Answer

Verified

415.5k+ views

**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.

Recently Updated Pages

what is the correct chronological order of the following class 10 social science CBSE

Which of the following was not the actual cause for class 10 social science CBSE

Which of the following statements is not correct A class 10 social science CBSE

Which of the following leaders was not present in the class 10 social science CBSE

Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE

Which one of the following places is not covered by class 10 social science CBSE

Trending doubts

How do you graph the function fx 4x class 9 maths CBSE

Which are the Top 10 Largest Countries of the World?

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Difference Between Plant Cell and Animal Cell

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE

The largest tea producing country in the world is A class 10 social science CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE