Question

At time \[t > 0\] the volume of a sphere is increasing at a rate of proportional to be reciprocal of its radius. At \[t = 0\] the radius of the sphere is \[1\] unit and \[t = 15\] the radius is \[2\] units. Find the radius of the sphere as a function of time \[t\].

Answer 1

Hint: In this problem, the volume of a sphere is increasing at a rate of proportional to be reciprocal of its radius. We have to find the radius of the sphere as a function of time
Using the function, we will find the relation between that radius and time and some constants. Then, using the condition we can find the values of the constants.

Complete step-by-step answer:
It is given that; at time \[t > 0\] the volume of a sphere is increasing at a rate of proportional to be reciprocal of its radius. At \[t = 0\] the radius of the sphere is \[1\] unit and \[t = 15\] the radius is \[2\] units.
We have to find the radius of the sphere.
Let us consider, \[V\] is the volume of the sphere and \[r\] is the radius of the sphere.
Since, the volume of a sphere is increasing at a rate of proportional to be reciprocal of its radius, the function can be written as,
\[\dfrac{{dV}}{{dt}} \propto \dfrac{1}{r}\]
It can be written as,
\[\dfrac{{dV}}{{dt}} = \dfrac{k}{r},k\] be any constant. … (1)
We know that, the volume of a sphere of radius \[r\] is \[V = \dfrac{4}{3}\pi {r^3}\]
Let us substitute the volume in equation (1) we get,
\[\dfrac{{d(\dfrac{4}{3}\pi {r^3})}}{{dt}} = \dfrac{k}{r}\]
Differentiate with respect to \[t\] we get,
\[\dfrac{4}{3}\pi .3{r^2}.\dfrac{{dr}}{{dt}} = \dfrac{k}{r}\]
Let us simplify the above equation, we get,
\[\pi{4} {r^3}dr = kdt\]
Since, \[k\] is constant.
Now let us integrate both sides we get,
\[\pi{4} \int {{r^3}} dr = k\int {dt} \]
Solving we get,
\[\pi{4} \dfrac{{{r^4}}}{4} = kt + C\]
So, we have,
\[\pi {r^4} = kt + C\]… (2)
We know that, when \[t = 0,r = 1\]
Substitute the values in equation (2) we get,
\[C = \pi \]
Then equation (2) becomes,
\[\pi {r^4} = kt + \pi \]… (3)
Again, we know that, when \[t = 15,r = 2\]
On substituting the values in (3) we get,
\[\pi {2^4} = 15k + \pi \]
On solving we get,
\[15k = 16\pi - \pi = 15\pi \]
So, \[k = \pi \]
Therefore, equation (3) becomes,
\[\pi {r^4} = \pi t + \pi \]
Now let us simplify the above equation, we get,
\[\pi {r^4} = \pi (t + 1)\]
Cancel the like terms that are in common and taking fourth root on both sides we get,
So, \[r = {(t + 1)^{\dfrac{1}{4}}}\]
Hence, the radius is \[r = {(t + 1)^{\dfrac{1}{4}}}\]

Note: We know that, the volume of a sphere of radius \[r\]is \[V = \dfrac{4}{3}\pi {r^3}\]while finding the radius we substitute the given values of t and r it is an important step which helps in finding the value of K and C.