The volume of a spherical balloon is increasing at a constant rate of $0.25m^3s^-1$ . Find the rate at which the radius is increasing at the instant when the volume is $10m^3$ ?

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Answer 1 · 2018-01-30T13:38:07

$0.0149 \ ms^-1$ (3sf )

Let us set up the following variables:

${ (t,"time elapsed", s), (r, "radius of the balloon at time "t, m), (V, "Volume of the balloon at time "t, m^3s^(-1)) :}$

Using the standard formula for the volume of a sphere, we have:

$V = 4/3pir^3$

If we differentiate wrt $r$ then we get:

$(dV)/(dr) = 4pir^2$

And applying the chain rule, we have:

$(dV)/(dt) = (dV)/(dr) * (dr)/(dt)$

And so we have:

$(dV)/(dt) = 4pir^2 (dr)/(dt)$

Now, the question tells us that the "volume is increasing at a constant rate of $0.25 \ m^3s^-1$ ", which mathematically means that:

$(dV)/(dt) = 0.25$

Therefore:

$4pir^2 (dr)/(dt) = 0.25 => (dr)/(dt)= 1/(16pir^2)$

And we are asked to find "rate at which the radius is increasing at the instant when the volume is $10 \ m^3$ . If $V=10$ , then:

$4/3pir^3 = 10$
$:. r^3 = 15/(2pi)$
$:. r = root(3)(15/(2pi))$

And with this value of $r$ we can calculate $(dr)/dt:$

$(dr)/(dt) = 1/(16pi \ (root(3)(15/(2pi)))^2)$
$\ \ \ \ \ = 0.0149 \ ms^-1$ (3sf )

The volume of a spherical balloon is increasing at a constant rate of 0.25m^3s^-10.25m^3s^-1. Find the rate at which the radius is increasing at the instant when the volume is 10m^310m^3?