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

The rate of change of the volume of the cylinder at that instant is -19,200π cubic kilometers per second.

Step-by-step explanation:

To find the rate of change of the volume of the cylinder, we need to use the formula for the volume of a cylinder, which is V = πr²h. We know that the radius of the base is decreasing at a rate of 12 kilometers per second. The height of the cylinder is fixed at 2.5 kilometers. At a certain instant, the radius is 40 kilometers. We can plug these values into the formula to find the rate of change of the volume.

V = π(40)²(2.5)

V = 50,275π

To find the rate of change of the volume, we can take the derivative of the equation with respect to time:

rac{dV}{dt} = rac{d}{dt}(50,275π)

Since the height is fixed, its derivative is zero. The derivative of the radius is -12 kilometers per second. Therefore, the rate of change of the volume of the cylinder at that instant is -12(40)²π = -19,200π cubic kilometers per second.

Answer 2

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)

`(dV)/(dt) = (d)/(dt)\pi r^2h`

As
`\pi, h` are constant:

`(dV)/(dt) = \pi h(d)/(dt) r^2\\\Rightarrow (dV)/(dt) = \pi h* 2 r(dr)/(dt) \\$Putting the values:$\\\Rigghtarrow(dV)/(dt) = 3.14 * 2.5* 2 * 40* 12 \\\Rigghtarrow(dV)/(dt) = 7536\ km^3/sec`