Question

If the volume of a spherical ball is increasing at the rate of \[4\pi cc/sec\], then the rate of increase of its radius (in cm/sec), when the volume is \[288\pi cc\], is
(A) \[\dfrac{1}{6}\]
(B) \[\dfrac{1}{9}\]
(C) \[\dfrac{1}{{36}}\]
(D) \[\dfrac{1}{{24}}\]

Answer 1

Hint: Here first we will find the radius of the sphere by using the formula of volume of sphere i.e.
\[V = \dfrac{4}{3}\pi {r^3}\] and then differentiate it to find the value of \[\dfrac{{dr}}{{dt}}\] which is the rate of change of radius.

Complete step-by-step answer:
It is given that:-
The volume of a spherical ball is increasing at the rate of\[4\pi cc/sec\]
This implies,
\[\dfrac{{dV}}{{dt}} = 4\pi cc/\sec \]…………………………(1)
Also, it is given that:-
The volume is \[288\pi cc\]
Hence,
\[V = 288\pi cc\]
Now we know that the volume of sphere is given by:-
\[V = \dfrac{4}{3}\pi {r^3}\]
Putting in the values we get:-
\[288\pi = \dfrac{4}{3}\pi {r^3}\]
Solving for \[{r^3}\] we get:-
\[{r^3} = \dfrac{{288\pi \times 3}}{{4\pi }}\]
Solving it further we get:-
\[{r^3} = 72 \times 3\]
Taking cube root of both the sides we get:-
\[\sqrt[3]{{{r^3}}} = \sqrt[3]{{72 \times 3}}\]
Solving it further we get:-
\[
  r = \sqrt[3]{{216}} \\
   \Rightarrow r = 6cm..........................\left( 2 \right) \\
 \]
Now we will differentiate the volume V with respect of t.
Hence on differentiating we get:-
\[\dfrac{d}{{dt}}\left( V \right) = \dfrac{d}{{dt}}\left( {\dfrac{4}{3}\pi {r^3}} \right)\]
Solving it further we get:-
\[\dfrac{{dV}}{{dt}} = \dfrac{4}{3}\pi \dfrac{d}{{dt}}\left( {{r^3}} \right)\]
Now we know that:-
\[\dfrac{d}{{dt}}\left( {{x^n}} \right) = n{x^{n - 1}}\]
Hence applying this formula we get:-
\[\dfrac{{dV}}{{dt}} = \dfrac{4}{3}\pi \left( {3{r^2}} \right).\dfrac{d}{{dt}}\left( r \right)\]
Simplifying it further we get:-
\[\dfrac{{dV}}{{dt}} = 4\pi \left( {{r^2}} \right).\dfrac{{dr}}{{dt}}\]
Putting the value of r from equation 2 we get:-
\[\dfrac{{dV}}{{dt}} = 4\pi {\left( 6 \right)^2}\dfrac{{dr}}{{dt}}\]
Solving it further we get:-
\[\dfrac{{dV}}{{dt}} = 4\left( {36} \right)\pi \dfrac{{dr}}{{dt}}\]
Equating this value with the value in equation1 we get:-
\[12\left( {36} \right)\pi \dfrac{{dr}}{{dt}} = 4\pi \]
Now evaluating for \[\dfrac{{dr}}{{dt}}\] we get:-
\[\dfrac{{dr}}{{dt}} = \dfrac{{4\pi }}{{4\left( {36} \right)\pi }}\]
Solving it further we get:-
\[
  \dfrac{{dr}}{{dt}} = \dfrac{1}{{36}} \\
   \Rightarrow \dfrac{{dr}}{{dt}} = \dfrac{1}{{36}}cm/\sec \\
 \]
Hence the rate of change of radius is \[\dfrac{1}{{36}}cm/\sec \]

Hence option C is the correct option.

Note: Students should note that when the rate of change is positive then it implies that the quantity is increasing at that rate while when the rate of change is negative then it implies that the quantity is decreasing at that rate.
Also, students must take care of calculations and the values substituted.