Question

The radius of the base of a cylinder is decreasing at a rate of 121212 kilometers per second. The height of the cylinder is fixed at 2.52.52, point, 5 kilometers. At a certain instant, the radius is 404040 kilometers. What is the rate of change of the volume of the cylinder at that instant (in cubic kilometers per second)

Answer 1

Answer:

7536 $km^3/sec$

Step-by-step explanation:

Given that:

Rate of decreasing of radius = 12 km/sec

Height of cylinder is fixed at = 2.5 km

Radius of cylinder = 40 km

To find:

The rate of change of Volume of the cylinder?

Solution:

First of all, let us have a look at the formula for volume of a cylinder.

$Volume = \pi r^2h$

Where $r$ is the radius and

$h$ is the height of cylinder.

As per question statement:

$r$ = 40 km (variable)

$h$ = 2.5 (constant)

$\dfrac{dV}{dt} = \dfrac{d}{dt}\pi r^2h$

As $\pi, h$ are constant:

$\dfrac{dV}{dt} = \pi h\dfrac{d}{dt} r^2\\\Rightarrow \dfrac{dV}{dt} = \pi h\times 2 r\dfrac{dr}{dt} \\ Putting the values: \\\Rigghtarrow\dfrac{dV}{dt} = 3.14 \times 2.5\times 2 \times 40\times 12 \\\Rigghtarrow\dfrac{dV}{dt} = 7536\ km^3/sec$