Question

If volume of a spherical ball is increasing at the rate of 4pie cc/sec, then the rate of increase of its radius, when the volume is 288pie cc

Answer 1

if volume of a spherical ball is increasing at the rate of 4pie cc/sec then the rate would be 4pie cc/sec

Answer 2

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The volume of the spherical ball is increasing at the rate of 4πcc/sec. Find the rate of the radius and the surface area are changing when the volume is 288π cc.

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Volume of a spherical ball increase at the rate of = 4πcc/s

$\huge\mathrm\pink{\frac{dv}{dt} = 4}$

Now, we are required to find the rate of the increase of the surface area when the volume is

= 288π

$\huge\mathrm\pink{\frac{dA}{dt} = ?¿}$

Now :

Since the volume of the sphere is = 288 π

we can find the radius = !??

using the volume of a sphere is $\frac{4}{3}πr³$

➪ $\frac{4}{3}πr³$ = 288π

taking Reciprocal :

➪ r³ = 288 × $\frac{3}{4}$

➪ r³ = 216

➪ r = 6

Now :

To find $\frac{dv}{dt}$

☞︎︎︎ $\frac{dv}{dt}$ = $\frac{dv}{dr}$ × $\frac{dr}{dt}$

Differentiate the volume of the sphere gives us:

➪ $\frac{dv}{dt}$ = $\frac{4}{3}πr³$ ×3r²

denominator and numerator 3 get cancel :

➪ 4πr²

Now :

to find $\frac{dr}{dt}$ :

☞︎︎︎ $\frac{dv}{dt}$ = $\frac{dv}{dr}$ × $\frac{dr}{dt}$

➪ 4πr² = $\frac{4}{3}πr³$ × $\frac{dr}{dt}$

➪ $\cancel\frac{4π}{4πr²}$ × = $\frac{dr}{dt}$

➪ $\frac{dr}{dt}$ = $\frac{1}{r²}$

Now the next process :

to fine $\frac{dA}{dt}$

☞︎︎︎ $\frac{dA}{dt}$ = $\frac{dA}{dr}$ × $\frac{dr}{dt}$

Remember, the surface area of the sphere is

Remember, the surface area of the sphere is 4πr²

So :

☞︎︎︎ A = 4πr²

☞︎︎︎ $\frac{dA}{dt}$ = 8πr

We already found that r = 6 and $\frac{dr}{dt}$ = $\frac{1}{r²}$

Putting this all together,

☞︎︎︎ $\frac{dA}{dt}$ = $\frac{dA}{dr}$ × $\frac{dr}{dt}$

➪ $\frac{dA}{dt}$ = 8πr × $\frac{1}{r²}$

➪ $\frac{dA}{dt}$ = $\frac{8πr}{r²}$

r and r get cancel :

➪ $\frac{dA}{dt}$ = $\frac{8π}{r}$

➪ $\frac{dA}{dt}$ = $\frac{8π}{6}$

➪ $\frac{dA}{dt}$ = $\frac{4}{3}π$

$\huge\bf\boxed{\boxed{\underline{\red{Ans = \frac{4}{3}π}}}}$