Question

Particle Acceleration A particle moves along the \(x\) -axis with velocity \(d x / d t=f(x) .\) Show that the particle's acceleration is \(f(x) f^{\prime}(x)\)

Short answer

The acceleration of the particle is indeed the product of its velocity and the derivative of the velocity function, \(d²x/dt² = f'(x) * f(x)\).

Step-by-step solution

Define velocity and acceleration

Velocity, in this exercise, is defined as \(dx/dt = f(x)\). Acceleration, being the derivative of the velocity, can be defined as the derivative with respect to time (t) of the function \(f(x)\), i.e, \(d^2x/dt^2\).

Apply the chain rule

When applying a second derivative (as in the derivative of the velocity, which is acceleration) to a function of a function (like our \(f(x)\)), we have to use the chain rule. The chain rule states that \(derivative (outer function) times derivative (inner function)\). In this case, our outer function is the velocity \(f(x)\), and our inner function is \(x(t)\). So, according to the chain rule, \(d^2x/dt^2 = f'(x)*(dx/dt)\).

Insert velocity into derivative

We already know the velocity \(dx/dt\) equals \(f(x)\). Thus, substituting \(f(x)\) for \(dx/dt\) in our previous derivative gives us \(d^2x/dt^2\) = \(f'(x)*f(x)\).