24. (a)Note that the level of the coffee in the cone is notneeded until part (b).Step 1:V15volume of coffee in poty5depth of coffee in potStep 2:}ddVt1} 510 in3/minStep 3:We want to find the value of }ddyt}.Step 4:V159pyStep 5:}ddVt1} 59p}ddyt}Step 6:1059p}ddyt}}ddyt} 5 }91p0}<0.354 in./minThe level in the pot is increasing at the rate ofapproximately 0.354 in./min.(b)Step 1:V25volume of coffee in filterr5radius of surface of coffee in filterh5depth of coffee in filterStep 2:At the instant in question,}ddVt2}5210 in3/min and h55 in.Step 3:We want to find 2}ddht}.Step 4:Note that }hr}5 }36}, so r5 }h2}.Then V25 }13}pr2h5 }p1h23}.Step 5:}ddVt2} 5 }p4h2} }ddht}Step 6:2105 }p(45)2} }ddht}}ddht}52}58p}in./minNote that }ddht} ,0, so the rate at which the level isfallingis positive. The level in the cone is falling at therate of }58p}<0.509 in./min.25.Step 1:Q5rate of CO2exhalation (mL/min)D5difference between CO2concentration in blood pumped to the lungs and CO2concentration in blood returning from the lungs (mL/L)y5cardiac outputStep 2:
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