a snowball melts do that the surface area decreases at rate of 0.5 m^2/min. find the rate at which the radius decreases when the radius is 4cm
2007-03-08 02:40:25 · 5 answers · asked by morgan 2 in Science & Mathematics ➔ Mathematics
thanks for the big hint creepy now i know where i was going wrong
2007-03-08 02:46:15 · update #1
S=4pi*r^2
dS/dt = dS/dr*dr/dt
ds/dr= 8pi*r and dS/dt= 50cm^2/min so
dr/dt= 50cm^2/min / 32pi cm = 0.5cm/min
2007-03-08 04:28:41 · answer #1 · answered by santmann2002 7 · 0⤊ 0⤋
Ok, you have to do this :
Surface Area = S = 4*pi*r^2
dS / dt = -8*pi*r*(dr / dt)
minus because it's decreasing, and dr/dt, the problem is asking you for that.
0.5 = -8*pi*(4/100)*dr/dt
dr / dt = - 0.49 m^2 / min
Hope that helped you
2007-03-08 10:46:12 · answer #2 · answered by anakin_louix 6 · 1⤊ 0⤋
Surface area of a sphere: 4pi * r^2
d surface area / dt = .5
radius = sqrt(SA/(4pi))
2007-03-08 10:42:59 · answer #3 · answered by creepy_mitch 2 · 0⤊ 0⤋
Let A be area and r be radius:
dA/dt=0.5m2/min
d(pi*r^2)/dt=0.5
2*pi*r*dr/dt=0.5
dr/dt=0.25/(pi*r)
When r=4cm
dr/dt=198.94cm/min
2007-03-08 10:46:34 · answer #4 · answered by Shrey G 3 · 1⤊ 0⤋
2*pi*r*dr/dt=0.5
dr/dt=0.25/(pi*r)
When r=4cm
2007-03-08 10:47:30 · answer #5 · answered by Ralph 1 · 0⤊ 0⤋