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[SOLVED] differentiation energy

dwsmith

Well-known member
Feb 1, 2012
1,673
$\mathcal{E} = \frac{1}{2}m\dot{x} + \frac{1}{2}kx^2$
The derivative is
$$
\frac{d\mathcal{E}}{dt} = \frac{1}{2}m\ddot{x} + kx\dot{x}
$$
but the solution is suppose to be
$$
\frac{d\mathcal{E}}{dt} = \dot{x}(m\ddot{x} + kx).
$$
How?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
$\mathcal{E} = \frac{1}{2}m\dot{x} + \frac{1}{2}kx^2$
The derivative is
$$
\frac{d\mathcal{E}}{dt} = \frac{1}{2}m\ddot{x} + kx\dot{x}
$$
but the solution is suppose to be
$$
\frac{d\mathcal{E}}{dt} = \dot{x}(m\ddot{x} + kx).
$$
How?
Hi dwsmith! :)

The first part of your energy is the kinetic energy.
However, kinetic energy contains speed squared instead of just speed.
So your energy formula should be:

$\qquad \mathcal{E} = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}kx^2$
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Hi dwsmith! :)

The first part of your energy is the kinetic energy.
However, kinetic energy contains speed squared instead of just speed.
So your energy formula should be:

$\qquad \mathcal{E} = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}kx^2$
The book has a typo then.