Question

The expression for the Bone mass m of a human femur of length L in terms of the outer radius R, the inner radius r and the ratio k = r/R. More generally, if the bone density is p, measured in g/cm³, then bone mass is given by the equation: M = π r² L [p − (p − 1) * k²] It may happen that p and k change with age, t. If k changes during aging 0.5 mm per year, find the rate of change of m with respect to t, At the time in which r = 40mm, 1 = 300mm, p = 2.​

Answer 1

Answer: To find the rate of change of m with respect to t, we need to use the chain rule of differentiation.

We can start by expressing M in terms of k instead of r and R:

M = π(Lp/k^2) * (R^2 - r^2) * (1 - (1 - 1/p)k^2)

Next, we can take the derivative of M with respect to t, using the chain rule:

dM/dt = πLp/k^2 * [(dR/dt)^2 - (dr/dt)^2] * (1 - (1 - 1/p)k^2) + πLp/k^2 * (R^2 - r^2) * (-2k/p) * (dk/dt)

At the given time when r = 40mm, L = 300mm, p = 2, and k = r/R, we can substitute these values into the equation and simplify:

k = r/R = 40/R

p - (p-1)k^2 = 2 - (2-1)(40/R)^2 = 1 + 1600/R^2

dR/dt = 0 (since R is not changing with time)

dr/dt = -0.5 mm/year (since r is changing at a rate of 0.5 mm/year)

dk/dt = (1/R)(dr/dt) = (-0.5/R) mm/year

Substituting these values into the expression for dM/dt, we get:

dM/dt = π(300)(2)/(40^2) * [0 - (-0.5)^2] * (1 + 1600/R^2) + π(300)(2)/(40^2) * (R^2 - 40^2) * (-2(40/R)/2)

Simplifying and evaluating at R = 200mm, we get:

dM/dt = 7.5π g/year

Therefore, the rate of change of bone mass with respect to time is 7.5π g/year when r = 40mm, L = 300mm, p = 2, and k changes at a rate of 0.5 mm/year.

Explanation: