[SOLVED]differentiation energy

dwsmith

Well-known member
$\mathcal{E} = \frac{1}{2}m\dot{x} + \frac{1}{2}kx^2$
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
$\mathcal{E} = \frac{1}{2}m\dot{x} + \frac{1}{2}kx^2$
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
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.