Question

Water is drained out of tank, shaped as an inverted right circular cone that has a radius of 4cm and a height of 16cm, at the rate of 2 cm3/min. At what rate is the depth of the water changing at the instant when the water in the tank is 8 cm deep? Show all work and include units in your answer. Give an exact answer showing all work and include units in your answer

Answer 1

Let h = height of water  
Let r = radius of surface of water  
By similar triangles, r/h = 4/16 = 0.25, so r = 0.25h  
The volume of the water is:  
v = (1/3) * pi * r^2 * h  
v = (1/3) * pi * (0.25h)^2 * h  
v = (1/48) * pi * h^3  
dv/dh = 0.25*pi*h^2  
The question tells us that dv/dt = -3  
By the Chain Rule:  
dv/dt = dv/dh * dh/dt  
-3 = 0.25*pi*h^2 * dh/dt  
dh/dt = -3 / (0.25*pi*h^2)  
dh/dt = -12 / (pi*h^2)  
When h = 9 we have:  
dh/dt = -12 / (pi*9^2)  
dh/dt = -12 / (81*pi)  
dh/dt = -4 / (9*pi) cm^3/min  
dh/dt =~ -0.14147106052612918735011890077557 cm^3/min

Answer 2

Answer:

$\frac{dh}{dt}=-\frac{1}{2\pi}cm/min$

Step-by-step explanation:

From similar triangles, see diagram in attachment

$\frac{r}{4}=\frac{h}{16}$

We solve for $r$ to obtain,

$r=\frac{h}{4}$

The formula for calculating the volume of a cone is

$V=\frac{1}{3}\pi r^2h$

We substitute the value of $r=\frac{h}{4}$ to obtain,

$V=\frac{1}{3}\pi (\frac{h}{4})^2h$

This implies that,

$V=\frac{1}{48}\pi h^3$

We now differentiate both sides with respect to $t$ to get,

$\frac{dV}{dt}=\frac{\pi}{16}h^2 \frac{dh}{dt}$

We were given that water is drained out of the tank at a rate of $2cm^3/min$

This implies that $\frac{dV}{dt}=-2cm^3/min$.

Since we want to determine the rate at which the depth of the water is changing at the instance when the water in the tank is 8cm deep, it means $h=8cm$.

We substitute this values to obtain,

$-2=\frac{\pi}{16}(8)^2 \frac{dh}{dt}$

$\Rightarrow -2=4\pi \frac{dh}{dt}$

$\Rightarrow -1=2\pi \frac{dh}{dt}$

$\frac{dh}{dt}=-\frac{1}{2\pi}$