Question 9

To build a beautiful trough with a semicircular cross section in celebration of п day on March 14, I tooka п-by-п piece of metal, bent it in semicircle, and welded two semicircles of radius 1 meter to each of the open sides. To keep up the celebration of day, I am pouring water into my trough at a rate of п cubicmeters per second, and at this moment there is cubic meters of water in my trough.

How quickly is the height of the water changing, in meters/sec?

This problem is quite a bit more challenging than the corresponding problem on the real exam; it might help to think about how to use trigonometry on this problem.

Fig. 1

The volume of the full trough is half the volume of a cylinder:

A line parallel to the top of the semicircle constitutes the current volume of water in the trough.

Let h be the height of the amount of water from the bottom of the trough to the waterline and let H be the remaining height from the waterline to the top of the trough. Thus r = h + H and H = r – h.

A line from the midpoint at the top of the trough to the point on the semicircle where the waterline meets creates a right triangle of height H, and an angle from top of the trough to the waterline of .

The area above the waterline is the area of the two right triangle of height H and the area of the two arcsections subtended by angle .

Area of triangle = Area of Arcsection =

The area of the cross-section of water in the trough at any time is, thus, the total area of the semicircle minus twice the area of the right triangle of height H and twice the area of the arcsection of angle ,

A = – – = – –

and because r = 1 meter; A = – bH –

where b is the base of the right triangle from the top of the waterline to the edge of the circle in Fig. 1.

[When empty, H = r, but b = 0, so bH = 0 and = , so A = 0. When full, and H are 0, so A = = ]

Now write b, H and is term of h, using fact the r = 1 meter.

h + H = r; h + H = 1; H = 1 – h

and therefore

and also so

So now the volume of the water in the trough at any height is: V = A * length = A* gives

1) V = ( – bH – )

2) V = ( – (1 – h) – arcsin(1 – h))

Using the initial volume condition V = equate this to expression 1)

= ( – bH – )

= – bH –

3) + = bH +

[Here is where mathematical insight and inspiration comes into play to make this problem simple.]

Notice in 3) that the term is the value of an angle (in radians), therefore equate: =

Therefore must equal bH. Remember, we are trying to find h (the height at this given volume of water). There are two ways to do this, the hard way and the easy way! Let's do the easy way first.

Since we know = , then from Fig. 1, b = cos = cos( ) = 1/2

4) = bH = H/2 = (1 – h)/2

= 1 – h

h = 1 – = 1 – 0.87 = 0.13

Now let's find h the hard(er) way. Using and H = (1 – h) we get 4) to be:

5) = bH = (1 – h)

3/16 = (1 – h)2 (2h – h2)

3/16 = 2h – 5h2 + 4h3 – h4

Newton's Method can now be used to find h from y(h) (a 4th degree polynomial with 4 roots).

Let y(h) = 2h – 5h2 + 4h3 – h4 – 3/16 = 0

and then y'(h) = 2 – 10h + 12h2 – 4h3

and to find h iterate: hn+1 = hn – y(hn)/y'(hn)

Below is Ruby code to perform Newton's Method with h0 that give correct root value of h.

def y(h); 2*h ­ 5*h*h + 4*h**3 ­ h**4 ­ 3.0/16 end                                 

def yy(h); 2 ­ 10*h + 12*h*h ­ 4*h**3 end

h = 0.0;  10.times { h = h ­ y(h)/yy(h) }; h  => 0.13397459621556132 

h = 0.25; 10.times { h = h ­ y(h)/yy(h) }; h  => 0.13397459621556135

h = 1 – (3**0.5)/2                            => 0.1339745962155614

[Other h0 values < 2 give all 4 roots of y(h): h1,2,3,4 = 1± 0.5 (0.5,1.5) and 1± (0.13,1.87) but h must be < r=1, so h = 1.5 or 1.87 are invalid and h = 0.5 would reduce the value of from .]

The really interesting question may be how would you find h if you didn't see (have the insight) to equate = and then = bH ? (This is frequently done with complex numbers, where with anexpression like (3 – x) + (5 + y)i = 7 + 4i you equate the real and imaginary parts of both sides to solve for x and y). The answer is YES!, we can find h using Newton's Method.

We can “brute force” our way to finding h from 3) using Newton's Method.

+ = bH +

+ = (1 – h) + arcsin(1 – h)

Let y(h) = – h + arcsin(1 – h) – ( + ) = 0

and then y'(h) = – – +

y'(h) = = = =

and to find h iterate: hn+1 = hn – y(hn)/y'(hn)

The Ruby code to do this is below:

include Math

def y(h); (2*h ­h*h)**0.5 ­ h*(2*h ­h*h)**0.5 + asin(1­h) ­ (PI/3 +(3**0.5)/4) end

def yy(h); ­2*(2*h ­h*h)**0.5 end 

h =0.0;   5.times {h= h ­ y(h)/yy(h)}; h    => (NaN+NaN*i) 

h =0.1;   5.times {h= h ­ y(h)/yy(h)}; h    => 0.13397459621556132

h =0.25;  5.times {h= h ­ y(h)/yy(h)}; h    => 0.13397459621556143

Thus we see, we can find the height h for any initial volume of water by “brute force” using Newton's Method. In fact, this becomes the generic process for solving for h for any initial volume amount without the need of applying any mathematical insight to the form or value of the volume expression.

Now that we have determined h we are ready to find dh/dt for the given dV/dt, using 2) V(h).

V = [ – (1 – h) – arcsin(1 – h)]

V = – [(1 – h) + arcsin(1 – h)]

dV/dt = d h{ – [(1 – h) + arcsin(1 – h)] } /dt

= 0 – d h[ – h + arcsin(1 – h)] /dt

-1 = d h[ – h + arcsin(1 – h)] /dt

-1 = [ – – + ]

-1 = [ – – – ]

-1 =

-1 = = =

= = = 0.99999999 = 1 meter/sec