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Mathematics 8 Online
OpenStudy (anonymous):

Suppose a spherical balloon grows in such a way that after t seconds, its volume is V = 4 sqrt(t) cm3 I found that v=52 but i cant find How fast is the volume changing after 169 seconds?

OpenStudy (anonymous):

i know its the derv

OpenStudy (anonymous):

first derivative of V

OpenStudy (anonymous):

V=4t^(1/2) V'=4(1/2)t^(-1/2)=2t^(-1/2)

OpenStudy (anonymous):

\[V'=\frac{2}{\sqrt{t}}\]

OpenStudy (anonymous):

how do i solve that?

OpenStudy (anonymous):

\[V'(169)=\frac{2}{\sqrt{169}}=\frac{2}{13}\]

OpenStudy (anonymous):

ok thanx

OpenStudy (anonymous):

you don't plug in 169 until you find the derivative

OpenStudy (anonymous):

wait so i do 2/sqrt (t) then plug in 169 for t?

OpenStudy (anonymous):

do= deriv

OpenStudy (anonymous):

V'(t)=2/sqrt(t) <= plug in 169 now

OpenStudy (anonymous):

ok

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