Home | Blog | Wiki | About

Tags: / math practical /

An example from Polya's 'How to solve it'.

I recently read a part of How to Solve it by the Hungarian mathematician George Polya (Pólya György) and found a nice example of how basic math can be used in every-day life to infer practically useful information.

It's named 'A rate problem' and appears in Part I.

In this blog post, I try to retell it. 😊

---

Consider a vessel in the shape of a hollow inverted right-circular cone with horizontal base. The base of this cone is the opening of the vessel.

Measurements of the cone are:

• base radius: a.
• height: b

Suppose water is entering into this vessel at a rate r.

We need to find the rate at which the water level is rising inside the vessel when the water level is at a height y from the pointed end of the cone.

The overall setup looks something like:

            a          
       +----------+           
       |          |           

  +--- +---------------------+
  |     \         .         /                      
  |      \        .        /                        
  |       \       .       /                    
  |        \-------------/  ------+            
  |         \     .     /         |            
b |          \    .    /          |              
  |           \   .   /           | y = current water level
  |            \  .  /            |            
  |             \ . /             |            
  |              \./              |                           
  +------------   V --------------+ 

Let's write down the data we know:

• base radius of the conical vessal: a.
• height of the conical vessel: b
• rate at which water is flowing into the vessel: r
• current water level: y

What we need to find is:

• rate at which y is increasing as water is flowing at the rate r.

Now the water level changes as time goes by since more water is entering the vessel without the vessel losing any of the water that's already inside it.

So we can say the y is a function of time t.

ie, the rate at which y increasing means rate at which the water level y is changing with respect to time.

Then what we need to find is:

 dy
────
 dt

(Because derivative of a function gives with respect to a variable gives the rate of change of that function with respect to that variable.)

All right, we have now got a notation/name for what we need to find out.

Now we need to find a relation between the information that we know and this dy/dt.

Let's have a look at the water that's already inside the vessel, even as more water is entering the vessel at rate r.

We can see that the water that's already inside the vessel assumes the shape of a cone as well.

Let x be the radius of the base of this cone.

            a        x 
       +----------+-----+     
       |          |     |     

  +--- +---------------------+
  |     \         .         /                      
  |      \        .        /                        
  |       \       .       /                    
  |        \-------------/  ------+            
  |         \     .     /         |            
b |          \    .    /          |              
  |           \   .   /           | y = current water level
  |            \  .  /            |            
  |             \ . /             |            
  |              \./              |                           
  +------------   V --------------+ 

Volume of a right circular cone is given by:

 πr²h
──────
  3   

where r is radius of base and h is height.

So the volume of water that's already inside the vessel when the water level is y is

     πx²y
V = ──────
      3   

Like in the case of the water level, the volume of water inside the vessel increases as time goes by. So V is a function of t as well.

So the rate at which volume of water changes is:

 dV
────
 dt

But we already know that water is entering the vessel at a rate r.

So

     dV
r = ────
     dt

Now, for a moment, consider the 3D vessel in 2D.

So it is like:

            a        x 
       +----------+------+     
       |          |      |     
                         .     
       B          P      .   C
  +--- +----------+----------+
  |     \         |      .  /                      
  |      \        |      . /                        
  |       \      O|      ./                    
  |        +------+------+  ------+            
  |       M \     |     / N       |            
b |          \    |    /          |              
  |           \   |   /           | y = current water level
  |            \  |  /            |            
  |             \ | /             |            
  |              \|/              |                           
  +------------   V --------------+ 

                  A

where the vessel corresponds to ΔABC and the cone formed by the water that's already inside the vessel corresponds to ΔAMN.

Let's have a closer look at these two triangles.

∠BAC = ∠MAN   (same angle)
∠ABC = ∠AMN   (since BC ‖ MN)

So, ΔABC and ΔAMN are similar triangles (by AA similarity criterion).

ie,, ΔABC ~ ΔAMN

This means that the ratio of any two corresponding sides would be the same.

   OM     AO
  ──── = ────
   PB     AP


    x     y
=> ─── = ───
    a     b


         a.y 
=>  x = ──────  
          b

Cool! We got a relation between the base radius of the water cone and a, y & b.

Now recall that

         πx²y
   V  = ──────
          3


         πx²y     π(a.y/b)²y 
=> V  = ─────  = ───────────── 
          3           3      


         πa²y²y
=> V  = ───────
          3b²

         3b²V
=> y³ = ──────
          πa² 

Differentiating on both sides with respect to t,

     dy      3b²     dV
3y².──── = ────── * ────
     dt      πa²     dt

(∵ a, b, h, etc are all constants.)


    dy        3b²       dV  
=> ──── = ────────── * ────
    dt     3y²*2πa²     dt


    dy        b²       dV
=> ──── = ───────── * ────
    dt     y²*2πa²     dt

And that's it! We have got dy/dt, which is what we were after.

Let's try an example case.

Suppose the values of constants were like:

• a = 4m
• b = 3m
• r = dV/dt = 2ms⁻¹
• y = 1m

Then the rate at which the water level is rising would be:

    dy         3²         
=> ──── = ────────── * 2   
    dt     1²*2π*4²       


               3²   
        = ─────────────
           1² * π * 4² 

               9    
        = ─────────────
              16π 

So, if the conical vessel is of height 3m with base radius 4m, the water level inside it rises at a rate of 9/16π ms⁻¹ when the water level is 1m..