
A balloon, which always remains spherical, has a variable diameter $\dfrac{3}{2}(2x + 1)$. Find the rate of change of its volume with respect to $x$.
Answer
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Hint: Use the formula of volume of sphere $v$=$\dfrac{4}{3}\pi {r^3}$ and then we find the volume of the balloon which is in spherical shape with respect to X.
Complete Step-by-Step solution:
So here $d$ be the diameter of the balloon which is given with the value $d$=$\dfrac{3}{2}(2x + 1)$
Let ${\text{r}}$ be the radius of balloon
We know that radius is half of the diameter so ${\text{r}}$ =$\dfrac{d}{2}$$ \to $ equation $1$
So to get the ${\text{r}}$ value let us substitute $d$ value in equation $1$
Then, ${\text{r}}$=$\dfrac{3}{4}(2x + 1)$
Here we have to find the volume of balloon, where balloon is in spherical shape
We know that volume of sphere $v$=$\dfrac{4}{3}\pi {r^3}$
So from this we can say that,
Volume of Sphere= Volume of balloon
So, volume of balloon $V$ =$\dfrac{4}{3} \times \pi \times {\left( {\dfrac{3}{4}} \right)^3}{(2x + 1)^3}$ [ $\because $ radius of balloon=$\dfrac{3}{4}(2x + 1)$ ]
$V$ =$\dfrac{4}{3} \times \pi \times \dfrac{{27}}{{64}}{(2x + 1)^3}$
$V$=$\dfrac{9}{{16}}\pi {(2x + 1)^3}$
So from this we got the volume of balloon as $V$=$\dfrac{9}{{16}}\pi {(2x + 1)^3}$
But here we have to find the rate of change volume of balloon with respect to $x$
To get the rate of change of volume we have to differentiate the volume of balloon ( $V$) with respect to $x$=$\dfrac{{dV}}{{dx}}$
$\dfrac{{dV}}{{dx}}$$ = \dfrac{9}{{16}} \times \pi \times {(2x + 1)^3}$
$\dfrac{{dV}}{{dx}}$$ = \dfrac{9}{{16}} \times \pi \times 3 \times {(2x + 1)^2} \times 2$
$\dfrac{{dV}}{{dx}}$=$\dfrac{{27}}{8}\pi {(2x + 1)^2}$
Hence the rate of change of volume of balloon $\dfrac{{dV}}{{dx}}$=$\dfrac{{27}}{8}\pi {(2x + 1)^2}$
NOTE: Whenever there are indirect values in the solution we have found the direct values by simplifying to get the answer. Example in above problem we need radius of balloon to find the volume of it but diameter of balloon is given so here we found the radius of balloon by using diameter where we know that radius is half of the diameter ${\text{r}}$=$\dfrac{d}{2}$
Complete Step-by-Step solution:
So here $d$ be the diameter of the balloon which is given with the value $d$=$\dfrac{3}{2}(2x + 1)$
Let ${\text{r}}$ be the radius of balloon
We know that radius is half of the diameter so ${\text{r}}$ =$\dfrac{d}{2}$$ \to $ equation $1$
So to get the ${\text{r}}$ value let us substitute $d$ value in equation $1$
Then, ${\text{r}}$=$\dfrac{3}{4}(2x + 1)$
Here we have to find the volume of balloon, where balloon is in spherical shape
We know that volume of sphere $v$=$\dfrac{4}{3}\pi {r^3}$
So from this we can say that,
Volume of Sphere= Volume of balloon
So, volume of balloon $V$ =$\dfrac{4}{3} \times \pi \times {\left( {\dfrac{3}{4}} \right)^3}{(2x + 1)^3}$ [ $\because $ radius of balloon=$\dfrac{3}{4}(2x + 1)$ ]
$V$ =$\dfrac{4}{3} \times \pi \times \dfrac{{27}}{{64}}{(2x + 1)^3}$
$V$=$\dfrac{9}{{16}}\pi {(2x + 1)^3}$
So from this we got the volume of balloon as $V$=$\dfrac{9}{{16}}\pi {(2x + 1)^3}$
But here we have to find the rate of change volume of balloon with respect to $x$
To get the rate of change of volume we have to differentiate the volume of balloon ( $V$) with respect to $x$=$\dfrac{{dV}}{{dx}}$
$\dfrac{{dV}}{{dx}}$$ = \dfrac{9}{{16}} \times \pi \times {(2x + 1)^3}$
$\dfrac{{dV}}{{dx}}$$ = \dfrac{9}{{16}} \times \pi \times 3 \times {(2x + 1)^2} \times 2$
$\dfrac{{dV}}{{dx}}$=$\dfrac{{27}}{8}\pi {(2x + 1)^2}$
Hence the rate of change of volume of balloon $\dfrac{{dV}}{{dx}}$=$\dfrac{{27}}{8}\pi {(2x + 1)^2}$
NOTE: Whenever there are indirect values in the solution we have found the direct values by simplifying to get the answer. Example in above problem we need radius of balloon to find the volume of it but diameter of balloon is given so here we found the radius of balloon by using diameter where we know that radius is half of the diameter ${\text{r}}$=$\dfrac{d}{2}$
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