Respuesta :
Answer:
Therefore the depth of the water is changing at the instant when the water in the tank is 9 cm deep at rate [tex]\frac{4}{27\pi}[/tex] [tex]cm^3[/tex]/ min.
Step-by-step explanation:
Given that,
Radius of the cone(r)= 6 cm
Height of the cone (h)= 12 cm
[tex]\therefore \frac rh =\frac 6{12}[/tex]
[tex]\Rightarrow h=2r[/tex]
[tex]\Rightarrow r=\frac h2[/tex]
The volume of the cone is (V) [tex]=\frac13 \pi r^2 h[/tex]
[tex]\therefore V=\frac13 \pi r^2 h[/tex]
Putting [tex]r=\frac h2[/tex]
[tex]\therefore V=\frac13 \pi (\frac h2)^2 h[/tex]
[tex]\Rightarrow V=\frac1 {12} \pi h^3[/tex]
Differentiating with respect to t
[tex]\frac{dV}{dt}=\frac 1{12}\pi . (3h^2) \frac{dh}{dt}[/tex]
[tex]\Rightarrow \frac{dV}{dt}=\frac{ 1}{4}\pi . h^2\frac{dh}{dt}[/tex] ....(1)
Given that water is drained out of tank at the rate 3 [tex]cm^3[/tex]/ min.
It means the rate change of volume is 3 [tex]cm^3[/tex]/ min that is [tex]\frac{dV}{dt}=3 \ cm^3/min[/tex]
Putting the value of [tex]\frac{dV}{dt}[/tex] in equation (1)
[tex]\therefore 3=\frac{ 1}{4}\pi . h^2\frac{dh}{dt}[/tex]
[tex]\Rightarrow \frac{dh}{dt}=\frac{3\times 4}{\pi h^2}[/tex]
[tex]\Rightarrow \frac{dh}{dt}=\frac{12}{\pi h^2}[/tex]
To find the rate of the depth of water changing at 9 cm depth, we need to put h=9 cm in the above equation.
[tex]\therefore \frac{dh}{dt}|_{h=9}=\frac{12}{\pi .9^2}[/tex]
[tex]=\frac{4}{27\pi}[/tex] [tex]cm^3[/tex]/ min.
Therefore the depth of the water is changing at the instant when the water in the tank is 9 cm deep at rate [tex]\frac{4}{27\pi}[/tex] [tex]cm^3[/tex]/ min.
