Real-World Applications പ്രയോഗങ്ങൾ
Whenever one quantity is worked out from another by multiplying, adding and taking powers – a bill from the kilometres, an area from a width, a height from the time – the rule is a polynomial. Move the sliders in each lab and watch the polynomial do the work.
- A rule as a polynomial: p(x) = 4x + 10 – the perimeter when every side grows by x (അളവുകളുടെ ബീജഗണിതം) – page 5.
- Area and volume rules: a(x) = x2 + 5x + 6, v(x) = x3 + 6x2 + 11x + 6 – page 6.
- Polynomial (ബഹുപദം), degree (കൃത്യങ്കം), coefficients (ഗുണകങ്ങൾ) – page 8.
- Values p(0), p(1), p(10) – page 10.
1. A border around a photo
A photo 30 cm long and 20 cm wide gets a mount (border) of width x cm on every side. The whole frame is (30 + 2x) by (20 + 2x), so the area of the border alone is (30 + 2x)(20 + 2x) − 600 = 4x2 + 100x – a second degree polynomial. The same rule works for a path around a garden or a lawn.
In the exam this looks like: “A square ground of side x m has a path 1 m wide around it. Write the area of the path. Is it a polynomial?” → (x + 2)2 − x2 = 4x + 4, a first degree polynomial.
2. The best open box from a card
This is exercise 1, question 2 made bigger. Cut squares of side x from the corners of an L × W card and fold up the sides. The volume is v(x) = x(L − 2x)(W − 2x), a third degree polynomial. Cutting too little gives a flat tray; cutting too much gives a tall thin box. Somewhere in between is the biggest box.
It starts with the textbook card, 7 cm × 5 cm: v(½) = 12, v(1) = 15, v(1½) = 12, and the best cut is about 0.96 cm (volume about 15.02 cubic cm). The green dot on the graph marks the biggest box; the red dot is your cut.
In the exam this looks like: “From a 12 cm × 8 cm card squares of side x are cut… Write v(x) and find v(1).” → v(x) = 4x3 − 40x2 + 96x, v(1) = 60.
3. Fencing with a fixed length of rope
A farmer has L metres of fencing for a rectangular plot. If one side is x, the other is L/2 − x, and the area is a(x) = x(L/2 − x). Two different values of x that add up to L/2 give the same area (the same plot turned around). The biggest plot is the square.
In the exam this looks like: “Rectangles are made with a 1 m rope. Why is a(10) = a(40)?” → a(x) = 50x − x2; sides 10 and 40 make the same rectangle.
4. A ball thrown straight up
A ball thrown up at 20 metres per second is at height h(t) = 20t − 5t2 metres after t seconds (taking the pull of gravity as 10 m/s²). It is a second degree polynomial in t: the 20t is how far the throw alone would carry it, the 5t2 is what gravity takes away.
In the exam this looks like: “h(t) = 20t − 5t2. Find h(1) and h(3). What do you notice?” → both are 15: the ball passes 15 m going up and again coming down.
5. Taxi fares: comparing two plans
Taxi A charges ₹30 to start and ₹15 per km: c(x) = 15x + 30. Taxi B charges ₹10 to start and ₹20 per km: d(x) = 20x + 10. Both are first degree polynomials – the coefficient of x is the rate, the constant term is the fixed charge (it is c(0)!).
In the exam this looks like: “Write the fare for x km as a polynomial and find its degree. For what distance are the two fares equal?” → 15x + 30 = 20x + 10 at x = 4 km.
6. Every number is a polynomial in disguise
4237 = 4 × 103 + 2 × 102 + 3 × 10 + 7, which is p(10) for p(x) = 4x3 + 2x2 + 3x + 7. Computers count in base 2: the binary number 1011 is q(2) for q(x) = x3 + x + 1, which is 8 + 2 + 1 = 11. And p(1) = 4 + 2 + 3 + 7 = 16 is the digit sum.
In the exam this looks like: “For p(x) = 4x3 + 2x2 + 3x + 7 find p(1) and p(10).” → 16 and 4237.
Check yourself
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