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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 $t$.

User 10482
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1 Answers1

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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} } $

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.

Etg
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