Question #16386

May 27, 2017

a) 27.5 kg b) 31.504 kg c) ${M}^{'} \left(t\right) = 0.3 - 0.002 t$ d) $0.272 {\text{ kg" " day}}^{-} 1$

Explanation:

a) This is the initial mass on May 25th, where t = 0 days after May 25th.

$M \left(0\right) = 27.5 \text{ kg}$

b) Substitute t = 14 into the expression for body mass

$M \left(14\right) = 27.5 + 0.3 \left(14\right) - 0.001 {\left(14\right)}^{2} = 27.5 + 4.2 - 0.196 = 31.504 \text{ kg}$

c) The rate of change at any time is given by the gradient of the tangent to the graph at any point. This is the given by the derivative of the body mass function and is found by using the power law

Derivative or $\text{dM"/"dt}$ $\text{ of }$ ${t}^{n} = n \cdot {t}^{n - 1}$

$R = \text{dM"/"dt} = 0.3 - 0.002 t$

d) To get the rate of change at t = 14, simply substitute 14 into the derivative function

${M}^{'} \left(14\right) = 0.3 - 0.002 \left(14\right) = 0.3 - 0.028 = 0.272 \text{ kg per day}$