A golf ball of mass 4.6 g is struck by a golf club at a speed of 50 m/s. The ball has inital velocity ,v, in metres per second, of v(m) = 83/(m+0.046), where m is the mass, in grams, of the golf club. Describe the rate of change of the initial velocity as the mass of the club increases. Show your work.

How does the mass of the club increase?

To describe the rate of change of the initial velocity as the mass of the club increases, we need to calculate the derivative of the initial velocity function with respect to the mass of the club.

Given the initial velocity function, v(m) = 83/(m + 0.046), where m is the mass of the golf club in grams, we need to find dv/dm, which represents the rate of change of v with respect to m.

To calculate the derivative, we'll use the quotient rule for differentiation. The quotient rule states that if we have a function f(x) = g(x)/h(x), then the derivative of f(x) is given by:

f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / (h(x))^2

In our case, g(x) = 83 and h(x) = m + 0.046. Let's differentiate each function:

g'(x) = 0 (since 83 is a constant)
h'(x) = 1 (the derivative of m with respect to m is 1)

Now we can substitute these values into the quotient rule formula:

v'(m) = (0 * (m + 0.046) - 83 * 1) / (m + 0.046)^2
= -83 / (m + 0.046)^2

Therefore, the rate of change of the initial velocity as the mass of the club increases is given by v'(m) = -83 / (m + 0.046)^2.

To interpret this result, we can note that the derivative is negative (-) and that it depends on the square of the quantity (m + 0.046) in the denominator. This means that as the mass of the club increases, the initial velocity decreases at an increasingly faster rate.