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If the radius of a sphere is increasing at a rate of 4 cm per second, how fast is the volume increasing when the diameter is 80 cm?

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Last updated date: 20th Jun 2024
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Answer
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Hint: We are given that radius of sphere is changing at 4cm/sec we have to find the rate of change of volume.
To do so, we will first learn what rate of change means, then we use differentiation to solve, we will be needing volume of sphere $v=\dfrac{4}{3}\pi {{r}^{3}}$ along with the relation between radius and diameter given as $r=\dfrac{d}{2}$ or $d=2r$ . we will also need $d\left( kx \right)=kd\left( x \right)$ and $d\left( {{x}^{n}} \right)=n{{x}^{n-1}}$ .

Complete step by step answer:
We are given that we have a sphere whose diameter is 80cm; we are also given that the radius of the sphere is increasing at a rate of 4cm/sec. we have to find the rate of change of volume.
Now for sphere of radius r, the volume is given as $\dfrac{4}{3}\pi {{r}^{3}}$
As we have that rate of change of radius is 4cm/sec
It means that radius of sphere is changing with time per second it increase 4 cm
So it means we have –
$\dfrac{dr}{dt}=4$ ………………………………. (1)
Now we are asked to find the rate of change of sphere.
It means we have to differentiate the volume with respect to time and have to see that how it is changing
As we know volume is gives as –
$v=\dfrac{4}{3}\pi {{r}^{3}}$
Now differentiating both side, we get –
$\dfrac{dv}{dt}=\dfrac{d\left( \dfrac{4}{3}\pi {{r}^{3}} \right)}{dt}$
As we know $d\left( kt \right)=kd\left( t \right)$
So,
$\dfrac{dv}{dt}=\dfrac{4}{3}\pi \dfrac{d\left( {{r}^{3}} \right)}{dt}$
Now as we know –
$\dfrac{d\left( {{x}^{n}} \right)}{dt}=n{{x}^{n-1}}\dfrac{dx}{dt}$
So using this we get –
$\dfrac{dv}{dt}=\dfrac{4}{3}\pi 3{{r}^{2}}\dfrac{dr}{dt}$
Now using $\dfrac{dr}{dt}=4$using (1)
And as diameter is 80cm so $r=\dfrac{d}{2}=\dfrac{80}{2}=40cm$
Using $\dfrac{dr}{dt}=4$and r=40 above, we get –
$\dfrac{dv}{dt}=\dfrac{4}{3}\pi \times 3\times 40\times 40\times 4$
By simplifying, we get –
$\dfrac{dv}{dt}=25600\pi $
Hence, volume is changing at a rate of $25600\pi $ 4cm/sec.

Note:
Remember we here do not have to find the volume, we have to find here the change in volume with time as it is given that radius is changing with time so we have to understand how changing in radius affects the volume. So not just use r=40 and $\dfrac{4}{3}\pi {{r}^{3}}$ to find volume.
We need differentiation to find an actual answer.