Answer:
(a) \(T=17.4\sqrt[4]{m}\)
(b) 0.057 sec/month.
Step-by-step explanation:
Let T is the circulation time of a mammal in seconds and m is the body mass in kilograms.
It is given that the circulation time of a mammal is proportional to the fourth root of the body mass of the mammal.
\(T\propto \sqrt[4]{m}\)
\(T=k\sqrt[4]{m}\)
where k is constant of proportionality.
(a) The proportionality constant is 17.4. So, the circulation time of a mammal is
\(T=17.4\sqrt[4]{m}\)
(b)
The above equation can be written as
\(T=17.4m^{\frac{1}{4}}\)
Differentiate with respect to time t.
\(\frac{dT}{dt}=17.4(\frac{1}{4}m^{\frac{1}{4}-1}\frac{dm}{dt})\)
\(\frac{dT}{dt}=4.35m^{-\frac{3}{4}}\frac{dm}{dt}\)
Substitute m=38 and \(\frac{dm}{dt}=0.2\) in the above equation.
\(\frac{dT}{dt}=4.35(38)^{-\frac{3}{4}}(0.2)\)
\(\frac{dT}{dt}=0.0568436\)
\(\frac{dT}{dt}\approx 0.057\)
Therefore, the rate of change of the circulation time of the child is 0.057 sec/month.
