Tenth Grade Mathematics: Real-World Mathematical Modeling with Quadratics

The real world is rarely strictly linear. When physical velocities change, when geometric dimensions are multiplied to calculate area, or when unit prices influence customer demand, the mathematical relationships that emerge are quadratic. Mathematical modeling is the process of translating a physical situation into a quadratic equation, solving it mathematically, and evaluating the solutions against physical reality (such as rejecting negative measurements for time or length).

The Geometry and Mensuration Model

Geometric problems frequently involve the dimensions of rectangles, triangles, and right-angled configurations governed by the Pythagorean theorem:

Right-Angled Triangle Model (Pythagorean Theorem):
                 |                 |     Altitude (h) |  \ Hypotenuse (c)        a^2 + b^2 = c^2
                 |                    +----+
                Base (b)

In rectangle problems, if the area is A and the perimeter is P, the length l and width w satisfy:

l + w = P / 2    and    l * w = A

Substituting w = P/2 - l into the area equation produces a quadratic equation for length: l^2 - (P/2)*l + A = 0.

The Uniform Motion and Speed Adjustment Model

A classic physical model involves a journey of fixed distance D where a change in speed delta_v results in a change in travel time delta_t:

Kinematic Relations:
Let standard speed = v km/h, Distance = D km.
Original Time: t1 = D / v
Increased Speed: (v + s) km/h  ===>  New Time: t2 = D / (v + s)

Time Difference Equation:
t1 - t2 = delta_t   ===>   D / v - D / (v + s) = delta_t

Clearing the algebraic fractions results in a standard quadratic equation in the velocity v.

Worked Speed Example: Train Transit

A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Formulation:
Let original speed = x km/h. Distance = 360 km.
Original time = 360 / x hours.
Increased speed = (x + 5) km/h.
New time = 360 / (x + 5) hours.

Equation:
360 / x - 360 / (x + 5) = 1

Multiply by x*(x + 5):
360*(x + 5) - 360*x = x*(x + 5)
360*x + 1800 - 360*x = x^2 + 5*x
x^2 + 5*x - 1800 = 0

Factor:
Find two numbers with product -1800 and sum 5: 45 and -40.
(x + 45) * (x - 40) = 0
x = -45  or  x = 40.

Physical Context Check:
Speed cannot be a negative quantity, so x = -45 is physically inadmissible.
Conclusion: The original speed of the train is 40 km/h.

The Rate of Flow: Water Taps and Cisterns

When two water taps fill a cistern together, their combined flow rate per unit time is the sum of their individual rates:

Taps Model:
If Tap A takes t hours alone, its rate = 1 / t cistern per hour.
If Tap B takes (t - k) hours alone, its rate = 1 / (t - k) cistern per hour.
Together filling in T hours:
                      1/t + 1/(t - k) = 1/T

Chapter Exercises

Exercise 1: The hypotenuse of a right triangle is 25 cm. If one of the remaining two sides is 5 cm longer than the other, find the lengths of the other two sides.

Exercise 2: An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.

Exercise 3: Two water taps together can fill a tank in 9 and 3/8 hours (75/8 hours). The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Solutions and Explanations

Solution 1: Let the length of the shorter side be x cm. Then the longer side has length (x + 5) cm. The hypotenuse is 25 cm. By the Pythagorean theorem: x^2 + (x + 5)^2 = 25^2 ==> x^2 + x^2 + 10x + 25 = 625 ==> 2x^2 + 10x - 600 = 0. Divide by 2: x^2 + 5x - 300 = 0. We factor this quadratic: find two numbers with product -300 and sum 5; these are 20 and -15. (x + 20)*(x - 15) = 0 ==> x = 15 or x = -20. Since a geometric length cannot be negative, we discard x = -20. Thus, x = 15 cm. The shorter side is 15 cm, and the longer side is 15 + 5 = 20 cm.

Solution 2: Let the average speed of the passenger train be x km/h. Then the average speed of the express train is (x + 11) km/h. Distance is 132 km. Time taken by passenger train is 132/x hours, and time taken by express train is 132/(x + 11) hours. Given that the difference in time is 1 hour: 132/x - 132/(x + 11) = 1 ==> 132[(x + 11 - x) / (x(x + 11))] = 1 ==> 132(11) = x^2 + 11x ==> 1452 = x^2 + 11x ==> x^2 + 11x - 1452 = 0. We find factors of -1452 that differ by 11: 44 and -33 (since 44 * (-33) = -1452 and 44 - 33 = 11). Thus: (x + 44)*(x - 33) = 0 ==> x = 33 or x = -44. Rejecting the negative speed, x = 33 km/h. Therefore, the average speed of the passenger train is 33 km/h, and the average speed of the express train is 33 + 11 = 44 km/h.

Solution 3: Let the smaller tap take x hours to fill the tank alone. Then the larger tap takes (x - 10) hours alone. In 1 hour, the smaller tap fills 1/x of the tank, and the larger tap fills 1/(x - 10) of the tank. Together they fill the tank in 75/8 hours, meaning in 1 hour they fill 1 / (75/8) = 8/75 of the tank. Thus: 1/x + 1/(x - 10) = 8/75 ==> [(x - 10 + x) / (x(x - 10))] = 8/75 ==> (2x - 10) / (x^2 - 10x) = 8/75. Cross-multiply: 75(2x - 10) = 8(x^2 - 10x) ==> 150x - 750 = 8x^2 - 80x ==> 8x^2 - 230x + 750 = 0. Divide by 2: 4x^2 - 115x + 375 = 0. Factor this quadratic: product is 4 * 375 = 1500; sum is -115. The numbers are -100 and -15. 4x^2 - 100x - 15x + 375 = 0 ==> 4x(x - 25) - 15(x - 25) = 0 ==> (x - 25)(4x - 15) = 0. Thus x = 25 or x = 15/4 = 3.75. If x = 3.75 hours, then the larger tap would take x - 10 = 3.75 - 10 = -6.25 hours, which is physically impossible. Therefore, x = 25 hours. The smaller tap takes 25 hours, and the larger tap takes 25 - 10 = 15 hours to fill the tank separately.