The radius increasing by 0.0145 feet/sec
Step-by-step explanation:
The given is:
- Helium is being pumped into a Sheridan balloon at a rate of 5 cubic feet per second
- The formula of the volume of the balloon is V = [tex]\frac{4}{3}[/tex] π r³ , where r is the radius of the balloon
∵ Helium is being pumped into a Sheridan balloon at a rate of
5 cubic feet per second
∴ [tex]\frac{dV}{dt}[/tex] = 5 feet³/sec
To find the rate of increasing of the radius find [tex]\frac{dV}{dr}[/tex]
∵ V = [tex]\frac{4}{3}[/tex] π r³
Differentiate V with respect to r
∴ [tex]\frac{dV}{dr}=(\frac{4}{3})\pi (3)r^{3-1}[/tex]
∴ [tex]\frac{dV}{dr}=4\pi r^{2}[/tex]
We need to find the increasing of the radius after 2 minutes
Let us find the volume after 2 minutes
∵ [tex]\frac{dV}{dt}[/tex] = 5 feet³/sec
∴ V = 5t feet³
∵ 1 minute = 60 seconds
∴ 2 minutes = 60 × 2 = 120 seconds
∴ V = 5(120) = 600 feet³
Now we can find r after 2 minutes by equating the rule of the volume by 600
∵ [tex]\frac{4}{3}[/tex] π r³ = 600
Divide both sides by [tex]\frac{4}{3}[/tex] π
∴ r³ = [tex]\frac{450}{\pi }[/tex]
Take ∛ for both sides
∴ [tex]r=\sqrt[3]{\frac{450}{\pi }}[/tex]
Substitute the value of r in [tex]\frac{dV}{dr}=4\pi r^{2}[/tex]
∴ [tex]\frac{dV}{dr}=4\pi (\sqrt[3]{\frac{450}{\pi }})^{2}[/tex]
∴ [tex]\frac{dV}{dr}[/tex] = 344.021
Now divide [tex]\frac{dV}{dt}[/tex] by [tex]\frac{dV}{dr}[/tex] to find [tex]\frac{dr}{dt}[/tex] the rate of increasing of the radius
∵ [tex]\frac{dV}{dt}[/tex] ÷ [tex]\frac{dV}{dr}[/tex] = 5 ÷ 344.021
∴ [tex]\frac{dV}{dt}[/tex] × [tex]\frac{dr}{dV}[/tex] = 0.0145
∴ [tex]\frac{dr}{dt}[/tex] = 0.0145 feet/sec
The radius increasing by 0.0145 feet/sec
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