cdot \dot{\vec{r}} \cdot \dot{\vec{r}})}{\partial t} = 0 $

$\dfrac{\partial KE}{\partial {\vec{r}} } =\dfrac{\partial ( \dfrac{1}{2}\cdot m \cdot \dot{\vec{r}} \cdot \dot{\vec{r}})}{\partial \vec{r}} = 0 $

$\dfrac{\partial KE}{\partial {\dot{\vec{r}}} } =\dfrac{\partial ( \dfrac{1}{2}\cdot m \cdot \dot{\vec{r}} \cdot \dot{\vec{r}})}{\partial \dot{\vec{r}}} = m \cdot \dot{\vec{r}\cdot } \ddot{\vec{r}} $

Therefore, from what you wrote we can conclude that $ \dfrac{dKE}{dt} = m \cdot \dot{\vec{r}\cdot } \ddot{\vec{r}}$. Simply because there is no $t$ nor ${\vec{r}}$ dependence explicitly in KE function.

$\endgroup$
  • $\begingroup$ Yes, the partial derivatives are confusing to me. Isn't $\dot{\vec r}$ a function of $t$ and $\vec {r}$ i.e. $\dot{\vec r}(t,\vec r)$? for example let $z = x+2y$ and $f(z) = 2z$. If I were to take $\frac{\partial f(z)}{\partial x}$, it would be zero with the "not explicitly dependent" argument, but clearly evaluates to $2$ if one substitutes for $z$ with $x+2y$ first. $\endgroup$ – User 10482 Sep 19 at 17:06
  • $\begingroup$ I suggest you to look up this link: physics.stackexchange.com/questions/9122/… There are pretty good explanations about what you are asking. I don't think I can come up with a better one. $\endgroup$ – Ekrem Sep 19 at 17:21
  • $\begingroup$ I understand it in theory but when I try an actual example (like I gave in the first comment) it does not work. Can you explain using that (or similar) example? $\endgroup$ – User 10482 Sep 19 at 18:00
  • $\begingroup$ Well, you can think as it is the definition of the partial derivative. In your example $f(z) = 2z$ is a function mapping $\mathbb{R} \rightarrow \mathbb{R}$. But $z$ also a function of $x$ and $y$ i.e. $z(x,y) = x + 2y$. If you substitute this function into $f(z)$, then you get another new function mapping $\mathbb{R}^2 \rightarrow \mathbb{R}$. Let's call this function$g(x,y)$. i.e. $g(x,y)=2x + 4y$. Therefore $\dfrac{\partial g(x,y)}{\partial x} = 2$ but $\dfrac{\partial f(z)}{\partial x} = 0 $. $\endgroup$ – Ekrem Sep 19 at 19:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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