Question

the radius of a cylinder is increasion at a rate of 1 meter per hour, and the height of the cylinder is decreasing at a rate of 4 meters per hour. at a certain instant, the base radius is 5 meters and the height is 8 meters. what is the rate of change of the volume of the cylinder at the instant? (Note the formula for volume of a cylinder is V r h)

Answer 1

Answer:

The volume is decreasing at a rate 20 cubic meters per hour.

Step-by-step explanation:

We are given the following in the question:

Rate of change of radius =

$\displaystyle\frac{dr}{dt} = 1 \text{ meter per hour}$

Rate of change of height =

$\displaystyle\frac{dr}{dt} = -4 \text{ meter per hour}$

At an instant,

radius, r = 5 meters

Height, h = 8 meters

Volume of cylinder , V=

$\pi r^2 h$

where r is the radius of cylinder and h is the height of cylinder.

Rate of change of volume of cylinder =

$\displaystyle\frac{dV}{dt} = \frac{d(\pi r^2 h)}{dt} = \pi\Big(2r\frac{dr}{dt}h + r^2\frac{dh}{dt}\Big)$

Putting the value, we get,

$\displaystyle\frac{dV}{dt} = \pi\Big(2(5)(1)(8) + (5)^2(-4)\Big) = -20\text{ cubic meters per hour}$

Thus, the volume is decreasing at a rate 20 cubic meters per hour.

Answer 2

The rate of change of the volume at that particular instant is - 62.8 m^3/h.

How to get the rate of change of the volume?

First, we can write the dimensions as:

• radius = R = (5m + 1m/h*t)
• height = H = (8m - 4 m/h*t)

Where t is the time in hours.

Then the volume of the cylinder will be:

V = pi*R^2*H = 3.14*(5m + 1m/h*t)^2*(8m - 4 m/h*t)

To get the rate of change, we need to differentiate it with respect to t, we will get:

V' = 3.14*(2*(5m + 1m/h*t)*(1m/h)*(8m - 4 m/h*t) + (5m + 1m/h*t)^2*(-4m/h))

V' = 3.14*( (2 m/h)*(5m + 1m/h*t)*(8m - 4 m/h*t) - (4m/h)*(5m + 1m/h*t)^2)

V' = 3.14*(2m/h)*( (5m + 1m/h*t)*(8m - 4 m/h*t) - 2*(5m + 1m/h*t)^2)

As you can see the rate of change depends on t, but we want the rate of change at this instant, then we use t = 0, replacing that on the above equation we get:

V'(0) = 3.14*(2m/h)*( (5m + 1m/h*0)*(8m - 4 m/h*0) - 2*(5m + 1m/h*0)^2)

V'(0) = 3.14*(2m/h)*( (5m)*(8m ) - 2*(5m)^2)=  -62.8 m^3/h

If you want to learn more about rates of change, you can read:

https://brainly.com/question/24313700