9. Quadratics: Applications in Geometry Physics Etc
Problems may come from several sources.
- Solving Systems of Equations - one quadratic, one linear.
- Examples from Physics.
- Constant Velocity Motion
- Quadratic in Time implies Constant Acceleration
- Constant speed and constant acceleration motion (enriched topic)
- Examples from Economics (do, but view with suspicion)
Apart from lip service to applications, mastery of quadratics is needed for calculus and beyond in science, engineering, mathematics and other quantitative disciplines based on calculus (or special functions such as logarithms and exponentials.)
Remark: Applications in economics of quadratics exist, and you may meet them, but those I have met seem more unreal, contrived, or artificial than the physic applications.
Remark: The quadratic formula may be used to solve ax2+bx + c = 0 directly. Or, factoring by inspection and factoring by completing the square and using the difference of two squares can be use to say ax2+bx + c = a(x-r)(x-s) for some real numbers r and s
Problem Type: Intersection of a line and a parabola.
The intersection is found by solving a systems of Equations - one quadratic, one linear.
The intersection of a line y = Ax + B and parabola y = ax2+bx + c may be found by solving Ax +B = ax2+bx + c. The latter yields ax2+(b-B)x + (c-B) = 0 which can be solved for x by the most convenient you see, say by inspection, by completing the square or by the quadratic formula. The latter quadratic in x may have two, one or no solutions. For each x solving the quadratic, there is a y = Ax+B to be computed in order to obtain the coordinates (x,y) of an intersection point.
Example: Find the intersection of the
straight line y = 3x-3 and
the quadratic y = 3x2-6x+3
At the intersection points, if any, the right hand sides of the equations must give the same value for y. Thus comparison of the two sides gives the equation
3x-3 = 3x2-6x+3
linear on one side and quadratic on the left. Add -3x + 3 to both sides
3x -3 = 3x2 - 6x + 3
-3x + 3 = -3x + 3 +
0 = 3x2 - 9x + 6
That implies that the first coordinate of any intersection point must satisfy the quadratic equation:
0 = 3x2 - 9x + 6
The latter can be solved with the aid of the quadratic formula or by factorization. The latter route may give the least amount of work: Let us try it.
0 = 3x2 - 9x + 6
= 3(x2 - 3x + 2) - take out common factor 3
= 3(x-2)(x-1) since 2 has two possible factorizations
2 = (2)(1) and 2 =(-2)(-1)
Here we are fortunate that -2 - 1 = 3.
That gives the factorization.
Now 0 = 3(x-2)(x-1) suggests x = 1 and x = 2 provide the x-coordinates of the intersection points. Let's compute the y-coordinates for each x-value and verify that the two expressions y = 3x-3 and y = 3x2-6x+3 give the same values for y.
| x | 1 | 2 |
| 3x -3 | 3 -3 = 0 | 3(2)-3 = 6-3 = 3 |
| 3x2-6x+3 | 3-6+3 = 0 | 3(2)2-6(2)+3 = 12-12+3 = 3 |
|
Therefore |
(x,y) = (1, 0) gives one point of intersection, and |
(x,y) = (2, 3) also gives an intersection point. |
Remark: In my scratch work, I made an error in the evaluation of at x=2 and had to reconsider my derivation of the solution and after a short delay, saw my error. Knowledge of how does not guarantee calculations are error-free, but practice may help you and I correct more quickly from errors or inconsistencies in our solutions.
Remark: The above problem did not ask us to graph the straight line and quadratic in the region about their intersection points. However, a graph follows.
Problem Type: Projectile Motion
Let t denote time. Let y denote height. Then quadratics y = at2+bt+c may be used to describe or approximate the height of thrown or free-falling projectiles such as bullets, rocks and balls when air resistance is negligible or neglected. For such projectiles, equations of the form x = pt+q may describe the projective movement in a horizontal direction.
If we express t in terms of x, we see that time t is given by a linear expression in x. That expression can be use to eliminate t in y = at2+bt+c to obtain a quadratic relation y = Ax2+Bx+C between the y and x coordinates of the projectile. So we conclude, the projectile follows a quadratic path in the xy plane. In the foregoing, the upper case letters A, B and C The letters A, B and C depend on the coefficients p, q, a, b and c. The do not have the same meaning or same value as the lower case letters a, b and c unless x = t in a unit-free description of the physical situation.
In practice, you may meet
- Vertical projectile motion - the position of a falling object subject to the constant pull of gravity at or near the earth's surface can be described using quadratics expressions y = at2+bt + c with time t in place of horizontal coordinate x as the independent variable. The direct use of this equation is to calculate coordinate y given the value of time t. One indirect use of this equation gives the value of y and asks for the value or possible values of t. You will need to solve a quadratic equation for t and if there are two numerical solutions, decide which one is required or selected by the information at hand. Further indirect uses of the formula may give you values of y and t, clearly or not, and ask you find the values of the coefficients a, b and c, before using y = at2+bt + c directly, or indirectly again.
- Projectile Motion in the Plane: Here y = at2+bt + c and x = Bt + C describes a falling body in the vertical xy plane near the earths surface. You may be asked to analyze these equations forwards and backwards.
- Free Sliding Object on a slanted plane. y = at2+bt + c and x = At2+ Bt + C but simplifications may follow, will follow, from using a slanted coordinate system with x or y coordinate in the plane. The equations for situation B or C may reappear. This might be an enriched problem in a senior high school physics course.
| Problems of type B: A cannon ball leaves the mouth of a cannon
with an initial horizontal velocity of 800 meters per second in a
direction (say the x-direction) and an initial vertical velocity of 600
meters per second (say the y-direction). (i) How will the ball be
when the ball is 4000 meters horizontally from the initial position.
(ii) When and where will the ball hit the ground? (iii) What is the
maximum height of the ball? Assume the position of the ball as it
leaves the cannon mouth is almost ground level, say y = 2 meters.
Solution: We assume air resistance is negligible for this cannon ball projectile, that it flies in a vertical plane with upward direction is positive. Then the x and y coordinates of the projectile are given by two formulas from physics, namely
where t = elapse time since the projectile was in its initial position, where g = 9.8 meters per second square = acceleration of a free-falling object due to gravity at the earth's surface, where (x0 , y0) give the initial position of the projectile; and where (vx , vy) = the initial velocity of the projectile. The latter means vx = initial horizontal velocity and vy = initial vertical velocity Substitution (Use) of data in equations. We will take the initial position x0 = 0 and use vx = initial horizontal velocity = 800 meters per second. So
Now (½)(9.8) = 4.7 gives
(i) Now we want to find the value of y when x = 4000 meter. The latter condition implies
Therefore t = 5 seconds when x = 4000 meters. That implies the value of y is given by the formula
evaluated at t = 5 seconds. That yields
That completes the solution to part (i). In part (ii), the question is when will y(t) = 0 after t = 0. That requires the solution of the equation
or equivalently
The positive solution T+ = t/sec of this equation follows from the quadratic with the aid of a calculator. That completes part (ii). Calculate the negative solution T- as well for use in part (iii).
For part (iii), the high point of the trajectory
occurs on the axis of symmetry t = h = -b/2a. The latter can be computed directly. The latter can be compute directly, or you can use symmetry to observe h = (T+ + T-) is half-way between the two zeroes of y = y(t). From the value t = h, the maximum value of y = y(t) can be obtained, again with the aid of a calculator.
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Remark 2: The description of projectile motion is provides a war-like qualitative idea of the flight of projectiles. Suffice it to say, I do not like the connection of mathematics to the arts of war, past and present. Mathematics skills and concepts have been driven by various motivations in consumer life, business, construction, science (planetary movements included), technology and war. Projectile motion provides the application of quadratics most easily visualized and so most useful for the development of mathematical skills - ouch.