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A spherical party balloon is being inflated with helium pumped in at a rate of 4 cubic feet per minute. How fast is the radius growing at the instant when the radius has reached 2 ft

Respuesta :

abdullahfarooqi

Answer:

0.159 ft/min

Explanation:

v= 4/3πr³

dv/dt = 8

dv/dt = 4/3π( 3r²) ( dr/dt )

plugging the values

  8=4/3π( 3r²) ( dr/dt )

dr/dt= 1/(6.28)=0.159 ft per min approx

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