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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?
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i know its the derv
first derivative of V
V=4t^(1/2) V'=4(1/2)t^(-1/2)=2t^(-1/2)
\[V'=\frac{2}{\sqrt{t}}\]
how do i solve that?
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\[V'(169)=\frac{2}{\sqrt{169}}=\frac{2}{13}\]
ok thanx
you don't plug in 169 until you find the derivative
wait so i do 2/sqrt (t) then plug in 169 for t?
do= deriv
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V'(t)=2/sqrt(t) <= plug in 169 now
ok
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