RRproblem5



Frances Loeb's work:

5 a) V= 1/3 π r^2 h R= 46h/6 V= 1/3 π (46h/6)^2 h I think you mean 45h/6 for a substitution for r (just a typo!) V= 18.75 π h^3 Agree with this!! Dv/dt = 18.75 π 3h^2 dh/dt Agree with this!! -50 = 56.25 π (5)^2 dh/dt Here you really wanted to plug in 6 for h, not 5! -50/ 1406.25π = dh/dt dh/dt= -.01132 m/min So my answer for this is different b) V= 1/3 π r^2 h h= 6r/45 V= 1/3 π r^2 6r/45 V= 22.5 π r^3 whoops over here you did 45/2 instead of 2/45 Dv/dt= 22.5 π 3r^2 dr/dt -50= 22.5 π 3(37.5)^2 dr/dt dr/dt= -1.6767e-4