Respuesta :
Answer: Level of the liquid dropping at 28.28 inch/second when the liquid is 2 inches deep.
Step-by-step explanation:
Since we have given that
Height = 9 inches
Diameter = 6 inches
Radius = 3 inches
So, [tex]\dfrac{r}{h}=\dfrac{3}{9}=\dfrac{1}{3}\\\\r=\dfrac{1}{3}h[/tex]
Volume of cone is given by
[tex]V=\dfrac{1}{3}\pi r^2h\\\\V=\dfrac{1}{3}\pi \dfrac{1}{9}h^2\times h\\\\V=\dfrac{1}{27}\pi h^3[/tex]
By differentiating with respect to time t, we get that
[tex]\dfrac{dv}{dt}=\dfrac{1}{27}\pi \times 3\times h^2\dfrac{dh}{dt}=\dfrac{1}{9}\pi h^2\dfrac{dh}{dt}[/tex]
Now, the liquid drips out the bottom of the filter at the constant rate of 4 cubic inches per second, ie [tex]\dfrac{dv}{dt}=-4\ in^3[/tex]
and h = 2 inches deep.
[tex]-4=\dfrac{1}{9}\times \pi\times (2)^2\dfrac{dh}{dt}\\\\-9\pi =\dfrac{dh}{dt}\\\\-28.28=\dfrac{dh}{dt}[/tex]
Hence, level of the liquid dropping at 28.28 inch/second when the liquid is 2 inches deep.
The level of the liquid drops at a rate of 28.28 inches per second when the liquid is 2 inches deep.
What is differentiation?
It is the rate of change of a function with respect to the variuable.
A filter filled with liquid is in the shape of a vertex-down cone with a height of 9 inches and a diameter of 6 inches at its open (upper) end.
If the liquid drips out the bottom of the filter at the constant rate of 4 cubic inches per second.
Then find the level of the liquid dropping when the liquid is 2 inches deep.
The relation between the radius and the height of the cone will be
[tex]\rm \dfrac{r}{h} = \dfrac{3}{9} = \dfrac{1}{3}[/tex]
Then
[tex]\rm r = \dfrac{h}{3}[/tex]
The volume of the cone is given by
[tex]\rm V = \dfrac{1}{3} \pi r^2 h\\\\V = \dfrac{1}{3} \pi \dfrac{1}{9} h^2 * h\\\\V = \dfrac{1}{27} \pi h^3[/tex]
By differentiating with respect to the time (t), we get
[tex]\rm \dfrac{dV}{dt} = \dfrac{1}{27} \pi *3*h^2 \dfrac{dh}{dt}[/tex]
It is given that the rate of volume rate is [tex]\rm \dfrac{dV}{dt} = -4 \ \ in^2[/tex] and h = 2 inches deep. Then
[tex]-4 = \dfrac{1}{9} \pi * 2^2 * \dfrac{dh}{dt}\\\\\\\dfrac{dh}{dt} = -28.28[/tex]
Thus, the level of the liquid drops at a rate of 28.28 inches per second when the liquid is 2 inches deep.
More about the differentiation link is given below.
https://brainly.com/question/24062595
