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anonymous
 5 years ago
A sphercal cloud is expanding at the rate of 400 cubric meters/day. At what rate is the cloud's radius increasing when the radius is 10 meters? at what rate is the cloud's surface are increasing when the radius is 20 meters?
anonymous
 5 years ago
A sphercal cloud is expanding at the rate of 400 cubric meters/day. At what rate is the cloud's radius increasing when the radius is 10 meters? at what rate is the cloud's surface are increasing when the radius is 20 meters?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This is a related rate problem right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We are looking for the instantaneous rate of change of the cloud's radius (how fast the cloud's radius is increasing right when the radius is 10 or 20 meters) It's an infinitesimally small period of time and during that time the surface area and radius change an infinitesimally small amount. since the surface area doesn't really change at that instant you could say that the rate of change of the volume is "spread" over the surface of the sphere. The equation for the surface area of a sphere is 4pi r^2 (4*pi*r*r) when the sphere has a radius of 10m it's surface area will be 4*pi*10*10=400*pi meters squared likewise when the sphere has a radius of 20 meters it's SA will be 1600pi meters squared the rate of change of the volume will be spread over the surface area so (400 meters^3) / (400*pi meters^2) or 1/pi meters per day when the cloud has a radius of 20 meters the rate of change of the radius will be 400 meters^3) / (1600*pi meters^2) or 1/4pi meters per day @imranmeah91 I don't know the terms used for these types of problems

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0damn, surface area, not radius. sorry about that.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you look at the equations for surface area and volume of a sphere there are some similarities SA=4*pi*r^2 V=4/3*pi*r^3 thus SA= 3/r*V since volume is increasing at 20 meters cubed per day, surface area is increasing at 3/r times that rate so when the radius of the cloud is 10 meters the surface area is increasing by 3/r or 3/10m * (20 meters cubed per day) or 6 meters squared per day when the cloud has a radius of 20 meters it's surface area is increasing by 3/(20m)* the rate of change in volume or 3/20m * 20 meters cubed / day or 3 meters squared per day.
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