Question

A spherical balloon is inflated so that its volume is increasing at the rate of 3 ft3/min. How fast is the diameter of the balloon increasing when the radius is 1 ft

Answer 1

When an spherical balloon volume is increasing at the rate of $3ft^3/min$ then the diameter of the balloon is increasing $\frac{3}{2\pi }ft /min$

How can we find the rate of change of balloon's diameter ?

The volume of a spherical balloon is $v=\frac{4}{3} \pi r^3$

In form of diameter we can write as

$v=\frac{4}{3} \pi (\frac{D}{2} )^3\\=\frac{1}{6} \pi D^3$

Now we will differentiate both sides wrt to $t$ we get

$\frac{dv}{dt} =\frac{1}{6} \pi 3D^2 \frac{dD}{dt} \\\frac{dD}{dt} =\frac{2}{\pi D^2} \frac{dv}{dt} \\\\when r=1\\D=2ft$

Given in the question $\frac{dv}{dt} =3ft^3/min$

thus when we substitute the values we get

$\frac{dD}{dt} =\frac{2}{\pi *2^2} (3)\\\frac{dD}{dt}=\frac{3}{2\pi } ft/min$

Learn more about the differentiation here:

brainly.com/question/28046488

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