# How fast does a tank drain?
1.(1)Suppose the tank is cylindrical with height 2cm and radius 1m and the hole is circular with radius 2cm. If we take g = 10 m/s^2, show that h satisfies the differential equation:
$$\frac{dh}{dt}=-0.0004\sqrt{20h}$$

### sol(1)

Recall: V=$\pi$$r^2$h
$$\frac{dV}{dt}=\pi r^2\frac{dh}{dt}=\pi\frac{dh}{dt}$$
$$\frac{dV}{dt}=-a\sqrt{2gh}=-\pi r^2\sqrt{2*32*h}=-0.0001\pi*8*\sqrt{h}$$
(while r=the radius of the hole)

Therefore,
$$\frac{dV}{dt}=\pi\frac{dh}{dt}=-0.0008\pi\sqrt{h}$$
#### => $$\frac{dh}{dt}=-0.0004\sqrt{20h}$$

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1.(2)Solve this equation to find the height of the water at time t, assuming the tank is full at time t = 0.

### sol(2)

Given initial condition is h(0)=2m
Solve the differential equation obtained in sol(1)
$$\frac{dh}{dt}=-0.0004\sqrt{20h}$$
$$\frac{dh}{\sqrt{20h}}=-0.0004dt$$
$$\int\frac{dh}{\sqrt{20h}}=-0.0004\int{dt}$$
$$\frac{\sqrt{h}}{\sqrt{5}}=-0.0004t+C$$
Apply the initial condition H(0)=2m
$$\sqrt{2}=0+\sqrt{5}C \space\space=>\space\space C=\frac{\sqrt{2}}{\sqrt{5}}$$
$$\sqrt{h}=-0.0004\sqrt{5}t+\frac{\sqrt{2}}{\sqrt{5}}$$
$$h=({-0.0004\sqrt{5}t+\frac{\sqrt{2}}{\sqrt{5}}} )^2$$
Hence the height of the water at time t is
> $$h=({-0.0004\sqrt{5}t+\frac{\sqrt{2}}{\sqrt{5}}} )^2$$

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1.(3)How long will it take for the water to drain completely?

### sol(3)

If the tank is empty then **h=0**
Let **h=0** be the result of (2)
$$({-0.0004\sqrt{5}t+\frac{\sqrt{2}}{\sqrt{5}}} )^2=0$$
$$-0.0004\sqrt{5}t+\frac{\sqrt{2}}{\sqrt{5}}=0$$
$$t=\frac{\sqrt{2}}{0.002}\approx707.106s$$

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2.(1)Suppose that a hole is drilled in the side of a cylindrical bottle and the height h of the water (above the hole) decreases from 10 cm to 3 cm in 68 seconds.Use equation2 to find an expression h(t).Evaluate h(t) for t = 10, 20, 30, 40, 50, and 60.

### sol(1)

To find out the solution of the above equation, we have to seperate the variables
$$\frac{dh}{dt}=k\sqrt{h}$$ into
$$h^{-1/2}dh=kdt$$
then integrate both sides and add the constants on both sides
$$\int h^{-1/2}dh +C1=\int kdt+C2$$
$$2h^{1/2}=kt+C$$
$$h^{1/2}=1/2(kt+C)$$
Hence$$h(t)=1/4(kt+C)^2$$
Value of h(t)for **t=10,20,30,40,50,60 sec**
$$h(10)=1/4(10k+C)^2\space\space h(20)=1/4(20k+C)^2\space\space h(30)=1/4(30k+C)^2\space\space h(40)=1/4(40k+C)^2\space\space h(50)=1/4(50k+C)^2\space\space h(60)=1/4(60k+C)^2\space\space$$

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3. In many parts of the world, the water for sprinkler systems in large hotels and hospitals is supplied by gravity from cylindrical tanks on or near the roofs of the buildings.Suppose such a tank has radius 3 m and the diameter of the outlet is 6 cm. An engineer has to guarantee that the water pressure will be at least 104kPa for a period of 10 minutes.What height should the engineer specify for the tank in order to make such a guarantee?

### sol

Recall: Pressure = Pa *g *d
After 10 min=600sec
$$P(t=600s)=10*h*600>=104kPa$$
=>$$h(600)>=104000/10$$
### Recall Torricelli's Law:
$$\frac{dV}{dt}=-a\sqrt{2gh}$$
$$and \space\space V=9\pi h$$
$$\frac{dV}{dt}=9\pi\frac{dh}{dt}$$
=>
$$9\pi\frac{dh}{dt}=-a\sqrt{2gh}$$
where g=10 a=area of the hole
$$9\pi\frac{dh}{dt}=-9\pi\sqrt{2*10*h}$$
$$\frac{dh}{dt}=-\sqrt{2*10*h}$$
$$\int \frac{dh}{\sqrt{h}}=\frac{-dt}{\sqrt{20}}=\frac{-t}{\sqrt{20}}+C1$$
$$\int h^{-1/2}dh=2\sqrt{h}+C2$$
=>
$$2\sqrt{h}=\frac{-t}{\sqrt{20}}+C$$
$$\sqrt{h}=\frac{-t}{2\sqrt{20}}+C/2$$
### Let k=C/2
=>
$$h(t)=(k-\frac{t}{2\sqrt{20}})^2$$
$$h(600)=(k-\frac{600}{2\sqrt{20}})^2>=10400/10$$
$$k>=1040+\frac{600}{2\sqrt{20}}$$
$$h(0)=(k-\frac{0}{2\sqrt{20}})^2=k^2$$
=>
$$k^2>=(1040-\frac{600}{2\sqrt{20}})^2\approx1174.168$$
Hence the tank must be at least 1174.168m tall.

4.(1)Suppose the tank has the shape of a sphere with radius 2m and is initially half full of water. If the radius of the circular hole is 1cm and we take g=10m/s^2,show that h satisfies the differential equation
$$(4h-h^2)\frac{dh}{dt}=-0.0001\sqrt{20h}$$

Recall:

$$A(h)\frac{dh}{dt}=-a\sqrt{2gh}$$
where A(h)=pi r^2,r=f(h)
$$\pi r^2\frac{dh}{dt}=-\pi(1/100)^2\sqrt{2*10*h}$$
### Recall: r=f(h)&Equation of a circle
$$r^2=4-(h-2)^2=4h-h^2$$
Hence
$$(4h-h^2)\frac{dh}{dt}=-0.0001\sqrt{20h}$$
4.(2)How long will it take for the water to drain completely?

We want to find the t such that h(t)=0m.
$$(4h-h^2)\frac{dh}{dt}=-0.0001\sqrt{20h}$$
$$(4h^{1/2}-h^{3/2})dh=-0.0001\sqrt{20}dt$$
$$\int(4h^{1/2}-h^{3/2})dh=\int-0.0001\sqrt{20}dt$$
$$\frac{8}{3}h^{3/2}-\frac{2}{5}h^{5/2}=-0.0001\sqrt{20}t+C$$
Note that h(0)=2
$$\frac{8}{3}2^{3/2}-\frac{2}{5}2^{5/2}=0+C$$
$$C=\frac{56}{15}\sqrt{2}$$
When h(t)=0
$$0-0=-0.0001\sqrt{20}t+\frac{56}{15}\sqrt{2}$$
$$t=\frac{56000}{15}\frac{\sqrt{2}}{\sqrt20}\approx11806s$$