A balloon is being filled with helium at the rate if $4 \frac{{ft}^{3}}{min}$ $4"ft"^3/"min"$ . What is the rate, in $\frac{{ft}^{2}}{min}$ $"ft"^2/"min"$ , at which the surface area is increasing when the volume is $\frac{32 π}{3} {ft}^{3}$ $(32pi)/3"ft"^3$ ?

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Answer 1 · 2018-07-31T07:29:48

The answer is $=4(ft^2)/min$

The volume of a balloon is

$V=4/3pir^3$

The volume is $V=32/3pi$

Therefore,

The radius is given by

$32/3pi=4/3pir^3$

$r^3=8$

$r=2$

So,

Differentiating wrt $t$

$(dV)/dt=4/3pi*3r^2(dr)/dt=4pir^2(dr)/dt$

As,

$(dV)/dt=4$

$4pir^2(dr)/dt=4$

$(dr)/dt=1/(pir^2)$

The surface area is

$A=4pir^2$

Differentiating wrt $t$

$(dA)/dt=8pirdr/dt=8pi*2*1/(pi2^2)=4(ft^2)/min$

A balloon is being filled with helium at the rate if 4"ft"^3/"min"4ft3min4"ft"^3/"min". What is the rate, in "ft"^2/"min"ft2min"ft"^2/"min", at which the surface area is increasing when the volume is (32pi)/3"ft"^332π3ft3(32pi)/3"ft"^3?