Gas is escaping from a spherical balloon at the rate of 2cm³/ min. How fast is the surface area changing when the radius is 13cm?[V=4/3(πr³) and s=4πr²].
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Step-by-step explanation:
dr/dt = 2cm^3 / min
r = 13 cm
s = 4πr^2
ds/dt = 4π ×2r× dr/dt
ds/dt = 4×π× 2× 13× 2
ds/dt = 204π cm^2/min
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