[H220151129IAAV] Integration for an Eclipsed Ellipse

Problem

Introduction

     Here we have a challenging H2 Mathematics question that tests students on the technique of substitution, finding area and volume of solid of revolution.  Note that the diagram is not drawn to scale and the line actually intersects the ellipse (oval shape) at the point (-2, 2).  Also the region  R  is the shaded area, but the label “R” is put outside of it.  Students should do their due diligence in ascertaining the intersection point themselves.

Solution

Remarks

     Since the word “exact” is not used in the instructions to part (iii), one can also use the Graphing Calculator to obtain the approximate answer  3.14  to 3 significant figures.  Part (iii) can also be done by another method of integration called the Shell Method, but this is not in the H2 syllabus.  Nevertheless, the student who uses it would not be penalised, unless explicitly forbidden in the rubric.

     The word “ellipse” means an oval shape, and is not to be confused with “eclipse” which means to occlude or hide (e.g. eclipse of the sun, eclipse of the moon).  By the way, the moon orbits around the earth in an ellipse and planets revolve around the sun in ellipses (ovals).
     Although not part of the syllabus, it may be advantageous to know that the area of a right ellipse is  pab,  where  a  and  b  are the semi-axes.  Imagine a circle of radius  a  being stretched by  ^b/_a.  Then its area  pa²  will be multiplied by the same factor  ^b/_a.  So this is not too difficult actually.  With the formula, one can verify one’s answer obtained by integration.

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H05. Work backwards        [ e.g. calculating new limits for substitution ]

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

Suitable Levels

* GCE ‘A’ Level H2 Mathematics

* IB Mathematics HL

* AP Calculus BC

* University / College calculus

* other syllabuses that involve applications of integration

* whoever is interested

[EM_20150412SCA] Is your Airport Design career “taking off”?

Question

An Airbus 380 has constant acceleration of  1 m/s2.  Its takeoff velocity is 280 km/h.  How long must the runway be at a minimum to allow the plane to take off?

Introduction

     A practical question for airport design, perhaps?  I present two solutions.  The first uses a graphical method (speed-time graph) which is in the (“Elementary”) Mathematics syllabus and the other uses formulas for motion under constant acceleration taught in Physics.  Whichever method is used, remember to convert from km/h to m/s.  The target velocity is  ⁷⁰⁰/₉ m/s,  and the time to achieve this is  ⁷⁰⁰/₉ s  starting from rest,  since the acceleration is  1 m/s².

Solution 1 [“Elementary” Mathematics, speed-time graph]

     The speed-time graph is very useful because it is able show the acceleration (as the gradient or slope of a straight line) and at the same time the area under the graph gives the numerical value of distance travelled.  If the acceleration is constant, we usually we get a trapezium.  But since the aeroplane starts from rest, we get a triangle (see diagram below).  All we need to do is to calculate the area under the graph and get the answer.
     It is interesting to observe that if we used the average speed  ⁷⁰⁰/₁₈ m/s,  we would also get the answer because the area under the graph (yellowish green rectangle) is the same as the area of the triangle.  This trick works for constant acceleration, but it may not work in other situations.

Solution 2 [Physics, constant acceleration]

     We use the important formulas  v = u + at  and  s = ut + ½at².  In our example,  u = 0  because the initial velocity is zero (the airplane starts from rest).  This makes our calculations very easy.  If we compare the two methods, you find that the calculations are very similar, and we get the same answer.  Remember that speed = |velocity|  the magnitude of velocity.  In this relatively easy problem, the velocity means the same as speed because we are going in a straight line and in one direction only.  In other situations, this may not be so.

Heuristics Used

H02. Use a diagram / model

H05. Work backwards

H09. Restate the problem in another way

H13* Use Equation / write a Mathematical Sentence

Suitable Levels
* GCE ‘O’ Level “Elementary” Mathematics
* GCE ‘O’ Level Physics
* other syllabuses that acceleration, speed and distance
* precocious kids who always want to learn more

[OlympiadGRN20150406] Grid Ratio Ninja

Perhaps you are an expert at Fruit Ninja, slashing fruits mercilessly with precision using your much-feared finger.  Now can you do the mathematical equivalent below?

Question

Solution

     We would be getting nowhere fast if we took a random approach.  We are only allowed two slashes.  Let us look at the question for more clues.  We have a  5  by  6  grid, giving a total of  30  cells.  The total number of parts in the ratio is  1 + 2 + 3 + 4 = 10.  10  parts correspond to  30 cells,  so each part is  3  cells.  So the number of cells in each portion are  3,  6,  9  and  12.

     Note that if we pair up  3  with  12  and  6  with  9,  we have  15 cells each.  We can just slash the grid down the middle to achieve this.

     We now need ratios of  1 : 4  and  2 : 3  for the remaining parts.  We certainly can achieve this if we were allowed two more slashes and if we used horizontal lines, like this:

However, we only have one more slash.  Can we make this slash do the work of the two horizonal slashes?  Yes!  Replace the two horizontal slashes with a single slanting slash that passes through the mid-points of the horizontal slashes.  This gives four trapezia.  The reason why it works is because for each portion, an extra triangular area is added but it is offset by the removal of a triangle of the same area, like this:-

We can check that the ratios of the areas of the portions are as required.  Done!

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H05. Work backwards

H08. Make suppositions

H09. Restate the problem in another way

H11. Solve part of the problem