A vertical cylinder is leaking water at a rate of 4m3/sec. If the cylinder has a height of 10m and a radius of 2m, at what rate is the height of the water changing when the height is 3m?

Respuesta :

Answer:

Therefore the rate change of height is  [tex]\frac{1}{\pi}[/tex] m/s.

Step-by-step explanation:

Given that a vertical cylinder is leaking water at rate of 4 m³/s.

It means the rate change of volume is 4 m³/s.

[tex]\frac{dV}{dt}=4 \ m^3/s[/tex]

The radius of the cylinder remains constant with respect to time, but the height of the water label changes with respect to time.

The height of the cylinder be h(say).

The volume of a cylinder is [tex]V=\pi r^2 h[/tex]

                                                 [tex]=( \pi \times 2^2\times h)\ m^3[/tex]

[tex]\therefore V= 4\pi h[/tex]

Differentiating with respect to t.

[tex]\frac{dV}{dt}=4\pi \frac{dh}{dt}[/tex]

Putting the value [tex]\frac{dV}{dt}[/tex]

[tex]\Rightarrow 4\pi \frac{dh}{dt}=4[/tex]

[tex]\Rightarrow \frac{dh}{dt}=\frac{4}{4\pi}[/tex]

[tex]\Rightarrow \frac{dh}{dt}=\frac{1}{\pi}[/tex]

The rate change of height does not depend on the height.

Therefore the rate change of height is  [tex]\frac{1}{\pi}[/tex] m/s.

The height of the water is changing at a rate of approximately 0.318 meters per second.

The volume formula of the cylinder ([tex]V[/tex]), in cubic meters, is described below:

[tex]V = \pi\cdot r^{2}\cdot h[/tex] (1)

Where:

  • [tex]r[/tex] - Radius of the cylinder, in meters.
  • [tex]h[/tex] - Height of the cylinder, in meters.

According to the statement, the leakage is represented by a decrease in the height of the water. By differential calculus and under the assumption that no rate of change exists for the radius, we have the expression for the rate of change in the height of the water within the cylinder ([tex]\dot h[/tex]), in meters per second:

[tex]\dot h = \frac{\dot V}{\pi\cdot r^{2}}[/tex] (2)

Where [tex]\dot V[/tex] is the rate change in the volume occupied by the water, in cubic meters per second.

If we know that [tex]\dot V = 4\,\frac{m^{3}}{s}[/tex] and [tex]r = 2\,m[/tex], then the rate of change in the height is:

[tex]\dot h = \frac{4\,\frac{m^{3}}{sec} }{\pi\cdot (2\,m)^{2}}[/tex]

[tex]\dot h \approx 0.318 \,\frac{m}{s}[/tex]

The height of the water is changing at a rate of approximately 0.318 meters per second.

We kindly invite to check this question on rates of change: https://brainly.com/question/18904995