The volume of air in a certain climber's lungs (in cubic centimeters) is modeled by the equation: Upper V equals 350 sin (52 pi t )plus 800 where t is the time in minutes. Find the maximum volume of air in the climber's lungs, and determine how many breaths the climber takes per minute.
Accepted Solution
A:
Answer:a) The maximun volume of air in the climber's lungs is [tex]1150 cm^{3}[/tex]b) The climber takes 26 breaths per minuteStep-by-step explanation:a) The function is given by [tex]V(t)=350sin(52\pi t)+800[/tex], where [tex]V(t)[/tex] is in [tex]cm^{3}[/tex], and [tex]t[/tex] is in minutes.So the first step is to derive it, by the chain rule we obtain [tex]V'(t)=(52\pi)350 cos(52 \pi t)[/tex].After that, we make the expression equals to zero, [tex]0=(52\pi)350 cos(52 \pi t)[/tex], then [tex]0=cos(52 \pi t)[/tex]. This leads to [tex]cos^{-1}(0)=52 \pi t[/tex].[tex]\frac{\pi}{2} =52 \pi t[/tex], and clearing it for [tex]t=\frac{\pi}{2*52\pi}=\frac{1}{104}[/tex].The second step is to evaluate the original expression for this value so, [tex]V(\frac{1}{104})=350sin(52\pi (\frac{1}{104}))+800=350sin (\frac{\pi}{2})+800=350+800=1150cm^{3}[/tex].b) As we can see this function have the form [tex]A*sin(\omega t)[/tex], where [tex]\omega=52\pi[/tex] is the angular frequency, so every [tex]2 \pi[/tex] radians we will have a breath, therefore Breaths Per Minute=[tex]BPM=\frac{\omega}{2\pi}=\frac{52\pi}{2 \pi} =26[/tex].