Respuesta :
Answer:
When the pile is 2 feet high, the height is changing at a rate of 0.14 feet per minute.
Step-by-step explanation:
The formula for the volume of a cone is:
[tex]V=\frac{1}{3} \pi {r}^{2} h[/tex]
Where:
r: radius of the base
h: height of the cone
We've been told that the diameter of the base is three times the height of the cone, which can be written like this:
[tex]diameter=2r=3h[/tex]
[tex]r=\frac{3}{2} h[/tex]
Then, replacing this in the volume formula:
[tex]V=\frac{1}{3} \pi {(\frac{3}{2}h)}^{2} h[/tex]
[tex]V=\frac{3}{4}\pi h^3[/tex]
Here, we've found a formula for the volume in terms of the height.
Given that the height changes in time, then the derivative of volume with respect of the time (the rate of change of the volume in time) will be:
[tex]\frac{dV}{dt} =\frac{dV}{dh}*\frac{dh}{dt}[/tex]
[tex]\frac{dV}{dt} =\frac{d}{dh}(\frac{3}{4}\pi h^3)*\frac{dh}{dt}[/tex]
[tex]\frac{dV}{dt} =(\frac{3}{4}\pi 3h^2)*\frac{dh}{dt}[/tex]
[tex]\frac{dV}{dt} =(\frac{9}{4}\pi h^2)*\frac{dh}{dt}[/tex]
Then the rate of change of the height will be:
[tex]\frac{dh}{dt} =\frac{4}{9} \frac{1}{\pi {h}^2} \frac{dV}{dt}[/tex]
So, the rate of change of the height when the pile is 2 ft high can be found using this last formula (dV/dt is the rate of change of the volume, which is 4 cubic feet per minute)
[tex]\frac{dh}{dt} =\frac{4}{9} \frac{1}{\pi {(2 ft)}^2} (4\frac{ft^3}{min})=0.14 \frac{ft}{min}[/tex]
So, when the pile is 2 feet high, the height is changing at a rate of 0.14 feet per minute.
Using implicit differentiation, it is found that the height of the pile is changing at a rate of 0.1415 feet per minute when the pile is 2 feet high.
The formula for the volume of a cone of height h and radius r is given by:
[tex]V = \frac{\pi r^2h}{3}[/tex]
In this problem, the diameter is three times the height. Since the radius is half the diameter:
[tex]2r = 3h[/tex]
[tex]r = \frac{3h}{2}[/tex]
Then
[tex]V = \frac{\pi (\frac{3h}{2})^2h}{3}[/tex]
[tex]V = \frac{3\pi h^3}{4}[/tex]
Applying implicit differentiation, the rate of change is:
[tex]\frac{dV}{dt} = \frac{9\pi h^2}{4}\frac{dh}{dt}[/tex]
- 2 feet high, thus [tex]h = 2[/tex].
- Rate of 4 cubic feet per minute, thus [tex]\frac{dV}{dt} = 4[/tex]
Then
[tex]\frac{dV}{dt} = \frac{9\pi h^2}{4}\frac{dh}{dt}[/tex]
[tex]4 = \frac{9\pi (2)^2}{4}\frac{dh}{dt}[/tex]
[tex]\frac{dh}{dt} = \frac{4}{9\pi}[/tex]
[tex]\frac{dh}{dt} = 0.1415[/tex]
The height of the pile is changing at a rate of 0.1415 feet per minute when the pile is 2 feet high.
A similar problem is given at https://brainly.com/question/18322401
