Question

air is being pumped into a spherical balloon at the rate of 7 cubic centimeters per second. What is the rate of change of the radius at the instant the volume equals 367?

Answer 1

The rate of change of radius of the spherical balloon is 0.0282 cm per sec .

Explanation:

Given as :

The rate at which air is filled inside balloon = 7 cm³ per sec

Let The radius = r cm

And The volume of balloon = v = 367 cm³

let The change of radius =
`\frac{\mathrm{d} r}{\mathrm{d} t}`

∵ volume = 367 cm³

∵ volume of sphere =
`(4)/(3) \pi` r³

i.e 367 =
`(4)/(3)` ×
`(22)/(7)` × r³

or, r³ =
`(7707)/(88)`

So, r³ = 87.57

∴ r =
`\sqrt[3]{87.57}` = 4.44 cm

I.e r = 4.44 cm

Now,
`\frac{\mathrm{d} v}{\mathrm{d} t}` =
`\frac{\mathrm{d} (4)/(3)\Pi r^(3)}{\mathrm{d} t}`

Or,
`\frac{\mathrm{d} v}{\mathrm{d} t}` = 4
`\pi`r²×
`\frac{\mathrm{d} r}{\mathrm{d} t}`

Or, 7 = 4
`\pi`(4.44)²×
`\frac{\mathrm{d} r}{\mathrm{d} t}`

Or, 7 = 4 × 3.14 × 19.71 ×
`\frac{\mathrm{d} r}{\mathrm{d} t}`

Or, 7 = 247.55 ×
`\frac{\mathrm{d} r}{\mathrm{d} t}`

∴
`\frac{\mathrm{d} r}{\mathrm{d} t}` =
`(7)/(247.55)`

Or,
`\frac{\mathrm{d} r}{\mathrm{d} t}` = 0.0282 cm per sec

Hence, The rate of change of radius of the spherical balloon is 0.0282 cm per sec . Answer

Answer 2

`(dr)/(dt)= \frac{7}{4\pi ((1101)/(4\pi ))^{(2)/(3)}}`

Explanation:

we know that volume of the spherical balloon =
`(4)/(3)* \pi * r^(3)`

given
`(dv)/(dt)` = 7

differentiating v with respect to time

`(dv)/(dt)=4\pi r^(2)(dr)/(dt)`

given
`367 = (4)/(3)\pi r^(3)`

`r = ((1101)/(4\pi ))^{(1)/(3)}`

`(dr)/(dt)= \frac{7}{4\pi ((1101)/(4\pi ))^{(2)/(3)}}`