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)
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.
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x
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1
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2
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3x -3
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3 -3 = 0
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3(2)-3 = 6-3
= 3
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3x2-6x+3
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3-6+3 = 0
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3(2)2-6(2)+3 =
12-12+3 =
3
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Therefore
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(x,y) = (1, 0) gives one
point of intersection, and
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(x,y) = (2, 3) also gives an intersection point.
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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.
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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
x = x(t) = x0 + vxt and
y = y(t) = y0 + vyt - ½gt2
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
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x = x(t) = x0 + vxt = 0 +
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800 m
sec
|
t =
|
800 m
sec
|
t
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Now (½)(9.8) = 4.7 gives
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y = y(t) = y0 + vyt - ½gt2 =
2 m +
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600 m
sec
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t
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- 4.7
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m
sec2
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t2
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(i) Now we want to find the value of y when x = 4000 meter. The
latter condition implies
800 m
sec
|
t
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= 4000m
|
|
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t =
|
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sec
800 m
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4000 m
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Therefore t = 5 seconds when x = 4000 meters. That
implies the value of y is given by the formula
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y = y(t) = 2 m +
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600 m
sec
|
t
|
- 4.7
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m
sec2
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t2
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evaluated at t = 5 seconds. That yields
y = [2 + (600)(5) - 4.7 (5)2] m =
[3002 - 4.7 (25)] m
= 2884.5 meters when x = 4000 meters.
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
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0 = y = 2 m +
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600 m
sec
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t
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- 4.7
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m
sec2
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t2
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or equivalently
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0 = 2 +
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600[
|
t
sec
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] - 4.7[
|
t
sec
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] 2
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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).
Exercise: Compute T+ ,
T- and then the t-coordinate h below of the
maximum height with the aid of a calculator.
For part (iii), the high point of the trajectory
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y = 2 m +
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600 m
sec
|
t
|
- 4.7
|
m
sec2
|
t2
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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.
Remark 1: In mathematics courses, I would advise students
to, try to delay or postpone the use of calculators and hence the
appearance of approximate calculations in a solution as much as
possible. The objective is to obtain an exact solution - one in
which there are no approximations.
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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.
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