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Find the rate of increase (with respect to r) in the surface area (S=4πr2) of a spherical balloon when: (A) r=2 inches → Rate of increase = (B) r=4 inches → Rate of increase = (C) r=6 inches → Rate of increase =
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rate of change = first derivative since we are findign the rate of change with respect to radius, we find \(\displaystyle {dS\over dr}\)
what will be the derivative?
S = 4pi*r^2 dS/dt = 4pi*2*r*dr/dt
not with "t" just with "r"
there is no variable "t" in the system. the question said .. with respect to radius (r)
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ok so what do i need to do?
@electrokid
\[ {dS\over dr}=4\pi(2r)=8\pi r \]
this is what is asked
ok so the answer is 8pi r
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yep. that is the rate of increase of S with respect to r
ok, thank you so much
yw
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