Respuesta :
[tex]T=17.4\sqrt[4]{m}[/tex]
Differentiate with respect to time [tex]s[/tex]:
[tex]\frac{dT}{ds}=\frac{17.4}{4m^{3/4}}\frac{dm}{ds}[/tex]
⇒[tex]\frac{dT}{ds}=\frac{4.35}{m^{3/4}}\frac{dm}{ds}[/tex]
Plug in [tex]m=38[/tex] and [tex]\frac{dm}{ds}=0.2[/tex]:
[tex]\frac{dT}{ds}=\frac{4.35}{38^{3/4}}*0.2[/tex]
⇒[tex]\frac{dT}{ds}[/tex]≈[tex]0.057[/tex] sec/month
Differentiate with respect to time [tex]s[/tex]:
[tex]\frac{dT}{ds}=\frac{17.4}{4m^{3/4}}\frac{dm}{ds}[/tex]
⇒[tex]\frac{dT}{ds}=\frac{4.35}{m^{3/4}}\frac{dm}{ds}[/tex]
Plug in [tex]m=38[/tex] and [tex]\frac{dm}{ds}=0.2[/tex]:
[tex]\frac{dT}{ds}=\frac{4.35}{38^{3/4}}*0.2[/tex]
⇒[tex]\frac{dT}{ds}[/tex]≈[tex]0.057[/tex] sec/month
Answer:
(a) [tex]T=17.4\sqrt[4]{m}[/tex]
(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.
[tex]T\propto \sqrt[4]{m}[/tex]
[tex]T=k\sqrt[4]{m}[/tex]
where k is constant of proportionality.
(a) The proportionality constant is 17.4. So, the circulation time of a mammal is
[tex]T=17.4\sqrt[4]{m}[/tex]
(b)
The above equation can be written as
[tex]T=17.4m^{\frac{1}{4}}[/tex]
Differentiate with respect to time t.
[tex]\frac{dT}{dt}=17.4(\frac{1}{4}m^{\frac{1}{4}-1}\frac{dm}{dt})[/tex]
[tex]\frac{dT}{dt}=4.35m^{-\frac{3}{4}}\frac{dm}{dt}[/tex]
Substitute m=38 and [tex]\frac{dm}{dt}=0.2[/tex] in the above equation.
[tex]\frac{dT}{dt}=4.35(38)^{-\frac{3}{4}}(0.2)[/tex]
[tex]\frac{dT}{dt}=0.0568436[/tex]
[tex]\frac{dT}{dt}\approx 0.057[/tex]
Therefore, the rate of change of the circulation time of the child is 0.057 sec/month.