Question

helium is being pumped into a sheridan balloon at a rate of 5 cubic feet per second. how fast is the radius increasing after 2 minutes? v=(4/3)(pi)r^3

Answer 1

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 = $\frac{4}{3}$ 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

∴ $\frac{dV}{dt}$ = 5 feet³/sec

To find the rate of increasing of the radius find $\frac{dV}{dr}$

∵ V = $\frac{4}{3}$ π r³

Differentiate V with respect to r

∴ $\frac{dV}{dr}=(\frac{4}{3})\pi (3)r^{3-1}$

∴ $\frac{dV}{dr}=4\pi r^{2}$

We need to find the increasing of the radius after 2 minutes

Let us find the volume after 2 minutes

∵  $\frac{dV}{dt}$ = 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

∵  $\frac{4}{3}$ π r³ = 600

Divide both sides by  $\frac{4}{3}$ π

∴ r³ = $\frac{450}{\pi }$

Take ∛ for both sides

∴ $r=\sqrt[3]{\frac{450}{\pi }}$

Substitute the value of r in  $\frac{dV}{dr}=4\pi r^{2}$

∴ $\frac{dV}{dr}=4\pi (\sqrt[3]{\frac{450}{\pi }})^{2}$

∴ $\frac{dV}{dr}$ = 344.021

Now divide $\frac{dV}{dt}$ by $\frac{dV}{dr}$ to find $\frac{dr}{dt}$ the rate of increasing of the radius

∵  $\frac{dV}{dt}$  ÷  $\frac{dV}{dr}$  = 5 ÷ 344.021

∴  $\frac{dV}{dt}$  × $\frac{dr}{dV}$ = 0.0145

∴ $\frac{dr}{dt}$ = 0.0145 feet/sec

The radius increasing by 0.0145 feet/sec

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