Course:MATH102/Question Challenge/2010 December 05

Question

When a person breathes, the volume of air in the lungs may be modelled by a function of the form

$\displaystyle V(t)=C+A\sin(\omega t+\phi ),$

where V is the volume in millilitres and t is time in seconds. Suppose the minimum and maximum volumes are 1400 and 3400 ml respectively, and the maximum rate of change of V is 1200 ml/sec.

What is the period of V(t)?

Solutions

Solution

$V(t)=C+A\sin(\omega t+\phi )$ We are told that the minimum and maximum volumes are 1400 and 3400 ml. From this we find the mean to be (1400+3400)/2= 4800/2= 2400. We also find the amplitude to be (3400-1400)/2=2000/2=1000. Hence $V(t)=2400+1000\sin(\omega t+\phi )$ The rate of change is $V'(t)$, and we calculate this to be $V'(t)=-1000\cos(\omega t+\phi )\cdot \omega$ The max rate of change will occur when the cosine term is -1, at which time $V'=1000\omega$. We also know that this rate is so 1200 ml/sec, so $V'=1000\omega =1200$ ml/sec. Hence $\omega ={\frac {1200}{1000}}=1.2$ /sec The period is given by: $T={\frac {2\pi }{\omega }}={\frac {2\pi }{1.2}}$ sec.