Saturday, February 26, 2011

River Boat Speed

Although Johnny is doing the CIE IGCSE Additional Mathematics programme, he is confused at times by questions in regular Mathematics which his teacher deems “not so hard,” such as the following:

Captain Jack has a motorboat that travels at 20 km per hour in still water. Given that the current in the river is 5 km per hour, the boat’s downstream speed (ignoring friction) will be 20 + 5 = 25 km/h while its speed in the upstream direction will be 20 – 5 = 15 km/h. Suppose that Jack has just enough fuel for 3 hours. How far can he go downstream before he has to turn around to get back home?”

A first attempt

The first physics formula Johnny learnt was ‘Distance = Speed × Time’ or D = S × T. For brevity he decides to use subscripts to identity total, average, downstream or upstream variables, T, A, D, and U, respectively. Thus his formula for the boat’s downstream distance is DD = SD × TD = 25 TD. Similarly his upstream distance formula is DU = SU × TU = 15 TU. Johnny also knows average speed is total distance divided by the total time, giving him the equation SA = DT ÷ TT. He reasons that the average speed, SA, is (15 + 25)/2 = 20 and consequently the total distance must be 20 × 3 = 60 km.

DD + DU = DT
SD × TD + SU × TU = DT
25 TD + 15 TU = 60
25 TD + 15 (3 – TD) = 60
25 TD + 45 – 15 TD = 60
10 TD = 15
TD = 1.5 hours

Consequently Johnny tells his group the downstream distance will be 25 km/h × 1.5 hours = 37.5 km.

Disagreement
Mary is in the same discussion group. She says to Johnny: That can’t be right. I used ‘Trial and Improvement.’ The two distance variables must be equal, DD = DU. Given that the total time is 3 hours, I wrote TU = 3 – TD. Obviously the downstream time cannot be less than 0 hours and your own calculations demonstrate that TD must be less than or equal to 1.5 hours. Using the vocabulary of functions, the domain of the downstream distance is 0 < TD < 1.5 hours.
Mary’s trials are shown below.

 Trial TD TU = 3 – TD DD = 25TD DU = 15TU Error 1st 1.5 hours 1.5 hours 35.5 km 22.5 km 15 km 2nd 0 hours 3 hours 0 km 45 km -45 km 3rd 1 hour 2 hours 25 km 30 km -5 km 4th 1.25 hour 1.75 hour 31.25 km 26.25 km +5 km 5th 1.125 hour 1.875 hour 28.125 km 28.125 km 0 km

Johnny replies:Your table certainly looks convincing. But the algebra I used is perfectly logical. How can there be a mistake?
At this point, as the teacher, I confirm that Mary is correct and ask the group to find the mistake in Johnny’s algebra.

Algebra revisited

Johnny’s experience tells him it is extremely difficult to proof-read for errors so he decides to take a slightly different approach. After discussion with Mary, he now realizes that he does not actually need the total distance. He only requires that:
DD = DU
25 TD = 15 TU
25 TD = 15 (3 – TD)
25 TD = 45 – 15 TD
40 TD = 45
TD = 45 ÷ 40 = 1.125 hours
DD = 25 km/h × 1.125 h = 28.125 km

While he is able to confirm Mary’s answer obtained by ‘Trial and Improvement,’ nevertheless, he is somewhat dissatisfied. Algebra required rather more mental gymnastics and, at least initially, appeared to steer him in the wrong direction. He asks me “Is algebra worth the effort?”
My honest answer is, “We have no need of algebra if we are only seeking the answer to one particular question. Algebra’s real advantage is that it is very easily generalized. Keeping the total time at 3 hours but supposing the current is some other constant than ‘5’, you can readily modify the previous solution as follows.
DD = DU
SD × TD = SU × TU
SD × TD = SU (3 – TD)
SD × TD = 3 SU – SU × TD
(SD + SU) TD = 3 SU
TD = 3 SU ÷ (SD + SU)
Would it be significantly harder to suppose we had fuel for H hours instead of 3 hours?

ICT

An approach very similar to Trial and Improvement uses the computer programme Geogebra to model the situation:

Sorry, the GeoGebra Applet could not be started. Please make sure that Java 1.4.2 (or later) is installed and active in your browser (Click here to install Java now)

Where did Johnny go wrong?

Like many students, Johnny is over-reliant on the arithmetic mean. For the purpose of determining the average speed, the Harmonic Mean is called for. How do you obtain the Harmonic Mean of two numbers a and b

 Procedure Algebra Captain Jack’s speeds Start with a and b a, b 15, 25 Take the reciprocal of each number 1/a and 1/b 1/15, 1/25 Rewrite with a common denominator in lowest terms b/ab, a/ab 1/15 = 5/75, 1/25 = 3/75 Find the arithmetic mean (a + b)/2/ab (5+3)/2/75 = 4/75 Take the reciprocal of the previous line 2ab/(a+b) 75/4

The total distance = (harmonic average of speed) × Total time = 75/4 × 3 = 225/4 = 56.25 km.
The downstream distance must be half the upstream distance or 28.125 km, as previously calculated.
(Please note the harmonic mean is not an explicit part of the IGCSE Maths Syllabus and consequently it is unlikely to be found on the examiner’s official mark scheme.)
Conclusion

As Grandma used to say, “There’s more than one way “to skin a cat.” (Cat befrienders are advised to replace the offending text with “poach an egg.”) Today’s message is, “There’s more than one way to solve a problem.” In the ‘real world,’ team skills and independent verification of results are important. Used correctly, algebra is indeed logical, but you should appreciate that an apparently small misunderstanding will lead to a false conclusion.  There is more to a mathematics education than the syllabus.

Tuesday, February 15, 2011

"In this question, i is a unit vector due east and j is a unit vector due north. At 0900 hours a ship sails from the point P with position vector (2i + 3j) km relative to an origin O. The ship sails north-east with a speed of 15 √2 km h–1.

i) Find, in terms of i and j, the velocity of the ship.     [2]

ii) Show that the ship will be at the point with position vector (24.5i + 25.5j) km at 1030 hours.       [1]

iii) Find, in terms of i, j and t, the position of the ship t hours after leaving P.                [2]

At the same time as the ship leaves P, a submarine leaves the point Q with position vector(47i – 27j) km.  The submarine proceeds with a speed of 25 km h–1 due north to meet the ship.

iv) Find, in terms of i and j, the velocity of the ship relative to the submarine.                  [2]

v) Find the position vector of the point where the submarine meets the ship.                [2]
CIE ADD MATHS summer 2008 paper 1 question 10"

The velocity of the ship:

North-east has a compass bearing (angle with the y-axis or North axis) of 045°. Therefore the angle with the x-axis (i.e., East axis) is also 45° and the velocity vector forms an isosceles right triangle.  Recall that the
sine and cosine of 45° are both 1/√2. Keep this exact value: do not use the approximation 0.707.

Since ship and sub both start with ‘s’, I have somewhat arbitrarily decided to use for the names of the boats their respective starting points in the equations which follow. ‘P’ subscripts denote the ship; ‘Q’ subscripts denote the submarine.

VP = 15 √2 (1/√2) i +  15 √2 (1/√2) j

VP = 15  i +  15 j

The position of the ship t hours after leaving P:

Displacement of the ship from origin, O  = ship’s original position + velocity * time

DP = P +t vP
DP = (2 i + 3 j) + (15 t i + 15 t j)
DP = (2 + 15 t) i + (3 + 15 t) j

The  ship will be at the point (24.5i + 25.5j) km at 1030 hours

Note that I did part iii) first and then returned to part ii) as a special case.

At 1030, one and a half hours have elapsed since 0900, so substitute t = 1.5 into the equation above.

DP = (2 + 22.5) i + (3 + 22.5) j

DP =  24.5 i + 25.5 j

The velocity of the ship relative to the submarine.

The velocity of P relative to Q is
vP vQ = (15  i + 15  j) – (0 i + 25 j)
= 15  i – 10 j

The position vector of the point where the submarine meets the ship.
For the ship to meet the sub, the two position vectors (or using the vocabulary of physics, their displacements) must be equal for the same value of time, t.
DP = DQ
When vectors are equal, all their components are equal. Thus, the coefficients of i and of j for both displacements are equal.
Equating the coefficients of i we get:
2 + 15 t  = 47
15 t = 45
t = 3
Likewise, equating the coefficients of j we get:
3 + 15 t = 25 t 27
30= 10 t
t = 3
As the t values are equal, we can be sure the ship meets the sub at 1200 noon.  I.e., noon is 3 hours after the starting time of 0900. (As an aside; had the solutions produced different values for t, we would have concluded the system was inconsistent, i.e., that the ship and the sub did not meet.)
The ship and sub meet at:
DP = DQ =  47 i + (25(3) 27) j
= 47 i + 48 j

An alternate method:

The velocity of submarine relative to itself is vQ v= 0. ( A bolded zero is a zero vector.)

Hence the relative position of Q remains as 47 i – 27 j.  We can 'pretend' that Q is not moving.  We do this all the time: we pretend the earth is not whizzing through space at thousands of km per second!  For the ship to meet the sub, its relative position must also be 47 i – 27 j.

We can calculate the Relative Position of P (with respect to Q) at time t using:
Relative Position of P  = the ship’s original position + Relative velocity * time

47 i – 27 j = (2 i + 3 j) + (15 t i – 10 t j)
47 i – 2 i – 15 t i = 3 j + 27 j – 10 t j
45 i– 15 t i = 30 j – 10 t j
(45 – 15 t) i = (30 – 10 t) j

Since i and j are in different directions this vector equation can only have the solution  0 = 0.
With scalars, the equation 1x = 1y implies x = y. However vector equations such as1i = 1j or 2i = 2j, declare an impossibility, since perceptible movement East is never equal to any amount of movement North. However we do allow a movement of 0 units East to equal a movement of 0 units North.

We require that (45 – 15 t) i = 0 i .  Hence t = 3.
We also require that (30 – 10 t) j = 0 j .  It is confirmed that t = 3.

(Had we found two different values of t, the system of equations would have been inconsistent-implying no meeting of the ship and sub. As the system is consistent the ship and the sub do meet.)

The actual (not relative) position of ship P at 12 noon is found is found just as we found its position at 1030.

Displacement of the ship P from origin, O  = ship’s original position + velocity * time

DP = vP
DP = (2 i + 3 j )+ (15 t i + 15 t j)
DP = (2 + 15 (3)) i + (3 + 15 (3)) j
DP = 47 i + 48 j

Monday, February 14, 2011

1 a) Express 2x2 – 8x + 5 in the form a(x + b)2 + c where a, b and c are integers. CIE ADD MATHS summer 2002, paper 1, question 11

Recall that (x + b) 2 = x 2 + 2bx + b 2. A common mistake is (x + b) 2 = x2 + b 2.  As a simple example, suppose x = b = 1. Then (1 + 1) 21 2 + 1 2.

Multiply (x + b) 2 through by a and add c: a(x + b) 2+ c = ax 2 + 2abx + ab 2 + c.

By polynomial identity, the squared terms of both forms can be equated, likewise the linear terms are equal, and finally the constant terms are also equal:

2x2 = ax 2     -->     a = 2
– 8x = 2abx     --> b = -2
5 = ab 2 + c    -->  c = -3

Thus 2x2 – 8x + 5  = 2(x + (-2))2 + (-3)

This sort of problem can also be solved by “completing the square,” which will be the subject of another blog. The equation y = 2x2 – 8x + 5 is written in expanded form, while y = 2(x – 2) 2 – 3 is the same parabola written in vertex form. As the squared term (x – 2) 2 cannot be negative, its minimum is 0 and this occurs when x = 2. Then, y = 2(0) 2 – 3 = -3. More generally a parabola when written as y = a(x – h) 2+ k has vertex (h, k). As an aside, the CIE examiner is deliberately muddying the waters by using the parameters a, b, and c in his writing of the vertex form. These particular letters are more commonly associated with the expanded form and its accompanying quadratic formula.

b) The function f is defined by f : x 2x2 – 8x + 5 for the domain 0 ≤ x ≤ 5.
i) Find the range of f.

From the expanded form (ax 2+bx +c) the y-intercept of the parabola is 5, the parabola is open
upwards (as a = 2 > 0) and from the vertex form (a(x - h)2 + k) its vertex is (2, -3). A rough sketch follows:

The minimum y-value is -3. The left end-point maximum is y = 5. The right endpoint of the domain is x = 5 and the corresponding value of y is 2(5– 2) 2 – 3 = 18 – 3 = 15. Thus the range of function f is -3 <= y <=15.

ii) Explain why f does not have an inverse.

Notice the horizontal line y = y1. For one value of y there exist two values of x. The function is not 1-1. Therefore it cannot have an inverse.

The function g is defined by g : x 2x2– 8x + 5 for the domain x k.

iii) Find the smallest value of k for which g has an inverse.

As seen above the graph of g should begin at the vertex and go upwards and right indefinitely. The domain is 2 ≤ x < infinity . Hence k = 2.

iv) For this value of k, find an expression for g–1

g(x) = y = 2x2– 8x + 5    Given
x = 2y2– 8y + 5    To find the inverse, swap ‘x’ and ‘y’.
x = 2(y – 2) 2 – 3     Switch to the (already known) vertex form, for easier simplification.
x + 3 = 2(y – 2) 2
½(x+3) = (y – 2) 2
sqrt{½(x+3)} =  sqrt { (y – 2)2 }   Square-root both sides; on the right, cancel the square and the square-root.
2 + sqrt {½(x+3)} = y = g–1
g–1(k) = g–1(2) = 2 +sqrt { ½(2+3)}    We were asked to find an expression; there is no requirement to simplify.