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A spherical balloon is inflated with helium at the rate of 100pie ft^3/min/ a) how fast is the balloon's radius increasing at the interest the radius is 5 feet? b) how fast is the surface area increasing at that instant.

Respuesta :

taskmasters

The problem at hand is on related rates, which could be solved using differential calculus

The volume of a sphere: V = (4/3)(pi)(r^3)

Differentiating with respect to time,

(dV/dt) = 4(pi)(r^2)(dr/dt)

Substituting the given,

(dV/dt) = 100 , r = 5

100 = 4(pi)(5^2)(dr/dt)

Solving for (dr/dt)

dr/dt= 0.318 ft/s

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