What Is the Derivative of Kinetic Energy?

Title: What Is the Derivative of Kinetic Energy? Category: physics

The phrase “derivative of kinetic energy” is incomplete unless you say what you are differentiating with respect to. For an object of constant mass, kinetic energy is $K=\frac{1}{2}mv^2$, where $v$ is its speed. Its derivative with respect to time describes how quickly kinetic energy changes; its derivative with respect to velocity is related to momentum.

Differentiate with respect to time: power

If the object’s velocity changes with time, use the chain rule. In one-dimensional motion, writing velocity as $v(t)$,

$$\frac{dK}{dt}=\frac{d}{dt}\left(\frac{1}{2}mv^2\right)=mv\frac{dv}{dt}=mva.$$

Here $a=dv/dt$. In vector form, kinetic energy depends on the squared speed, $K=\frac{1}{2}m\mathbf{v}\cdot\mathbf{v}$, so

$$\frac{dK}{dt}=m\mathbf{v}\cdot\mathbf{a}.$$

For constant mass and Newton’s second law, $\mathbf{F}_{\mathrm{net}}=m\mathbf{a}$. Therefore,

$$\frac{dK}{dt}=\mathbf{F}_{\mathrm{net}}\cdot\mathbf{v}=P_{\mathrm{net}},$$

where $P_{\mathrm{net}}$ is the net power delivered to the object. Power is the rate of energy transfer, measured in watts (joules per second). The dot product matters: a force perpendicular to the velocity can change the direction of motion without changing the kinetic energy at that instant.

Differentiate with respect to velocity: momentum

In one dimension, treating $v$ as the signed velocity,

$$\frac{dK}{dv}=mv.$$

In three dimensions, the corresponding statement is about the gradient with respect to the velocity vector:

$$\nabla_{\mathbf{v}}K=m\mathbf{v}=\mathbf{p},$$

where $\mathbf{p}$ is linear momentum. If you use speed as a nonnegative scalar, $dK/dv=mv$ points along the direction of increasing speed; it is not by itself the full vector momentum.

Worked example

A $2\,\mathrm{kg}$ object moves in a straight line at $4\,\mathrm{m/s}$ and has acceleration $2\,\mathrm{m/s^2}$ in the direction of motion. Its kinetic energy is

$$K=\frac{1}{2}(2)(4^2)=16\,\mathrm{J}.$$

The time derivative at that instant is

$$\frac{dK}{dt}=mva=(2)(4)(2)=16\,\mathrm{W}.$$

This agrees with the force-times-velocity calculation: the net force is $ma=4\,\mathrm{N}$, so $P=Fv=(4)(4)=16\,\mathrm{W}$. If instead you differentiate with respect to velocity, you get $dK/dv=mv=(2)(4)=8\,\mathrm{kg\,m/s}$, the object’s momentum in this one-dimensional example.

Derivative with respect to position

For one-dimensional motion, if velocity depends on position, the chain rule gives

$$\frac{dK}{dx}=\frac{dK}{dt}\frac{dt}{dx}=mva\frac{1}{v}=ma=F_{\mathrm{net}},$$

when $v\ne 0$ and acceleration and force are along the same axis. This is the differential form of the work–energy theorem: the change in kinetic energy along a path equals the work done by the net force. In multiple dimensions, the path-based statement is $dK=\mathbf{F}_{\mathrm{net}}\cdot d\mathbf{r}$; it is generally not meaningful to treat $K$ as a function of position alone unless the motion and velocity are specified.

Common mistakes

• Leaving out the variable. $dK/dt$ and $dK/dv$ are different quantities and have different units.
• Forgetting the chain rule. The time derivative is not just $mv$; that is the derivative with respect to scalar velocity. For time, multiply by $dv/dt$.
• Confusing speed with velocity. Speed is scalar, while velocity and momentum are vectors. Use the vector dot product for power.
• Assuming the result holds unchanged for variable mass. If mass changes, differentiating $K=\frac{1}{2}mv^2$ adds a term: $dK/dt=\frac{1}{2}v^2(dm/dt)+mv(dv/dt)$ in one dimension. Applying Newton’s law to a variable-mass system requires additional care.

So, if a mechanics problem asks for the derivative without specifying a variable, check its context. A time derivative usually asks for power; a derivative with respect to velocity usually points to momentum.