A body of mass \(m\) moves along a trajectory \(\vec{r}(t)\) in three-dimensional space with constant kinetic energy, What geometric relationship has to exist between the body's velocity vector, \(\vec{v}(t),\) and its acceleration vector, \(\vec{a}(t),\) in order to accomplish this

For a body of mass \(m\) moving with velocity \(\vec{v}(t)\), its kinetic energy, \(K\), can be written as: K = \frac{1}{2}m\|\vec{v}(t)\|^2.

To determine when the kinetic energy is constant, we need to find when its derivative with respect to time is zero. We differentiate \(K\) with respect to \(t\): \frac{dK}{dt} = \frac{d}{dt}\left(\frac{1}{2}m\|\vec{v}(t)\|^2\right).

Applying the chain rule, we get: \frac{dK}{dt} = m\|\vec{v}(t)\|\frac{d\|\vec{v}(t)\|}{dt}. Now, we notice that the derivative of \(\|\vec{v}(t)\|\) with respect to \(t\) can be related to acceleration. Since the acceleration \(\vec{a}(t) = \frac{d\vec{v}(t)}{dt}\), we see that: \frac{d\|\vec{v}(t)\|}{dt} = \frac{\vec{v}(t) \cdot \vec{a}(t)}{\|\vec{v}(t)\|}, where \(\cdot\) denotes the dot product.

Substituting this result into the expression for \(\frac{dK}{dt}\), we get: \frac{dK}{dt} = m\vec{v}(t) \cdot \vec{a}(t).

Recall that we want to find the relationship that will make the derivative of kinetic energy equal to zero. Therefore, for constant kinetic energy, we must have: \frac{dK}{dt} = m\vec{v}(t) \cdot \vec{a}(t) = 0. This means that the velocity vector \(\vec{v}(t)\) and the acceleration vector \(\vec{a}(t)\) should be orthogonal or perpendicular to each other, that is, their dot product should be zero: \vec{v}(t) \cdot \vec{a}(t) = 0. This geometric relationship indicates that the angle between the velocity and acceleration vectors is 90 degrees, making their dot product zero. To maintain constant kinetic energy along the trajectory, the velocity and acceleration vectors must always be perpendicular to each other.