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The radius, r, in inches, of a spherical balloon is related to the volume, V, by r(V) = cubed root(3V/4pi). Air is pumped into the balloon, so the volume after t seconds is given by V(t) = 10 + 20t.

 

a. Find the composite function r(V(t))

b. Find the exact time when the radius reaches 10.

 

For a, I got: 

 

r(V(t)) = r(cubed root[3(10+20t)/4pi] or r(cubed root(30+60t/4pi)

 

For b, I know that r is 10, (10(cubed root(30+60t/4pi)) but I'm not sure how to solve it. 

 Sep 19, 2020
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The radius, r, in inches, of a spherical balloon is related to the volume, V, by r(V) = cubed root(3V/4pi). Air is pumped into the balloon, so the volume after t seconds is given by V(t) = 10 + 20t.

 

This is what I get.

 

\(r(V)=\left(\frac{3V}{4\pi}\right)^{1/3}\qquad V(t)=10+20t\\~\\ r(V(t))=\left(\frac{3(10+20t)}{4\pi}\right)^{1/3}\\~\\ r(V(t))=\left(\frac{3(5+10t)}{2\pi}\right)^{1/3}\\~\\ r(V(t))=10\qquad when\\~\\ \left(\frac{3(5+10t)}{2\pi}\right)^{1/3}=10\\~\\ \frac{3(5+10t)}{2\pi}=1000\\ 5+10t=\frac{2000\pi}{3}\\ 10t=\frac{2000\pi-15}{3}\\ t=\frac{2000\pi-15}{30}\\ t=\frac{400\pi-3}{6} \)

 

 

 

 

LaTex:

r(V)=\left(\frac{3V}{4\pi}\right)^{1/3}\qquad V(t)=10+20t\\~\\
r(V(t))=\left(\frac{3(10+20t)}{4\pi}\right)^{1/3}\\~\\
r(V(t))=\left(\frac{3(5+10t)}{2\pi}\right)^{1/3}\\~\\
r(V(t))=10\qquad when\\~\\
\left(\frac{3(5+10t)}{2\pi}\right)^{1/3}=10\\~\\
\frac{3(5+10t)}{2\pi}=1000\\
5+10t=\frac{2000\pi}{3}\\
10t=\frac{2000\pi-15}{3}\\
t=\frac{2000\pi-15}{30}\\
t=\frac{400\pi-3}{6}

 Sep 19, 2020

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