Air is escaping a spherical balloon. When the radius of the balloon is 30cm, it is decreasing at a rate of 5cm/min. At this moment, what is the rate of change of the volume of the balloon with respect to time..?
answer: 18 000PIcm^3/min..
I'm not sure how you get that..please explain..thanks..Calculus...Chain Rule..?v=(4/3)r^3Pi; dv/dr=(4/3)*3r^2Pi=4r^2Pi
dr/dt=-5
dv/dt=(dv/dr)(dr/dt)=4r^2Pi * (-5)
r=30,
so, the answer is 4*30^2*Pi * (-5) = -18000Pi
%26quot;-%26quot; means decreasing.Calculus...Chain Rule..?Volume for a sphere is
V=(4/3) 蟺 r^3
the derivative of that is
dV/dt=4蟺 r^2
dr/dt= -5
so we need to put that next to our derivative of the radius because that's a different quantity than volume (implicit derivation)
dV/dt=4 蟺 r^2(dr/dt)
4 蟺 (30^2)(-5)
dV/dt= -18,000蟺 cm^3/minCalculus...Chain Rule..?V = (4/3) * pi * r^3
dV / dt = (4/3) * 3 * pi * r^2 * dr/dt = 4 * pi * r^2 * dr/dt
dr/dt = -5 cm/min
r = 30 cm
dV / dt = 4 * pi * (30)^2 * (-5)
dV / dt = (3600 * -5) * pi
dV / dt = -18000 * pi cm^3 / minCalculus...Chain Rule..?Vs = 4/3 pi r^3
dV/dt = 4 pi r^2 dr/dt
r = 30 cm
dr/dt = -5 cm/min
dV/dt = 4 pi (30)^2 (-5) = -20* 900 pi = -18000 pi cm^3/min
