Question

An edge of variable cube in increasing at the rate of 10 cm/s . How fast the volume of the cube will increase when the edge is 5 cm long.

Answer 1

v=a^3
dv/dt=3a^2(da/dt)
dv/dt=3a^2(+10)
at a=5cm
dv/dt=3(25)(10)
therefore volume will be increasing by 3(25)(10)cm^3/s

Answer 2

Concept:

Here, we will apply the concept of derivation.

Given:

An edge of a variable cube is increasing at the rate of 10 cm/s.

To find:

How fast the volume of the cube will increase when the edge is 5 cm long.

Solution:

We know that the volume of a cube is = $a^{3}$

Now, in this question, we have to determine the change in the volume of the cube, which means, we have to determine the derivative of the volume with respect to the change in time.

Let the time be t

$v=a^{3} \\\frac{dv}{dt} = 3a^{2}\frac{da}{dt} \\\frac{dv}{dt} = 3a^{2} (10)$

We know that the edge of a variable cube is increasing at the rate of 10 cm/s, so we can replace $\frac{da}{dt} with 10$

So, the edge is 5cm long, a=5

$\frac{dv}{dt}=3a^{2}*10 \\ \\\frac{dv}{dt} =3*5*5*10\\\frac{dv}{dt}=750cm$

∴The volume is increasing at a rate of $750cm^{3} /s$

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