1
$\begingroup$

We are using the Euler equation to calculate the minimum:

Euler equation: $-\frac{d}{dt}\hat{L}_{x'}(t) + \hat{L}_x(t) = 0$

We have the following $L = 12tx + x'^2$ ($x$ is a function of $t$)

Now calculating these derivatives my book says it equals $-2x'' + 12t = 0$

Can anybody please explain how they came to this answer?

Thanks in advance.

$\endgroup$
1
$\begingroup$

Hints. Rewrite as follows: $$-\frac{d}{dt} \frac{\partial L}{\partial x'} + \frac{\partial L}{\partial x} = 0 \qquad ; \qquad L = 12tx + (x')^2$$ What is $\frac{\partial L}{\partial x'}$ , what is $-\frac{d}{dt} \frac{\partial L}{\partial x'}$ then, and, at last, what is $\frac{\partial L}{\partial x}$ ?
If you can assemble these, then you're finished.

$\endgroup$
0
$\begingroup$

We first differentiate with respect to $t$, that is, $x''(t)$, then we differentiate with respect to $x'$, that is $1$, using the chain rule we get $\frac{d}{dt}\frac{d}{dx}\hat{L} = 1\cdot x'' = x''$

$\endgroup$

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.